PROBLEM WORKBOOK - Langlo...

PROBLEMWORKBOOK

Cover Photo: Lawrence Manning/CORBIS

Cover Design: Jason Wilson

Copyright © by Holt, Rinehart and Winston

All rights reserved. No part of this publication may be reproduced or transmittedin any form or by any means, electronic or mechanical, including photocopy,recording, or any information storage and retrieval system, without permission inwriting from the publisher.

Teachers using HOLT PHYSICS may photocopy blackline masters in completepages in sufficient quantities for classroom use only and not for resale.

Printed in the United States of America

ISBN 0-03-057337-8

1 2 3 4 5 6 095 05 04 03 02 01

Holt PhysicsProblem WorkbookThis workbook contains additional worked-out samples and practiceproblems for each of the problem types from the Holt Physics text.

Contributing Writers

Boris M. KorsunskyPhysics Instructor

Science Department

Northfield Mount Hermon School

Northfield, MA

Angela BerensteinScience Writer

Urbana, IL

John StokesScience Writer

Socorro, NM

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Contents iii

Section Title Page

Sample and Practice 1A Metric Prefixes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

Sample and Practice 2A Average Velocity and Displacement . . . . . . . . . . . . . . . . . . . . . . . 3

Sample and Practice 2B Average Acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

Sample and Practice 2C Displacement with Constant Acceleration. . . . . . . . . . . . . . . . 7

Sample and Practice 2D Velocity and Displacement with ConstantAcceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

Sample and Practice 2E Final Velocity After Any Displacement . . . . . . . . . . . . . . . . . . 12

Sample and Practice 2F Falling Object . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

Sample and Practice 3A Finding Resultant Magnitude and Direction . . . . . . . . . . . 16

Sample and Practice 3B Resolving Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

Sample and Practice 3C Adding Vectors Algebraically . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

Sample and Practice 3D Projectiles Launched Horizontally. . . . . . . . . . . . . . . . . . . . . . . 22

Sample and Practice 3E Projectiles Launched at an Angle . . . . . . . . . . . . . . . . . . . . . . . . 24

Sample and Practice 3F Relative Velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

Sample and Practice 4A Net External Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

Sample and Practice 4B Newton’s Second Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

Sample and Practice 4C Coefficients of Friction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

Sample and Practice 4D Overcoming Friction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

Sample and Practice 5A Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

Sample and Practice 5B Kinetic Energy. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

Sample and Practice 5C Work-Kinetic Energy Theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . 44

Sample and Practice 5D Potential Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

Sample and Practice 5E Conservation of Mechanical Energy . . . . . . . . . . . . . . . . . . . . . 50

Sample and Practice 5F Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

Sample and Practice 6A Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

Sample and Practice 6B Force and Impulse . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

Sample and Practice 6C Stopping Distance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

Sample and Practice 6D Conservation of Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

Sample and Practice 6E Perfectly Inelastic Collision . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

Sample and Practice 6F Kinetic Energy in PerfectlyInelastic Collisions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

Sample and Practice 6G Elastic Collisions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

Sample and Practice 7A Angular Displacement. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

Sample and Practice 7B Angular Speed . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

Contents

Contentsiv

Section Title Page

Sample and Practice 7C Angular Acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

Sample and Practice 7D Angular Kinematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

Sample and Practice 7E Tangential Speed . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

Sample and Practice 7F Tangential Acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

Sample and Practice 7G Centripetal Acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80

Sample and Practice 7H Force That Maintains Circular Motion . . . . . . . . . . . . . . . . . . 81

Sample and Practice 7I Gravitational Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

Sample and Practice 8A Torque . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

Sample and Practice 8B Rotational Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

Sample and Practice 8C Newton's Second Law for Rotation. . . . . . . . . . . . . . . . . . . . . . . 91

Sample and Practice 8D Conservation of Angular Momentum . . . . . . . . . . . . . . . . . . . 94

Sample and Practice 8E Conservation of Mechanical Energy . . . . . . . . . . . . . . . . . . . . . 96

Sample and Practice 9A Buoyant Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98

Sample and Practice 9B Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

Sample and Practice 9C Pressure as a Function of Depth . . . . . . . . . . . . . . . . . . . . . . . . 102

Sample and Practice 9D Bernoulli’s Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104

Sample and Practice 9E Gas Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106

Sample and Practice 10A Temperature Conversion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108

Sample and Practice 10B Conservation of Energy. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109

Sample and Practice 10C Calorimetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111

Sample and Practice 10D Heat of Phase Change . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113

Sample and Practice 11A Work Done on or by a Gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115

Sample and Practice 11B The First Law of Thermodynamics . . . . . . . . . . . . . . . . . . . . . 116

Sample and Practice 11C Heat-Engine Efficiency . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118

Sample and Practice 12A Hooke’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119

Sample and Practice 12B Simple Harmonic Motion of a Simple Pendulum . . . . 121

Sample and Practice 12C Simple Harmonic Motion of a Mass-Spring System . . 122

Sample and Practice 12D Wave Speed . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123

Sample and Practice 13A Intensity of Sound Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124

Sample and Practice 13B Harmonics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125

Sample and Practice 14A Electromagnetic Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126

Sample and Practice 14B Concave Mirrors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127

Sample and Practice 14C Convex Mirrors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129

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Contents v

Section Title Page

Sample and Practice 15A Snell’s Law. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131

Sample and Practice 15B Lenses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132

Sample and Practice 15C Critical Angle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135

Sample and Practice 16A Interference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136

Sample and Practice 16B Diffraction Gratings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138

Sample and Practice 17A Coulomb’s Law. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140

Sample and Practice 17B The Superposition Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142

Sample and Practice 17C Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145

Sample and Practice 17D Electric Field Strength . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148

Sample and Practice 18A Electrical Potential Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151

Sample and Practice 18B Potential Difference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153

Sample and Practice 18C Capacitance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156

Sample and Practice 19A Current . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159

Sample and Practice 19B Resistance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160

Sample and Practice 19C Electric Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161

Sample and Practice 19D Cost of Electrical Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162

Sample and Practice 20A Resistors in Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163

Sample and Practice 20B Resistors in Parallel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165

Sample and Practice 20C Equivalent Resistance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167

Sample and Practice 20D Current in and Potential DifferenceAcross a Resistor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171

Sample and Practice 21A Particle in a Magnetic Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174

Sample and Practice 21B Force on a Current-Carrying Conductor . . . . . . . . . . . . . . 176

Sample and Practice 22A Induced emf and Current . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178

Sample and Practice 22B Induction in Generators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180

Sample and Practice 22C rms Currents and Potential Differences . . . . . . . . . . . . . . . 182

Sample and Practice 22D Transformers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184

Sample and Practice 23A Quantum Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186

Sample and Practice 23B The Photoelectric Effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 188

Sample and Practice 23C De Broglie Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190

Sample and Practice 25A Binding Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191

Sample and Practice 25B Nuclear Decay. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193

Sample and Practice 25C Measuring Nuclear Decay . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195

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Problem 1A 1

NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 1AMETRIC PREFIXES

P R O B L E MIn Hindu chronology, the longest time measure is a para. One para equals 311 040 000 000 000 years. Calculate this value in megahours and innanoseconds. Write your answers in scientific notation.

S O L U T I O NGiven: 1 para = 311 040 000 000 000 years

Unknown: 1 para = ? Mh

1 para = ? ns

Express the time in years in terms of scientific notation. Then build conversion

factors from the relationships given in Table 1-3.

1 para = 3.1104 × 1014 years

365

1

.2

y

5

ea

d

r

ays ×

1

24

da

h

y ×

1 ×1 M

10

h6 h

365

1

.2

y

5

ea

d

r

ays ×

1

24

da

h

y ×

36

1

0

h

0 s ×

1 ×1

1

n

0

s−9s

Convert from years to megahours by multiplying the time by the first conversion

expression.

1 para = 3.1104 × 1014 years × 365

1

.2

y

5

ea

d

r

ays ×

1

24

da

h

y ×

1 ×1 M

10

h6 h

=

Convert from years to nanoseconds by multiplying the time by the second con-

version expression.

1 para = 3.1104 × 1014 years × 365

1

.2

y

5

ea

d

r

ays ×

1

24

da

h

y ×

36

1

0

h

0 s ×

1 ×1

1

n

0

s−9 s

= 9.8157 × 1030 ns

2.7266 × 1012 Mh

ADDITIONAL PRACTICE

1. One light-year is the distance light travels in one year. This distance is

equal to 9.461 × 1015 m. After the sun, the star nearest to Earth is Alpha

Centauri, which is about 4.35 light-years from Earth. Express this dis-

tance in

a. megameters.

b. picometers.

Holt Physics Problem Workbook2

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2. It is estimated that the sun will exhaust all of its energy in about ten

billion years. By that time, it will have radiated about 1.2 × 1044 J (joules)

of energy. Express this amount of energy in

a. kilojoules.

b. nanojoules.

3. The smallest living organism discovered so far is called a mycoplasm. Its

mass is estimated as 1.0 × 10–16 g. Express this mass in

a. petagrams.

b. femtograms.

c. attograms.

4. The “extreme” prefixes that are officially recognized are yocto, which in-

dicates a fraction equal to 10–24, and yotta, which indicates a factor equal

to 1024. The maximum distance from Earth to the sun is 152 100 000 km.

Using scientific notation, express this distance in

a. yoctometers (ym).

b. yottameters (Ym).

5. In 1993, the total production of nuclear energy in the world was

2.1 × 1015 watt-hours, where a watt is equal to one joule (J) per second.

Express this number in

a. joules.

b. gigajoules.

6. In Einstein’s special theory of relativity, mass and energy are equivalent.

An expression of this equivalence can be made in terms of electron volts

(units of energy) and kilograms, with one electron volt (eV) being equal

to 1.78 × 10–36 kg. Using this ratio, express the mass of the heaviest

mammal on earth, the blue whale, which has an average mass of

1.90 × 105 kg, in

a. mega electron volts.

b. tera electron volts.

7. The most massive star yet discovered in our galaxy is one of the stars

in the Carina Nebula, which can be seen from Earth’s Southern

Hemisphere and from the tropical latitudes of the Northern Hemisphere.

The star, designated as Eta Carinae, is believed to be 200 times as massive

as the sun, which has a mass of nearly 2 × 1030 kg. Find the mass of Eta

Carinae in

a. milligrams.

b. exagrams.

8. The Pacific Ocean has a surface area of about 166 241 700 km2 and an

average depth of 3940 m. Estimate the volume of the Pacific Ocean in

a. cubic centimeters.

b. cubic millimeters.

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Problem 2A 3

NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 2AAVERAGE VELOCITY AND DISPLACEMENT

P R O B L E MThe fastest fish, the sailfish, can swim 1.2 × 102 km/h. Suppose you have a friend who lives on an island 16 km away from the shore. If you send a message using a sailfish as a messenger, how long will it take for themessage to reach your friend?

S O L U T I O NGiven: vavg = 1.2 × 102 km/h

∆x = 16 km

Unknown: ∆t = ?

Use the definition of average speed to find ∆t.

vavg = ∆∆

x

t

Rearrange the equation to calculate ∆t.

∆t = v

∆

av

x

g

∆t = =

= 8.0 min

2.0

1

k

6

m

k

/

m

min

16 km

1.2 × 102 k

h

m 60

1

m

h

in

ADDITIONAL PRACTICE

1. The Sears Tower in Chicago is 443 m tall. Joe wants to set the world’s

stair climbing record and runs all the way to the roof of the tower. If Joe’s

average upward speed is 0.60 m/s, how long will it take Joe to climb from

street level to the roof of the Sears Tower?

2. An ostrich can run at speeds of up to 72 km/h. How long will it take an

ostrich to run 1.5 km at this top speed?

3. A cheetah is known to be the fastest mammal on Earth, at least for short

runs. Cheetahs have been observed running a distance of 5.50 × 102 m

with an average speed of 1.00 × 102 km/h.

a. How long would it take a cheetah to cover this distance at this speed?

b. Suppose the average speed of the cheetah were just 85.0 km/h.

What distance would the cheetah cover during the same time inter-

val calculated in (a)?


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4. A pronghorn antelope has been observed to run with a top speed of

97 km/h. Suppose an antelope runs 1.5 km with an average speed of

85 km/h, and then runs 0.80 km with an average speed of 67 km/h.

a. How long will it take the antelope to run the entire 2.3 km?

b. What is the antelope’s average speed during this time?

5. Jupiter, the largest planet in the solar system, has an equatorial radius of

about 7.1 × 104 km (more than 10 times that of Earth). Its period of ro-

tation, however, is only 9 h, 50 min. That means that every point on

Jupiter’s equator “goes around the planet” in that interval of time. Calcu-

late the average speed (in m/s) of an equatorial point during one period

of Jupiter’s rotation. Is the average velocity different from the average

speed in this case?

6. The peregrine falcon is the fastest of flying birds (and, as a matter of fact,

is the fastest living creature). A falcon can fly 1.73 km downward in 25 s.

What is the average velocity of a peregrine falcon?

7. The black mamba is one of the world’s most poisonous snakes, and with

a maximum speed of 18.0 km/h, it is also the fastest. Suppose a mamba

waiting in a hide-out sees prey and begins slithering toward it with a

velocity of +18.0 km/h. After 2.50 s, the mamba realizes that its prey can

move faster than it can. The snake then turns around and slowly returns

to its hide-out in 12.0 s. Calculate

a. the mamba’s average velocity during its return to the hideout.

b. the mamba’s average velocity for the complete trip.

c. the mamba’s average speed for the complete trip.

8. In the Netherlands, there is an annual ice-skating race called the “Tour of

the Eleven Towns.” The total distance of the course is 2.00 × 102 km, and

the record time for covering it is 5 h, 40 min, 37 s.

a. Calculate the average speed of the record race.

b. If the first half of the distance is covered by a skater moving with

a speed of 1.05v, where v is the average speed found in (a), how

long will it take to skate the first half? Express your answer in hours

and minutes.

Problem 2B 5

NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 2BAVERAGE ACCELERATION

P R O B L E MIn 1977 off the coast of Australia, the fastest speed by a vessel on the waterwas achieved. If this vessel were to undergo an average acceleration of1.80 m/s2, it would go from rest to its top speed in 85.6 s. What was thespeed of the vessel?

S O L U T I O N

Given: aavg = 1.80 m/s2

∆t = 85.6 s

vi = 0 m/s

Unknown: vf = ?

Use the definition of average acceleration to find vf.

aavg = ∆∆

v

t =

vf

∆–

t

vi

Rearrange the equation to calculate vf.

vf = aavg ∆t + vi

vf = 1.80 m

s285.6 s + 0 m

s

= 154 m

s

= 154 m

s 3.60

1

×h

103 s110

k3m

m

= 554 k

h

m

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ADDITIONAL PRACTICE

1. If the vessel in the sample problem accelerates for 1.00 min, what will

its speed be after that minute? Calculate the answer in both meters per

second and kilometers per hour.

2. In 1935, a French destroyer, La Terrible, attained one of the fastest

speeds for any standard warship. Suppose it took 2.0 min at a constant

acceleration of 0.19 m/s2 for the ship to reach its top speed after start-

ing from rest. Calculate the ship’s final speed.

3. In 1934, the wind speed on Mt. Washington in New Hampshire

reached a record high. Suppose a very sturdy glider is launched in this

wind, so that in 45.0 s the glider reaches the speed of the wind. If the


NAME ______________________________________ DATE _______________ CLASS ____________________

glider undergoes a constant acceleration of 2.29 m/s2, what is the

wind’s speed? Assume that the glider is initially at rest.

4. In 1992, Maurizio Damilano, of Italy, walked 29 752 m in 2.00 h.

a. Calculate Damilano’s average speed in m/s.

b. Suppose Damilano slows down to 3.00 m/s at the midpoint in his

journey, but then picks up the pace and accelerates to the speed

calculated in (a). It takes Damilano 30.0 s to accelerate. Find the

magnitude of the average acceleration during this time interval.

5. South African frogs are capable of jumping as far as 10.0 m in one hop.

Suppose one of these frogs makes exactly 15 of these jumps in a time

interval of 60.0 s.

a. What is the frog’s average velocity?

b. If the frog lands with a velocity equal to its average velocity and

comes to a full stop 0.25 s later, what is the frog’s average

acceleration?

6. In 1991 at Smith College, in Massachusetts, Ferdie Adoboe ran

1.00 × 102 m backward in 13.6 s. Suppose it takes Adoboe 2.00 s to

achieve a velocity equal to her average velocity during the run. Find

her average acceleration during the first 2.00 s.

7. In the 1992 Summer Olympics, the German four-man kayak team cov-

ered 1 km in just under 3 minutes. Suppose that between the starting

point and the 150 m mark the kayak steadily increases its speed from

0.0 m/s to 6.0 m/s, so that its average speed is 3.0 m/s.

a. How long does it take to cover the 150 m?

b. What is the magnitude of the average acceleration during that

part of the course?

8. The highest speed ever achieved on a bicycle was reached by John

Howard of the United States. The bicycle, which was accelerated by

being towed by a vehicle, reached a velocity of +245 km/h. Suppose

Howard wants to slow down, and applies the brakes on his now freely

moving bicycle. If the average acceleration of the bicycle during brak-

ing is –3.0 m/s2, how long will it take for the bicycle’s velocity to de-

crease by 20.0 percent?

9. In 1993, bicyclist Rebecca Twigg of the United States traveled 3.00 km

in 217.347 s. Suppose Twigg travels the entire distance at her average

speed and that she then accelerates at –1.72 m/s2 to come to a complete

stop after crossing the finish line. How long does it take Twigg to come

to a stop?

10. During the Winter Olympic games at Lillehammer, Norway, in 1994,

Dan Jansen of the United States skated 5.00 × 102 m in 35.76 s. Sup-

pose it takes Jansen 4.00 s to increase his velocity from zero to his

maximum velocity, which is 10.0 percent greater than his average ve-

locity during the whole run. Calculate Jansen’s average acceleration

during the first 4.00 s.

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Problem 2C 7

NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 2CDISPLACEMENT WITH CONSTANT ACCELERATION

P R O B L E MIn England, two men built a tiny motorcycle with a wheel base (the dis-tance between the centers of the two wheels) of just 108 mm and a wheel’smeasuring 19 mm in diameter. The motorcycle was ridden over a distanceof 1.00 m. Suppose the motorcycle has constant acceleration as it travelsthis distance, so that its final speed is 0.800 m/s. How long does it take themotorcycle to travel the distance of 1.00 m? Assume the motorcycle is ini-tially at rest.

S O L U T I O NGiven: vf = 0.800 m/s

vi = 0 m/s

∆x = 1.00 m

Unknown: ∆t = ?

Use the equation for displacement with constant acceleration.

∆x = 12

(vi + vf )∆t

Rearrange the equation to calculate ∆t.

∆t = v

2

f

∆+

x

vi

∆t = = 0

2

.

.

8

0

0

0

0 s

= 2.50 s

(2)(1.00 m)0.800

m

s + 0

m

s

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ADDITIONAL PRACTICE

1. In 1993, Ileana Salvador of Italy walked 3.0 km in under 12.0 min. Sup-

pose that during 115 m of her walk Salvador is observed to steadily in-

crease her speed from 4.20 m/s to 5.00 m/s. How long does this increase

in speed take?

2. In a scientific test conducted in Arizona, a special cannon called HARP

(High Altitude Research Project) shot a projectile straight up to an alti-

tude of 180.0 km. If the projectile’s initial speed was 3.00 km/s, how long

did it take the projectile to reach its maximum height?

3. The fastest speeds traveled on land have been achieved by rocket-

powered cars. The current speed record for one of these vehicles is

about 1090 km/h, which is only 160 km/h less than the speed of

sound in air. Suppose a car that is capable of reaching a speed of


NAME ______________________________________ DATE _______________ CLASS ____________________

1.09 × 103 km/h is tested on a flat, hard surface that is 25.0 km long. The

car starts at rest and just reaches a speed of 1.09 × 103 km/h when it

passes the 20.0 km mark.

a. If the car’s acceleration is constant, how long does it take to make

the 20.0 km drive?

b. How long will it take the car to decelerate if it goes from its maxi-

mum speed to rest during the remaining 5.00 km stretch?

4. In 1990, Dave Campos of the United States rode a special motorcycle

called the Easyrider at an average speed of 518 km/h. Suppose that at

some point Campos steadily decreases his speed from 100.0 percent to

60.0 percent of his average speed during an interval of 2.00 min. What is

the distance traveled during that time interval?

5. A German stuntman named Martin Blume performed a stunt called “the

wall of death.” To perform it, Blume rode his motorcycle for seven

straight hours on the wall of a large vertical cylinder. His average speed

was 45.0 km/h. Suppose that in a time interval of 30.0 s Blume increases

his speed steadily from 30.0 km/h to 42.0 km/h while circling inside the

cylindrical wall. How far does Blume travel in that time interval?

6. An automobile that set the world record for acceleration increased speed

from rest to 96 km/h in 3.07 s. How far had the car traveled by the time

the final speed was achieved?

7. In a car accident involving a sports car, skid marks as long as 290.0 m

were left by the car as it decelerated to a complete stop. The police report

cited the speed of the car before braking as being “in excess of 100 mph”

(161 km/h). Suppose that it took 10.0 seconds for the car to stop. Esti-

mate the speed of the car before the brakes were applied. (REMINDER:

Answer should read, “speed in excess of . . .”)

8. Col. Joe Kittinger of the United States Air Force crossed the Atlantic

Ocean in nearly 86 hours. The distance he traveled was 5.7 × 103 km.

Suppose Col. Kittinger is moving with a constant acceleration during

most of his flight and that his final speed is 10.0 percent greater than his

initial speed. Find the initial speed based on this data.

9. The polar bear is an excellent swimmer, and it spends a large part of its

time in the water. Suppose a polar bear wants to swim from an ice floe to

a particular point on shore where it knows that seals gather. The bear

dives into the water and begins swimming with a speed of 2.60 m/s. By

the time the bear arrives at the shore, its speed has decreased to 2.20 m/s.

If the polar bear’s swim takes exactly 9.00 min and it has a constant de-

celeration, what is the distance traveled by the polar bear?

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Problem 2D 9

NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 2DVELOCITY AND DISPLACEMENT WITH CONSTANT ACCELERATION

P R O B L E MSome cockroaches can run as fast as 1.5 m/s. Suppose that two cock-roaches are separated by a distance of 60.0 cm and that they begin to runtoward each other at the same moment. Both insects have constant accel-eration until they meet. The first cockroach has an acceleration of0.20 m/s2 in one direction, and the second one has an acceleration of0.12 m/s2 in the opposite direction. How much time passes before the two insects bump into each other?

S O L U T I O NGiven: a1 = 0.20 m/s2 (first cockroach’s acceleration)

vi,1 = 0 m/s (first cockroach’s initial speed)

a2 = 0.12 m/s2 (second cockroach’s acceleration)

vi,2 = 0 m/s (second cockroach’s initial speed)

d = 60.0 cm = 0.60 m (initial distance between the insects)

Unknown: ∆x1 = ? ∆x2 = ? ∆t = ?

Choose an equation(s) or situation: Use the equation for displacement with

constant acceleration for each cockroach.

∆x1 = vi,1∆t + 12

a1∆t2

∆x2 = vi,2∆t + 12

a2∆t2

The distance the second cockroach travels can be expressed as the difference be-

tween the total distance that initially separates the two insects and the distance

that the first insect travels.

∆x2 = d – ∆x1

Rearrange the equation(s) to isolate the unknown(s): Substitute the expression

for the first cockroach’s displacement into the equation for the second cockroach’s

displacement using the equation relating the two displacements to the initial dis-

tance between the insects.

∆x2 = d – ∆x1 = d – vi,1∆t + 12

a1∆t2= vi,2∆t + 1

2a2∆t2

The equation can be rewritten to express ∆t in terms of the known quantities. To

simplify the calculation, the terms involving the initial speeds, which are both

zero, can be removed from the equations.

d – 12

a1∆t2 = 12

a2∆t2

∆t = a12

+d a2

1. DEFINE

2. PLAN

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NAME ______________________________________ DATE _______________ CLASS ____________________

Substitute the values into the equation(s) and solve:

∆t = = =

The final speeds for the first and second cockroaches are 0.38 m/s and 0.23 m/s,

respectively. Both of these values are well below the maximum speed for cock-

roaches in general.

1.9 s1.2 m0.32 m/s2

(2)(0.60 m)0.20

m

s2 + 0.12 m

s2

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ADDITIONAL PRACTICE

1. In 1986, the first flight around the globe without a single refueling was

completed. The aircraft’s average speed was 186 km/h. If the airplane

landed at this speed and accelerated at −1.5 m/s2, how long did it take

for the airplane to stop?

2. In 1976, Gerald Hoagland drove a car over 8.0 × 102 km in reverse. For-

tunately for Hoagland and motorists in general, the event took place on

a special track. During this drive, Hoagland’s average velocity was

about –15.0 m/s. Suppose Hoagland decides during his drive to go for-

ward. He applies the brakes, stops, and then accelerates until he moves

forward at same speed he had when he was moving backward. How

long would the entire reversal process take if the average acceleration

during this process is +2.5 m/s2?

3. The first permanent public railway was built by George Stephenson

and opened in Cleveland, Ohio, in 1825. The average speed of the

trains was 24.0 km/h. Suppose a train moving at this speed accelerates

–0.20 m/s2 until it reaches a speed of 8.0 km/h. How long does it take

the train to undergo this change in speed?

4. The winding cages in mine shafts are used to move workers in and out

of the mines. These cages move much faster than any commercial ele-

vators. In one South African mine, speeds of up to 65.0 km/h are at-

tained. The mine has a depth of 2072 m. Suppose two cages start their

downward journey at the same moment. The first cage quickly attains

the maximum speed (an unrealistic situation), then proceeds to de-

scend uniformly at that speed all the way to the bottom. The second

cage starts at rest and then increases its speed with a constant accelera-

tion of magnitude 4.00 × 10–2 m/s2. How long will the trip take for

each cage? Which cage will reach the bottom of the mine shaft first?

5. In a 1986 bicycle race, Fred Markham rode his bicycle a distance of

2.00 × 102 m with an average speed of 105.4 km/h. Markham and the

bicycle started the race with a certain initial speed.

a. Find the time it took Markham to cover 2.00 × 102 m.

b. Suppose a car moves from rest under constant acceleration. What

is the magnitude of the car’s acceleration if the car is to finish the

race at exactly the same time Markham finishes the race?

3. CALCULATE

4. EVALUATE

Problem 2D 11

NAME ______________________________________ DATE _______________ CLASS ____________________

6. Some tropical butterflies can reach speeds of up to 11 m/s. Suppose a

butterfly flies at a speed of 6.0 m/s while another flying insect some dis-

tance ahead flies in the same direction with a constant speed. The but-

terfly then increases its speed at a constant rate of 1.4 m/s2 and catches

up to the other insect 3.0 s later. How far does the butterfly travel dur-

ing the race?

7. Mary Rife, of Texas, set a women’s world speed record for sailing. In

1977, her vessel, Proud Mary, reached a speed of 3.17 × 102 km/h. Sup-

pose it takes 8.0 s for the boat to decelerate from 3.17 × 102 km/h to

2.00 × 102 km/h. What is the boat’s acceleration? What is the displace-

ment of the Proud Mary as it slows down?

8. In 1994, a human-powered submarine was designed in Boca Raton,

Florida. It achieved a maximum speed of 3.06 m/s. Suppose this sub-

marine starts from rest and accelerates at 0.800 m/s2 until it reaches

maximum speed. The submarine then travels at constant speed for an-

other 5.00 s. Calculate the total distance traveled by the submarine.

9. The highest speed achieved by a standard nonracing sports car is

3.50 × 102 km/h. Assuming that the car accelerates at 4.00 m/s2, how

long would this car take to reach its maximum speed if it is initially at

rest? What distance would the car travel during this time?

10. Stretching 9345 km from Moscow to Vladivostok, the Trans-Siberian

railway is the longest single railroad in the world. Suppose the train is ap-

proaching the Moscow station at a velocity of +24.7 m/s when it begins a

constant acceleration of –0.850 m/s2. This acceleration continues for 28 s.

What will be the train’s final velocity when it reaches the station?

11. The world’s fastest warship belongs to the United States Navy. This ves-

sel, which floats on a cushion of air, can move as fast as 1.7 × 102 km/h.

Suppose that during a training exercise the ship accelerates +2.67 m/s2,

so that after 15.0 s its displacement is +6.00 × 102 m. Calculate the

ship’s initial velocity just before the acceleration. Assume that the ship

moves in a straight line.

12. The first supersonic flight was performed by then Capt. Charles

Yeager in 1947. He flew at a speed of 3.00 × 102 m/s at an altitude of

more than 12 km, where the speed of sound in air is slightly less than

3.00 × 102 m/s. Suppose Capt. Yeager accelerated 7.20 m/s2 in 25.0 s to

reach a final speed of 3.00 × 102 m/s. What was his initial speed?

13. Peter Rosendahl rode his unicycle a distance of 1.00 × 102 m in 12.11 s.

If Rosendahl started at rest, what was the magnitude of his acceleration?

14. Suppose that Peter Rosendahl began riding the unicycle with a speed of

3.00 m/s and traveled a distance of 1.00 × 102 m in 12.11s. What would

the magnitude of Rosendahl’s acceleration be in this case?

15. In 1991, four English teenagers built an electric car that could attain a

speed 30.0 m/s. Suppose it takes 8.0 s for this car to accelerate from

18.0 m/s to 30.0 m/s. What is the magnitude of the car’s acceleration?

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NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 2EFINAL VELOCITY AFTER ANY DISPLACEMENT

P R O B L E MIn 1970, a rocket-powered car called Blue Flame achieved a maximumspeed of 1.00 ( 103 km/h (278 m/s). Suppose the magnitude of the car’sconstant acceleration is 5.56 m/s2. If the car is initially at rest, what is thedistance traveled during its acceleration?

S O L U T I O NGiven: vi = 0 m/s

vf = 278 m/s

a = 5.56 m/s2

Unknown: ∆x = ?

Choose an equation(s) or situation: Use the equation for the final velocity after

any displacement.

vf2 = vi

2 + 2a∆x

Rearrange the equation(s) to isolate the unknown(s):

∆x = vf

2

2

−a

vi2


∆x = =

Using the appropriate kinematic equation, the time of travel for Blue Flame is

found to be 50.0 s. From this value for time the distance traveled during the ac-

celeration is confirmed to be almost 7 km. Once the car reaches its maximum

speed, it travels about 16.7 km/min.

6.95 × 103 m278

m

s

2

− 0 m

s

2

(2)5.56 m

s2

1. DEFINE

2. PLAN

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3. CALCULATE

4. EVALUATE

ADDITIONAL PRACTICE

1. In 1976, Kitty Hambleton of the United States drove a rocket-engine

car to a maximum speed of 965 km/h. Suppose Kitty started at rest

and underwent a constant acceleration with a magnitude of 4.0 m/s2.

What distance would she have had to travel in order to reach the maxi-

mum speed?

2. With a cruising speed of 2.30 × 103 km/h, the French supersonic pas-

senger jet Concorde is the fastest commercial airplane. Suppose the

landing speed of the Concorde is 20.0 percent of the cruising speed. If

the plane accelerates at –5.80 m/s2, how far does it travel between the

time it lands and the time it comes to a complete stop?

Problem 2E 13

NAME ______________________________________ DATE _______________ CLASS ____________________

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3. The Boeing 747 can carry more than 560 passengers and has a maxi-

mum speed of about 9.70 × 102 km/h. After takeoff, the plane takes a

certain time to reach its maximum speed. Suppose the plane has a con-

stant acceleration with a magnitude of 4.8 m/s2. What distance does the

plane travel between the moment its speed is 50.0 percent of maximum

and the moment its maximum speed is attained?

4. The distance record for someone riding a motorcycle on its rear wheel

without stopping is more than 320 km. Suppose the rider in this un-

usual situation travels with an initial speed of 8.0 m/s before speeding

up. The rider then travels 40.0 m at a constant acceleration of

2.00 m/s2. What is the rider’s speed after the acceleration?

5. The skid marks left by the decelerating jet-powered car The Spirit of

America were 9.60 km long. If the car’s acceleration was –2.00 m/s2,

what was the car’s initial velocity?

6. The heaviest edible mushroom ever found (the so-called “chicken of

the woods”) had a mass of 45.4 kg. Suppose such a mushroom is at-

tached to a rope and pulled horizontally along a smooth stretch of

ground, so that it undergoes a constant acceleration of +0.35 m/s2. If

the mushroom is initially at rest, what will its velocity be after it has

been displaced +64 m?

7. Bengt Norberg of Sweden drove his car 44.8 km in 60.0 min. The

feature of this drive that is interesting is that he drove the car on two

side wheels.

a. Calculate the car’s average speed.

b. Suppose Norberg is moving forward at the speed calculated in

(a). He then accelerates at a rate of –2.00 m/s2. After traveling

20.0 m, the car falls on all four wheels. What is the car’s final

speed while still traveling on two wheels?

8. Starting at a certain speed, a bicyclist travels 2.00 × 102 m. Suppose the

bicyclist undergoes a constant acceleration of 1.20 m/s2. If the final

speed is 25.0 m/s, what was the bicyclist’s initial speed?

9. In 1994, Tony Lang of the United States rode his motorcycle a short dis-

tance of 4.0 × 102 m in the short interval of 11.5 s. He started from rest

and crossed the finish line with a speed of about 2.50 × 102 km/h. Find

the magnitude of Lang’s acceleration as he traveled the 4.0 × 102 m

distance.

10. The lightest car in the world was built in London and had a mass of

less than 10 kg. Its maximum speed was 25.0 km/h. Suppose the driver

of this vehicle applies the brakes while the car is moving at its maxi-

mum speed. The car stops after traveling 16.0 m. Calculate the car’s

acceleration.


NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 2FFALLING OBJECT

P R O B L E MThe famous Gateway to the West Arch in St. Louis, Missouri, is about 192 m tall at its highest point. Suppose Sally, a stuntwoman, jumps off thetop of the arch. If it takes Sally 6.4 s to land on the safety pad at the baseof the arch, what is her average acceleration? What is her final velocity?

S O L U T I O NGiven: vi = 0 m/s

∆y = –192 m

∆t = 6.4 s

Unknown: a = ?

vf = ?

Choose an equation(s) or situation: Both the acceleration and the final speed

are unknown. Therefore, first solve for the acceleration during the fall using the

equation that requires only the known variables.

∆y = vi ∆t + 1

2a ∆t 2

Then the equation for vf that involves acceleration can be used to solve for vf .

vf = vi + a∆t


a = 2(∆y

∆–

t

v2i ∆t)

vf = vi + a∆t


=

vf = 0 m

s + –9.4

m

s2(6.4s) =

Sally’s downward acceleration is less than the free-fall acceleration at Earth’s sur-

face (9.81 m/s2). This indicates that air resistance reduces her downward acceler-

ation by 0.4 m/s2. Sally’s final speed, 60 m\s2, is such that, if she could fall at this

speed at the beginning of her jump with no acceleration, she would travel a dis-

tance equal to the arch’s height in just a little more than 3 s.

–6.0 × 101 m

s

–9.4 m

s2a = (2)(–192 m)–0

m

s(6.4s)

(6.4 s)2

1. DEFINE

2. PLAN

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3. CALCULATE

4. EVALUATE

Problem 2F 15

NAME ______________________________________ DATE _______________ CLASS ____________________

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ADDITIONAL PRACTICE

1. The John Hancock Center in Chicago is the tallest building in the United

States in which there are residential apartments. The Hancock Center is

343 m tall. Suppose a resident accidentally causes a chunk of ice to fall

from the roof. What would be the velocity of the ice as it hits the ground?

Neglect air resistance.

2. Brian Berg of Iowa built a house of cards 4.88 m tall. Suppose Berg

throws a ball from ground level with a velocity of 9.98 m/s straight up.

What is the velocity of the ball as it first passes the top of the card house?

3. The Sears Tower in Chicago is 443 m tall. Suppose a book is dropped

from the top of the building. What would be the book’s velocity at a

point 221 m above the ground? Neglect air resistance.

4. The tallest roller coaster in the world is the Desperado in Nevada. It has a

lift height of 64 m. If an archer shoots an arrow straight up in the air and

the arrow passes the top of the roller coaster 3.0 s after the arrow is shot,

what is the initial speed of the arrow?

5. The tallest Sequoia sempervirens tree in California’s Redwood National

Park is 111 m tall. Suppose an object is thrown downward from the top

of that tree with a certain initial velocity. If the object reaches the ground

in 3.80 s, what is the object’s initial velocity?

6. The Westin Stamford Hotel in Detroit is 228 m tall. If a worker on the

roof drops a sandwich, how long does it take the sandwich to hit the

ground, assuming there is no air resistance? How would air resistance af-

fect the answer?

7. A man named Bungkas climbed a palm tree in 1970 and built himself a

nest there. In 1994 he was still up there, and he had not left the tree for

24 years. Suppose Bungkas asks a villager for a newspaper, which is

thrown to him straight up with an initial speed of 12.0 m/s. When

Bungkas catches the newspaper from his nest, the newspaper’s velocity is

3.0 m/s, directed upward. From this information, find the height at

which the nest was built. Assume that the newspaper is thrown from a

height of 1.50 m above the ground.

8. Rob Colley set a record in “pole-sitting” when he spent 42 days in a bar-

rel at the top of a flagpole with a height of 43 m. Suppose a friend want-

ing to deliver an ice-cream sandwich to Colley throws the ice cream

straight up with just enough speed to reach the barrel. How long does it

take the ice-cream sandwich to reach the barrel?

9. A common flea is recorded to have jumped as high as 21 cm. Assuming

that the jump is entirely in the vertical direction and that air resistance is

insignificant, calculate the time it takes the flea to reach a height of 7.0 cm.


NAME ______________________________________ DATE _______________ CLASS ____________________

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Holt Physics

Problem 3AFINDING RESULTANT MAGNITUDE AND DIRECTION

Cheetahs are, for short distances, the fastest land animals. In the courseof a chase, cheetahs can also change direction very quickly. Suppose acheetah runs straight north for 5.0 s, quickly turns, and runs 3.00 × 102 mwest. If the magnitude of the cheetah’s resultant displacement is 3.35 ×102 m, what is the cheetah’s displacement and velocity during the firstpart of its run?

S O L U T I O NGiven: ∆t1 = 5.0 s

∆x = 3.00 s × 102 m

d = 3.35 × 102 m

Unknown: ∆y = ? vy = ?

Diagram:

1. DEFINE

2. PLAN

3. CALCULATE

P R O B L E M

∆y = ?

∆ x = 3.00 × 102 m

d = 3.35 × 102 m

N

Choose the equation(s) or situation: Use the Pythagorean theorem to subtract

one of the displacements at right angles from the total displacement, and thus

determine the unknown component of displacement.

d2 = ∆x2 + ∆y2

Use the equation relating displacement to constant velocity and time, and use the

calculated value for ∆y and the given value for ∆t to solve for v.

∆v =


∆y 2 = d 2 – ∆x2

∆y =√

d 2– ∆x 2

vy =

Substitute the values into the equation(s) and solve: Because the value for ∆y

is a displacement magnitude, only the positive root is used (∆y > 0).

∆y =√

(3.35× 102 m)2 − (3.00 × 102 m)2

=√

1.12 × 105 m2− 9.00× 104 m2

∆y

∆t

∆y

∆t

Problem 3A 17

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4. EVALUATE

=√

2.2× 104 m

=

vy = 1.5

5

×.0

10

s

2 m =

The cheetah has a top speed of 30 m/s, or 107 km/h. This is equal to about

67 miles/h.

3.0 × 101 m/s, north

1.5 × 102 m, north

ADDITIONAL PRACTICE

1. An ostrich cannot fly, but it is able to run fast. Suppose an ostrich runs

east for 7.95 s and then runs 161 m south, so that the magnitude of the

ostrich’s resultant displacement is 226 m. Calculate the magnitude of the

ostrich’s eastward component and its running speed.

2. The pronghorn antelope, found in North America, is the best long-

distance runner among mammals. It has been observed to travel at an av-

erage speed of more than 55 km/h over a distance of 6.0 km. Suppose the

antelope runs a distance of 5.0 km in a direction 11.5° north of east,

turns, and then runs 1.0 km south. Calculate the resultant displacement.

3. Kangaroos can easily jump as far 8.0 m. If a kangaroo makes five such

jumps westward, how many jumps must it make northward to have a

northwest displacement with a magnitude of 68 m? What is the angle of

the resultant displacement with respect to north?

4. In 1926, Gertrude Ederle of the United States became the first woman to

swim across the English channel. Suppose Ederle swam 25.2 km east

from the coast near Dover, England, then made a 90° turn and traveled

south for 21.3 km to a point east of Calais, France. What was Ederle’s re-

sultant displacement?

5. The emperor penguin is the best diver among birds: the record dive is

483 m. Suppose an emperor penguin dives vertically to a depth of 483 m

and then swims horizontally a distance of 225 m. What angle would the

vector of the resultant displacement make with the water’s surface? What

is the magnitude of the penguin’s resultant displacement?

6. A killer whale can swim as fast as 15 m/s. Suppose a killer whale swims in

one direction at this speed for 8.0 s, makes a 90° turn, and continues

swimming in the new direction with the same speed as before. After a cer-

tain time interval, the magnitude of the resultant displacement is 180.0 m.

Calculate the amount of time the whale swims after changing direction.

7. Woodcocks are the slowest birds: their average speed during courtship

displays can be as low as 8.00 km/h. Suppose a woodcock flies east for

15.0 min. It then turns and flies north for 22.0 min. Calculate the magni-

tude of the resultant displacement and the angle between the resultant

displacement and the woodcock’s initial displacement.

ADDITIONAL PRACTICE

1. A common flea can jump a distance of 33 cm. Suppose a flea makes five

jumps of this length in the northwest direction. If the flea’s northward

displacement is 88 cm, what is the flea’s westward displacement.


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Holt Physics

Problem 3BRESOLVING VECTORS

Certain iguanas have been observed to run as fast as 10.0 m/s. Suppose aniguana runs in a straight line at this speed for 5.00 s. The direction of mo-tion makes an angle of 30.0° to the east of north. Find the value of theiguana’s northward displacement.

S O L U T I O NGiven: v = 10.0 m/s

t = 5.00 s

q = 30.0°

Unknown: ∆y = ?

Diagram:

1. DEFINE

2. PLAN

3. CALCULATE

4. EVALUATE

P R O B L E M

Ν

30.0° d = v∆t∆yθ =

Choose the equation(s) or situation: The northern component of the vector is

equal to the vector magnitude times the cosine of the angle between the vector

and the northward direction.

∆y = d(cos q)

Use the equation relating displacement with constant velocity and time, and sub-

stitute it for d in the previous equation.

d = v∆t

∆y = v∆t(cos q)


∆y = 10.0 m

s(5.00 s)(cos 30.0°)

=

The northern component of the displacement vector is smaller than the displace-

ment itself, as expected.

43.3 m, north

Problem 3B 19

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2. The longest snake ever found was a python that was 10.0 m long. Sup-

pose a coordinate system large enough to measure the python’s length is

drawn on the ground. The snake’s tail is then placed at the origin and the

snake’s body is stretched so that it makes an angle of 60.0° with the posi-

tive x-axis. Find the x and y coordinates of the snake’s head. (Hint: The y-

coordinate is positive.)

3. A South-African sharp-nosed frog set a record for a triple jump by trav-

eling a distance of 10.3 m. Suppose the frog starts from the origin of a

coordinate system and lands at a point whose coordinate on the y-axis is

equal to −6.10 m. What angle does the vector of displacement make with

the negative y-axis? Calculate the x component of the frog.

4. The largest variety of grasshopper in the world is found in Malaysia.

These grasshoppers can measure almost a foot (0.305 m) in length and

can jump 4.5 m. Suppose one of these grasshoppers starts at the origin of

a coordinate system and makes exactly eight jumps in a straight line that

makes an angle of 35° with the positive x-axis. Find the grasshopper’s

displacements along the x- and y-axes. Assume both component dis-

placements to be positive.

5. The landing speed of the space shuttle Columbia is 347 km/h. If the shut-

tle is landing at an angle of 15.0° with respect to the horizontal, what are

the horizontal and the vertical components of its velocity?

6. In Virginia during 1994 Elmer Trett reached a speed of 372 km/h on his

motorcycle. Suppose Trett rode northwest at this speed for 8.7 s. If the

angle between east and the direction of Trett’s ride was 60.0°, what was

Trett’s displacement east? What was his displacement north?

7. The longest delivery flight ever made by a twin-engine commercial jet

took place in 1990. The plane covered a total distance of 14 890 km from

Seattle, Washington to Nairobi, Kenya in 18.5 h. Assuming that the plane

flew in a straight line between the two cities, find the magnitude of the

average velocity of the plane. Also, find the eastward and southward

components of the average velocity if the direction of the plane’s flight

was at an angle of 25.0° south of east.

8. The French bomber Mirage IV can fly over 2.3 × 103 km/h. Suppose this

plane accelerates at a rate that allows it to increase its speed from 6.0 ×102 km/h to 2.3 × 103 km/h. in a time interval of 120 s. If this accelera-

tion is upward and at an angle of 35° with the horizontal, find the accel-

eration’s horizontal and vertical components.


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P R O B L E M

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Holt Physics

Problem 3CADDING VECTORS ALGEBRAICALLY

The record for the longest nonstop closed-circuit flight by a model air-plane was set in Italy in 1986. The plane flew a total distance of 1239 km.Assume that at some point the plane traveled 1.25 × 103 m to the east, then1.25 × 103 m to the north, and finally 1.00 × 103 m to the southeast. Calcu-late the total displacement for this portion of the flight.

S O L U T I O NGiven: d1 = 1.25 × 103 m d2 = 1.25 × 103 m d3 = 1.00 × 103 m

Unknown: ∆xtot = ? ∆ytot = ? d = ? q = ?

Diagram:

1. DEFINE

2. PLAN

d3 = 1.00 × 103 m

d2 = 1.25 × 103 m

d1 = 1.25 × 103 m

∆ xtot

∆ytotdtot

45.0°

N

3. CALCULATE

Choose the equation(s) or situation: Orient the displacements with respect to

the x-axis of the coordinate system.

q1 = 0.00° q2 = 90.0° q3 = −45.0°Use this information to calculate the components of the total displacement along

the x-axis and the y-axis.

∆xtot = ∆x1 + ∆x2 + ∆x3

= d1(cos q1) + d2(cos q2) + d3(cos q3)

∆ytot = ∆y1 + ∆y2 + ∆y3

= d1(sin q1) + d2(sin q2) + d3(sin q3)

Use the components of the total displacement, the Pythagorean theorem, and the

tangent function to calculate the total displacement.

d =√

(∆xtot)2 + (∆ytot)2 q = tan−1 Substitute the values into the equation(s) and solve:

∆xtot = (1.25 × 103 m)(cos 0°) + (1.25 × 103 m)(cos 90.0°)

+(1.00 × 103 m)[cos (−45.0°)]

= 1.25 × 103 m + 7.07 × 102 m

= 1.96 × 103 m

∆ytot = (1.25 × 103 m)(sin 0°) + (1.25 × 103 m)(sin 90.0°)

+(1.00 × 103 m)[sin (−45.0°)]

= 1.25 × 103 m + 7.07 × 102 m

= 0.543 × 103 m

d =√

(1.96× 103 m)2 + (0.543× 103 m)2

∆ytot∆xtot

Problem 3C 21

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ADDITIONAL PRACTICE

1. For six weeks in 1992, Akira Matsushima, from Japan, rode a unicycle

more than 3000 mi across the United States. Suppose Matsushima is rid-

ing through a city. If he travels 250.0 m east on one street, then turns

counterclockwise through a 120.0° angle and proceeds 125.0 m north-

west along a diagonal street, what is his resultant displacement?

2. In 1976, the Lockheed SR-71A Blackbird set the record speed for any air-

plane: 3.53 × 103 km/h. Suppose you observe this plane ascending at this

speed. For 20.0 s, it flies at an angle of 15.0° above the horizontal, then

for another 10.0 s its angle of ascent is increased to 35.0°. Calculate the

plane’s total gain in altitude, its total horizontal displacement, and its re-

sultant displacement.

3. Magnor Mydland of Norway constructed a motorcycle with a wheelbase

of about 12 cm. The tiny vehicle could be ridden at a maximum speed

11.6 km/h. Suppose this motorcycle travels in the directions d1 and d2

shown in the figure below. Calculate d1 and d2, and determine how long

it takes the motorcycle to reach a net displacement of 2.0 × 102 m to the

right?

4. The fastest propeller-driven aircraft is the Russian TU-95/142, which can

reach a maximum speed of 925 km/h. For this speed, calculate the plane’s

resultant displacement if it travels east for 1.50 h, then turns 135° north-

west and travels for 2.00 h.

5. In 1952, the ocean liner United States crossed the Atlantic Ocean in less

than four days, setting the world record for commercial ocean-going ves-

sels. The average speed for the trip was 57.2 km/h. Suppose the ship

moves in a straight line eastward at this speed for 2.50 h. Then, due to a

strong local current, the ship’s course begins to deviate northward by

30.0°, and the ship follows the new course at the same speed for another

1.50 h. Find the resultant displacement for the 4.00 h period.

d =√

3.84 × 106 m2+ 2.95× 105 m2 =√

4.14 × 106 m2

d =

q = tan−1 q =

The magnitude of the total displacement is slightly larger than that of the total

displacement in the eastern direction alone.

15.5° north of east

0.543 × 103 m1.96 × 103 m

2.03 × 103 m

4. EVALUATE


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Holt Physics

Problem 3DPROJECTILES LAUNCHED HORIZONTALLY

A movie director is shooting a scene that involves dropping s stuntdummy out of an airplane and into a swimming pool. The plane is 10.0 mabove the ground, traveling at a velocity of 22.5 m/s in the positive x di-rection. The director wants to know where in the plane’s path the dummyshould be dropped so that it will land in the pool. What is the dummy’shorizontal displacement?

S O L U T I O N

Given: ∆y = −10.0 m g = 9.81 m/s2 vx = 22.5 m/s

Unknown: ∆t = ? ∆x = ?

Diagram: The initial velocity vector of the

stunt dummy only has a horizontal

component. Choose the coordinate

system oriented so that the positive y

direction points upward and the pos-

itive x direction points to the right.

Choose the equation(s) or situation: The dummy drops with no initial vertical

velocity. Because air resistance is neglected, the dummy’s horizontal velocity re-

mains constant.

∆y = − 12

g∆t2

∆x = vx∆t


= ∆t 2

∆t = where ∆y is negative

First find the time it takes for the dummy to reach the ground.

∆t = = (2

−)

9

(

.

−8

1

1

0

m

.0

/

m

s2)

= 1.43 s

Find out how far horizontally the dummy can travel during this period of time.

∆x = vx∆t = (22.5 m/s)(1.43 s)

=

The stunt dummy will have to drop from the plane when the plane is at a hori-

zontal distance of 32.2 m from the pool. The distance is within the correct order

of magnitude, given the other values in this problem.

∆x 32.2 m

2∆y−g

2∆y−g

2∆y−g

1. DEFINE

2. PLAN

3. CALCULATE

4. EVALUATE

vx,i = vx = 22.5 m/s

ay = – g

y

x

Problem 3D 23

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ADDITIONAL PRACTICE

1. Florence Griffith-Joyner of the United States set the women’s world

record for the 200 m run by running with an average speed of 9.37 m/s.

Suppose Griffith-Joyner wants to jump over a river. She runs horizontally

from the river’s higher bank at 9.37 m/s and lands on the edge of the op-

posite bank. If the difference in height between the two banks is 2.00 m,

how wide is the river?

2. The longest banana split ever made was 7.320 km long (needless to say,

more than one banana was used). If an archer were to shoot an arrow

horizontally from the top of Mount Everest, which is located 8848 m

above sea level, would the arrow’s horizontal displacement be larger than

7.32 km? Assume that the arrow cannot be shot faster than 100.0 m/s,

that there is no air resistance, and that the arrow lands at sea level.

3. The longest shot on a golf tournament was made by Mike Austin in 1974.

The ball went a distance of 471 m. Suppose the ball was shot horizontally

off a cliff at 80.0 m/s. Calculate the height of the cliff.

4. Recall Elmer Trett, who in 1994 reached a speed of 372 km/h on his mo-

torcycle. Suppose Trett drives off a horizontal ramp at this speed and

lands a horizontal distance of 40.0 m away from the edge of the ramp.

What is the height of the ramp? Neglect air resistance.

5. A Snorkel fire engine is designed for putting out fires that are well above

street level. The engine has a hydraulic lift that lifts the firefighter and a

system that delivers pressurized water to the firefighter. Suppose that the

engine cannot move closer than 25 m to a building that has a fire on its

sixth floor, which is 25 m above street level. Also assume that the water

nozzle is stuck in the horizontal position (an improbable situation). If

the horizontal speed of the water emerging from the hose is 15 m/s, how

high above the street must the firefighter be lifted in order for the water

to reach the fire?

6. The longest stuffed toy ever manufactured is a 420 m snake made by

Norwegian children. Suppose a projectile is thrown horizontally from a

height half as long as the snake and the projectile’s horizontal displace-

ment is as long as the snake. What would be the projectile’s initial speed?

7. Libyan basketball player Suleiman Nashnush was the tallest basketball

player ever. His height was 2.45 m. Suppose Nashnush throws a basket-

ball horizontally from a level equal to the top of his head. If the speed of

the basketball is 12.0 m/s when it lands, what was the ball’s initial speed?

(Hint: Consider the components of final velocity.)

8. The world’s largest flowerpot is 1.95 m high. If you were to jump hori-

zontally from the top edge of this flowerpot at a speed of 3.0 m/s, what

would your landing velocity be?


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Holt Physics

Problem 3EPROJECTILES LAUNCHED AT AN ANGLE

The narrowest strait on earth is Seil Sound in Scotland, which lies be-tween the mainland and the island of Seil. The strait is only about 6.0 mwide. Suppose an athlete wanting to jump “over the sea” leaps at an angleof 35° with respect to the horizontal. What is the minimum initial speedthat would allow the athlete to clear the gap? Neglect air resistance.

S O L U T I O NGiven: ∆x = 6.0 m

q = 35°g = 9.81 m/s

Unknown: vi = ?

Diagram:

1. DEFINE

2. PLAN

∆ x

v

= 6.00 m

θ = 35°

Choose the equation(s) or situation: The horizontal component of the athlete’s

velocity, vx, is equal to the initial speed multiplied by the cosine of the angle, q,

which is equal to the magnitude of the horizontal displacement, ∆x, divided by the

time interval required for the complete jump.

vx = vi cos q = ∆∆

x

t

At the midpoint of the jump, the vertical component of the athlete’s velocity, vy ,

which is the upward vertical component of the initial velocity, vi sin q, minus the

downward component of velocity due to free-fall acceleration, equals zero. The

time required for this to occur is half the time necessary for the total jump.

vy = vi sin q − g ∆2

t = 0

vi sin q = g

2

∆t

Rearrange the equation(s) to isolate the unknown(s): Express ∆t in the second

equation in terms of the displacement and velocity component in the first equation.

vi sin q = 2

gvi c

∆o

x

s q

vi2 =

2 sin

g

q∆x

cos q

Problem 3E 25

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1. In 1993, Wayne Brian threw a spear a record distance of 201.24 m. (This

is not an official sport record because a special device was used to “elon-

gate” Brian’s hand.) Suppose Brian threw the spear at a 35.0° angle with

respect to the horizontal. What was the initial speed of the spear?

2. April Moon set a record in flight shooting (a variety of long-distance

archery). In 1981 in Utah, she sent an arrow a horizontal distance of

9.50 × 102 m. What was the speed of the arrow at the top of the flight if the

arrow was launched at an angle of 45.0° with respect to the horizontal?

3. In 1989 during overtime in a high school basketball game in Erie, Penn-

sylvania, Chris Eddy threw a basketball a distance of 27.5 m to score and

win the game. If the shot was made at a 50.0° angle above the horizontal,

what was the initial speed of the ball?

4. In 1978, Geoff Capes of the United Kingdom won a competition for

throwing 5 lb bricks; he threw one brick a distance of 44.0 m. Suppose

the brick left Capes’ hand at an angle of 45.0° with respect to the

horizontal.

a. What was the initial speed of the brick?

b. What was the maximum height reached by the brick?

c. If Capes threw the brick straight up with the speed found in (a),

what would be the maximum height the brick could achieve?

5. In 1991, Doug Danger rode a motorcycle to jump a horizontal distance

of 76.5 m. Find the maximum height of the jump if his angle with re-

spect to the ground at the beginning of the jump was 12.0°.

6. Michael Hout of Ohio can run 110.0 meter hurdles in 18.9 s at an aver-

age speed of 5.82 m/s. What makes this interesting is that he juggles three

balls as he runs the distance. Suppose Hout throws a ball up and forward

at twice his running speed and just catches it at the same level. At what

angle, q , must the ball be thrown? (Hint: Consider horizontal displace-

ments for Hout and the ball.)

ADDITIONAL PRACTICE

vi = 2sin

g q∆ x

cosq

Substitute the values into the equation(s) and solve: Select the positive root for vi .

vi = =

By substituting the value for vi into the original equations, you can determine the

time for the jump to be completed, which is 0.92 s. From this, the height of the

jump is found to equal 1.0 m.

7.9 m

s

9.81m

s2(6.0 m)

(2)(sin 35°)(cos 35°)

4. EVALUATE

3. CALCULATE


NAME ______________________________________ DATE _______________ CLASS ____________________

7. A scared kangaroo once cleared a fence by jumping with a speed of

8.42 m/s at an angle of 55.2° with respect to the ground. If the jump

lasted 1.40 s, how high was the fence? What was the kangaroo’s horizon-

tal displacement?

8. Measurements made in 1910 indicate that the common flea is an impres-

sive jumper, given its size. Assume that a flea’s initial speed is 2.2 m/s, and

that it leaps at an angle of 21° with respect to the horizontal. If the jump

lasts 0.16 s, what is the magnitude of the flea’s horizontal displacement?

How high does the flea jump?

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Problem 3A 27

NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 3FRELATIVE VELOCITY

P R O B L E MThe world’s fastest current is in Slingsby Channel, Canada, where thespeed of the water reaches 30.0 km/h. Suppose a motorboat crosses thechannel perpendicular to the bank at a speed of 18.0 km/h relative to thebank. Find the velocity of the motorboat relative to the water.

S O L U T I O NGiven: vwb = 30.0 km/h along the channel (velocity of the water, w,

with respect to the bank, b)

vmb = 18.0 km/h perpendicular to the channel (velocity of the

motorboat, m, with respect to the bank, b)

Unknown: vmw = ?

Diagram:

1. DEFINE

2. PLAN

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vwb

vmbvmw

Choose the equation(s) or situation: From the vector diagram, the resultant

vector (the velocity of the motorboat with respect to the bank, vmb) is equal to

the vector sum of the other two vectors, one of which is the unknown.

vmw = vmb + vwb

Use the Pythagorean theorem to calculate the magnitude of the resultant velocity,

and use the tangent function to find the direction. Note that because the vectors

vmb and vwb are perpendicular to each other, the product that results from mul-

tiplying one by the other is zero. The tangent of the angle between vmb and vmw

is equal to the ratio of the magnitude of vwb to the magnitude of vmb.

vmw2 = vmb

2 + vwb2

tanq = v

v

m

wb

b


vmw =√

vmb2 + vwb

2

q = tan−1 v

v

m

wb

b

Substitute the values into the equation(s) and solve: Choose the positive root

for vmw.

vmw = 18.0k

h

m2

+ 30.0 kh

m

2

= 35.0 k

h

m

3. CALCULATE


NAME ______________________________________ DATE _______________ CLASS ____________________

The angle between vmb and vmw is as follows:

q = tan−1 =

The motorboat must move in a direction 59° with respect to vmb and against the

current, and with a speed of 35.0 km/h in order to move 18.0 km/h perpendicu-

lar to the bank.

59.0° away from the oncoming current

30.0 k

h

m

18.0

k

h

m

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4. EVALUATE

ADDITIONAL PRACTICE

1. In 1933, a storm occurring in the Pacific Ocean moved with speeds

reaching a maximum of 126 km/h. Suppose a storm is moving north at

this speed. If a gull flies east through the storm with a speed of 40.0 km/h

relative to the air, what is the velocity of the gull relative to Earth?

2. George V Coast in Antarctica is the windiest place on Earth. Wind speeds

there can reach 3.00 × 102 km/h. If a research plane flies against the wind

with a speed of 4.50 × 102 km/h relative to the wind, how long does it

take the plane to fly between two research stations that are 250 km apart?

3. Turtles are fairly slow on the ground, but they are very good swimmers,

as indicated by the reported speed of 9.0 m/s for the leatherback turtle.

Suppose a leatherback turtle swims across a river at 9.0 m/s relative to

the water. If the current in the river is 3.0 m/s and it moves at a right

angle to the turtle’s motion, what is the turtle’s displacement with respect

to the river’s bank after 1.0 min?

4. California sea lions can swim as fast as 40.0 km/h. Suppose a sea lion be-

gins to chase a fish at this speed when the fish is 60.0 m away. The fish, of

course, does not wait, and swims away at a speed 16.0 km/h. How long

would it take the sea lion to catch the fish?

5. The spur-wing goose is one of the fastest birds in the world when it

comes to level flying: it can reach a speed of 90.0 km/h. Suppose two

spur-wing geese are separated by an unknown distance and start flying

toward each other at their maximum speeds. The geese pass each other

40.0 s later. Calculate the initial distance between the geese.

6. The fastest snake on Earth is the black mamba, which can move over a

short distance at 18.0 km/h. Suppose a mamba moves at this speed to-

ward a rat sitting 12.0 m away. The rat immediately begins to run away at

33.3 percent of the mamba’s speed. If the rat jumps into a hole just be-

fore the mamba can catch it, determine the length of time that the chase

lasts.

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Problem 4A 29

NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 4ANET EXTERNAL FORCE

P R O B L E MThe muscle responsible for closing the mouth is the strongest muscle in thehuman body. It can exert a force greater than that exerted by a man lifting amass of 400 kg. Richard Hoffman of Florida recorded the force of biting at4.33 103 N. If each force shown in the diagram below has a magnitudeequal to the force of Hoffman’s bite, determine the net force.

S O L U T I O NDefine the problem, and identify the variables.

Given: F1 = −4.33 × 103 N

F2 = 4.33 × 103 N

F3 = 4.33 × 103 N

Unknown: Fnet = ? qnet = ?Diagram:

q = –60.0°

F1 = –4.33 × 103 N

F2 = 4.33 × 103 N

F3 = 4.33 × 103 N

Select a coordinate system, and apply it to the free-body diagram: Let F1 lie

along the negative y-axis and F2 lie along the positive x-axis. Now F3 must be re-

solved into x and y components.

Find the x and y components of all vectors: As indicated in the sketch, the angle

between F3 and the x-axis is 60.0°. Because this angle is in the quadrant bounded

by the positive x and negative y axes, it has a negative value.

F3,x = F3(cos q) = (4.33 × 103 N) [cos (−60.0°)] = 2.16 × 103 N

F3,y = F3(sin q) = (4.33 × 103 N) [sin (−60.0°)] = −3.75 × 103 N

Find the net external force in both the x and y directions.

For the x direction: ΣFx = F2 + F3,x = Fx,net

ΣFx = 4.33 × 103 N + 2.16 × 103 N = 6.49 × 103 N

For the y direction: ΣFy = F1 + F3,y = Fy,net

ΣFy = (−4.33 × 103 N) + (−3.75 × 103 N) = −8.08 × 103 N

Find the net external force.

Use the Pythagorean theorem to calculate Fnet . Use qnet = tan−1 FF

x

y,

,

n

n

e

e

t

t to find

the angle between the net force and the x-axis.

Fnet =√

(Fx,net)2+ (Fy,net)Fnet =

√(6.49× 103 N)2 + (−8.08 × 103 N)2 =

√10.74× 107 N2

Fnet = 1.036 × 104 N

1. DEFINE

2. PLAN

3. CALCULATE


NAME ______________________________________ DATE _______________ CLASS ____________________

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ADDITIONAL PRACTICE

1. Joe Ponder, from North Carolina, once used his teeth to lift a pumpkin

with a mass of 275 kg. Suppose Ponder has a mass of 75 kg, and he

stands with each foot on a platform and lifts the pumpkin with an at-

tached rope. If he holds the pumpkin above the ground between the

platforms, what is the force exerted on his feet? (Draw a free-body dia-

gram showing all of the forces present on Ponder.)

2. In 1994, Vladimir Kurlovich, from Belarus, set the record as the world’s

strongest weightlifter. He did this by lifting and holding above his head

a barbell whose mass was 253 kg. Kurlovich’s mass at the time was

roughly 133 kg. Draw a free-body diagram showing the various forces

in the problem. Calculate the normal force exerted on each of

Kurlovich’s feet during the time he was holding the barbell.

3. The net force exerted by a woodpecker’s head when its beak strikes a

tree can be as large as 4.90 N, assuming that the bird’s head has a mass

of 50.0 g. Assume that two different muscles pull the woodpecker’s head

forward and downward, exerting a net force of 4.90 N. If the forces ex-

erted by the muscles are at right angles to each other and the muscle

that pulls the woodpecker’s head downward exerts a force of 1.70 N,

what is the magnitude of the force exerted by the other muscle? Draw a

free-body diagram showing the forces acting on the woodpecker’s head.

4. About 50 years ago, the San Diego Zoo, in California, had the largest go-

rilla on Earth: its mass was about 3.10 × 102 kg. Suppose a gorilla with this

mass hangs from two vines, each of which makes an angle of 30.0° with

the vertical. Draw a free-body diagram showing the various forces, and

find the magnitude of the force of tension in each vine. What would hap-

pen to the tensions if the upper ends of the vines were farther apart?

5. The mass of Zorba, a mastiff born in London, England, was measured

in 1989 to be 155 kg. This mass is roughly the equivalent of the com-

bined masses of two average adult male mastiffs. Suppose Zorba is

placed in a harness that is suspended from the ceiling by two cables that

are at right angles to each other. If the tension in one cable is twice as

large as the tension in the other cable, what are the magnitudes of the

two tensions? Assume the mass of the cables and harness to be negligi-

ble. Before doing the calculations, draw a free-body diagram showing

the forces acting on Zorba.

qnet = tan−1 −68.4.098××

1

1

0

03

3

N

N =

The net force is larger than the individual forces, but it is not quite three times as

large as any one force, which would be the case if all three forces were acting in

one direction only. The angle is negative to indicate that it is in the quadrant

below the positive x-axis, where the values along the y-axis are negative. The net

force is 1.036 × 104 N at an angle of 51.2° below the positive x-axis.

−51.2°

4. EVALUATE

Problem 4B 31

NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 4BNEWTON’S SECOND LAW

P R O B L E M

A 1.5 kg ball has an acceleration of 9.0 m/s2 to the left. What is the netforce acting on the ball?

S O L U T I O NGiven: m = 1.5 kg

a = 9.0 m/s2 to the left

Unknown: F = ?

Use Newton’s second law, and solve for F.

ΣF = ma

Because there is only one force,

ΣF = F

F = (1.5 kg)(9.0 m/s2) = 14 N

F = 14 N to the left

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ADDITIONAL PRACTICE

1. David Purley, a racing driver, survived deceleration from 173 km/h to

0 km/h over a distance of 0.660 m when his car crashed. Assume that

Purley’s mass is 70.0 kg. What is the average force acting on him during

the crash? Compare this force to Purley’s weight. (Hint: Calculate the

average acceleration first.)

2. A giant crane in Washington, D. C. was tested by lifting a 2.232 × 106 kg

load.

a. Find the magnitude of the force needed to lift the load with a net

acceleration of 0 m/s2.

b. If the same force is applied to pull the load up a smooth slope

that makes a 30.0° angle with the horizontal, what would be the

acceleration?

3. When the click beetle jumps in the air, its acceleration upward can be as

large as 400.0 times the acceleration due to gravity. (An acceleration

this large would instantly kill any human being.) For a beetle whose

mass is 40.00 mg, calculate the magnitude of the force exerted by the

beetle on the ground at the beginning of the jump with gravity taken

into account. Calculate the magnitude of the force with gravity ne-

glected. Use 9.807 m/s2 as the value for free-fall acceleration.


NAME ______________________________________ DATE _______________ CLASS ____________________

4. In 1994, a Bulgarian athlete named Minchev lifted a mass of 157.5 kg.

By comparison, his own mass was only 54.0 kg. Calculate the force act-

ing on each of his feet at the moment he was lifting the mass with an

upward acceleration of 1.00 m/s2. Assume that the downward force on

each foot is the same.

5. In 1967, one of the high school football teams in California had a tackle

named Bob whose mass was 2.20 × 102 kg. Suppose that after winning a

game the happy teammates throw Bob up in the air but fail to catch

him. When Bob hits the ground, his average upward acceleration over

the course of the collision is 75.0 m/s2. (Note that this acceleration has a

much greater magnitude than free-fall acceleration.) Find the average

force that the ground exerts on Bob during the collision.

6. The whale shark is the largest type of fish in the world. Its mass can be

as large as 2.00 × 104 kg, which is the equivalent mass of three average

adult elephants. Suppose a crane lifts a net with a 2.00 × 104 kg whale

shark off the ground. The net is steadily accelerated from rest over an

interval of 2.5 s until the net reaches a speed of 1.0 m/s. Calculate the

magnitude of the tension in the cable pulling the net.

7. The largest toad ever caught had a mass of 2.65 kg. Suppose a toad with

this mass is placed on a metal plate that is attached to two cables, as

shown in the figure below. If the plate is pulled upward so that it has a

net acceleration of 2.55 m/s2, what is magnitude of the tension in the

cables? (The plate’s weight can be disregarded.)

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q1 = 45°

Fg = 26 N

FT,1 FT, 2q

2 = 45°

8. In 1991, a lobster with a mass of 20.0 kg was caught off the coast of

Nova Scotia, Canada. Imagine this lobster involved in a friendly tug of

war with several smaller lobsters on a horizontal plane at the bottom of

the sea. Suppose the smaller lobsters are able to drag the large lobster,

so that after the large lobster has been moved 1.55 m its speed is

0.550 m/s. If the lobster is initially at rest, what is the magnitude of the

net force applied to it by the smaller lobsters? Assume that friction and

resistance due to moving through water are negligible.

9. A 0.5 mm wire made of carbon and manganese can just barely support

the weight of a 70.0 kg person. Suppose this wire is used to lift a 45.0 kg

load. What maximum upward acceleration can be achieved without

breaking the wire?

Problem 4B 33

NAME ______________________________________ DATE _______________ CLASS ____________________

10. The largest hydraulic turbines in the world have shafts with individual

masses that equal 3.18 × 105 kg. Suppose such a shaft is delivered to the

assembly line on a trailer that is pulled with a horizontal force of

81.0 kN. If the force of friction opposing the motion is 62.0 kN, what is

the magnitude of the trailer’s net acceleration? (Disregard the mass of

the trailer.)

11. An average newborn blue whale has a mass of 3.00 × 103 kg. Suppose

the whale becomes stranded on the shore and a team of rescuers tries

to pull it back to sea. The rescuers attach a cable to the whale and pull it

at an angle of 20.0° above the horizontal with a force of 4.00 kN. There

is, however, a horizontal force opposing the motion that is 12.0 percent

of the whale’s weight. Calculate the magnitude of the whale’s net accel-

eration during the rescue pull.

12. One end of the cable of an elevator is attached to the elevator car, and

the other end of the cable is attached to a counterweight. The counter-

weight consists of heavy metal blocks with a total mass almost the same

as the car’s. By using the counterweight, the motor used to lift and

lower the car needs to exert a force that is only about equal to the total

weight of the passengers in the car. Suppose the car with passengers has

a mass of 1.600 × 103 kg and the counterweight has a mass of 1.200 ×103 kg. Calculate the magnitude of the car’s net acceleration as it falls

from rest at the top of the shaft to the ground 25.0 m below. Calculate

the car’s final speed.

13. The largest squash ever grown had a mass of 409 kg. Suppose you want

to push a squash with this mass up a smooth ramp that is 6.00 m long

and that makes a 30.0° angle with the horizontal. If you push the

squash with a force of 2080 N up the incline, what is

a. the net force exerted on the squash?

b. the net acceleration of the squash?

c. the time required for the squash to reach the top of the ramp?

14. A very thin boron rod with a cross-section of 0.10 mm × 0.10 mm can

sustain a force of 57 N. Assume the rod is used to pull a block along a

smooth horizontal surface.

a. If the maximum force accelerates the block by 0.25 m/s2, find the

mass of the block.

b. If a second force of 24 N is applied in the direction opposite the

57 N force, what would be the magnitude of the block’s new

acceleration?

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NAME ______________________________________ DATE _______________ CLASS ____________________

15. A hot-air balloon with a total mass of 2.55 × 103 kg is being pulled

down by a crew tugging on a rope. The tension in the rope is 7.56 ×103 N at an angle of 72.3° below the horizontal. This force is aided in

the vertical direction by the balloon’s weight and is opposed by a buoy-

ant force of 3.10 × 104 N that lifts the balloon upward. A wind blowing

from behind the crew exerts a horizontal force of 920 N on the balloon.

a. What is the magnitude and direction of the net force?

b. Calculate the magnitude of the balloon’s net acceleration.

c. Suppose the balloon is 45.0 m above the ground when the crew be-

gins pulling it down. How far will the balloon travel horizontally by

the time it reaches the ground if the balloon is initially at rest?

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Problem 4C 35

NAME ______________________________________ DATE _______________ CLASS ____________________

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Holt Physics

Problem 4CCOEFFICIENTS OF FRICTION

P R O B L E MA 20.0 kg trunk is pushed across the floor of a moving van by a horizontalforce. If the coefficient of kinetic friction between the trunk and the flooris 0.255, what is the magnitude of the frictional force opposing the ap-plied force?


mk = 0.255

g = 9.81 m/s2

Unknown: Fk = ?

Use the equation for frictional force, substituting mg for the normal force Fn.

Fk = mkFn = mkmg

Fk = (0.255)(20.0 kg)(9.81 m/s2)

Fk = 50.0 N

ADDITIONAL PRACTICE

1. The largest flowers in the world are the Rafflesia arnoldii, found in

Malaysia. A single flower is almost a meter across and has a mass up to

11.0 kg. Suppose you cut off a single flower and drag it along the flat

ground. If the coefficient of kinetic friction between the flower and the

ground is 0.39, what is the magnitude of the frictional force that must

be overcome?

2. Robert Galstyan, from Armenia, pulled two coupled railway wagons a

distance of 7 m using his teeth. The total mass of the wagons was about

2.20 × 105 kg. Of course, his job was made easier by the fact that the

wheels were free to roll. Suppose the wheels are blocked and the coeffi-

cient of static friction between the rails and the sliding wheels is 0.220.

What would be the magnitude of the minimum force needed to move

the wagons from rest? Assume that the track is horizontal.

3. The steepest street in the world is Baldwin Street in Dunedin, New

Zealand. It has an inclination angle of 38.0° with respect to the hori-

zontal. Suppose a wooden crate with a mass of 25.0 kg is placed on

Baldwin Street. An additional force of 59 N must be applied to the crate

perpendicular to the pavement in order to hold the crate in place. If the

coefficient of static friction between the crate and the pavement is

0.599, what is the magnitude of the frictional force?


NAME ______________________________________ DATE _______________ CLASS ____________________

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4. Now imagine that a child rides a wagon down Baldwin Street. In order

to keep from moving too fast, the child has secured the wheels of the

wagon so that they do not turn. The wagon and child then slide down

the hill at a constant velocity. What is the coefficient of kinetic friction

between the tires of the wagon and the pavement?

5. The steepest railroad track that allows trains to move using their own

locomotion and the friction between their wheels and the track is lo-

cated in France. The track makes an angle of 5.2° with the horizontal.

Suppose the rails become greasy and the train slides down the track

even though the wheels are locked and held in place with blocks. If the

train slides down the tracks with a constant velocity, what is the coeffi-

cient of kinetic friction between the wheels and track?

6. Walter Arfeuille of Belgium lifted a 281.5 kg load off the ground using

his teeth. Suppose Arfeuille can hold just three times that mass on a

30.0° slope using the same force. What is the coefficient of static fric-

tion between the load and the slope?

7. A blue whale with a mass of 1.90 × 105 kg was caught in 1947. What is

the magnitude of the minimum force needed to move the whale along

a horizontal ramp if the coefficient of static friction between the ramp’s

surface and the whale is 0.460?

8. Until 1979, the world’s easiest driving test was administered in Egypt.

To pass the test, one needed only to drive about 6 m forward, stop, and

drive the same distance in reverse. Suppose that at the end of the 6 m

the car’s brakes are suddenly applied and the car slides to a stop. If the

force required to stop the car is 6.0 × 103 N and the coefficient of ki-

netic friction between the tires and pavement is 0.77, what is the mag-

nitude of the car’s normal force? What is the car’s mass?

9. The heaviest train ever pulled by a single engine was over 2 km long.

Suppose a force of 1.13 × 108 N is needed to overcome static friction in

the train’s wheels. If the coefficient of static friction is 0.741, what is the

train’s mass?

10. In 1994, a 3.00 × 103 kg pancake was cooked and flipped in Manchester,

England. If the pancake is placed on a surface that is inclined 31.0° with

respect to the horizontal, what must the coefficient of kinetic friction

be in order for the pancake to slide down the surface with a constant

velocity? What would be the magnitude of the frictional force acting on

the pancake?

Problem 4C 37

NAME ______________________________________ DATE _______________ CLASS ____________________

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Holt Physics

Problem 4DOVERCOMING FRICTION

P R O B L E MIn 1988, a very large telephone constructed by a Dutch telecommunica-tions company was demonstrated in the Netherlands. Suppose this tele-phone is towed a short distance by a horizontal force equal to 8670 N, sothat the telephone’s net acceleration is 1.30 m/s2. Given that the coeffi-cient of kinetic friction between the phone and the ground is 0.120, calcu-late the mass of the telephone.

S O L U T I O NGiven: Fapplied = 8670 N

anet = 1.30 m/s2

mk = 0.120

g = 9.81 m/s2

Unknown: m = ?

Choose the equation(s) or situation: Apply Newton’s second law to describe the

forces acting on the telephone.

Fnet = manet = Fapplied − Fk

The frictional force, Fk, depends on the normal force, Fn, exerted on the tele-

phone. For a horizontal surface, the normal force equals the telephone’s weight.

Fk = mkFn = mk(mg)

Substituting the equation for Fk into the equation for Fnet provides an equation

with all known and unknown variables.

manet = Fapplied − mkmg


manet + mkmg = Fapplied

m(anet + mkg) = Fapplied

m = an

F

e

a

t

p

+pli

med

kg

Substitute the values into the equations and solve:

m =

m =

m = =

Because the mass is constant, the sum of the acceleration terms in the denomina-

tor must be constant for a constant applied force. Therefore, the net acceleration

decreases as the coefficient of friction increases.

3.50 × 103 kg8670 N2.48 m/s2

8670 N1.30 m/s2 + 1.18 m/s2

8670 N1.30 m/s2 + (0.120)(9.81 m/s2)

1. DEFINE

2. PLAN

3. CALCULATE

4. EVALUATE


NAME ______________________________________ DATE _______________ CLASS ____________________

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1. Isaac Newton developed the laws of mechanics. Brian Newton put

those laws into application when he ran a marathon in about 8.5 h

while carrying a bag of coal. Suppose Brian Newton wants to remove

the bag from the finish line. He drags the bag with an applied horizon-

tal force of 130 N, so that the bag as a net acceleration of 1.00 m/s2.

If the coefficient of kinetic friction between the bag and the pavement

is 0.158, what is the mass of the bag of coal?

2. The most massive car ever built was the official car of the General Sec-

retary of the Communist Party in the former Soviet Union. Suppose

this car is moving down a 10.0° slope when the driver suddenly applies

the brakes. The net force acting on the car as it stops is –2.00 × 104 N. If

the coefficient of kinetic friction between the car’s tires and the pave-

ment is 0.797, what is the car’s mass? What is the magnitude of the

normal force that the pavement exerts on the car?

3. Suppose a giant hamburger slides down a ramp that has a 45.0° incline.

The coefficient of kinetic friction between the hamburger and the

ramp is 0.597, so that the net force acting on the hamburger is 6.99 ×103 N. What is the mass of the hamburger? What is the magnitude of

the normal force that the ramp exerts on the hamburger?

4. An extremely light, drivable car with a mass of only 9.50 kg was built in

London. Suppose that the wheels of the car are locked, so that the car

no longer rolls. If the car is pushed up a 30.0° slope by an applied force

of 80.0 N, the net acceleration of the car is 1.64 m/s2. What is the coef-

ficient of kinetic friction between the car and the incline?

5. Cleopatra’s Needle, an obelisk given by the Egyptian government to

Great Britain in the nineteenth century, is 20+ m tall and has a mass of

about 1.89 × 105 kg. Suppose the monument is lowered onto its side

and dragged horizontally to a new location. An applied force of

7.6 × 105 N is exerted on the monument, so that its net acceleration is

0.11 m/s2. What is the magnitude of the frictional force?

6. Snowfall is extremely rare in Dunedin, New Zealand. Nevertheless,

suppose that Baldwin Street, which has an incline of 38.0°, is covered

with snow and that children are sledding down the street. A sled and

rider move downhill with a constant acceleration. What would be the

magnitude of the sled’s net acceleration if the coefficient of kinetic fric-

tion between the snow and the sled’s runners is 0.100? Does the accel-

eration depend on the masses of the sled and rider?

7. The record speed for grass skiing was set in 1985 by Klaus Spinka, of

Austria. Suppose it took Spinka 6.60 s to reach his top speed after he

started from rest down a slope with a 34.0° incline. If the coefficient of

kinetic friction between the skis and the grass was 0.198, what was the

magnitude of Spinka’s net acceleration? What was his speed after 6.60 s?

ADDITIONAL PRACTICE

Problem 5A 39

NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 5AWORK AND ENERGY

P R O B L E MThe largest palace in the world is the Imperial Palace in Beijing, China.Suppose you were to push a lawn mower around the perimeter of a rec-tangular area identical to that of the palace, applying a constant horizon-tal force of 60.0 N. If you did 2.05 × 105 J of work, how far would you havepushed the lawn mower? If the Imperial Palace is 9.60 × 102 m long, howwide is it?

S O L U T I O NGiven: F = 60.0 N

W = 2.05 × 105 J

x = 9.60 × 102 m

Unknown: d = ?

y = ?

Use the equation for net work done by a constant force.

W = Fd(cos q)

To calculate the width, y, recall that the perimeter of an area equals the sum of

twice its width and twice its length.

d = 2x + 2y

Rearrange the equations to solve for d and y. Note that the force is applied in the

direction of the displacement, so q = 0°.

d = F(c

W

os q) =

(60

2

.

.

0

05

N

×)(

1

c

0

o

5

s

J

0°)

d =

y = d −

2

2x =

y = = 1.50 ×

2

103 m

y = 7.50 × 102 m

3.42 × 103m − 1.92 × 103 m

2

3.42 × 103 m − (2)(9.60 × 102 m)

2

3.42 × 103 m

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ADDITIONAL PRACTICE

1. Lake Point Tower in Chicago is the tallest apartment building in the

United States (although not the tallest building in which there are

apartments). Suppose you take the elevator from street level to the roof

of the building. The elevator moves almost the entire distance at con-

stant speed, so that it does 1.15 × 105 J of work on you as it lifts the en-


NAME ______________________________________ DATE _______________ CLASS ____________________

tire distance. If your mass is 60.0 kg, how tall is the building?

Ignore the effects of friction.

2. In 1985 in San Antonio, Texas, an entire hotel building was moved several

blocks on 36 dollies. The mass of the building was about 1.45 × 106 kg. If

1.00 × 102 MJ of work was done to overcome the force of resistance that

was just 2.00 percent of the building’s weight, how far was the building

moved?

3. A hummingbird has a mass of about 1.7 g. Suppose a hummingbird

does 0.15 J of work against gravity, so that it ascends straight up with a

net acceleration of 1.2 m/s2. How far up does it move?

4. In 1453, during the siege of Constantinople, the Turks used a cannon

capable of launching a stone cannonball with a mass of 5.40 × 102 kg.

Suppose a soldier dropped a cannonball with this mass while trying to

load it into the cannon. The cannonball rolled down a hill that made an

angle of 30.0° with the horizontal. If 5.30 × 104 J of work was done by

gravity on the cannonball as it rolled down a hill, how far did it roll?

5. The largest turtle ever caught in the United States had a mass of over

800 kg. Suppose this turtle were raised 5.45 m onto the deck of a re-

search ship. If it takes 4.60 × 104 J of work to lift the turtle this distance

at a constant velocity, what is the turtle’s weight?

6. During World War II, 16 huge wooden hangers were built for United

States Navy airships. The hangars were over 300 m long and had a maxi-

mum height of 52.0 m. Imagine a 40.0 kg block being lifted by a winch

from the ground to the top of the hangar’s ceiling. If the winch does 2.08

× 104 J of work in lifting the block, what force is exerted on the block?

7. The Warszawa Radio mast in Warsaw, Poland, is 646 m tall, making it

the tallest human-built structure. Suppose a worker raises some tools

to the top of the tower by means of a small elevator. If 2.15 × 105 J of

work is done in lifting the tools, what is the force exerted on them?

8. The largest mincemeat pie ever created had a mass of 1.02 × 103 kg.

Suppose that a pie with this mass slides down a ramp that is 18.0 m

long and is inclined to the ground by 10.0°. If the coefficient of kinetic

friction is 0.13, what is the net work done on the pie during its descent?

9. The longest shish kebab ever made was 881.0 m long. Suppose the meat

and vegetables need to be delivered in a cart from one end of this shish

kebab’s skewer to the other end. A cook pulls the cart by applying a

force of 40.00 N at an angle of 45.00° above the horizontal. If the force

of friction acting on the cart is 28.00 N, what is the net work done on

the cart and its contents during the delivery?

10. The world’s largest chandelier was created by a company in South

Korea and hangs in one of the department stores in Seoul, South Korea.

The chandelier’s mass is about 9.7 × 103 kg. Consider a situation in

which this chandelier is placed in a wooden crate whose mass is negli-

gible. The chandelier is then pulled along a smooth horizontal surface

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Problem 5A 41

NAME ______________________________________ DATE _______________ CLASS ____________________

by two forces that are parallel to the smooth surface, are at right angles

to each other, and are applied 45° to either side of the direction in

which the chandelier is moving. If each of these forces is 1.2 × 103 N,

how much work must be done on the chandelier to pull it 12 m?

11. The world’s largest flag, which was manufactured in Pennsylvania, has

a length of 154 m and a width of 78 m. The flag’s mass is 1.24 × 103 kg,

which may explain why the flag has never been flown from a flagpole.

Suppose this flag is being pulled by two forces: a force of 8.00 × 103 N

to the east and a force of 5.00 × 103 N that is directed 30.0° south of

east. How much work is done in moving the flag 20.0 m directly south?

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NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 5BKINETIC ENERGY

P R O B L E MSilvana Cruciata from Italy set a record in one-hour running by running18.084 km in 1.000 h. If Cruciata’s kinetic energy was 694 J, what was hermass?

S O L U T I O NGiven: ∆x = 18.084 km = 1.8084 × 104 m

∆t = 1.000 h = 3.600 × 103 s

KE = 694 J

Unknown: m = ?

Choose the equation(s) or situation: Use the definition of average velocity to

calculate Cruciata’s speed.

vavg = ∆∆

x

t

Use the equation for kinetic energy, using vavg for the velocity term, to solve for

m.

KE = 1

2 m vavg

2

Rearrange the equation(s) to isolate the unknown(s): Substitute the average

velocity equation into the equation for kinetic energy and solve for m.

m = v

2

a

K

vg

E2 = =

2K

∆E

x

∆2t 2


m = =

If the average speed is rounded to 5.0 m/s, and the kinetic energy is rounded to

700 J, the estimated mass is 56 kg, which is close to the calculated value.

55.0 kg(2)(694 J)(3.600 × 103 s)2

(1.8084 × 104 m)2

2KE

∆∆

x

t2

1. DEFINE

2. PLAN

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3. CALCULATE

4. EVALUATE

ADDITIONAL PRACTICE

1. In 1994, Leroy Burrell of the United States set what was then a new world

record for the men’s 100 m run. He ran the 1.00 × 102 m distance in 9.85 s.

Assuming that he ran with a constant speed equal to his average speed, and

his kinetic energy was 3.40 × 103 J, what was Burrell’s mass?

2. The fastest helicopter, the Westland Lynx, has a top speed of 4.00 ×102 km/h. If its kinetic energy at this speed is 2.10 × 107 J, what is the

helicopter’s mass?

Problem 5B 43

NAME ______________________________________ DATE _______________ CLASS ____________________

3. Dan Jansen of the United States won a speed-skating competition at the

1994 Winter Olympics in Lillehammer, Norway. He did this by skating

500 m with an average speed of 50.3 km/h. If his kinetic energy was

6.54 × 103 J, what was his mass?

4. In 1987, the fastest auto race in the United States was the Busch Clash in

Daytona, Florida. That year, the winner’s average speed was about

318 km/h. Suppose the kinetic energy of the winning car was 3.80 MJ.

What was the mass of the car and its driver?

5. In 1995, Karine Dubouchet of France reached a record speed in downhill

skiing. If Dubouchet’s mass was 51.0 kg, her kinetic energy would have

been 9.96 × 104 J. What was her speed?

6. Susie Maroney from Australia set a women’s record in long-distance

swimming by swimming 93.625 km in 24.00 h.

a. What was Maroney’s average speed?

b. If Maroney’s mass was 55 kg, what was her kinetic energy?

7. The brightest, hottest, and most massive stars are the brilliant blue stars

designated as spectral class O. If a class O star with a mass of 3.38 × 1031 kg

has a kinetic energy of 1.10 × 1042 J, what is its speed? Express your answer

in km/s (a typical unit for describing the speed of stars).

8. The male polar bear is the largest land-going predator. Its height when

standing on its hind legs is over 3 m and its mass, which is usually

around 500 kg, can be as large as 680 kg. In spite of this bulk, a running

polar bear can reach speeds of 56.0 km/h.

a. Determine the kinetic energy of a running polar bear, using the

maximum values for its mass and speed.

b. What is the ratio of the polar bear’s kinetic energy to the kinetic

energy of Leroy Burrell, as given in item 1?

9. Escape speed is the speed required for an object to leave Earth’s orbit. It

is also the minimum speed an incoming object must have to avoid being

captured and pulled into an orbit around Earth. The escape speed for a

projectile launched from Earth’s surface is 11.2 km/s. Suppose a meteor

is pulled toward Earth’s surface and, as a meteorite, strikes the ground

with a speed equal to this escape speed. If the meteorite has a diameter of

about 3 m and a mass of 2.3 × 105 kg, what is its kinetic energy at the

instant it collides with Earth’s surface?

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NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 5CWORK-KINETIC ENERGY THEOREM

P R O B L E MThe Great Pyramid of Khufu in Egypt, used to have a height of 147 m andsides that sloped at an angle of 52.0° with respect to the ground. Stoneblocks with masses of 1.37 × 104 kg were used to construct the pyramid.Suppose that a block with this mass at rest on top of the pyramid beginsto slide down the side. Calculate the block’s kinetic energy at ground levelif the coefficient of kinetic friction is 0.45.

S O L U T I O NGiven: m = 1.37 × 104 kg

h = 147 m

g = 9.81 m/s2

q = 52.0°mk = 0.45

vi = 0 m/s

Unknown: KEf = ?

Choose the equation(s) or situation: The net work done by the block as it slides

down the side of the pyramid can be expressed by using the definition of work in

terms of net force. Because the net force is parallel to the displacement, the net

work is simply the net force multiplied by the displacement. It can also be ex-

pressed in terms of changing kinetic energy by using the work–kinetic energy

theorem.

Wnet = Fnetd

Wnet = ∆KE

The net force on the block equals the difference between the component of the

force due to free-fall acceleration along the side of the pyramid and the frictional

force resisting the downward motion of the block.

Fnet = mg(sin q) − Fk = mg (sin q) − mkmg (cos q)

The distance the block travels along the side of the pyramid equals the height of

the pyramid divided by the sine of the angle of the side’s slope.

h = d(sin q)

d = sin

h

q

Because the block is initially at rest, its initial kinetic energy is zero, and the

change in kinetic energy equals the final kinetic energy.

∆KE = KEf − KEi = KEf

1. DEFINE

2. PLAN

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Problem 5C 45

NAME ______________________________________ DATE _______________ CLASS ____________________

Rearrange the equation(s) to isolate the unknown(s): Combining these equa-

tions yields the following expression for the final kinetic energy.

KEf = Fnetd = mg (sin q − mk cos q)sin

h

q

KEf = mgh1 − ta

mn

k

q


KEf = (1.37 × 104 kg)(9.81 m/s2)(147 m)1 − tan

0.

5

4

2

5

.0°

KEf = (1.37 × 104 kg)(9.81 m/s2)(147 m)(1.00 − 0.35)

KEf = (1.37 × 104 kg)(9.81 m/s2)(147 m)(0.65)

KEf =

Note that the net force, and thus the final kinetic energy, is about two-thirds of

what it would be if the side of the pyramid were frictionless.

1.3 × 107 J

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3. CALCULATE

4. EVALUATE

ADDITIONAL PRACTICE

1. The tops of the towers of the Golden Gate Bridge, in San Francisco, are

227 m above the water. Suppose a worker drops a 655 g wrench from the

top of a tower. If the average force of air resistance is 2.20 percent of the

force of free fall, what will the kinetic energy of the wrench be when it

hits the water?

2. Bonny Blair of the United States set a world record in speed skating when

she skated 5.00 × 102 m with an average speed of 12.92 m/s. Suppose

Blair crossed the finish line at this speed and then skated to a stop. If the

work done by friction over a certain distance was −2830 J, what would

Blair’s kinetic energy be, assuming her mass to be 55.0 kg.

3. The CN Tower in Toronto, Canada, is 553 m tall, making it the tallest

free-standing structure in the world. Suppose a chunk of ice with a mass

of 25.0 g falls from the top of the tower. The speed of the ice is 30.0 m/s

as it passes the restaurant, which is located 353 m above the ground.

What is the magnitude of the average force due to air resistance?

4. In 1979, Dr. Hans Liebold of Germany drove a race car 12.6 km with an

average speed of 404 km/h. Suppose Liebold applied the brakes to reduce

his speed. What was the car’s final speed if −3.00 MJ of work was done by

the brakes? Assume the combined mass of the car and driver to be 1.00 ×103 kg.

5. The summit of Mount Everest is 8848.0 m above sea level, making it the

highest summit on Earth. In 1953, Edmund Hillary was the first person

to reach the summit. Suppose upon reaching there, Hillary slid a rock

with a 45.0 g mass down the side of the mountain. If the rock’s speed is


NAME ______________________________________ DATE _______________ CLASS ____________________

27.0 m/s when it is 8806.0 m above sea level, how much work was done

on the rock by air resistance?

6. In 1990, Roger Hickey of California reached a speed 35.0 m/s on his

skateboard. Suppose it took 21 kJ of work for Roger to reach this speed

from a speed of 25.0 m/s. Calculate Hickey’s mass.

7. At the 1984 Winter Olympics, William Johnson of the United States

reached a speed of 104.5 km/h in the downhill skiing competition. Sup-

pose Johnson left the slope at that speed and then slid freely along a hori-

zontal surface. If the coefficient of kinetic friction between the skis and

the snow was 0.120 and his final speed was half of his initial speed, find

the distance William traveled.

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Problem 5D 47

NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 5DPOTENTIAL ENERGY

P R O B L E MIn 1993, Javier Sotomayor from Cuba set a record in the high jump byclearing a vertical distance of 2.45 m. If the gravitational potential energyassociated with Sotomayor at the top point of his trajectory was 1.59 103 J, what was his mass?

S O L U T I O NGiven: h = 2.45 m

g = 9.81 m/s2

PEg = 1.59 × 103 J

Unknown: m = ?

Use the equation for gravitational potential energy, and rearrange it to solve

for m.

PEg = mgh

m = P

g

E

h

g

m = = 66.2 kg(1.59 × 103 J)

9.81 ms2s2(2.45 m)

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ADDITIONAL PRACTICE

1. In 1992, Ukrainian Sergei Bubka used a short pole to jump to a height

of 6.13 m. If the maximum potential energy associated with Bubka was

4.80 kJ at the midpoint of his jump, what was his mass?

2. Naim Suleimanoglu of Turkey has a mass of about 62 kg, yet he can lift

nearly 3 times this mass. (This feat has earned Suleimanoglu the nick-

name of “Pocket Hercules.”) If the potential energy associated with a

barbell lifted 1.70 m above the floor by Suleimanoglu is 3.04 × 103 J,

what is the barbell’s mass?

3. In 1966, a special research cannon built in Arizona shot a projectile to a

height of 180 km above Earth’s surface. The potential energy associated

with the projectile when its altitude was 10.0 percent of the maximum

height was 1.48 × 107J. What was the projectile’s mass? Assume that

constant free-fall acceleration at this altitude is the same as at sea level.

4. The highest-caliber cannon ever built (though never used) is located in

Moscow, Russia. The diameter of the cannon’s barrel is about 89 cm,

and the cannon’s mass is 3.6 × 104 kg. Suppose this cannon were lifted

by airplane. If the potential energy associated with this cannon were


NAME ______________________________________ DATE _______________ CLASS ____________________

8.88 × 108 J, what would be its height above sea level? Assume that

constant free-fall acceleration at this altitude is the same as at sea level.

5. In 1987, Stefka Kostadinova from Bulgaria set a new women’s record in

high jump. It is known that the ratio of the potential energy associated

with Kostadinova at the top of her jump to her mass was 20.482 m2/s2.

What was the height of her record jump?

6. In 1992, David Engwall of California used a slingshot to launch a dart

with a mass of 62 g. The dart traveled a horizontal distance of 477 m.

Suppose the slingshot had a spring constant of 3.0 × 104 N/m. If the

elastic potential energy stored in the slingshot just before the dart was

launched was 1.4 × 102 J, how far was the slingshot stretched?

7. Suppose a 51 kg bungee jumper steps off the Royal Gorge Bridge, in

Colorado. The bridge is situated 321 m above the Arkansas River. The

bungee cord’s spring constant is 32 N/m, the cord’s relaxed length is

104 m, and its length is 179 m when the jumper stops falling. What is

the total potential energy associated with the jumper at the end of his

fall? Assume that the bungee cord has negligible mass.

8. Situated 4080 m above sea level, La Paz, Bolivia, is the highest capital in

the world. If a car with a mass of 905 kg is driven to La Paz from a loca-

tion that is 1860 m above sea level, what is the increase in potential

energy?

9. In 1872, a huge gold nugget with a mass of 286 kg was discovered in

Australia. The nugget was displayed for the public before it was melted

down to extract pure gold. Suppose this nugget is attached to the ceil-

ing by a spring with a spring constant of 9.50 × 103 N/m. The nugget is

released from a height of 1.70 m above the floor, and is caught when it

is no longer moving downward and is about to be pulled back up by

the elastic force of the spring.

a. If the spring stretches a total amount of 59.0 cm, what is the elas-

tic potential energy associated with the spring-nugget system?

b. What is the gravitational potential energy associated with the

nugget just before it is dropped?

c. What is the gravitational potential energy associated with the

nugget after the spring has stretched 59.0 cm?

d. What is the difference between the gravitational potential energy

values in parts (b) and (c)? How does this compare with your

answer for part (a)?

10. When April Moon set a record for flight shooting in 1981, the arrow

traveled a distance of 9.50 × 102 m. Suppose the arrow had a mass of

65.0 g, and that the angle at which the arrow was launched was 45.0°

above the horizontal.

a. What was the kinetic energy of the arrow at the instant it left the

bowstring? (Hint: Review Section 3E to determine the initial

speed of the arrow.)

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Problem 5D 49

NAME ______________________________________ DATE _______________ CLASS ____________________

b. If the bowstring was pulled back 55.0 cm from its relaxed posi-

tion, what was the spring constant of the bowstring? (Hint:

Assume that all of the elastic potential energy stored in the bow-

string is converted to the arrow’s initial kinetic energy.)

c. Assuming that air resistance is negligible, determine the maxi-

mum height that the arrow reaches. (Hint: Equate the arrow’s ini-

tial kinetic energy to the sum of the maximum gravitational

potential energy associated with the arrow and the arrow’s kinetic

energy at maximum height.)

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NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 5ECONSERVATION OF MECHANICAL ENERGY

P R O B L E MThe largest apple ever grown had a mass of about 1.47 kg. Suppose youhold such an apple in your hand. You accidentally drop the apple, thenmanage to catch it just before it hits the ground. If the speed of the appleat that moment is 5.42 m/s, what is the kinetic energy of the apple? Fromwhat height did you drop it?


g = 9.81 m/s2

v = 5.42 m/s

Unknown: KE = ? h = ?

Choose the equation(s) or situation: Use the equations for kinetic and

gravitational potential energy.

KE = 1

2 mv 2

PEg = mgh

The zero level for gravitational potential energy is the ground. Because the apple

ends its fall at the zero level, the final gravitational potential energy is zero.

PEg,f = 0

The initial gravitational potential energy is measured at the point from which the

apple is released.

PEg,i = mgh

The apple is initially at rest, so the initial kinetic energy is zero.

KEi = 0

The final kinetic energy is therefore:

KEf = 1

2 mv 2


PEg,i = (1.47 kg)(9.81 ms2)h

KEf = 1

2(1.47 kg)5.42

m

s2

Solving for KE yields the following result:

KE = KEf = 1

2(1.47 kg)5.42

m

s2

= 21.6 J

1. DEFINE

2. PLAN

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3. CALCULATE

Problem 5E 51

NAME ______________________________________ DATE _______________ CLASS ____________________

Now use the principle of conservation for mechanical energy and the calculated

quantity for KEf to evaluate h.

MEi = MEf

PEi + KEi = PEf + KEf

PEi + 0 J = 0 J + 21.6 J

mgh = 21.6 J

h = =

Note that the height of the apple can be determined without knowing the apple’s

mass. This is because the conservation equation reduces to the equation relating

speed, acceleration, and displacement: v2 = 2gh.

1.50 m21.6 J

(1.47 kg)9.81 ms2

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4. EVALUATE

ADDITIONAL PRACTICE

1. The largest watermelon ever grown had a mass of 118 kg. Suppose this

watermelon were exhibited on a platform 5.00 m above the ground. After

the exhibition, the watermelon is allowed to slide along to the ground

along a smooth ramp. How high above the ground is the watermelon at

the moment its kinetic energy is 4.61 kJ?

2. One species of eucalyptus tree, Eucalyptus regnens, grows to heights simi-

lar to those attained by California redwoods. Suppose a bird sitting on

top of one specimen of eucalyptus tree drops a nut. If the speed of the

falling nut at the moment it is 50.0 m above the ground is 42.7 m/s, how

tall is the tree? Do you need to know the mass of the nut to solve this

problem? Disregard air resistance.

3. In 1989, Michel Menin of France walked on a tightrope suspended under

a balloon nearly at an altitude of 3150 m above the ground. Suppose a

coin falls from Menin’s pocket during his walk. How high above the

ground is the coin when its speed is 60.0 m/s?

4. In 1936, Col. Harry Froboess of Switzerland jumped into the ocean from

the airship Graf Hindenburg, which was 1.20 × 102 m above the water’s

surface. Assuming Froboess had a mass of 72.0 kg, what was his kinetic

energy at the moment he was 30.0 m from the water’s surface? What was

his speed at that moment? Neglect the air resistance.

5. Suppose a motorcyclist rides a certain high-speed motorcycle. He reaches

top speed and then coasts up a hill. The maximum height reached by the

motorcyclist is 250.0 m. If 2.55 × 105 J of kinetic energy is dissipated by

friction, what was the initial speed of the motorcycle? The combined

mass of the motorcycle and motorcyclist is 250.0 kg.

6. The deepest mine ever drilled has a depth of 12.3 km (by contrast,

Mount Everest has height of 8.8 km). Suppose you drop a rock with a


NAME ______________________________________ DATE _______________ CLASS ____________________

mass of 120.0 g down the shaft of this mine. What would the rock’s ki-

netic energy be after falling 3.2 km? What would the potential energy as-

sociated with the rock be at that same moment? Assume no air resistance

and a constant free-fall acceleration.

7. Desperado, a roller coaster built in Nevada, has a vertical drop of 68.6 m.

The roller coaster is designed so that the speed of the cars at the end of

this drop is 35.6 m/s. Assume the cars are at rest at the start of the drop.

What percent of the initial mechanical energy is dissipated by friction?

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Problem 5F 53

NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 5FPOWER

P R O B L E MMartinus Kuiper of the Netherlands ice skated for 24 h with an averagespeed of 6.3 m/s. Suppose Kuiper’s mass was 65 kg. If Kuiper provided520 W of power to accelerate for 2.5 s, how much work did he do?

S O L U T I O NGiven: P = 520 W

∆t = 2.5 s

Unknown: W = ?

Use the equation for power and rearrange it to solve for work.

P = ∆W

t

W = P∆t = (520 W)(2.5 s) = 1300 J

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ADDITIONAL PRACTICE

1. The most powerful ice breaker in the world was built in the former So-

viet Union. The ship is almost 150 m long, and its nuclear engine gener-

ates 56 MW of power. How much work can this engine do in 1.0 h?

2. Reginald Esuke from Cameroon ran over 3 km down a mountain slope

in just 62.25 min. How much work was done if the power developed

during Esuke’s descent was 585.0 W?

3. The world’s tallest lighthouse is located in Japan and is 106 m tall. A

winch that provides 3.00 × 102 W of power is used to raise 14.0 kg of

equipment to the lighthouse top at a constant velocity. How long does it

take the equipment to reach the lighthouse top?

4. The first practical car to use a gasoline engine was built in London in

1826. The power generated by the engine was just 2984 W. How long

would this engine have to run to produce 3.60 × 104 J of work?

5. Dennis Joyce of the United States threw a boomerang and caught it at the

same location 3.0 min later. Suppose Joyce decided to work out while

waiting for the boomerang to return. If he expended 54 kJ of work, what

was his average power output during the workout?

6. In 1984, Don Cain threw a flying disk that stayed aloft for 16.7 s. Suppose

Cain ran up a staircase during this time, reaching a height of 18.4 m. If

his mass was 72.0 kg, how much power was needed for Cain’s ascent.


NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 6AMOMENTUM

P R O B L E MThe world’s most massive train ran in South Africa in 1989. Over 7 kmlong, the train traveled 861.0 km in 22.67 h. Imagine that the distance wastraveled in a straight line north. If the train’s average momentum was7.32 108 kg•m/s to the north, what was its mass?

S O L U T I O NGiven: ∆x = 861.0 km to the north

∆t = 22.67 h

pavg = 7.32 × 108 kg

s

•m to the north

Unknown: vavg = ? m = ?

Use the definition of average velocity to calculate vavg, and then substitute this

value for velocity in the definition of momentum to solve for mass.

vavg = ∆∆

x

t =

(2

(

2

8

.6

6

7

1.

h

0

)

×(3

1

6

0

0

3

0

m

s/

)

h) = 10.55

m

s to the north

pavg = mvavg

m = p

va

a

v

v

g

g = = 6.94 × 107 kg7.32 × 108

kg

s

•m

10.55 m

s

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ADDITIONAL PRACTICE

1. In 1987, Marisa Canofoglia, of Italy, roller-skated at a record-setting

speed of 40.3 km/h. If the magnitude of Canofoglia’s momentum was

6.60 × 102 kg•m/s, what was her mass?

2. In 1976, a 53 kg helicopter was built in Denmark. Suppose this heli-

copter flew east with a speed of 60.0 m/s and the total momentum of

the helicopter and pilot was 7.20 × 103 kg•m/s to the east. What was the

mass of the pilot?

3. One of the smallest planes ever flown was the Bumble Bee II, which had

a mass of 1.80 × 102 kg. If the pilot’s mass was 7.0 × 101 kg, what was

the velocity of both plane and pilot if their momentum was 2.08 × 104 kg•m/s to the west?

4. The first human-made satellite, Sputnik I, had a mass of 83.6 kg and a

momentum with a magnitude of 6.63 × 105 kg•m/s. What was the

satellite’s speed?

Problem 6A 55

NAME ______________________________________ DATE _______________ CLASS ____________________

5. Among the largest passenger ships currently in use, the Norway has

been in service the longest. The Norway is more than 300 m long, has a

mass of 6.9 × 107 kg, and can reach a top cruising speed of 33 km/h.

Calculate the magnitude of the ship’s momentum.

6. In 1994, a tower 22.13 m tall was built of Lego® blocks. Suppose a block

with a mass of 2.00 g is dropped from the top of this tower. Neglecting

air resistance, calculate the block’s momentum at the instant the block

hits the ground.

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NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 6BFORCE AND MOMENTUM

P R O B L E M

In 1993, a generator with a mass of 1.24 105 kg was flown from Ger-many to a power plant in India on a Ukrainian-built plane. This consti-tuted the heaviest single piece of cargo ever carried by a plane. Supposethe plane took off with a speed of 101 m/s toward the southeast and thenaccelerated to a final cruising speed of 197 m/s. During this acceleration,a force of 4.00 105 N in the southeast direction was exerted on the gen-erator. For how much time did the force act on the generator?


vi = 101 m/s to the southeast

vf = 197 m/s to the southeast

F = 4.00 × 105 N to the southeast

Unknown: ∆t = ?

Use the impulse-momentum theorem to determine the time the force acts on the

generator.

F ∆t = ∆p = mvf − mvi

∆t = mvf

F

− mvi

∆t =

∆t =

∆t = 1.1

4

9

.0

×0

1

×07

10

k5g•

N

m/s

∆t = 29.8 s

2.44 × 107 kg•m/s − 1.25 × 107 kg•m/s

4.00 × 105 N

(1.24 × 105 kg)(197 m/s) − (1.24 × 105 kg)(101 m/s)

4.00 × 105 N

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ADDITIONAL PRACTICE

1. In 1991, a Swedish company, Kalmar LMV, constructed a forklift truck

capable of raising 9.0 × 104 kg to a height of about 2 m. Suppose a mass

this size is lifted with an upward velocity of 12 cm/s. The mass is ini-

tially at rest and reaches its upward speed because of a net force of 6.0 ×103 N exerted upward. For how long is this force applied?

2. A bronze statue of Buddha was completed in Tokyo in 1993. The statue

is 35 m tall and has a mass of 1.00 × 106 kg. Suppose this statue were to

be moved to a different location. What is the magnitude of the impulse

that must act on the statue in order for the speed to increase from 0 m/s

Problem 6B 57

NAME ______________________________________ DATE _______________ CLASS ____________________

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to 0.20 m/s? If the magnitude of the net force acting on the statue is

12.5 kN, how long will it take for the final speed to be reached?

3. In 1990, Gary Stewart of California made 177 737 jumps on a pogo

stick. Suppose that the pogo stick reaches a height of 12.0 cm with each

jump and that the average net force acting on the pogo stick during the

contact with the ground is 330 N upward. What is the time of contact

with the ground between the jumps? Assume the total mass of Stewart

and the pogo stick is 65 kg. (Hint: The difference between the initial

and final velocities is one of direction rather than magnitude.)

4. The specially designed armored car that was built for Leonid Brezhnev

when he was head of the Soviet Union had a mass of about 6.0 × 103 kg.

Suppose this car is accelerated from rest by a force of 8.0 kN to the east.

What is the car’s velocity after 8.0 s?

5. In 1992, Dan Bozich of the United States drove a gasoline-powered go-

cart at a speed of 125.5 km/h. Suppose Bozich applies the brakes upon

reaching this speed. If the combined mass of the go-cart and driver is

2.00 × 102 kg, the decelerating force is 3.60 × 102 N opposite the cart’s

motion, and the time during which the deceleration takes place is

10.0 s. What is the final speed of Bozich and the go-cart?

6. The “human cannonball” has long been a popular—and extremely

dangerous—circus stunt. In order for a 45 kg person to leave the can-

non with the fastest speed yet achieved by a human cannonball, a 1.6 ×103 N force must be exerted on that person for 0.68 s. What is the

record speed at which a person has been shot from a circus cannon?

7. The largest steam-powered locomotive was built in the United States in

1943. It is still operational and is used for entertainment purposes. The

locomotive’s mass is 4.85 × 105 kg. Suppose this locomotive is traveling

northwest along a straight track at a speed of 20.0 m/s. What force

must the locomotive exert to increase its velocity to 25.0 m/s to the

northwest in 5.00 s?

8. With upward speeds of 12.5 m/s, the elevators in the Yokohama Land-

mark Tower in Yokohama, Japan, are among the fastest elevators in the

world. Suppose a passenger with a mass of 70.0 kg enters one of these

elevators. The elevator then goes up, reaching full speed in 4.00 s. Cal-

culate the net force that is applied to the passenger during the elevator’s

acceleration.

9. Certain earthworms living in South Africa have lengths as great as 6.0 m

and masses as great as 12.0 kg. Suppose an eagle picks up an earthworm

of this size, only to drop it after both have reached a height of 40.0 m

above the ground. By skillfully using its muscles, the earthworm man-

ages to extend the time during which it collides with the ground to

0.250 s. What is the net force that acts on the earthworm during its col-

lision with the ground? Assume the earthworm’s vertical speed when it

is initially dropped to be 0 m/s.


NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 6CSTOPPING DISTANCE

P R O B L E MThe largest nuts (and, presumably, the largest bolts) are manufactured inEngland. The nuts have a mass of 4.74 103 kg each, which is greaterthan any passenger car currently in production. Suppose one of thesenuts slides along a rough horizontal surface with an initial velocity of2.40 m/s to the right. If the force of friction acting on the nut is 6.8 103 N to the left, what is the change in the nut’s momentum after 1.1 s?How far does the nut travel during its change in momentum?

S O L U T I O N

Given: m = 4.74 × 103 kg

vi = 2.40 m/s to the right = +2.40 m/s

F = 6.8 × 103 N to the left = −6.8 × 103 N

∆t = 1.1 s

Unknown: ∆p = ? ∆x = ?

Use the impulse-momentum theorem to calculate the change in momentum.

Use the definition of momentum to find vf , and then use the equation for stop-

ping distance to solve for ∆x.

∆p = F∆t = (−6.8 × 103 N)(1.1 s)

∆p =

∆p = mvf − mvi

vf = ∆p +

m

mvi

vf =

vf = =

vf =

∆x = 12

(vi + vf)∆t = 12

(2.40 m/s + 0.82 m/s)(1.1 s) = 12

(3.22 m/s)(1.1 s)

∆x = 1.8 m/s to the right

0.82 m/s to the right

3900 kg•m/s4.74 × 103 kg

−7.5 ×103 kg•m/s + 1.14 × 104 kg•m/s

4.74 × 103 kg

−7.5 × 103 kg•m/s + (4.74 × 103 kg)(2.40 m/s)

4.74 × 103 kg

−7.5 × 103 kg•m/s to the right, or 7.5 × 103 kg•m/s to the left

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ADDITIONAL PRACTICE

1. The most powerful tugboats in the world are built in Finland. These

boats exert a force with a magnitude of 2.85 × 106 N. Suppose one of

these tugboats is trying to slow a huge barge that has a mass of 2.0 × 107 kg and is moving with a speed of 3.0 m/s. If the tugboat exerts its

maximum force for 21 s in the direction opposite to that in which the

barge is moving, what will be the change in the barge’s momentum? How

far will the barge travel before it is brought to a stop?

Problem 6C 59

NAME ______________________________________ DATE _______________ CLASS ____________________

2. In 1920, a 6.5 × 104 kg meteorite was found in Africa. Suppose a me-

teorite with this mass enters Earth’s atmosphere with a speed of 1.0 km/s.

What is the change in the meteorite’s momentum if an average constant

force of –1.7 × 106 N acts on the meteorite for 30.0 s? How far does the

meteorite travel during this time?

3. The longest canoe in the world was constructed in New Zealand. The

combined mass of the canoe and its crew of more than 70 people was

2.03 × 104 kg. Suppose the canoe is rowed from rest to a velocity of

5.00 m/s to the east, at which point the crew takes a break for 20.3 s. If a

constant average retarding force of 1.20 × 103 N to the west acts on the

canoe, what is the change in the momentum of the canoe and crew? How

far does the canoe travel during the time the crew is not rowing?

4. The record for the smallest dog in the world belongs to a terrier who had

a mass of only 113 g. Suppose this dog runs to the right with a speed of

2.00 m/s when it suddenly sees a mouse. The dog becomes scared and

uses its paws to bring itself to rest in 0.80 s. What is the force required to

stop the dog? What is the dog’s stopping distance?

5. In 1992, an ice palace estimated to be 4.90 × 106 kg was built in Min-

nesota. Despite this sizable mass, this structure could be moved at a con-

stant velocity because of the small force of friction between the ice blocks

of its base and the ice of the lake upon which it was constructed. Imagine

moving the entire palace with a speed of 0.200 m/s on this very smooth,

icy surface. Once the palace is no longer being pushed, it coasts to a stop

in 10.0 s. What is the average force of kinetic friction acting on the

palace? What is the palace’s stopping distance?

6. Steel Phantom is a roller coaster in Pennsylvania that, like the Desperado

in Nevada, has a vertical drop of 68.6 m. Suppose a roller-coaster car

with a mass of 1.00 × 103 kg travels from the top of that drop without

friction. The car then decelerates along a horizontal stretch of track until

it comes to a stop. How long does it take the car to decelerate if the con-

stant force acting on it is –2.24 × 104 N? How far does the car travel

along the horizontal track before stopping? Assume the car’s speed at the

peak of the drop is zero.

7. Two Japanese islands are connected by a long rail tunnel that extends

horizontally underwater. Imagine a communication system in which a

small rail car with a mass of 100.0 kg is launched by a type of cannon in

order to transport messages between the two islands. Assume a rail car

from one end of the tunnel has a speed of 4.50 × 102 m/s, which is just

large enough for a constant frictional force of –188 N to cause the car to

stop at the other end of the tunnel. How long does it take for the car to

travel the length of the tunnel? What is the length of the tunnel?

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NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 6DCONSERVATION OF MOMENTUM

P R O B L E MKangaroos are good runners that can sustain a speed of 56 km/h (15.5 m/s). Suppose a kangaroo is sitting on a log that is floating in a lake.When the kangaroo gets scared, she jumps off the log with a velocity of15 m/s toward the bank. The log moves with a velocity of 3.8 m/s awayfrom the bank. If the mass of the log is 250 kg, what is the mass of thekangaroo?

S O L U T I O NGiven: vk,i = initial velocity of kangaroo = 0 m/s

vl,i = initial velocity of log = 0 m/s

vk,f = final velocity of kangaroo = 15 m/s toward bank = +15 m/s

vl,f = final velocity of log = 3.8 m/s away from bank = −3.8 m/s

ml = mass of log = 250 kg

Unknown: mk = mass of kangaroo = ?

Choose the equation(s) or situation: Because the momentum of the kangaroo-

log system is conserved and therefore remains constant, the total initial momen-

tum of the kangaroo and log will equal the total final momentum of the kanga-

roo and log.

mk vk,i + ml vl,i = mk vk, f + ml vl, f

The initial velocities of the kangaroo and log are zero, and therefore the initial

momentum for each of them is zero. It thus follows that the total final momen-

tum for the kangaroo and log must also equal zero. The momentum-conservation

equation reduces to the following:

mk vk, f + ml vl, f = 0

Rearrange the equation(s) and isolate the unknown(s):

mk = −m

vk

lv

,f

l,f


mk =

mk =

Because the log is about four times as massive as the kangaroo, its velocity is

about one-fourth as large as the kangaroo’s velocity.

63 kg

−(250 kg)(−3.8 m/s)

15 m/s

1. DEFINE

2. PLAN

3. CALCULATE

4. EVALUATE

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Problem 6D 61

NAME ______________________________________ DATE _______________ CLASS ____________________

1. The largest single publication in the world is the 1112-volume set of

British Parliamentary Papers for 1968 through 1972. The complete set has

a mass of 3.3 × 103 kg. Suppose the entire publication is placed on a cart

that can move without friction. The cart is at rest, and a librarian is sit-

ting on top of it, just having loaded the last volume. The librarian jumps

off the cart with a horizontal velocity relative to the floor of 2.5 m/s to

the right. The cart begins to roll to the left at a speed of 0.05 m/s. Assum-

ing the cart’s mass is negligible, what is the librarian’s mass?

2. The largest grand piano in the world is really grand. Built in London, it

has a mass of 1.25 × 103 kg. Suppose a pianist finishes playing this piano

and pushes herself from the piano so that she rolls backwards with a

speed of 1.4 m/s. Meanwhile, the piano rolls forward so that in 4.0 s it

travels 24 cm at constant velocity. Assuming the stool that the pianist is

sitting on has a negligible mass, what is the pianist’s mass?

3. With a mass of 114 kg, Baby Bird is the smallest monoplane ever flown.

Suppose the Baby Bird and pilot are coasting along the runway when the

pilot jumps horizontally to the runway behind the plane. The pilot’s ve-

locity upon leaving the plane is 5.32 m/s backward. After the pilot jumps

from the plane, the plane coasts forward with a speed of 3.40 m/s. If the

pilot’s mass equals 60.0 kg, what is the velocity of the plane and pilot be-

fore the pilot jumps?

4. The September 14, 1987, issue of the New York Times had a mass of

5.4 kg. Suppose a skateboarder picks up a copy of this issue to have a look

at the comic pages while rolling backward on the skateboard. Upon realiz-

ing that the New York Times doesn’t have a “funnies” section, the skate-

boarder promptly throws the entire issue in a recycling container. The

newspaper is thrown forward with a speed of 7.4 m/s. When the skater

throws the newspaper away, he rolls backward at a speed of 1.4 m/s. If

the combined mass of the skateboarder and skateboard is assumed to be

50.0 kg, what is the initial velocity of the skateboarder and newspaper?

5. The longest bicycle in the world was built in New Zealand in 1988. It is

more than 20 m in length, has a mass of 3.4 × 102 kg, and can be ridden

by four people at a time. Suppose four people are riding the bike south-

east when they realize that the street turns and that the bike won’t make

it around the corner. All four riders jump off the bike at the same time

and with the same velocity (9.0 km/h to the northwest, as measured rela-

tive to Earth). The bicycle continues to travel forward with a velocity of

28 km/h to the southeast. If the combined mass of the riders is 2.6 × 102 kg, what is the velocity of the bicycle and riders immediately before

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ADDITIONAL PRACTICE


NAME ______________________________________ DATE _______________ CLASS ____________________

6. The largest frog ever found was discovered in Cameroon in 1989. The

frog’s mass was nearly 3.6 kg. Suppose this frog is placed on a skateboard

with a mass of 3.0 kg. The frog jumps horizontally off the skateboard to

the right, and the skateboard rolls freely in the opposite direction with a

speed of 2.0 m/s relative to the ground. If the frog and skateboard are ini-

tially at rest, what is the initial horizontal velocity of the frog?

7. In 1994, a pumpkin with a mass of 449 kg was grown in Canada. Sup-

pose you want to push a pumpkin with this mass along a smooth, hori-

zontal ramp. You give the pumpkin a good push, only to find yourself

sliding backwards at a speed of 4.0 m/s. How far will the pumpkin slide

3.0 s after the push? Assume your mass to be 60.0 kg.

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Problem 6E 63

NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 6EPERFECTLY INELASTIC COLLISIONS

P R O B L E MThe Chinese giant salamander is one of the largest of salamanders. Sup-pose a Chinese giant salamander chases a 5.00 kg carp with a velocity of3.60 m/s to the right and the carp moves with a velocity of 2.20 m/s in thesame direction (away from the salamander). If the speed of the salaman-der and carp immediately after the salamander catches the carp is 3.50 m/s to the right, what is the salamander’s mass?

S O L U T I O NGiven: mc = mass of carp = 5.00 kg

vs,i = initial velocity of salamander = 3.60 m/s to the right

vc,i = initial velocity of carp = 2.20 m/s to the right

vf = final velocity = 3.50 m/s to the right

Unknown: ms = mass of salamander = ?

Use the equation for a perfectly inelastic collision and rearrange it to solve for ms.

msvs,i + mcvc,i = (ms + mc)vf

ms = mc

v

v

s

f

,

−

i −m

vc

f

vc,i

ms =

ms =

ms = 6

0

.5

.1

k

0

g

m

•m

/s

/s

ms = 65 kg

17.5 kg•m/s − 11.0 kg•m/s

0.10 m/s

(5.00 kg)(3.50 m/s) − (5.00 kg)(2.20 m/s)

3.60 m/s − 3.50 m/s

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1. Zorba, an English mastiff with a mass of 155 kg, jumps forward horizon-

tally at a speed of 6.0 m/s into a boat that is floating at rest. After the

jump, the boat and Zorba move with a velocity of 2.2 m/s forward. Cal-

culate the boat’s mass.

2. Yvonne van Gennip of the Netherlands ice skated 10.0 km with an aver-

age speed of 10.8 m/s. Suppose van Gennip crosses the finish line at her

average speed and takes a huge bouquet of flowers handed to her by a

fan. As a result, her speed drops to 10.01 m/s. If van Gennip’s mass is

63.0 kg, what is the mass of the bouquet?

ADDITIONAL PRACTICE


NAME ______________________________________ DATE _______________ CLASS ____________________

3. The world’s largest guitar was built by a group of high school students in

Indiana. Suppose that this guitar is placed on a light cart. The cart and

guitar are then pushed with a velocity of 4.48 m/s to the right. One of the

students tries to slow the cart by stepping on it as it passes by her. The

new velocity of the cart, guitar, and student is 4.00 m/s to the right. If the

student’s mass is 54 kg, what is the mass of the guitar? Assume the mass

of the cart is negligible.

4. The longest passenger buses in the world operate in Zaire. These buses

are more than 30 m long, have two trailers, and have a total mass of 28 ×103 kg. Imagine a safety test involving one of these buses and a truck

with a mass of 12 × 103 kg. The truck with an unknown velocity hits a

bus that is at rest so that the two vehicles move forward together with a

speed of 3.0 m/s. Calculate the truck’s velocity prior to the collision.

5. Sumo wrestlers must be very heavy to be successful in their sport, which

involves pushing the rival out of the ring. One of the greatest sumo

champions, Akebono, had a mass of 227 kg. The heaviest sumo wrestler

ever, Konishiki, at one point had a mass of 267 kg. Suppose these two

wrestlers are opponents in a match. Akebono moves left with a speed of

4.0 m/s, while Konishiki moves toward Akebono with an unknown

speed. After the wrestlers undergo an inelastic collision, both have a ve-

locity of zero. From this information, calculate Konishiki’s velocity be-

fore colliding with Akebono.

6. Louis Borsi, of London, built a drivable car that had a mass of 9.50 kg

and could move as fast as 24.0 km/h. Suppose the inventor falls out of

this car and the car proceeds driverless to the north at its maximum

speed. The inventor’s young daughter, who has a mass of 32.0 kg,

“catches” the car by jumping northward from a nearby stairway. The ve-

locity of the car and girl is 11.0 km/h to the north. What was the velocity

of Borsi’s daughter as she jumped in the car?

7. In 1990, Roger Hickey of California attained a speed of 89 km/h while

standing on a skateboard. Suppose Hickey is riding horizontally at his

stand-up speed when he catches up to another skateboarder, who is

moving at 69 km/h in the same direction. If the second skateboarder

steps sideways onto Hickey’s skateboard, the two riders move forward

with a new speed. Calculate this speed, assuming that both skateboarders

have equal, but unknown, masses and that the mass of the skateboard is

negligible.

8. The white shark is the largest carnivorous fish in the world. The mass of

a white shark can be as great as 3.0 × 103 kg. In spite of (or perhaps be-

cause of) the mass and ferocity of the shark, it is prized by commercial

and sports fishers alike. Suppose Joe, who is one of these fishers, goes to a

cliff that overlooks the ocean. To see if the sharks are biting, Joe drops a

2.5 × 102 kg fish as bait into the ocean below. As it so happens, a 3.0 ×103 kg white shark is prowling the ocean floor just below the cliff. The

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Problem 6E 65

NAME ______________________________________ DATE _______________ CLASS ____________________

shark sees the bait, which is sinking straight down at a speed of

3.0 m/s. The shark swims upward with a speed of 1.0 m/s to swallow the

bait. What is the velocity of the shark right after it has swallowed the

bait?

9. The heaviest cow on record had a mass of 2.267 × 103 kg and lived in

Maine at the beginning of the twentieth century. Imagine that during an

agricultural exhibition, the cow’s owner puts the cow on a railed cart that

has a mass of 5.00 × 102 kg and pushes the cow and cart left to the stage

with a speed of 2.00 m/s. Another farmer puts his cow, which has a mass

of 1.800 × 103 kg, on an identical cart and pushes it toward the stage

from the opposite direction with a speed of 1.40 m/s. The carts collide

and stick together. What is the velocity of the cows after the collision?

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NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 6FKINETIC ENERGY IN PERFECTLY INELASTIC COLLISIONS

P R O B L E M

Alaskan moose can be as massive as 8.00 × 102 kg. Suppose two feudingmoose, both of which have a mass of 8.00 × 102 kg, back away and thenrun toward each other. If one of them runs to the right with a speed of8.0 m/s and the other runs to the left with a speed of 6.0 m/s, whatamount of kinetic energy will be dissipated after their inelastic collision?

S O L U T I O N

Given: m1 = mass of first moose = 8.00 × 102 kg

m2 = mass of second moose = 8.00 × 102 kg

v1, i = initial velocity of first moose = 8.0 m/s to the right

v2, i = initial velocity of second moose = 6.0 m/s to the left

= –6.0 m/s to the right

Unknown: ∆KE = ?

Use the equation for a perfectly inelastic collision.

m1v1, i + m2 v2, i = (m1 + m2)vf

(8.00 × 102 kg)(8.0 m/s) + (8.00 × 102 kg)(–6.0 m/s)

= (2)(8.00 × 102 kg)vf

6.4 × 103 kg•m/s – 4.8 × 103 kg•m/s = (16.0 × 102 kg)vf

vf = 1.6

16

×.0

1

×03

1

k

0

g2

•

k

m

g

/s = 1.0 m/s to the right

Use the equation for kinetic energy to calculate the kinetic energy of each moose

before the collision and the final kinetic energy of the two moose combined.

Initial kinetic energy:

KEi = KE1,i + KE2,i = 12

m1v1,i2 + 1

2 m2v2,i

2

KEi = 12

(8.00 × 102 kg)(8.0 m/s)2 + 12

(8.00 × 102 kg)(−6.0 m/s)2

KEi = 2.6 × 104 J + 1.4 × 104 J = 4.0 × 104 J

Final kinetic energy:

KEf = KE1,f + KE2,f = 12

(m1 + m2)vf2

KEf =

KEf = 8.0 × 102 J

Change in kinetic energy:

∆KE = KEf − KEi = 8.0 × 102 J − 4.0 × 104 J =

By expressing ∆KE as a negative number, we indicate that the energy has left the

system to take a form other than mechanical energy.

−3.9 × 104 J

(2)(8.00 × 102 kg)(1.0 m/s)2

2

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Problem 6F 67

NAME ______________________________________ DATE _______________ CLASS ____________________

1. The hog-nosed bat is the smallest mammal on Earth: it is about the same

size as a bumblebee and has an average mass of 2.0 g. Suppose a hog-

nosed bat with this mass flies at 2.0 m/s when it detects a bug with a

mass of 0.20 g flying directly toward it at 8.0 m/s. What fraction of the

total kinetic energy is dissipated when it swallows the bug?

2. The heaviest wild lion ever measured had a mass of 313 kg. Suppose this

lion is walking by a lake when it sees an empty boat floating at rest near

the shore. The curious lion jumps into the boat with a speed of 6.00 m/s,

causing the boat with the lion in it to move away from the shore with a

speed of 2.50 m/s. How much kinetic energy is dissipated in this inelastic

collision.

3. The cheapest car ever commercially produced was the Red Bug Back-

board, which sold in 1922 in the United States for about $250. The car’s

mass was only 111 kg. Suppose two of these cars are used in a stunt crash

for an action film. If one car’s initial velocity is 9.00 m/s to the right and

the other car’s velocity is 5.00 m/s to the left, how much kinetic energy is

dissipated in the crash?

4. In 1986, four high school students built an electric car that could reach a

speed of 106.0 km/h. The mass of the car was just 60.0 kg. Imagine two

of these cars used in a stunt show. One car travels east with a speed of

106.0 km/h, and the other car travels west with a speed of 75.0 km/h. If

each car’s driver has a mass of 50.0 kg, how much kinetic energy is dissi-

pated in the perfectly inelastic head-on collision?

5. The Arctic Snow Train, built for the U.S. Army, has a mass of 4.00 × 105 kg

and a top speed of 32.0 km/h. Suppose such a train moving at its top

speed is hit from behind by another snow train with a mass of 1.60 × 105 kg and a speed of 45.0 km/h in the same direction. What is the

change in kinetic energy after the trains’ perfect inelastic collision?

6. There was a domestic cat in Australia with a mass of 21.3 kg. Suppose

this cat is sitting on a skateboard that is not moving. A 1.80 × 10–1 kg

treat is thrown to the cat. When the cat catches the treat, the cat and

skateboard move with a speed of 6.00 × 10–2 m/s. How much kinetic en-

ergy is dissipated in the process? Assume one-dimensional motion.

7. In 1985, a spider with a mass of 122 g was caught in Surinam, South

America. (Recall that the smallest dog in the world had a smaller mass.)

Suppose a spider with this mass runs at a certain unknown speed when it

collides inelastically with another spider, which has a mass of 96.0 g and

is at rest. Find the fraction of the kinetic energy that is dissipated in the

perfect inelastic collision. Assume that the resting spider is on a low-

friction surface. Do you need to know the first spider’s velocity to calcu-

late the fraction of the dissipated kinetic energy?Cop

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NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 6GELASTIC COLLISIONS

P R O B L E MAmerican juggler Bruce Sarafian juggled 11 identical balls at one time in1992. Each ball had a mass of 0.20 kg. Suppose two balls have an elastic head-on collision during the act. The first ball moves away from the collision witha velocity of 3.0 m/s to the right, and the second ball moves away with a veloc-ity of 4.0 m/s to the left. If the first ball’s velocity before the collision is 4.0 m/sto the left, what is the velocity of the second ball before the collision?

S O L U T I O NGiven: m1 = m2 = 0.20 kg

v1, i = initial velocity of ball 1 = 4.0 m/s to the left

= −4.0 m/s to the right

v1, f = final velocity of ball 1 = 3.0 m/s to the right

v2, f = final velocity of ball 2 = 4.0 m/s to the left

= –4.0 m/s to the right

Unknown: v2, i = initial velocity of ball 2 = ?

Choose the equation(s) or situation: Use the equation for the conservation of

momentum to determine the initial velocity of ball 2. Because both balls have

identical masses, the mass terms cancel.

m1v1, i + m2v2, i = m1v1, f + m2 v2, f

v1, i + v2, i = v1, f + v2, f


v2, i = v1, f + v2, f − v1, i


v2,i = 3.0 m/s − 4.0 m/s − (−4.0 m/s)

v2,i =

Confirm your answer by making sure that kinetic energy is also conserved.

12

m1v1,i2 + 1

2m2v2,i

2 = 12

m1v1,f2 + 1

2m2v2,f

2

v1,i2 + v2,i

2 = v1,f2 + v2,f

2

(−4.0 m/s)2 + (3.0 m/s)2 = (3.0 m/s)2 + (−4.0 m/s)2

16 m2/s2 + 9.0 m2/s2 = 9.0 m2/s2 + 16 m2/s2

25 m2/s2 = 25 m2/s2


1. DEFINE

2. PLAN

3. CALCULATE

4. EVALUATE

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ADDITIONAL PRACTICE

1. The moon’s orbital speed around Earth is 3.680 × 103 km/h. Suppose the

moon suffers a perfectly elastic collision with a comet whose mass is 50.0

percent that of the moon. (A partially inelastic collision would be a much

Problem 6G 69

NAME ______________________________________ DATE _______________ CLASS ____________________

more realistic event.) After the collision, the moon moves with a speed of

–4.40 × 102 km/h, while the comet moves away from the moon at 5.740

× 103 km/h. What is the comet’s speed before the collision?

2. The largest beet root on record had a mass of 18.40 kg. The largest cab-

bage on record had a mass of 56.20 kg. Imagine these two vegetables

traveling in opposite directions. The cabbage, which travels 5.000 m/s to

the left, collides with the beet root. After the collision, the cabbage has a

velocity of 6.600 × 10–2 m/s to the left, and the beet root has a velocity of

10.07 m/s to the left. What is the beet root’s velocity before the perfectly

elastic collision?

3. The first astronaut to walk in outer space without being tethered to a

spaceship was Capt. Bruce McCandless. In 1984, he used a jet backpack,

which cost about $15 million to design, to move freely about the exterior

of the space shuttle Challenger. Imagine two astronauts working in outer

space. Suppose they have equal masses and accidentally run into each

other. The first astronaut moves 5.0 m/s to the right before the collision

and 2.0 m/s to the left afterwards. If the second astronaut moves 5.0 m/s

to the right after the perfectly elastic collision, what was the second astro-

naut’s initial velocity?

4. Speeds as high as 273 km/h have been recorded for golf balls. Suppose a

golf ball whose mass is 45.0 g is moving to the right at 273 km/h and

strikes another ball that is at rest. If after the perfectly elastic collision the

golf ball moves 91 km/h to the left and the other ball moves 182 km/h to

the right, what is the mass of the second ball?

5. Jana Novotna of what is now the Czech Republic has the strongest serve

among her fellow tennis players. In 1993, she sent the ball flying with a

speed of 185 km/h. Suppose a tennis ball moving to the right at this speed

hits a moveable target of unknown mass. After the one-dimensional, per-

fectly elastic collision, the tennis ball bounces to the left with a speed of

80.0 km/h. If the tennis ball’s mass is 5.70 × 10–2 kg, what is the target’s

mass? (Hint: Use the conservation of kinetic energy to solve for the sec-

ond unknown quantity.)

6. Recall the two colliding snow trains in item 6 of the previous section.

Suppose now that the collision between the two trains is perfectly elastic

instead of inelastic. The train with a 4.00 × 105 kg and a velocity of

32.0 km/h to the right is struck from behind by a second train with a

mass of 1.60 × 105 kg and a velocity of 36.0 km/h to the right. If the first

train’s velocity increases to 35.5 km/h to the right, what is the final veloc-

ity of the second train after the collision?

7. A dump truck used in Canada has a mass of 5.50 × 105 kg when loaded

and 2.30 × 105 kg when empty. Suppose two such trucks, one loaded and

one empty, crash into each other at a monster truck show. The trucks are

supplied with special bumpers that make a collision almost perfectly

elastic. If the trucks hit each other at equal speeds of 5.00 m/s and the

less massive truck recoils to the right with a speed of 9.10 m/s, what is the

velocity of the full truck after the collision?

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NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 7AANGULAR DISPLACEMENT

P R O B L E M

The diameter of the outermost planet, Pluto, is just 2.30 103 km. If aspace probe were to orbit Pluto near the planet’s surface, what would thearc length of the probe’s displacement be after it had completed eight orbits?

S O L U T I O N

Given: r = 2.30 ×

2

103 km = 1.15 × 103 km

∆q = 8 orbits = 8(2p rad) = 16p rad

Unknown: ∆s = ?

Use the angular displacement equation and rearrange to solve for ∆s.

∆q = ∆r

s

∆s = r ∆q = (1.15 × 103 km)(16p rad) = 5.78 × 104 km

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ADDITIONAL PRACTICE

1. A neutron star can have a mass three times that of the sun and a radius

as small as 10.0 km. If a particle travels +15.0 rad along the equator of a

neutron star, through what arc length does the particle travel? Does the

particle travel in the clockwise or counterclockwise direction from the

viewpoint of the neutron star’s “north” pole?

2. John Glenn was the first American to orbit Earth. In 1962, he circled

Earth counterclockwise three times in less that 5 h aboard his spaceship

Friendship 7. If his distance from the Earth’s center was 6560 km, what

arc length did Glenn and his spaceship travel through?

3. Jupiter’s diameter is 1.40 × 105 km. Suppose a space vehicle travels

along Jupiter’s equator with an angular displacement of 1.72 rad.

a. Through what arc length does the space vehicle move?

b. How many orbits around Earth’s equator can be completed if the

vehicle travels at the surface of the Earth through this arc length

around Earth? Use 6.37 × 103 km for Earth’s radius.

4. In 1981, the space shuttle Columbia became the first reusable spacecraft

to orbit Earth. The shuttle’s total angular displacement as it orbited

Earth was 225 rad. How far from Earth’s center was Columbia if it

moved through an arc length of 1.50 × 106 km? Use 6.37 × 103 km for

Earth’s radius.

5. Mercury, the planet closest to the sun, has an orbital radius of 5.8 × 107 km.

Find the angular displacement of Mercury as it travels through an arc

length equal to the radius of Earth’s orbit around the sun (1.5 × 108 km).

Problem 7A 71

NAME ______________________________________ DATE _______________ CLASS ____________________

6. From 1987 through 1988, Sarah Covington-Fulcher ran around the

United States, covering a distance of 1.79 × 104 km. If she had run that

distance clockwise around Earth’s equator, what would her angular dis-

placement have been? (Earth’s average radius is 6.37 × 103 km).

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NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 7BANGULAR SPEED

P R O B L E MIn 1975, an ultrafast centrifuge attained an average angular speed of 2.65 104 rad/s. What was the centrifuge’s angular displacement after 1.5 s?

S O L U T I O NGiven: wavg = 2.65 × 104 rad/s

∆t = 1.5 s

Unknown: ∆q = ?

Use the angular speed equation and rearrange to solve for ∆q.

wavg = ∆∆

qt

∆q = wavg ∆t = (2.65 × 104 rad/s)(1.5 s)

∆q = 4.0 × 104 rad

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ADDITIONAL PRACTICE

1. The largest steam engine ever built was constructed in 1849. The en-

gine had one huge cylinder with a radius of 1.82 m. If a beetle were to

run around the edge of the cylinder with an average angular speed of

1.00 × 10–1 rad/s, what would its angular displacement be after 60.0 s?

What arc length would it have moved through?

2. The world’s largest planetarium dome is 30 m in diameter. What would

your angular displacement be if you ran around the perimeter of this

dome for 120 s with an average angular speed of 0.40 rad/s?

3. The world’s most massive magnet is located in a research center in

Dubna, Russia. This magnet has a mass of over 3.0 × 107 kg and a ra-

dius of 30.0 m. If you were to run around this magnet so that you trav-

eled 5.0 × 102 m in 120 s, what would your average angular speed be?

4. A floral clock in Japan has a radius of 15.5 m. If you ride a bike around

the clock, making 16 revolutions in 4.5 min, what is your average angu-

lar speed?

5. A revolving globe with a radius of 5.0 m was built between 1982 and

1987 in Italy. If the globe revolves with the same average angular speed

as Earth, how long will it take for a point on the globe’s equator to

move through 0.262 rad?

6. A water-supply tunnel in New York City is 1.70 × 102 km long and has

a radius of 2.00 m. If a beetle moves around the tunnel’s perimeter with

an average angular speed of 5.90 rad/s, how long will it take the beetle

to travel a distance equal to the tunnel’s length?

Problem 7C 73

NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 7CANGULAR ACCELERATION

P R O B L E MA self-propelled Catherine wheel (a spinning wheel with fireworks alongits rim) with a diameter of 20.0 m was built in 1994. Its maximum angu-lar speed was 0.52 rad/s. How long did the wheel undergo an angular ac-celeration of 2.6 10–2 rad/s2 in order to reach its maximum angularspeed?

S O L U T I O NGiven: w1 = 0 rad/s

w2 = 0.52 rad/s

aavg = 2.6 × 10−2 rad/s2

Unknown: ∆t = ?

Use the angular acceleration equation and rearrange to solve for ∆t.

aavg = ∆∆wt

∆t = a∆

a

w

vg =

w2

a−

av

w

g

1 = = 2.0 × 101 s0.52 rad/s − 0 rad/s

2.6 × 10−2 rad/s2

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ADDITIONAL PRACTICE

1. Peter Rosendahl of Sweden rode a unicycle with a wheel diameter of

2.5 cm. If the wheel’s average angular acceleration was 2.0 rad/s2, how

long would it take for the wheel’s angular speed to increase from

0 rad/s to 9.4 rad/s?

2. Jupiter has the shortest day of all of the solar system’s planets. One ro-

tation of Jupiter occurs in 9.83 h. If an average angular acceleration of

–3.0 × 10–8 rad/s2 slows Jupiter’s rotation, how long does it take for

Jupiter to stop rotating?

3. In 1989, Dave Moore of California built the Frankencycle, a bicycle

with a wheel diameter of more than 3 m. If you ride this bike so that

the wheels’ angular speed increases from 2.00 rad/s to 3.15 rad/s in

3.6 s, what is the average angular acceleration of the wheels?

4. In 1990, David Robilliard rode a bicycle on the back wheel for more

than 5 h. If the wheel’s initial angular speed was 8.0 rad/s and Robilliard

tripled this speed in 25 s, what was the average angular acceleration?

5. Earth takes about 365 days to orbit once around the sun. Mercury, the

innermost planet, takes less than a fourth of this time to complete one

revolution. Suppose some mysterious force causes Earth to experience

an average angular acceleration of 6.05 × 10–13 rad/s2, so that after


NAME ______________________________________ DATE _______________ CLASS ____________________

12.0 days its angular orbital speed is the same as that of Mercury. Cal-

culate this angular speed and the period of one orbit.

6. The smallest ridable tandem bicycle was built in France and has a

length of less than 40 cm. Suppose this bicycle accelerates from rest so

that the average angular acceleration of the wheels is 0.800 rad/s2. What

is the angular speed of the wheels after 8.40 s?

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Problem 7D 75

NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 7DANGULAR KINEMATICS

P R O B L E MIn 1990, a pizza with a radius of 18.7 m was baked in South Africa. Sup-pose this pizza was placed on a rotating platform. If the pizza acceleratedfrom rest at 5.00 rad/s2 for 25.0 s, what was the pizza’s final angular speed?

S O L U T I O NGiven: wi = 0 rad/s

a = 5.00 rad/s2

∆t = 25.0 s

Unknown: wf = ?

Use the rotational kinematic equation relating final angular speed to initial angu-

lar speed, angular acceleration, and time.

wf = wi + a∆t

wf = 0 rad/s + (5.00 rad/s2)(25.0 s)

wf = 125 rad/s

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ADDITIONAL PRACTICE

1. In 1987, Takayuki Koike of Japan rode a unicycle nonstop for 160 km in

less than 7 h. Suppose at some point in his trip Koike accelerated

downhill. If the wheel’s angular speed was initially 5.0 rad/s, what

would the angular speed be after the wheel underwent an angular ac-

celeration of 0.60 rad/s2 for 0.50 min?

2. The moon orbits Earth in 27.3 days. Suppose a spacecraft leaves the

moon and follows the same orbit as the moon. If the spacecraft has a

constant angular acceleration of 1.0 × 10–10 rad/s2, what is its angular

speed after 12 h of flight? (Hint: At takeoff, the spaceship has the same

angular speed as the moon.)

3. African baobab trees can have circumferences of up to 43 m. Imagine

riding a bicycle around a tree this size. If, starting from rest, you

travel a distance of 160 m around the tree with a constant angular ac-

celeration of 5.0 × 10–2 rad/s2, what will your final angular speed be?

4. In 1976, Kathy Wafler produced an unbroken apple peel 52.5 m in

length. Suppose Wafler turned the apple with a constant angular

acceleration of –3.2 × 10–5 rad/s2, until her final angular speed was

0.080 rad/s. Assuming the apple was a sphere with a radius of 8.0 cm,

calculate the apple’s initial angular speed.


NAME ______________________________________ DATE _______________ CLASS ____________________

5. In 1987, a giant hanging basket of flowers with a mass of 4000 kg was

constructed. The radius of the basket was 3.0 m. Suppose this basket

was placed on the ground and an admiring spectator ran around it to

see every detail again and again. At first the spectator’s angular speed

was 0.820 rad/s, but he steadily decreased his speed to 0.360 rad/s by

the time he had traveled 20.0 m around the basket. Find the spectator’s

angular acceleration.

6. One of the largest scientific devices in the world is the particle accelera-

tor at Fermilab, in Batavia, Illinois. The accelerator consists of a giant

ring with a radius of 1.0 km. Suppose a maintenance engineer drives

around the accelerator, starting at an angular speed of 5.0 × 10–3 rad/s

and accelerating at a constant rate until one trip is completed in

14.0 min. Find the engineer’s angular acceleration.

7. With a diameter of 207.0 m, the Superdome in New Orleans, Louisiana,

is the largest “dome” in the world. Imagine a race held along the Super-

dome’s outside perimeter. An athlete whose initial angular speed is

7.20 × 10–2 rad/s runs 12.6 rad in 4 min 22 s. What is the athlete’s con-

stant angular acceleration?

8. In 1986, Fred Markham rode a bicycle that was pulled by an automo-

bile 200 m in 6.83 s. Suppose the angular speed of the bicycle’s wheels

increased steadily from 27.0 rad/s to 32.0 rad/s. Find the wheels’ angu-

lar acceleration.

9. Consider a common analog clock. At midnight, the hour and minute

hands coincide. Then the minute hand begins to rotate away from the

hour hand. Suppose you adjust the clock by pushing the hour hand

clockwise with a constant acceleration of 2.68 × 10–5 rad/s2. What is the

angular displacement of the hour hand after 120.0 s? (Note that the un-

accelerated hour hand makes one full rotation in 12 h.)

10. Herman Bax of Canada grew a pumpkin with a circumference of 7 m.

Suppose an ant crawled around the pumpkin’s “equator.” The ant

started with an angular speed of 6.0 × 10–3 rad/s and accelerated

steadily at a rate of 2.5 × 10–4 rad/s2 until its angular speed was tripled.

What was the ant’s angular displacement?

11. Tal Burt of Israel rode a bicycle around the world in 77 days. If Burt

could have ridden along the equator, his average angular speed would

have been 9.0 × 10–7 rad/s. Now consider an object moving with this an-

gular speed. How long would it take the object to reach an angular speed

of 5.0 × 10–6 rad/s if its angular acceleration was 7.5 × 10–10 rad/s2?

12. A coal-burning power plant in Kazakhstan has a chimney that is nearly

500 m tall. The radius of this chimney is 7.1 m at the base. Suppose a

factory worker takes a 500.0 m run around the base of the chimney. If

the worker starts with an angular speed of 0.40 rad/s and has an angu-

lar acceleration of 4.0 × 10–3 rad/s2, how long will the run take?

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Problem 7E 77

NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 7ETANGENTIAL SPEED

P R O B L E MIn about 45 min, Nicholas Mason inflated a weather balloon using onlylung power. If a fly, moving with a tangential speed of 5.11 m/s, were tomake exactly 8 revolutions around this inflated balloon in 12.0 s, whatwould the balloon’s radius be?

S O L U T I O NGiven: vt = 5.11 m/s

∆q = 8 rev

∆t = 12.0 s

Unknown: w = ? r = ?

Use ∆q and ∆t to calculate the fly’s angular speed. Then rearrange the tangential

speed equation to determine the balloon’s radius.

w = ∆∆qt

= = 4.19 rad/s

vt = rw

r = wvt =

4

5

.1

.1

9

1

r

m

ad

/

/

s

s = 1.22 m

(8 rev)2p r

r

a

e

d

v

12.0 s

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ADDITIONAL PRACTICE

1. The world’s tallest columns, which stand in front of the Education

Building in Albany, New York, are each 30 m tall. If a fly circles a col-

umn with an angular speed of 4.44 rad/s, and its tangential speed is

4.44 m/s, what is the radius of the column?

2. The longest dingoproof wire fence stretches across southeastern Aus-

tralia. Suppose this fence were to have a circular shape. A rancher driv-

ing around the perimeter of the fence with a tangential speed of

16.0 m/s has an angular speed of 1.82 × 10–5 rad/s. What is the fence’s

radius and length (circumference)?

3. The smallest self-sustaining gas turbine has a tiny wheel that can rotate

at 5.24 × 103 rad/s. If the wheel rim’s tangential speed is 131 m/s, what

is the wheel’s radius?

4. Earth’s average tangential speed around the sun is about 29.7 km/s. If

Earth’s average orbital radius is 1.50 × 108 km, what is its angular or-

bital speed in rad/s?


NAME ______________________________________ DATE _______________ CLASS ____________________

5. Two English engineers designed a ridable motorcycle that was less than

12 cm long. The front wheel’s diameter was only 19.0 mm. Suppose this

motorcycle was ridden so that the front wheel had an angular speed of

25.6 rad/s. What would the tangential speed of the front wheel’s rim

have been?

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Problem 7F 79

NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 7FTANGENTIAL SPEED

P R O B L E MIn 1988, Stu Cohen made his kite perform 2911 figure eights in just 1 h. Ifthe kite made a circular “loop” with a radius of 1.5 m and had a tangentialacceleration of 0.6 m/s2, what was the kite’s angular acceleration?

S O L U T I O N

Given: r = 1.5 m at = 0.6 m/s2

Unknown: a = ?

Apply the tangential acceleration equation, solving for angular acceleration.

at = ra

a = a

rt =

0.

1

6

.5

m

m

/s2

= 0.4 rad/s2

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1. The world’s largest aquarium, at the EPCOT center in Orlando, Florida,

has a radius of 32 m. Bicycling around this aquarium with a tangential

acceleration of 0.20 m/s2, what would your angular acceleration be?

2. Dale Lyons and David Pettifer ran the London marathon in less than

4 h while bound together at one ankle and one wrist. Suppose they

made a turn with a radius of 8.0 m. If their tangential acceleration was

−1.44 m/s2, what was their angular acceleration?

3. In 1991, Timothy Badyna of the United States ran 10 km backward in

just over 45 min. Suppose Badyna made a turn, so that his angular

speed decreased 2.4 × 10–2 rad/s in 6.0 s. If his tangential acceleration

was –0.16 m/s2, what was the radius of the turn?

4. Park Bong-tae of South Korea made 14 628 turns of a jump rope in

1.000 h. Suppose the rope’s average angular speed was gained from rest

during the first turn and that the rope during this time had a tangential

acceleration of 33.0 m/s2. What was the radius of the rope’s circular path?

5. To “crack” a whip requires making its tip move at a supersonic speed.

Kris King of Ohio achieved this with a whip 56.24 m long. If the tip of

this whip moved in a circle and its angular speed increased from

6.00 rad/s to 6.30 rad/s in 0.60 s, what would the magnitude of the tip’s

tangential acceleration be?

6. In 1986 in Japan, a giant top with a radius of 1.3 m was spun. The top

remained spinning for over an hour. Suppose these people accelerated

the top from rest so that the first turn was completed in 1.8 s. What was

the tangential acceleration of points on the top’s rim?

ADDITIONAL PRACTICE

Holt Physics

Problem 7GCENTRIPETAL ACCELERATION

P R O B L E MCalculate the orbital radius of the Earth, if its tangential speed is 29.7 km/sand the centripetal acceleration acting on Earth is 5.9 10–3 m/s2.

S O L U T I O N

Given: vt = 29.7 km/s ac = 5.9 × 10−3 m/s2

Unknown: r = ?

Use the centripetal acceleration equation written in terms of tangential speed.

Rearrange the equation to solve for r.

ac = v

rt2

r = v

at

c

2

= (2

5

9

.9

.7

××1

1

0

0−3

3

m

m

/

/

s

s2)2

= 1.5 × 1011 m = 1.5 × 108 km


NAME ______________________________________ DATE _______________ CLASS ____________________

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ADDITIONAL PRACTICE

1. The largest salami in the world, made in Norway, was more than 20 m

long. If a hungry mouse ran around the salami’s circumference with a

tangential speed of 0.17 m/s, the centripetal acceleration of the mouse

was 0.29 m/s2. What was the radius of the salami?

2. The largest steerable single-dish radio telescope is located at the branch

of the Max Planck Institute in Bonn, Germany. Suppose this telescope

rotates about its axis with the same angular speed as Earth. The cen-

tripetal acceleration of the points at the edge of the telescope is 2.65 ×10−7 m/s2. What is the radius of the telescope dish?

3. In 1994, Susan Williams of California blew a bubble-gum bubble

with a diameter of 58.4 cm. If this bubble were rigid and the centripetal

acceleration of the equatorial points of the bubble were 8.50 × 10−2 m/s2,

what would the tangential speed of those points be?

4. An ostrich lays the largest bird egg. A typical diameter for an ostrich

egg at its widest part is 12 cm. Suppose an egg of this size rolls down a

slope so that the centripetal acceleration of the shell at its widest part is

0.28 m/s2. What is the tangential speed of that part of the shell?

5. A waterwheel built in Hamah, Syria, rotates continuously. The wheel’s

radius is 20.0 m. If the wheel rotates once in 16.0 s, what is the magni-

tude of the centripetal acceleration of the wheel’s edge?

6. In 1995, Cathy Marsal of France cycled 47.112 km in 1.000 hour. Calcu-

late the magnitude of the centripetal acceleration of Marsal with re-

spect to Earth’s center. Neglect Earth’s rotation, and use 6.37 × 103 km as

Earth’s radius.

Problem 7H 81

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Holt Physics

Problem 7HFORCE THAT MAINTAINS CIRCULATION

P R O B L E MThe royal antelope of western Africa has an average mass of only 3.2 kg.Suppose this antelope runs in a circle with a radius of 30.0 m. If a force of8.8 N maintains this circular motion, what is the antelope’s tangentialspeed?


r = 30.0 m

Fc = 8.8 N

Unknown: vt = ?

Use the equation for the force that maintains circular motion, and rearrange it to

solve for tangential speed.

Fc = m

r

vt2

vt = F

mcr = (8.8 N

3.

)

2

(3

k0

g

.0m) = 82

m

s2

2

vt = 9.1 m/s

1. Gregg Reid of Atlanta, Georgia, built a motorcycle that is over 4.5 m

long and has a mass of 235 kg. The force that holds Reid and his motor-

cycle in a circular path with a radius 25.0 m is 1850 N. What is Reid’s

tangential speed? Assume Reid’s mass is 72 kg.

2. With an average mass of only 30.0 g, the mouse lemur of Madagascar is

the smallest primate on Earth. Suppose this lemur swings on a light

vine with a length of 2.4 m, so that the tension in the vine at the bot-

tom point of the swing is 0.393 N. What is the lemur’s tangential speed

at that point?

3. In 1994, Mata Jagdamba of India had very long hair. It was 4.23 m long.

Suppose Mata conducted experiments with her hair. First, she deter-

mined that one hair strand could support a mass of 25 g. She then at-

tached a smaller mass to the same hair strand and swung it in the hori-

zontal plane. If the strand broke when the tangential speed of the mass

reached 8.1 m/s, how large was the mass?

4. Pat Kinch used a racing cycle to travel 75.57 km/h. Suppose Kinch

moved at this speed around a circular track. If the combined mass of

Kinch and the cycle was 92.0 kg and the average force that maintained

his circular motion was 12.8 N, what was the radius of the track?

ADDITIONAL PRACTICE

5. In 1992, a team of 12 athletes from Great Britain and Canada rappelled

446 m down the CN Tower in Toronto, Canada. Suppose an athlete

with a mass of 75.0 kg, having reached the ground, took a joyful swing

on the 446 m-long rope. If the speed of the athlete at the bottom point

of the swing was 12 m/s, what force maintained the athlete’s circular

motion? What was the tension in the rope? Neglect the rope’s mass.


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Problem 7I 83

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Holt Physics

Problem 7IGRAVITATIONAL FORCE

P R O B L E M

The sun has a mass of 2.0 1030 kg and a radius of 7.0 105 km. Whatmass must be located at the sun’s surface for a gravitational force of470 N to exist between the mass and the sun?

S O L U T I O N

Given: m1 = 2.0 × 1030 kg

r = 7.0 × 105 km = 7.0 × 108 m

G = 6.673 × 10−11 N•m2/kg2

Fg = 470 N

Unknown: m2 = ?

Use Newton’s universal law of gravitation, and rearrange it to solve for the second

mass.

Fg = G m1

r2m2

m2 = =

m2 = 1.7 kg

(470 N)(7.0 × 108 m)2

6.673 × 10−11 N

k

•

g

m2

2

(2.0 × 1030 kg)

Fg r2

G m1

ADDITIONAL PRACTICE

1. Deimos, a satellite of Mars, has an average radius of 6.3 km. If the gravi-

tational force between Deimos and a 3.0 kg rock at its surface is 2.5 ×10–2 N, what is the mass of Deimos?

2. A 3.08 × 104 kg meteorite is on exhibit in New York City. Suppose this

meteorite and another meteorite are separated by 1.27 × 107 m (a dis-

tance equal to Earth’s average diameter). If the gravitational force be-

tween them is 2.88 × 10−16 N, what is the mass of the second meteorite?

3. In 1989, a cake with a mass of 5.81 × 104 kg was baked in Alabama.

Suppose a cook stood 25.0 m from the cake. The gravitational force ex-

erted between the cook and the cake was 5.0 × 10−7 N. What was the

cook’s mass?

4. The largest diamond ever found has a mass of 621 g. If the force of

gravitational attraction between this diamond and a person with a

mass of 65.0 kg is 1.0 × 10–12 N, what is the distance between them?

5. The passenger liners Carnival Destiny and Grand Princess, built re-

cently, have a mass of about 1.0 × 108 kg each. How far apart must

these two ships be to exert a gravitational attraction of 1.0 × 10−3 N on

each other?

6. In 1874, a swarm of locusts descended on Nebraska. The swarm’s mass

was estimated to be 25 × 109 kg. If this swarm were split in half and the

halves separated by 1.0 × 103 km, what would the magnitude of the

gravitational force between the halves be?

7. Jupiter, the largest planet in the solar system, has a mass 318 times that

of Earth and a volume that is 1323 times greater than Earth’s. Calculate

the magnitude of the gravitational force exerted on a 50.0 kg mass on

Jupiter’s surface.


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Problem 8A 85

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Holt Physics

Problem 8ATORQUE

P R O B L E MA beam that is hinged near one end can be lowered to stop traffic at a rail-road crossing or border checkpoint. Consider a beam with a mass of12.0 kg that is partially balanced by a 20.0 kg counterweight. The counter-weight is located 0.750 m from the beam’s fulcrum. A downward force of1.60 102 N applied over the counterweight causes the beam to move up-ward. If the net torque on the beam is 29.0 N•m when the beam makes anangle of 25.0° with respect to the ground, how long is the beam’s longersection? Assume that the portion of the beam between the counterweightand fulcrum has no mass.

S O L U T I O NGiven: mb = 12.0 kg

mc = 20.0 kg

dc = 0.750 m

Fapplied = 1.60 × 102 N

tnet = 29.0 N•m

q = 90.0° − 25.0° = 65.0°

g = 9.81 m/s2

Unknown: l = ?

Diagram:

Choose the equation(s) or situation: Apply the definition of torque to each

force and add up the individual torques.

t = F d (sin q)

tnet = ta + tb + tc

where ta = counterclockwise torque produced by applied force = Fapplied dc (sin q)

tb = clockwise torque produced by weight of beam

= −mb g (sin q)

tc = counterclockwise torque produced by counterweight

= mc g dc (sin q)

tnet = Fapplied dc (sin q) − mb g (sin q) + mc g dc (sin q)

Note that the clockwise torque is negative, while the counterclockwise torques are

positive.

l2

l2

mb g

dc

θ

1. DEFINE

2. PLAN


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mb g = (Fapplied + mc g) dc − stinne

qt

l =


l =

l =

l =

l =

l =

l =

For a constant applied force, the net torque is greatest when q is 90.0° and de-

creases as the beam rises. Therefore, the beam rises fastest initially.

3.99 m

(2)(235 N•m)(12.0 kg)(9.81 m/s2)

(2)(2.67 × 102 N•m − 32.0 N•m]

(12.00 kg)(9.81 m/s2)

(2) [356 N)(0.750 m) − 32.0 N•m]

(12.0 kg)(9.81 m/s2)

(2) [(1.60 × 102 N + 196 N)(0.750 m) − 32.0 N•m]

(12.0 kg)(9.81 m/s2)

(2)([1.60 × 102 N + (20.0 kg)(9.81 m/s2)] (0.750 m) − 2s

9

i

.

n

0

6

N

5.

•

0

m

°

(12.0 kg)(9.81 m/s2)

2(Fapplied + mc g)dc − stinne

qt

mb g

l2

3. CALCULATE

4. EVALUATE

ADDITIONAL PRACTICE

1. The nests built by the mallee fowl of Australia can have masses as large as

3.00 × 105 kg. Suppose a nest with this mass is being lifted by a crane.

The boom of the crane makes an angle of 45.0° with the ground. If the

axis of rotation is the lower end of the boom at point A, the torque pro-

duced by the nest has a magnitude of 3.20 × 107 N•m. Treat the boom’s

mass as negligible, and calculate the length of the boom.

2. The pterosaur was the most massive flying dinosaur. The average mass

for a pterosaur has been estimated from skeletons to have been between

80.0 and 120.0 kg. The wingspan of a pterosaur was greater than 10.0 m.

Suppose two pterosaurs with masses of 80.0 kg and 120.0 kg sat on the

middle and the far end, respectively, of a light horizontal tree branch.

The pterosaurs produced a net counterclockwise torque of 9.4 kN •m

about the end of the branch that was attached to the tree. What was the

length of the branch?

Problem 8A 87

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3. A meterstick of negligible mass is fixed horizontally at its 100.0 cm mark.

Imagine this meterstick used as a display for some fruits and vegetables

with record-breaking masses. A lemon with a mass of 3.9 kg hangs from

the 70.0 cm mark, and a cucumber with a mass of 9.1 kg hangs from the

x cm mark. What is the value of x if the net torque acting on the meter-

stick is 56.0 N•m in the counterclockwise direction?

4. In 1943, there was a gorilla named N’gagi at the San Diego Zoo. Suppose

N’gagi were to hang from a bar. If N’gagi produced a torque of –1.3 ×104 N•m about point A, what was his weight? Assume the bar has negli-

gible mass.

5. The first—and, in terms of the number of passengers it could carry, the

largest—Ferris wheel ever constructed had a diameter of 76 m and held

36 cars, each carrying 60 passengers. Suppose the magnitude of the

torque, produced by a ferris wheel car and acting about the center of the

wheel, is –1.45 × 106 N•m. What is the car’s weight?

6. In 1897, a pair of huge elephant tusks were obtained in Kenya. One tusk

had a mass of 102 kg, and the other tusk’s mass was 109 kg. Suppose both

tusks hang from a light horizontal bar with a length of 3.00 m. The first

tusk is placed 0.80 m away from the end of the bar, and the second, more

massive tusk is placed 1.80 m away from the end. What is the net torque

produced by the tusks if the axis of rotation is at the center of the bar?

Neglect the bar’s mass.

7. A catapult, a device used to hurl heavy objects such as large stones, con-

sists of a long wooden beam that is mounted so that one end of it pivots

freely in a vertical arc. The other end of the beam consists of a large hol-

lowed bowl in which projectiles are placed. Suppose a catapult provides

an angular acceleration of 50.0 rad/s2 to a 5.00 × 102 kg boulder. This can

be achieved if the net torque acting on the catapult beam, which is 5.00 m

long, is 6.25 × 105 N•m.

a. If the catapult is pulled back so that the beam makes an angle of

10.0° with the horizontal, what is the magnitude of the torque pro-

duced by the 5.00 × 102 kg boulder?

b. If the force that accelerates the beam and boulder acts perpendicu-

larly on the beam 4.00 m from the pivot, how large must that force

be to produce a net torque of 6.25 × 105 N•m?

Holt Physics

Problem 8B


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ROTATIONAL EQUILIBRIUM

P R O B L E M

In 1960, a polar bear with a mass of 9.00 × 102 kg was discovered inAlaska. Suppose this bear crosses a 12.0 m long horizontal bridge thatspans a gully. The bridge consists of a wide board that has a uniform mass of 2.50 × 102 kg and whose ends are loosely set on either side of thegully. When the bear is two-thirds of the way across the bridge, what isthe normal force acting on the board at the end farthest from the bear?

S O L U T I O N

Given: mb = mass of bridge = 2.50 × 102 kg

mp = mass of polar bear = 9.00 × 102 kg

l = length of bridge = 12.0 m

g = 9.81 m/s2

Unknown: Fn,1 = ?

Diagram:

Choose the equation(s) or situation:

Apply the first condition of equilibrium: The unknowns in this problem are the

normal forces exerted upward by the ground on either end of the board. The known

quantities are the weights of the bridge and the polar bear. All of the forces are in the

vertical (y) direction.

Fy = Fn,1 + Fn,2 − mbg − mpg = 0

Because there are two unknowns and only one equation, the solution cannot be

obtained from the first condition of equilibrium alone.

Choose a point for calculating net torque: Choose the end of the bridge farthest

from the bear as the pivot point. The torque produced by Fn,1 will be zero.

Apply the second condition of equilibrium: The torques produced by the bridge’s

and polar bear’s weights are clockwise and therefore negative. The normal force on

the end of the bridge opposite the axis of rotation exerts a counterclockwise (positive)

torque.

tnet = −(mbg)db − (mpg)dp + Fn,2d2 = 0

12.0 m

Fn,1 Fn,2mp mb

1. DEFINE

2. PLAN

Problem 8B 89

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The lever arm for the bridge’s weight (db) is the distance from the bridge’s

center of mass to the pivot point, or half the bridge’s length. The lever arm for the

polar bear is two-thirds the bridge’s length. The lever arm for the normal force

farthest from the pivot equals the entire length of the bridge.

db = 12

l , dp = 23

l , d2 = l

The torque equation thus takes the following form:

tnet = − − + Fn,2l = 0

Rearrange the equation(s) to isolate the unknowns:

Fn,2 = + = m

2b +

2

3

mp g

Fn,1 = mb g + mp g − Fn,2


Fn,2 = 2.50 ×2

102 kg +

(2)(9.00

3

× 102 kg) (9.81 m/s2)

Fn,2 = (125 kg + 6.00 × 102 kg)(9.81 m/s2)

Fn,2 = (725 kg)(9.81 m/s2)

Fn,2 = 7.11 × 103 N

Fn,1 = (2.50 × 102 kg)(9.81 m/s2) + (9.00 × 102 kg)(9.81 m/s2)

− 7.11 × 103 N

Fn,1 = 2.45 × 103 N + 8.83 × 103 N − 7.11 × 103 N

Fn,1 =

The sum of the upward normal forces exerted on the ends of the bridge must

equal the weight of the polar bear and the bridge. (The individual normal forces

change as the polar bear moves across the bridge.)

(4.17 kN + 7.11 kN) = (2.50 × 102 kg + 9.00 × 102 kg)(9.81 m/s2)

11.28 kN = 11.28 × 103 N

4.17 × 103 N

2mp gl3l

mb gl2l

2mp gl3

mb gl2

ADDITIONAL PRACTICE

3. CALCULATE

4. EVALULATE

1. The heaviest sea sponge ever collected had a mass of 40.0 kg, but after

drying out, its mass decreased to 5.4 kg. Suppose two loads equal to the

wet and dry masses of this giant sponge hang from the opposite ends of a

horizontal meterstick of negligible mass and that a fulcrum is placed

70.0 cm from the larger of the two masses. How much extra force must

be applied to the end of the meterstick with the smaller mass in order to

provide equilibrium?

2. A Saguaro cactus with a height of 24 m and an estimated age of 150 years

was discovered in 1978 in Arizona. Unfortunately, a storm toppled it in

1986. Suppose the storm produced a torque of 2.00 × 105 N•m that

acted on the cactus. If the cactus could withstand a torque of only


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1.2 × 105 N•m, what minimum force could have been applied to the cac-

tus keep it standing? At what point and in what direction should this

force have been applied? Assume that the cactus itself was very strong

and that the roots were just pulled out of the ground.

3. In 1994, John Evans set a record for brick balancing by holding a load of

bricks with a mass of 134 kg on his head for 10 s. Another, less extreme,

method of balancing this load would be to use a lever. Suppose a board

with a length of 7.00 m is placed on a fulcrum and the bricks are set on

one end of the board at a distance of 2.00 m from the fulcrum. If a force

is applied at a right angle to the other end of the board and the force has

a direction that is 60.0° below the horizontal and away from the bricks,

how great must this force be to keep the load in equilibrium? Assume the

board has negligible mass.

4. In 1994, a vanilla ice lollipop with a mass of 8.8 × 103 kg was made in

Poland. Suppose this ice lollipop was placed on the end of a lever 15 m in

length. A fulcrum was placed 3.0 m from the lollipop so that the lever

made an angle of 20.0° with the ground. If the force was applied perpen-

dicular to the lever, what was the smallest magnitude this force could

have and still lift the lollipop? Neglect the mass of the lever.

5. The Galápagos fur seals are very small. An average adult male has a mass

of 64 kg, and a female has a mass of only 27 kg. Suppose one average

adult male seal and one average adult female seal sit on opposite ends of

a light board that has a length of 3.0 m. How far from the male seal

should the board be pivoted in order for equilibrium to be maintained?

6. Goliath, a giant Galápagos tortoise living in Florida, has a mass of 3.6 ×102 kg. Suppose Goliath walks along a heavy board above a swimming

pool. The board has a mass of 6.0 × 102 kg and a length of 15 m, and it is

placed horizontally on the edge of the pool so that only 5.0 m of it ex-

tends over the water. How far out along this 5.0 m extension of the board

can Goliath walk before he falls into the pool?

7. The largest pumpkin ever grown had a mass of 449 kg. Suppose this

pumpkin was placed on a platform that was supported by two bases

5.0 m apart. If the left base exerted a normal force of 2.70 × 103 N on the

platform, how far must the pumpkin have been from the platform’s left

edge? The platform had negligible mass.

8. In 1991, a giant stick of Brighton rock (a type of rock candy) was made

in England. The candy had a mass of 414 kg and a length of 5.00 m.

Imagine that the candy was balanced horizontally on a fulcrum. A child

with a mass of 40.0 kg sat on one end of the stick. How far must the ful-

crum have been from the child in order to maintain equilibrium?

Problem 8C 91

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Holt Physics

Problem 8CNEWTON’S SECOND LAW FOR ROTATION

P R O B L E MThe giant sequoia General Sherman in California has a mass of about 2.00 106 kg, making it the most massive tree in the world. Its height of 83.0 mis also impressive. Imagine a uniform bar with the same mass and length asthe tree. If this bar is rotated about an axis that is perpendicular to andpasses through the bar’s midpoint, how large an angular acceleration wouldresult from a torque of 4.60 107 N•m? (Note: Assume the bar is thin.)

S O L U T I O N

Given: M = 2.00 × 106 kg

l = 83.0 m

t = 4.60 × 107 N•m

Unknown: a × ?

Calculate the bar’s moment of inertia using the formula for a thin rod with the

axis of rotation at its center.

I = 1

1

2 M l 2 =

1

1

2 (2.00 × 106 kg)(83.0 m)2 = 1.15 × 109 kg•m2

Now use the equation for Newton’s second law for rotating objects. Rearrange the

equation to solve for angular acceleration.

t = Ia

a = tI

= (

(

1

4

.1

.6

5

0

××

1

1

0

09

7

k

N

g•

•

m

m2)

) = 4.00 × 10−2 rad/s2

ADDITIONAL PRACTICE

1. One of the largest Ferris wheels currently in existence is in Yokohama,

Japan. The wheel has a radius of 50.0 m and a mass of 1.20 × 106 kg. If

a torque of 1.0 × 109 N•m is needed to turn the wheel from a state of

rest, what would the wheel’s angular acceleration be? Treat the wheel as

a thin hoop.

2. In 1992, Jacky Vranken from Belgium attained a speed of more than

250 km/h on just the back wheel of a motorcycle. Assume that all of the

back wheel’s mass is located at its outer edge. If the wheel has a mass of

22 kg and a radius of 0.36 m, what is the wheel’s angular acceleration

when a torque of 5.7 N•m acts on the wheel?

3. In 1995, a fully functional pencil with a mass of 24 kg and a length of

2.74 m was made. Suppose this pencil is suspended at its midpoint and

a force of 1.8 N is applied perpendicular to its end, causing it to rotate.

What is the angular acceleration of the pencil?


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4. The turbines at the Grand Coulee Third Power Plant in the state of

Washington have rotors with a mass of 4.07 × 105 kg and a radius of

5.0 m each. What angular acceleration would one of these rotors have if

a torque of 5.0 × 104 N•m were applied? Assume the rotor is a uniform

disk.

5. J. C. Payne of Texas amassed a ball of string that had a radius of 2.00 m.

Suppose a force of 208 N was applied tangentially to the ball’s surface

in order to give the ball an angular acceleration of 3.20 × 10–2 rad/s2.

What was the ball’s moment of inertia?

6. The heaviest member of British Parliament ever was Sir Cyril Smith.

Calculate his peak mass by finding first his moment of inertia from the

following situation. If Sir Cyril were to have ridden on a merry-go-

round with a radius of 8.0 m, a torque of 7.3 × 103 N•m would have

been needed to provide him with an angular acceleration of 0.60 rad/s2.

7. In 1975, a centrifuge at a research center in England made a carbon-

fiber rod spin about its center so fast that the tangential speed of the

rod’s tips was about 2.0 km/s. The length of the rod was 15.0 cm. If it

took 80.0 s for a torque of 0.20 N•m to bring the rod to rest from its

maximum speed, what was the rod’s moment of inertia?

8. The largest tricycle ever built had rear wheels that were almost 1.70 m

in diameter. Neglecting the mass of the spokes, the moment of inertia

of one of these wheels is equal to that of a thin hoop rotated about its

symmetry axis. Find the wheel’s moment of inertia and its mass if a

torque of 125 N•m is applied to the wheel so that in 2.0 s the wheel’s

angular speed increases from 0 rad/s to 12 rad/s.

9. In 1990, a cherry pie with a radius of 3.00 m and a mass of 17 × 103 kg

was baked in Canada. Suppose the pie was placed on a light rotating

platform attached to a motor. If this motor brought the angular speed

of the pie from 0 rad/s to 3.46 rad/s in 12 s, what was the torque the

motor must have produced? Assume the mass of the platform was neg-

ligible and the pie was a uniformly solid disk.

10. In just over a month in 1962, a shaft almost 4.00 × 108 m deep and with

a radius of 4.0 m was drilled in South Africa. The mass of the soil taken

out was about 1.0 × 108 kg. Imagine a rigid cylinder with a mass, ra-

dius, and length equal to these values. If this cylinder rotates about its

symmetry axis so that it undergoes a constant angular acceleration

from 0 rad/s to 0.080 rad/s in 60.0 s, how large a torque must act on the

cylinder?

11. In 1993, a bowl in Canada was filled with strawberries. The mass of the

bowl and strawberries combined was 2390 kg, and the moment of iner-

tia about the symmetry axis was estimated to be 2.40 × 103 kg•m2. Sup-

pose a constant angular acceleration was applied to the bowl so that it

made its first two complete rotations in 6.00 s. How large was the

torque that acted on the bowl?

Problem 8C 93

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12. A steel ax with a mass of 7.0 × 103 kg and a length of 18.3 m was made

in Canada. If Paul Bunyan were to take a swing with such an ax, what

torque would he have to produce in order for the blade to have a tan-

gential acceleration of 25 m/s2? Assume that the blade follows a circle

with a radius equal to the ax handle’s length and that nearly all of the

mass is concentrated in the blade.

Holt Physics

Problem 8DCONSERVATION OF ANGULAR MOMENTUM

P R O B L E M

The average distance from Earth to the moon is 3.84 105 km. The aver-age orbital speed of the moon when it is at its average distance fromEarth is 3.68 103 km/h. However, in 1912 the average orbital speed was3.97 103 km/h, and in 1984 it was 3.47 103 km/h. Calculate the dis-tances that correspond to the 1912 and 1984 orbital speeds, respectively.

S O L U T I O N

Given: ravg = 3.84 × 105 km

vavg = 3.68 × 103 km/h

v1 = 3.97 × 103 km/h

v2 = 3.47 × 103 km/h

Unknown: r1 = ? r2 = ?

Choose the equation(s) or situation: Because there are no external torques, the

angular momentum of the Earth-moon system is conserved.

Lavg = L1 = L2

Iavg wavg = I1 w1 = I2 w2

If the moon is treated as a point mass revolving around a central axis, its moment

of inertia is simply mr2, and the conservation of momentum expression takes the

following form:

mmoon (vavg)2 v

ra

a

v

v

g

g = mmoon (r1)2 v

r1

1 = mmoon (r2)2 v

r2

2Because the mass of the moon is unchanged, the mass term cancels, and the

equation reduces to the following:

ravg vavg = r1 v1 = r2 v2


r1 = ravg

v1

vavg r2 = ravg

v2

vavg


r1 = =

r2 = =

Because angular momentum is conserved in the absence of external torques, the

tangential orbital speed of the moon is greater than its average value when the

moon is closer to Earth. Similarly, the smaller tangential orbital speed occurs

when the moon is farther from Earth.

4.07 × 105 km(3.84 × 105 km)(3.68 × 103 km/h)

(3.47 × 103 km/h)

3.56 × 105 km(3.84 × 105 km)(3.68 × 103 km/h)

(3.97 × 103 km/h)

1. DEFINE

2. PLAN

3. CALCULATE

4. EVALUATE


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Problem 8D 95

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1. Encke’s comet revolves around the sun in a period of just over three years

(the shortest period of any comet). The closest it approaches the sun is

4.95 × 107 km, at which time its orbital speed is 2.54 × 105 km/h. At what

distance from the sun would Encke’s comet have a speed equal to 1.81 ×105 km/h?

2. In 1981, Sammy Miller reached a speed of 399 km/h on a rocket-

powered ice sled. Suppose the sled, moving at its maximum speed, was

hooked to a post with a radius of 0.20 m by a light cord with an un-

known initial length. The rocket engine was then turned off, and the sled

began to circle the post with negligible resistance as the cord wrapped

around the post. If the speed of the sled after 20 turns was 456 km/h,

what was the length of the unwound cord?

3. Earth is not a perfect sphere, in part because of its rotation about its axis.

A point on the equator is in fact over 21 km farther from Earth’s center

than is the North pole. Suppose you model Earth as a solid clay sphere

with a mass of 25.0 kg and a radius of 15.0 cm. If you begin rotating the

sphere with a constant angular speed of 4.70 × 10–3rad/s (about the same

as Earth’s), and the sphere continues to rotate without the application of

any external torques, what will the change in the sphere’s moment of in-

ertia be when the final angular speed equals 4.74 × 10–3 rad/s?

4. In 1971, a model plane built in the Soviet Union by Leonid Lipinsky

reached a speed of 395 km/h. The plane was held in a circular path by a

control line. Suppose the plane ran out of gas while moving at its maxi-

mum speed and Lipinsky pulled the line in to bring the plane home

while it continued in a circular path. If the line’s initial length is 1.20 ×102 m and Lipinsky shortened the line by 0.80 m every second, what was

the plane’s speed after 32 s?

5. The longest spacewalk by a team of astronauts lasted more than 8 h. It

was performed in 1992 by three crew members from the space shuttle

Endeavour. Suppose that during the walk two astronauts with equal

masses held the opposite ends of a rope that was 10.0 m long. From the

point of view of the third astronaut, the other two astronauts rotated

about the midpoint of the rope with an angular speed of 1.26 rad/s. If

the astronauts shortened the rope equally from both ends, what was their

angular speed when the rope was 4.00 m long?

6. In a problem in the previous section, a cherry pie with a radius of 3.00 m

and a mass of 17 × 103 kg was rotated on a light platform. Suppose that

when the pie reached an angular speed of 3.46 rad/s there was no net

torque acting on it. Over time, the filling in the pie began to move out-

ward, changing the pie’s moment of inertia. Assume the pie acted like a

uniform, rigid, spinning disk with a mass of 16.80 × 103 kg combined

with a 0.20 × 103 kg particle. If the smaller mass shifted from a position

2.50 m from the center to one that was 3.00 m from the center, what was

the change in the angular speed of the pie?

ADDITIONAL PRACTICE

Holt Physics

Problem 8ECONSERVATION OF MECHANICAL ENERGY

P R O B L E MIn 1990, Eddy McDonald of Canada completed 8437 loops with a yo-yo inan hour, setting a world record. Assume that the yo-yo McDonald usedhad a mass of 6.00 10–2 kg. The yo-yo descended from a height of 0.600 mdown a vertical string and had a linear speed of 1.80 m/s by the time itreached the bottom of the string. If its final angular speed was 82.6 rad/s,what was the yo-yo’s moment of inertia?

S O L U T I O N

Given: m = 6.00 × 10−2 kg h = 0.600 m

vf = 1.80 m/s wf = 82.6 rad/s

g = 9.81 m/s2

Unknown: I = ?

Choose the equation(s) or situation: Apply the principle of conservation of

mechanical energy.

MEi = MEf

Initially, the system possesses only gravitational potential energy.

MEi = PEg = mgh

When the yo-yo reaches the bottom of the string, this potential energy has been

converted to translational and rotational kinetic energy.

MEf = KEtrans + KErot = 12

mvf2 + 1

2Iwf

2

Equate the initial and final mechanical energy.

mgh = 12

mvf2 + 1

2Iwf

2


12

Iwf2 = mgh – 1

2mvf

2 I = 2 mgh

w−

f2

mvf2

= m(2g

wh

f

−2

vf2)


I =

I = =

Assuming that the yo-yo can be approximated by a solid disk with a central axis of

rotation, the yo-yo’s moment of inertia is described by the equation I = 12

MR2.

From the calculated value for I, the yo-yo’s radius can be found to be 5.0 cm.

R = M

2I = = 0.050 m

(2)(7.6 × 10−5 kg •m2)

6.00 × 10−2 kg

7.6 × 10−5 kg•m2(6.00 × 10−2 kg)(8.6 m2/s2)

(82.6 rad/s)2

(6.00 × 10−2 kg)[(2) (9.81 m/s2)(0.600 m) − (1.80 m/s)2]

(82.6 rad/s)2

1. DEFINE

2. PLAN

3. CALCULATE

4. EVALUATE


NAME ______________________________________ DATE _______________ CLASS ____________________

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Problem 8E 97

NAME ______________________________________ DATE _______________ CLASS ____________________

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1. In 1993, a group of students in England made a giant yo-yo that was 10 ft

in diameter and had a mass of 407 kg. With the use of a crane, the yo-yo

was launched from a height of 57.0 m above the ground. Suppose the lin-

ear speed of the yo-yo at the end of its descent was 12.4 m/s. If the angu-

lar speed at the end of the descent was 28.0 rad/s, what was the yo-yo’s

moment of inertia?

2. In 1994, a bottle over 3 m in height and with a radius at its base of 0.56 m

was made in Australia. Treat the bottle as a thin-walled cylinder rotating

about its symmetry axis, which has the same rotational properties as a

thin hoop rotating about its symmetry axis. What is the linear speed that

the bottle acquires after rolling down a slope with a height of 5.0 m? Do

you need to know the mass of the bottle?

3. In 1988, a cheese with a mass of 1.82 × 104 kg was made in Wisconsin.

Suppose the cheese had a cylindrical shape. The cheese was set to roll

along a horizontal road with an undetermined speed. The road then

went uphill, and the cheese rolled up the hill until its vertical displace-

ment was 1.2 m, at which point it came to a stop. Assuming that there

was no slipping between the rim of the cheese and the ground, calculate

the initial linear speed of the cheese.

4. In 1982, a team of ten people rolled a cylindrical barrel with a mass of

64 kg for almost 250 km without stopping. Imagine that at some point

during the trip the barrel was stopped at the crest of a steep hill. The bar-

rel was accidentally released and rolled down the hill. If the linear speed

of the barrel at the bottom of the slope was 12.0 m/s, how high was the

hill? Assume that the barrel’s moment of inertia was equal to 0.80 mr 2.

5. A potato with a record-breaking mass of 3.5 kg was grown in 1994. Sup-

pose a child saw this potato and decided to pretend it was a soccer ball.

The child kicked the potato so that it rolled without slipping at a speed of

5.4 m/s. The potato rolled up a slope with a 30.0° incline. Assuming that

the potato could be approximated as a uniform, solid sphere with a ra-

dius of 7.0 cm, what was the distance along the slope that the potato

rolled before coming to a stop?

6. In 1992, an artificial egg with a mass of 4.8 × 103 kg was made in Aus-

tralia. Assume that the egg is a solid sphere with a radius of 2.0 m. Calcu-

late the minimum height of a slope that the egg rolls down if it is to

reach an angular speed of 5.0 rad/s at the bottom of the slope. What is

the translational kinetic energy of the egg at the bottom of the slope?

7. An onion grown in 1994 had a record-breaking mass of 5.55 kg. Assume

that this onion can be approximated by a uniform, solid sphere. Suppose

the onion rolled down an inclined ramp that had a height of 1.40 m.

What was the onion’s rotational kinetic energy? Assume that there was

no slippage between the ramp and the onion’s surface.

ADDITIONAL PRACTICE


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NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 9ABUOYANT FORCE

P R O B L E MThe highest natural concentration of salts in water are found in the evaporating remnants of old oceans, such as the Dead Sea in Israel. Sup-pose a swimmer with a volume of 0.75 m3 is able to float just beneath thesurface of water with a density of 1.02 103 kg/m3. How much extramass can the swimmer carry and be able to float just beneath the surfaceof the Dead Sea, which has a density of 1.22 103 kg/m3?

S O L U T I O N

Given: V = 0.75 m3

r1 = 1.02 × 103 kg/m3

r2 = 1.22 × 103 kg/m3

Unknown: m′ = ?

Choose the equation(s) or situation: In both bodies of water, the buoyant force

equals the weight of the floating object.

FB,1 = Fg,1

r1Vg = mg

FB,2 = Fg,2

r2Vg = (m + m′)g = r1Vg + m′g


m′ = (r2 − r1)V


m′ = (1.22 × 103 kg/m3 − 1.02 × 103 kg/m3)(0.75 m3)

m′ = (0.20 × 103 kg/m3)(0.75 m3)

m′ =

The mass that can be supported by buoyant force increases with the difference in

fluid densities.

150 kg

1. DEFINE

2. PLAN

3. CALCULATE

4. EVALUATE

ADDITIONAL PRACTICE

1. The heaviest pig ever raised had a mass of 1158 kg. Suppose you placed

this pig on a raft made of dry wood. The raft completely submerged in

water so that the raft’s top surface was just level with the surface of the

lake. If the raft’s volume was 3.40 m3, what was the mass of the raft’s dry

wood? The density of fresh water is 1.00 × 103 kg/m3.

2. La Belle, one of four ships that Robert La Salle used to establish a French

colony late in the seventeenth century, sank off the coast of Texas. The

ship’s well-preserved remains were discovered and excavated in the

1990s. Among those remains was a small bronze cannon, called a min-

ion. Suppose the minion’s total volume is 4.14 × 10−2 m3. What is the

Problem 9A 99

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NAME ______________________________________ DATE _______________ CLASS ____________________

minion’s mass if its apparent weight in sea water is 3.115 × 103 N? The

density of sea water is 1.025 × 103 kg/m3.

3. William Smith built a small submarine capable of diving as deep as

30.0 m. The submarine’s volume can be approximated by that of a cylin-

der with a length of 3.00 m and a cross-sectional area of 0.500 m2. Sup-

pose this submarine dives in a freshwater river and then moves out to

sea, which naturally consists of salt water. What mass of fresh water must

be added to the ballast to keep the submarine submerged? The density

of fresh water is 1.000 × 103 kg/m3, and the density of sea water is

1.025 × 103 kg/m3.

4. The largest iceberg ever observed had an area of 3.10 × 104 km2, which is

larger than the area of Belgium. If the top and bottom surfaces of the ice-

berg were flat and the thickness of the submerged part was 0.84 km, how

large was the buoyant force acting on the iceberg? The density of sea

water equals 1.025 × 103 kg/m3.

5. A cannon built in 1868 in Russia could fire a cannonball with a mass of

4.80 × 102 kg and a radius of 0.250 m. When suspended from a scale and

submerged in water, a cannonball of this type has an apparent weight of

4.07 × 103 N. How large is the buoyant force acting on the cannonball?

The density of fresh water is 1.00 × 103 kg/m3.

6. The tallest iceberg ever measured stood 167 m above the water. Suppose

that both the top and the bottom of this iceberg were flat and the thick-

ness of the submerged part was estimated to be 1.50 km. Calculate the

density of the ice. The density of sea water equals 1.025 × 103 kg/m3.

7. The Russian submarines of the “Typhoon” class are the largest sub-

marines in the world. They have a length of 1.70 × 102 m and an average

diameter of 13.9 m. When submerged, they displace 2.65 × 107 kg of sea

water. Assume that these submarines have a simple cylindrical shape. If a

“Typhoon”-class submarine has taken on enough ballast that it descends

with a net acceleration of 2.00 m/s2, what is the submarine’s density dur-

ing its descent?

8. To keep Robert La Salle’s ship La Belle well preserved, shipbuilders are re-

constructing the ship in a large tank filled with fresh water. Polyethylene

glycol, or PEG, will then be slowly added to the water until a 30 percent

PEG solution is formed. Suppose La Belle displaces 6.00 m3 of liquid

when submerged. If the ship’s apparent weight decreases by 800 N as the

PEG concentration increases from 0 to 30 percent, what is the density of

the final PEG solution? The density of fresh water is 1.00 × 103 kg/m3.

NAME ______________________________________ DATE _______________ CLASS ____________________


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Holt Physics

Problem 9BPRESSURE

P R O B L E MThe largest helicopter in the world, which was built in Russia, has a massof 1.03 103 kg. If you placed this helicopter on a large piston of a hy-draulic lift, what force would need to be applied to the small piston inorder to slowly lift the helicopter? Assume that the weight of the heli-copter is distributed evenly over the large piston’s area, which is 1.40 102 m2. The area of the small piston is 0.80 m2.


A1 = 1.40 × 102 m2

A2 = 0.80 m2

g = 9.81 m/s2

Unknown: F2 = ?

Use the equation for pressure to equate the two opposing pressures in terms of

force and area.

P = A

F P1 = P2

A

F1

1 =

A

F2

2

F2 = F1 A

A2

1 = mg

A

A2

1

F2 = (1.03 × 105 kg)(9.81 m/s2)1.4

0

0

.8

×0

1

m

02

2

m2F2 = 5.8 × 103 N

ADDITIONAL PRACTICE

1. Astronauts and cosmonauts have used pressurized spacesuits to explore

the low-pressure regions of space. The pressure inside one of these suits

must be close to that of Earth’s atmosphere at sea level so that the space

explorer may be safe and comfortable. The pressure on the outside of the

suit is a fraction of 1.0 Pa. Clearly, pressurized suits are made of ex-

tremely sturdy material that can tolerate the stress from these pressure

differences. If the average interior surface area of a pressurized spacesuit

is 3.3 m2, what is the force exerted on the suit’s material? Assume that the

pressure outside the suit is zero and that the pressure inside the suit is

1.01 × 105 Pa.

2. A strange idea to control volcanic eruptions is developed by a daydream-

ing engineer. The engineer imagines a giant piston that fits into the

NAME ______________________________________ DATE _______________ CLASS ____________________

Problem 9B 101

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volcano’s shaft, which leads from Earth’s surface down to the magma

chamber. The piston controls an eruption by exerting pressure that is

equal to or greater than the pressure of the hot gases, ash, and magma

that rise from the magma chamber through the shaft. The engineer as-

sumes that the pressure of the volcanic material is 4.0 × 1011 Pa, which is

the pressure in Earth’s interior. If the material rises into a cylindrical

shaft with a radius of 50.0 m, what force is needed on the other side of

the piston to balance the pressure of the volcanic material?

3. The largest goat ever grown on a farm had a mass of 181 kg; on the other

hand, the smallest “pygmy” goats have a mass of only about 16 kg. Imag-

ine an agricultural show in which a large goat with a mass of 181 kg

exerts a pressure on a hydraulic-lift piston that is equal to the pressure

exerted by three pygmy goats, each of which has a mass of 16.0 kg. The

area of the piston on which the large goat stands is 1.8 m2. What is the

area of the piston on which the pygmy goats stand?

4. The greatest load ever raised was the offshore Ekofisk complex in the

North Sea. The complex, which had a mass of 4.0 × 107 kg, was raised

6.5 m by more than 100 hydraulic jacks. Imagine that his load could have

been raised using a single huge hydraulic lift. If the load had been placed

on the large piston and a force of 1.2 × 104 N had been applied to the

small piston, which had an area of 5.0 m2, what must the large piston’s

area have been?

5. The pressure that can exist in the interior of a star due to the weight of

the outer layers of hot gas is typically several hundred billion times

greater than the pressure exerted on Earth’s surface by Earth’s atmos-

phere. Suppose a pressure equal to that estimated for the sun’s interior

(2.0 × 1016 Pa) acts on a spherical surface within a star. If a force of

1.02 N × 1031 N produces this pressure, what is the area of the surface?

What is the sphere’s radius r? (Recall that a sphere’s surface area equals

4pr2.)

6. The eye of a giant squid can be more than 35 cm in diameter—the

largest eye of any animal. Giant squid also live at depths greater than a

mile below the ocean’s surface. At a depth of 2 km, the outer half of a

giant squid’s eye is acted on by an external force of 4.6 × 106 N. Assum-

ing the squid’s eye has a diameter of 38 cm, what is the pressure on the

eye? (Hint: Treat the eye as a sphere.)

7. The largest tires in the world, which are used for certain huge dump

trucks, have diameters of about 3.50 m. Suppose one of these tires has a

volume of 5.25 m3 and a surface area of 26.3 m2. If a force of 1.58 × 107 N

acts on the inner area of the tire, what is the absolute pressure inside the

tire? What is the gauge pressure on the tire’s surface?

Holt Physics

Problem 9CPRESSURE AS A FUNCTION OF DEPTH

P R O B L E MIn 1969, a whale dove and remained underwater for nearly 2 h. Evidenceindicates that the whale reached a depth of 3.00 km, where the whale sus-tained a pressure of 3.03 107 Pa. Estimate the density of sea water.

S O L U T I O NGiven: h = 3.00 km = 3.00 × 103 m

P = 3.03 × 107 Pa

Po = 1.01 × 105 Pa

g = 9.81 m/s2

Unknown: r = ?

Use the equation for fluid pressure as a function of depth, and rearrange it to solve

for density.

P = Po + rgh

r = P

g

−h

Po

r =

r =

r = 1.03 × 103 kg/m3

3.02 × 107 Pa(9.81 m/s2)(3.00 × 103 m)

(3.03 × 107 Pa) − (1.01 × 105 Pa)

(9.81 m/s2)(3.00 × 103 m)


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NAME ______________________________________ DATE _______________ CLASS ____________________

ADDITIONAL PRACTICE

1. In 1994, a 16.8 m tall oil-filled barometer was constructed in Belgium.

Suppose the barometer column was 80.0 percent filled with oil. What

was the density of the oil if the pressure at the bottom of the column was

2.22 × 105 Pa and the air pressure at the top of the oil column was 1.01 ×105 Pa?

2. One of the lowest atmospheric pressures ever measured at sea level was

8.88 × 104 Pa, which existed in hurricane Gilbert in 1988. This same

pressure can be found at a height 950 m above sea level. Use this infor-

mation to estimate the density of air.

3. In 1993, Francisco Ferreras of Cuba held his breath and took a dive that

lasted more than 2 min. The maximum pressure Ferreras experienced

was 13.6 times greater than atmospheric pressure. To what depth did

Ferreras dive? The density of sea water is 1.025 × 103 kg/m3.

4. A penguin can endure pressures as great as 4.90 × 106 Pa. What is the

maximum depth to which a penguin can dive in sea water?

NAME ______________________________________ DATE _______________ CLASS ____________________

Problem 9C 103

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5. In 1942, the British ship Edinburgh, which was carrying a load of 460

gold ingots, sank off the coast of Norway. In 1981, all of the gold was re-

covered from a depth of 245 m by a team of 12 divers. What was the

pressure exerted by the ocean’s water at that depth?

6. In 1960, a bathyscaph descended 10 916 m below the ocean’s surface.

What was the pressure exerted on the bathyscaph at that depth?


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NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 9DBERNOULLI’S EQUATION

P R O B L E MThe widest road tunnel in the world is located in California. The tunnelhas a cross-sectional area of about 4.00 × 102 m2. On the other hand, theThree Rivers water tunnel in Georgia has a cross-sectional area of only8.0 m2. Imagine connecting together two tunnels with areas equal tothese along a flat region and setting fresh water to flow at 4.0 m/s in thenarrower tunnel. If the pressure in the wider tunnel is 1.10 × 105 Pa, whatis the pressure in the narrower tunnel?

S O L U T I O N1. DEFINE

2. PLAN

Given: A1 = 8.0 m2 A2 = 4.00 × 102 m2

r = 1.00 × 103 kg/m3

v1 = 4.0 m/s

P2 = 1.10 × 105 Pa

Unknown: P1 = ?

Choose the equation(s) or situation: Because this problem involves fluid flow, it

requires the application of Bernoulli’s equation.

P1 + 12

rv12 + rgh1 = P2 + 1

2rv2

2 + rgh2

The flow of water is horizontal, so h1 and h2 are equal.

P1 + 12

rv12 = P2 + 1

2rv2

2

To find the speed of the flowing water in the wider tunnel, use the continuity

equation.A1v1 = A2v2

Substitute this equation for v2 into Bernoulli’s equation.

P1 + 12

rv12 = P2 + 1

2r

A

A1

2 v12


P1 = P2 + 12

rv12

A

A1

22

−1Substitute the values into the equation(s) and solve:

P1 = 1.10 × 105 Pa

+ 12

(1.00 × 103 kg/m3)(4.0 m/s)24.00

8.

×0

1

m

02

2

m22−1

P1 = 1.10 × 105 Pa + (8.0 × 103 N•m/m3)(4.0 × 10−4 − 1)

P1 = 1.10 × 105 Pa − (8.0 × 103 Pa)(0.9996)

P1 = 1.10 × 105 Pa − 8.0 × 103 Pa

P1 = 1.02 × 105 Pa

3. CALCULATE

NAME ______________________________________ DATE _______________ CLASS ____________________

Problem 9D 105

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4. EVALUATE

ADDITIONAL PRACTICE

1. The Chicago system of sewer tunnels has a total length of about 200 km.

The tunnels are all at the same level and vary in diameter from less than

3 m to 10 m. Consider a connection between a wide tunnel and a narrow

tunnel. The pressure in the wide tunnel is 12 percent greater than the

pressure in the narrow one, and the speeds of flowing water in the wide

and the narrow tunnels are 0.60 m/s and 4.80 m/s, respectively. Based on

this information, find the pressure in the wide tunnel.

2. A certain New York City water-supply tunnel is almost 170 km long, is all

at the same level, and has a diameter of 4.10 m. Suppose water flows

through the tunnel at a speed of 3.0 m/s until it reaches a narrow section

where the tunnel’s diameter is 2.70 m. The pressure in the narrow section

is 82 kPa. Use the continuity equation to find the water’s speed in the

narrow section of the tunnel. Then find the pressure in the wide portion

of the tunnel.

3. An enormous open vat owned by a British cider company has a volume

of nearly 7000 m3. If a small hole is drilled halfway down the side of this

vat when it is full of cider, the cider will hit the ground 19.7 m away from

the bottom of the vat. How tall is the vat?

4. The tallest cooling tower in the world is in Germany. Consider a water

pipe coming down from the top of the tower. The pipe is punctured near

the bottom, which causes the water to flow from the puncture hole with

a speed of 59 m/s. How high is the tower? Assume the pressure inside the

pipe is equal to the pressure outside the pipe.

5. A water tower built in Oklahoma has a capacity of 1893 m3 and is about

66.0 m high. Suppose a little hole with a cross-sectional area of 10.0 cm2

is drilled near the bottom of the tower’s water tank. At what speed does

the water initially flow through the hole? Assume that the air pressure in-

side and outside the tank is the same.

6. The Nurek Dam, in Tajikistan, is the tallest dam in the world—its height

is more than 300 m. Suppose the difference in water levels on the oppo-

site sides of the dam is 3.00 × 102 m. If a small crack appears in the dam

near the lower water level, at what speed will the water stream leave the

crack? Assume the air pressure on either side of the dam is the same.

7. The world’s largest litter bin has a volume of more than 40 m3 and is

6.0 m tall. If this bin is filled with water and then a hole with an area of

0.16 cm2 is drilled near the bottom, at what speed will the water leave the

bin? Assume that the water level in the bin drops slowly and that the bin

is open to the air.

The pressure of the water increases when it flows into the wider tunnel. The

speed of the water’s flow decreases, as indicated by the continuity equation.

v2 = (4.0 m/s)(8/400) = 8 × 10−2 m/s

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Holt Physics

Problem 9ETHE IDEAL GAS LAW

P R O B L E MA hot-air balloon named Double Eagle V traveled a record distance ofmore than 8000 km from Japan to California in 1981. The volume of theballoon was 1.13 104 m3. If the balloon contained 2.84 1029 gas parti-cles that had an average temperature of 355 K, what was the absolutepressure of the gas in the balloon?

S O L U T I O NGiven: V = 1.13 × 104 m3

N = 2.84 × 1029 particles

T = 355 K

kB = 1.38 × 10−23 J/K

Unknown: P = ?

Choose the equation(s) or situation: To find the pressure of the gas, use the

ideal gas law.

PV = NkBT


P = Nk

VBT


P =

P =

The pressure inside the balloon is about 20 percent greater than standard air

pressure. This pressure corresponds to the higher temperature of the air in the

balloon. The hot air’s temperature is also nearly 20 percent greater than 300 K.

1.23 × 105 Pa

(2.84 × 1029 particles)(1.38 × 10−23 J/K)(355 K)

1.13 × 104 m3

1. DEFINE

2. PLAN

3. CALCULATE

4. EVALUATE

ADDITIONAL PRACTICE

1. The official altitude record for a balloon was set in 1961 by two American

officers piloting a high-altitude helium balloon with a volume of 3.4 ×105 m3. Assume that the temperature of the gas was 280 K. If the balloon

contained 1.4 × 1030 atoms of helium, find the absolute pressure in the

balloon at the maximum altitude of 35 km.

2. The estimated number of locusts that made up a swarm that infested

Nebraska in 1874 was 1.2 × 1013. This number was about 7000 times the

total human population of Earth back then and about 2000 times the total

human population today. It is, however, only a few billionths of the number

of molecules in a liter of gas. If a container with a volume of 1.0 × 10−3 m3

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is filled with 1.2 × 1013 gas molecules maintained at a temperature

of 300.0 K, what is the pressure of the gas in the container?

3. In terms of volume, the largest pyramid in the world is not in Egypt but

in Mexico. The Pyramid of the Sun at Teotihuacan has a volume of 3.3 ×106 m3. If you fill a balloon to this same volume with 1.5 × 1032 mole-

cules of nitrogen at a temperature of 360 K, what will the absolute pres-

sure of the gas be?

4. A snow palace more than 30 m high was built in Japan in 1994. Suppose

a container with the same volume as this snow palace is filled with 1.00 ×1027 molecules. If the temperature of the gas is 2.70 × 102 K and the gas

pressure is 36.2 Pa, what is the volume of the gas?

5. Suppose the volume of a balloon decreases so that the temperature of the

balloon decreases from 280 K to 240 K and its pressure drops from 1.6 ×104 Pa to 1.7 × 104 Pa. What is the new volume of the gas?

6. The longest navigable tunnel in the world was built in France. Suppose

the entire tunnel, which has a cross-sectional area of 2.50 × 102 m2, is

filled with air at a temperature of 3.00 × 102 K and a pressure of 101 kPa.

If the tunnel contains 4.34 × 1031 molecules, what is the volume and

length of the tunnel?

7. A balloon is filled with 7.36 × 104 m3 of hot air. If the pressure inside the

balloon is 1.00 × 105 Pa and there are 1.63 × 1030 particles of air inside,

what is the average temperature of the air inside the balloon?

8. In 1993, a group of American researchers drilled a 3053 m shaft in the

ice sheet of Greenland. Suppose the cross-sectional area of the shaft is

0.040 m2. If the air in the shaft consists of 3.6 × 1027 molecules at an

average pressure of 105 kPa, what is the air’s average temperature?

9. The cylinder of the largest steam engine had a radius of 1.82 m. Suppose

the length of the cylinder is six times the radius. Steam at a pressure of

2.50 × 106 Pa and a temperature of 495 K enters the cylinder when the

piston has reduced the volume in the cylinder to 3.00 m3. The piston is

then pushed outward until the volume of the steam in the cylinder is

57.0 m3. If the pressure of the steam after expansion is 1.01 × 105 Pa,

what is the temperature of the steam?


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Holt Physics

Problem 10A

P R O B L E M

S O L U T I O N

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TEMPERATURE CONVERSION

The temperature at the surface of the sun is estimated to be 5.97 × 103 K.Express this temperature in degrees Fahrenheit and in degrees Celsius.

Given: T = 5.97 × 103K

Unknown: TF = ? TC = ?

Use the Celsius-Fahrenheit and Celsius-Kelvin temperature conversion equations.

TC = T − 273.15 = (5.97 × 103 − 0.27 × 103)K =

TF = 95

TC + 32.0 = 95

(5.70 × 103) + 32.0°F = 1.03 × 104°F

5.70 × 103 K

ADDITIONAL PRACTICE

1. Usually, people die if their body temperature drops below 35°C. There

was one case, however, of a two-year-old girl who had been accidentally

locked outside in the winter. She survived, even though her body temper-

ature dropped as low as 14°C. Express this temperature in kelvins and in

degrees Fahrenheit.

2. In experiments conducted by the United States Air Force, subjects en-

dured air temperatures of 4.00 × 102°F. Express this temperature in

degrees Celsius and in kelvins.

3. The temperature of the moon’s surface can reach 117°C when exposed to

the sun and can cool to −163°C when facing away from the sun. Express

this temperature change in degrees Fahrenheit.

4. Because of Venus’ proximity to the sun and its thick, high-pressure

atmosphere, its temperature can rise to 860.0°F. Express this temperature

in degrees Celsius.

5. On January 22, 1943, the air temperature at Spearfish, South Dakota, rose

49.0°F in 2 min to reach a high temperature of 7.00°C. What were the ini-

tial and final temperatures in degrees Fahrenheit? What was the tempera-

ture in degrees Celsius before the temperature increase?

6. In 1916, Browning, Montana, experienced a temperature decrease of 56°C

during a 24 h period. The final temperature was −49°C. Express in kelvins

the temperatures at the beginning and the end of the 24 h period.

7. In 1980, Willie Jones of Atlanta, Georgia, was hospitalized with heat-

stroke, having a body temperature of 116°F. Fortunately, he survived.

Express Willie’s body temperature in kelvins.

Problem 10B 109

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Holt Physics

Problem 10BCONSERVATION OF ENERGY

The moon has a mass of 7.3 × 1022 kg and an average orbital speed of1.02 × 103 m/s. Suppose all of the moon’s kinetic energy goes into in-creasing the internal energy of a quantity of water at a temperature of100.0°C. If it takes 2.26 × 106 J to vaporize 1.00 kg of water that is initiallyat 100.0°C, what mass of this water can be vaporized?

S O L U T I O NGiven: mm = mass of moon = 7.3 × 1022 kg

vm = speed of moon = 1.02 × 103 m/s

k = energy needed to vaporize 1 kg of water = 2.26 × 106 J

Unknown: mw = mass of water vaporized = ?

Choose the equation(s) or situation: Use the equation for energy conservation,

including a term for the change in internal energy.

∆PE + ∆KE + ∆U = 0

The moon’s kinetic energy alone is taken into consideration. Therefore ∆PE

equals zero. Because all of the moon’s kinetic energy is transferred to the water’s

internal energy, the moon’s final kinetic energy is also zero.

∆KE + ∆U = KEf − KEi + ∆U = 0 − KEi + ∆U = 0

∆U = KEi = 12

mm(vm)2

To find the mass of water that will be vaporized, the change in internal energy

must be divided by the conversion constant, k.

mw = ∆k

U =

mm

2

(v

km)2


mw = =

The mass of the water vaporized is only 23 percent of the moon’s mass. This sug-

gests that a smaller object moving with the moon’s speed would vaporize a mass

of water only 0.23 times as large as its own mass. Because the water is already at

its boiling point, this result indicates that a considerable amount of energy goes

into the process of vaporization.

1.7 × 1022 kg(7.3 × 1022 kg)(1.02 × 103 m/s)2

2(2.26 × 106 J/kg)

1. DEFINE

2. PLAN

3. CALCULATE

4. EVALUATE

ADDITIONAL PRACTICE

1. A British-built Hovercraft—a vehicle that cruises on a cushion of air—

has a mass of 3.05 × 105 kg and can attain a speed of 120.0 km/h. Sup-

pose this vehicle slows down from 120.0 km/h to 90.0 km/h and that the


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change in its kinetic energy is used to raise the temperature of a quantity

of water by 10.0°C. Knowing that 4186 J is required to raise the tempera-

ture of 1.00 kg of water by 1.00°C, calculate the mass of the heated water.

2. The Westin Stamford Hotel in Detroit is 228 m tall. Suppose a piece of

ice, which initially has a temperature of 0.0°C, falls from the hotel roof

and crashes to the ground. Assuming that 50.0 percent of the ice’s me-

chanical energy during the fall and collision is absorbed by the ice and

that 3.33 × 105 J is required to melt 1.00 kg of ice, calculate the fraction

of the ice’s mass that would melt.

3. The French-built Concorde, the fastest passenger jet plane, is known to

travel with a speed as great as 2.333 × 103 km/h. Suppose the plane travels

horizontally at an altitude of 4.000 × 103 m and at maximum speed when

a fragment of metal breaks free from the plane. The metal has, of course,

the same horizontal speed as the plane, and when it lands on the ground,

it will have absorbed 1.00 percent of its total mechanical energy. If it takes

355 J to raise the temperature of 1.00 kg of this metal by 1.00°C, how

great a temperature change will the metal fragment experience from the

time it breaks free from the Concorde to the time it lands on the ground?

Ignore air resistance.

4. Mount Everest is the world’s highest mountain. Its height is 8848 m. Sup-

pose a steel alpine hook were to slowly slide off the summit of Everest

and fall all the way to the base of the mountain. If 20.0 percent of the

hook’s mechanical energy is absorbed by the hook as internal energy, cal-

culate the final temperature of the hook. Assume that the hook’s initial

temperature is −18.0°C and that the hook’s temperature increases by

1.00°C for every 448 J/kg that is added.

5. The second tallest radio mast in the world is located near Fargo, North

Dakota. The tower, which has an overall height of 629 m, was built in just

one month by a team of 11 workers. Suppose one of these builders left a

3.00 g copper coin on the top of the tower. During extremely windy

weather, the coin falls off the tower and reaches the ground at the tower’s

base with a speed of 42 m/s. If the coin absorbs 5.0 percent of its total me-

chanical energy, by how much does the coin’s internal energy increase?

6. In 1993, Russell Bradley carried a load of bricks with a total mass of

312 kg up a ramp that had a height of 2.49 m. Suppose Bradley puts the

load on the ramp and then pushes the load off the edge with a horizontal

speed of 0.50 m/s. If the bricks absorb half their total mechanical energy,

how much does their internal energy change?

7. Angel Falls, the highest waterfall in the world, is located in Venezuela. Es-

timate the height of the waterfall, assuming that the water that falls the

complete distance experiences a temperature increase of 0.230°C. In your

calculation, assume that the water absorbs 10.0 percent of its mechanical

energy and that 4186 J is needed to raise the temperature of 1.00 kg of

water by 1.00°C.

Problem 10C 111

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Holt Physics

Problem 10CCALORIMETRY

In 1990, the average rate for use of fresh water in the United States was approximately 1.50 × 107 kg each second. Suppose a group of teenagersbuild a really big calorimeter and that they place in it a mass of waterequal to the mass of fresh water consumed in 1.00 s. They then use a testsample of gold with a mass equal to that of the United States gold reservefor 1992. Initially, the gold has a temperature of 80.0°C and the water has atemperature of 1.00°C. If the final equilibrium temperature of the goldand water is 2.30°C, what is the mass of the gold?

S O L U T I O NGiven: Tf = 2.30°C mwater = mw = 1.50 × 107 kg

Tgold = Tg = 80.0°C Twater = Tw = 1.00°C

cp,gold = cp,g = 129 J/kg•°C cp,water = cp,w = 4186 J/kg•°C

Unknown: mgold = mg = ?

Choose the equation(s) or situation: Equate the energy removed from the gold

to the energy absorbed by the water.

energy removed from metal = energy absorbed by water

cp,gmg(Tg – Tf ) = cp,wmw(Tf − Tw)


mg =


mg =

mg = =

Although the masses of the gold sample and water are unrealistic, the tempera-

tures are entirely reasonable, indicating that temperature is an intrinsic variable

of matter that is independent of the quantity of a substance. The small increase

in the water’s temperature and the large decrease in the gold’s temperature is the

result of the water having both a larger mass and a larger specific heat capacity.

8.14 × 106 kg(4186 J/kg•°C)(1.50 × 107 kg)(1.30°C)

(129 J/kg•°C)(77.7°C)

(4186 J/kg•°C)(1.50 × 107 kg)(2.30°C − 1.00°C)

(129 J/kg•°C)(80.0°C − 2.30°C)

cp,wmw(Tf − Tw)

cp,g(Tg – Tf )

1. DEFINE

2. PLAN

3. CALCULATE

4. EVALUATE

ADDITIONAL PRACTICE

1. In 1992, the average rate of energy consumption in the United States was

about 2.8 × 109 W. Suppose all of the copper produced in the United

States in 1992 is placed in the giant calorimeter used in the sample


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problem. The quantity of energy transferred by heat from the copper to

the water is equal to the energy used in the United States during 1.2 s of

1992. If the initial temperature of the copper is 26°C, and the final tem-

perature is 21°C, what is the mass of the copper?

2. In 1992, a team of firefighters pumped 143 × 103 kg of water in less than

four days. What mass of wood can be cooled from a temperature of

280.0°C to one of 100.0°C using this amount of water? Assume that the

initial temperature of the water is 20.0°C and that all of the water has a

final equilibrium temperature of 100.0°C, but that none of the water is

vaporized. Use 1.700 × 103 J/kg•°C for the specific heat capacity of wood.

3. One of the nuclear generators at a power plant in Lithuania has a nomi-

nal power of 1.450 GW, making it the most powerful generator in the

world. Nippon Steel Corporation in Japan is the world’s largest steel pro-

ducer. Between April 1, 1993, and March, 31, 1994, Nippon Steel’s mills

produced 25.1 × 109 kg of steel. Suppose this entire quantity of steel is

heated and then placed in the giant calorimeter used in the sample prob-

lem. If the quantity of energy transferred by heat from the steel to the

water equals 1.00 percent of all the energy produced by the Lithuanian

generator in a year, what is the temperature change of the steel? Assume

that the specific heat capacities of steel and iron are the same.

4. In 1994, to commemorate the 200th anniversary of a beverage company,

a giant bottle was constructed and filled with 2.25 × 103 kg of the com-

pany’s lemonade. Suppose the lemonade has an initial temperature of

28.0°C when 9.00 × 102 kg of ice with a temperature of −18.0°C is added

to it. What is the lemonade’s temperature at the moment the temperature

of the ice reaches 0.0°C? Assume that the lemonade has the same specific

heat capacity as water.

5. The water in the Arctic Ocean has a total mass of 1.33 × 1019 kg. The

average temperature of the water is estimated to be 4 .000°C. What

would the temperature of the water in the Arctic Ocean be if the

energy produced in 1.000 × 103 y by the world’s largest power plant

(1.33 × 1010 W) were transferred by heat to it?

6. There is a little island off the shore of Brazil where the weather is ex-

tremely consistent. From 1911 to 1990, the lowest temperature on the is-

land was 18°C (64°F) and the highest temperature was 32°C (90°F). It is

known that the liquid in a standard can of soft drink absorbs 20.8 kJ of

energy when its temperature increases from 18.0°C to 32.0°C. If the soft

drink has a mass of 0.355 kg, what is its specific heat capacity?

7. The lowest temperature ever recorded in Alaska is −62°C. The highest

temperature ever recorded in Alaska is 38°C. Suppose a piece of metal

with a mass of 180 g and a temperature of −62.0°C is placed in a

calorimeter containing 0.500 kg of water with a temperature of 38.0°C. If

the final equilibrium temperature of the metal and water is 36.9°C, what

is the specific heat capacity of the metal? Use the calculated value of the

specific heat capacity and Table 10-4 to identify the metal.

Problem 10D 113

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Holt Physics

Problem 10DHEAT OF PHASE CHANGE

The world’s deepest gold mine, which is located in South Africa, is over 3 km deep. Every day, the mine transfers enough energy by heat to themine’s cooling systems to melt 3.36 × 107 kg of ice at 0.0°C. If the energyoutput from the mine is increased by 2.0 percent, to what final tempera-ture will the 3.36 × 107 kg of ice-cold water be heated?

S O L U T I O NGiven: mice = mwater = m = 3.36 × 107 kg

Ti = 0.0°C

Lf = 3.33 × 105 J/kg

cp,w = 4186 J/(kg•°C)

Q = energy added to water = (2.0 × 10−2)Q

Unknown: Tf = ?

Choose the equation(s) or situation: First, determine the amount of energy

needed to melt 3.36 × 107 kg of ice by using the equation for the heat of fusion.

Q = miceLf = mLf

The energy added to the now liquid ice (Q) can then be determined.

Q = (2.0 × 10−2)Q = (2.0 × 10−2)mLf

Finally, the energy added to the water equals the product of the water’s mass, spe-

cific heat capacity, and change in temperature.

Q = mcp,w(Tf − Ti) = (2.0 × 10−2)mLf


Tf = + Ti = (2.0 × 10−2) cp

L

,

f

w + Ti


Tf = (2.0 × 10−2) + 0.0°C =

Note that the result is independent of the mass of the ice and water. The amount

of energy needed to raise the water’s temperature by 1°C is a little more than

1 percent of the energy required to melt the ice.

1.6°C(3.33 × 10−5 J/kg)(4186 J/kg•°C)

(2.0 × 10−2)mLfmcp,w

1. DEFINE

2. PLAN

3. CALCULATE

4. EVALUATE

ADDITIONAL PRACTICE

1. Lake Superior contains about 1.20 × 1016 kg of water, whereas Lake Erie

contains only 4.8 × 1014 kg of water. Suppose aliens use these two lakes

for cooking. They heat Lake Superior to 100.0°C and freeze Lake Erie to


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0.0°C. Then they mix the two lakes together to make a “lake shake.” What

would be the final temperature of the mixture? Assume that the entire

energy transfer by heat occurs between the lakes.

2. The lowest temperature measured on the surface of a planetary body in the

solar system is that of Triton, the largest of Neptune’s moons. The surface

temperature on this distant moon can reach a low of −235°C. Suppose an

astronaut brings a water bottle containing 0.500 kg of water to Triton. The

water’s temperature decreases until the water freezes, then the temperature

of the ice decreases until it is in thermal equilibrium with Triton at a tem-

perature of −235°C. If the energy transferred by heat from the water to Tri-

ton is 471 kJ, what is the value of the water’s initial temperature?

3. Suppose that an ice palace built in Minnesota in 1992 is brought into con-

tact with steam with a temperature of 100.0°C. The temperature and mass

of the ice palace are 0.0°C and 4.90 × 106 kg, respectively. If all of the steam

liquefies by the time all of the ice melts, what is the mass of the steam?

4. In 1992, 1.804 × 106 kg of silver was produced in the United States. What mass

of ice must be melted so that this mass of liquid silver can solidify? Assume

that both substances are brought into contact at their melting temperatures.

The latent heat of fusion for silver is 8.82 × 104 J/kg.

5. The United States Bullion Depository at Fort Knox, Kentucky, contains

almost half a million standard mint gold bars, each with a mass of

12.4414 kg. Assuming an initial bar temperature of 5.0°C, each bar will

melt if it absorbs 2.50 MJ of energy transferred by heat. If the specific

heat capacity of gold is 129 J/kg •°C and the melting point of gold is

1063°C, calculate the heat of fusion of gold.

6. The world’s largest piggy bank has a volume of 7.20 m3. Suppose the bank

is filled with copper pennies and that the pennies occupy 80.0 percent of

the bank’s total volume. The density of copper is 8.92 × 103 kg/m3.

a. Find the total mass of the coins in the piggy bank.

b. Consider the mass found in (a). If these copper coins are brought

to their melting point, how much energy must be added to the

coins in order to melt 15 percent of their mass? The latent heat of

fusion for copper is 1.34 × 105 J/kg.

7. The total mass of fresh water on Earth is 3.5 × 1019 kg. Suppose all this

water has a temperature of 10.0°C. Suppose the entire energy output of

the sun is used to bring all of Earth’s fresh water to a boiling temperature

of 100.0°C, after which the water is completely vaporized.

a. How much energy must be added to the fresh water through heat in

order to raise its temperature to the boiling point and vaporize it?

b. If the rate at which energy is transferred from the sun is 4.0 × 1026 J/s,

how long will it take for the sun to provide sufficient energy for the

heating and vaporization process?

Problem 11A 115

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Holt Physics

Problem 11A

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WORK DONE ON OR BY A GAS

Suppose 2.4 × 108 J of work was required to inflate 1.6 million balloonsthat were released at one time in 1994. If each balloon was filled at a con-stant pressure that was 25 kPa in excess of atmospheric pressure, whatwas the change in volume for each balloon?

S O L U T I O NGiven: W = 2.4 × 108 J

P = 25 kPa = 2.5 × 104 Pa

n = number of balloons = 1.6 × 106

Unknown: ∆V = ?

Use the definition of work in terms of changing volume. The total volume

change involves 1.6 million balloons, so the volume change of one balloon must

be multiplied by n = 1.6 × 106.

W = nP∆V

∆V = n

W

P = = 6.0 × 10−3 m32.4 × 108 J(1.6 × 106)(2.5 × 104 N/m2)

ADDITIONAL PRACTICE

1. The largest glass bottle made by the method of glass-blowing was over

2 m tall. Suppose the net pressure used to expand the bottle to full vol-

ume was 5.1 kPa. If 3.6 × 103 J of work was done in expanding the bottle

from an initial volume of 0.0 m3, what was the final volume?

2. Russell Bradley carried 207 kg of bricks 3.65 m up a ladder. If the amount

of work required to perform that task is used to compress a gas at a con-

stant pressure of 1.8 × 106 Pa, what is the change in volume of the gas?

3. Nicholas Mason inflated a weather balloon using just the power of his

lungs. The balloon’s final radius was 1.22 m. If 642 kJ of work was done

to inflate the balloon, at what net pressure was the balloon inflated?

4. Calculate the pressure needed to inflate a sphere that has a volume equal

to that of the sun assuming that the work done was 3.6 × 1034 J. The

sun’s radius is 7.0 × 105 km.

5. In 1979, an extremely low pressure of 87 kPa was measured in a storm over

the Pacific Ocean. Suppose a gas is compressed at this pressure and its vol-

ume decreases by 25.0 × 10−3 m3. How much work is done by the gas?

6. Susan Williams, of California, blew a bubble-gum bubble with a radius

of 29.2 cm. If this were done with a constant net pressure of 25.0 kPa, the

work done could have been used to launch a model airplane. If the air-

plane’s mass were 160.0 g, what would have been the launch speed?


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ADDITIONAL PRACTICE

1. The heaviest snake ever found had a mass of 227 kg and measured

8.45 m in length. Suppose a sample of a gas with an initial internal en-

ergy of 42.0 kJ performs an amount of work equal to that needed to lift

the snake to a height equal to its length. If 4.00 kJ of energy is transferred

Holt Physics

Problem 11BTHE FIRST LAW OF THERMODYNAMICS

In 1992, residents of Arkansas consumed, on average, 11.4 L of gasolineper vehicle per day. If this amount of gasoline burns completely in a purecombustion reaction, it will release 4.3 × 108 J of energy. Suppose thisamount of energy is transferred by heat from a quantity of gas confinedin a very large cylinder. The cylinder, however, is equipped with a piston,and shortly after the energy is transferred by heat from the cylinder, workis done on the gas. The magnitude of the energy transferred by work isequal to one-third the magnitude of the energy transferred by heat. If theinitial internal energy of the gas is 1.00 × 109 J, what is the final internalenergy of the gas?

S O L U T I O NGiven: Ui = 1.00 × 109 J Q = −4.3 × 108 J

W = Q/3 = −(4.3 × 108 J)/3 = −1.4 × 108 J

Work is done on the gas, so work, W , has a negative value and increases the inter-

nal energy of the gas. Energy is transferred from the gas by heat, Q, which re-

duces the gas’ internal energy. Therefore, Q must have a negative value.

Unknown: Uf = ?

Choose the equation(s) or situation: Apply the first law of thermodynamics

using the definition of the change in internal energy (∆U = Uf − Ui) and the

values for the energy transferred by heat, Q, and work, W, to find the value for

the final internal energy.

∆U = Q − W

Uf − Ui = Q − W


Uf = Q − W + Ui


Uf = (−4.3 × 108 J) − (−1.4 × 108 J) + (1.00 × 109 J)

Uf =

The final internal energy is less than the initial internal energy because three

times as much energy was transferred away from the gas by heat as was trans-

ferred to the gas by work done on the gas.

7.1 × 108 J

1. DEFINE

2. PLAN

3. CALCULATE

4. EVALUATE

Problem 11B 117

NAME ______________________________________ DATE _______________ CLASS ____________________

to the gas by heat during the lifting process, what will be the final inter-

nal energy of the gas?

2. The most massive cannon ever built was made in Russia in 1868. This

cannon had a mass in excess of 1.40 × 105 kg and could fire cannonballs

with masses up to 4.80 × 102 kg. If the gunpowder in the cannon burned

very quickly (adiabatically), forming compressed gas in the barrel, then

the gas would expand and perform work on the cannonball. Suppose the

initial speed of one of these cannonballs is 2.00 × 102 m/s.

a. How much work is done by the expanding gas?

b. If the internal energy of the gas is 12.0 MJ when the cannonball

leaves the barrel, what is the initial internal energy of the gas im-

mediately after all of the powder burns?

3. The world’s largest jelly, which had a mass of about 4.00 × 104 kg, was

made in Australia in 1981.

a. How much energy must be transferred by heat from the jelly in order

for its temperature to decrease 20.0°C? Assume that the specific heat

capacity of the jelly equals that of water.

b. Suppose all this energy is transferred by heat to a sample of gas. At

the same time, the gas does 1.64 × 109 J of work on its surround-

ings. What is the net change in the internal energy of the gas?

4. The surface of Lake Ontario is 75.0 m above sea level. The mass of water

in the lake is about 1.64 × 1015 kg. Imagine an enormous sample of gas

that performs an amount of work capable of lifting Lake Ontario 75.0 m

and that also transfers to the lake enough energy by heat to vaporize the

entire lake. If the initial temperature of the lake water is 6.0°C and the

internal energy of the gas decreases by 90.0 percent, what is the final

internal energy of the gas?

5. An average elephant has a mass of 5.00 × 103 kg. Contrary to popular be-

lief, elephants are not slow; they can achieve speeds of up to 40.0 km/h.

Imagine a sample of gas that does an amount of work equal to the work

required for an average elephant to move from rest to its maximum

speed. If the initial internal energy of the gas, 2.50 × 105 J, is to be dou-

bled, how much energy must be transferred to the gas by heat?

6. The rate of nuclear energy production in the United States in 1992 was

about 5.9 × 109 J/s. Suppose one second’s worth of this energy is trans-

ferred by heat to an ideal gas. How much work must be done on or by

this gas so that the net increase in its internal energy is 2.6 × 109 J?

7. Dan Koko, a professional stuntman, jumped onto an air pad from a

height of almost 1.00 × 102 m. His impact speed was about 141 km/h.

a. If all of Koko’s kinetic energy was transferred by heat to the air in

the air pad after the inelastic collision with the air pad, how much

work was done?

b. Assuming that the initial internal energy of the air in the pad is

4.00 MJ, determine the percent increase in the internal energy of

this air after Koko’s jump. Assume that Koko’s mass was 76.0 kg.

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Holt Physics

Problem 11CHEAT-ENGINE EFFICIENCY

In 1989, Brendan Keenoy ran up 1760 steps in the CN Tower in Toronto,reaching a height of 342 m in 7 min 52 s. Suppose the amount of workdone by Keenoy is done by a heat engine. The engine’s input energy is 1.34 MJ, and its efficiency is 0.18. How much energy is exhausted from,and how much work is done by the engine?

S O L U T I O N

Given: Qh = 1.34 MJ = 1.34 × 106 J eff = 0.18

Unknown: Qc = ? Wnet = ?

Use the equation for the efficiency of a heat engine, expressed in terms of Qh and

Wnet . The net work equals the work done in climbing the tower (mgh).

Qc = Qh (1 − eff ) = (1.34 × 106 J)(1 − 0.18) = (1.34 × 106 J)(0.82)

Qc =

Wnet = Qh − Qc = 1.34 × 106 J − 1.1 × 106 J = 2.4 × 105 J

1.1 × 106 J

ADDITIONAL PRACTICE

1. The oldest working steam engine was designed in 1779 by James Watt.

Suppose this engine’s efficiency is 8.0 percent. How much energy must be

transferred by heat to the engine’s surroundings if 2.5 kJ is transferred by

heat into the engine? How much work is done?

2. In 1894, the first turbine-driven ship was designed. Suppose the ships

turbine’s had an efficiency of 16 percent. How much energy would have

been exhausted by the turbines if the input energy was 2.0 × 109 J and

the net power was 1.5 MW?

3. A steam engine built in 1812 still works at its original site in England. The

engine delivers 19 kW of net power. If the engine’s efficiency is 6.0 percent,

how much energy must be transferred by heat to the engine in 1.00 h?

4. The first motorcycle was built in Germany in 1885, and it could reach a

speed of 19 km/h. The output power of the engine was 370 W. If the engine’s

efficiency was 0.19, what was the input energy after 1.00 min?

5. A fairly efficient steam engine was built in 1840. It required burning only

0.80 kg of coal to perform 2.6 MJ of net work. Calculate the engine’s effi-

ciency if burning coal releases 32.6 MJ of energy per kilogram of coal.

6. The world’s tallest mobile crane can lift 3.00 × 104 kg to a height of

1.60 × 102 m. What is the efficiency of a heat engine that does the same

task while losing 3.60 × 108 J of energy by heat to the surroundings?

Problem 12A 119

NAME ______________________________________ DATE _______________ CLASS ____________________

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Holt Physics

Problem 12AHOOKE’S LAW

The pygmy shrew has an average mass of 2.0 g. If 49 of these shrews areplaced on a spring scale with a spring constant of 24 N/m, what is thespring’s displacement?

S O L U T I O NGiven: m = mass of one shrew = 2.0 g = 2.0 × 10−3 kg

n = 49

g = 9.81 m/s2

k = 24 N/m

Unknown:

Choose the equation(s) or situation: When the shrews are attached to the

spring, the equilibrium position changes. At the new equilibrium position, the

net force acting on the shrews is zero. So the spring force (given by Hooke’s law)

must be equal to and opposite the weight of the shrews.

Fnet = 0 = Felastic + Fg

Felastic = −kx

Fg = −mtotg = −nmg

−kx − nmg = 0


x = −n

k

mg


x =

x =

Forty-nine shrews of 2.0 g each provide a total mass of about 0.1 kg, or a weight

of just under 1 N. From the value of the spring constant, a force of 1 N should

displace the spring by 1/24 of a meter, or about 4 cm. This indicates that the final

result is consistent with the rest of the data.

−4.0 × 10−2 m

−(49)(2.0 × 10−3 kg)(9.81 m/s2)

(24 N/m)

1. DEFINE

2. PLAN

3. CALCULATE

4. EVALUATE

P R O B L E M

ADDITIONAL PRACTICE

1. The largest meteorite of lunar origin reportedly has a mass of 19 g. If the

meteorite placed on a scale whose spring constant is 83 N/m, what is the

compression of the spring?


NAME ______________________________________ DATE _______________ CLASS ____________________

2. In 1952, a great rainfall hit the island of Reunion in the Indian Ocean. In

less than 24 h, 187 kg of rain fell on each square meter of soil. If a 187 kg

mass is placed on a scale that has a spring constant of 1.53 × 104 N/m,

how far is the spring compressed?

3. The largest tigers, and therefore the largest members of the cat family, are

the Siberian tigers. Male Siberian tigers are reported to have an average

mass of about 389 kg. By contrast, a variety of very small cat that is na-

tive to India has an average adult mass of only 1.5 kg. Suppose this small

cat is placed on a spring scale, causing the spring to be extended from its

equilibrium position by 1.2 mm. How far would the spring be extended

if a typical male Siberian tiger were placed on the same scale?

4. The largest known crab is a giant spider crab that had a mass of 18.6 kg.

The distance from the end of one of this crab’s claws to the end of the

other claw measured about 3.7 m. If this particular giant spider crab

were hung from an elastic band so that the elongation of the band was

equal to the crab’s claw span, what would be the spring constant of the

elastic band?

5. The CN Tower in Toronto, Canada, is 533 m tall, making it the world’s

tallest free-standing structure. Suppose an unusually long bungee cord is

attached to the top of the CN Tower. The equilibrium length of the cord

is equal to one-third the height of the tower. When a test mass of 70.0 kg

is attached, the cord stretches to a length that equals two-thirds of the

tower’s height. From this information, determine the spring constant of

the bungee cord.

6. The largest ruby in the world may be found in New York. This ruby is

109 mm long, 91 mm wide, and 58 mm thick, making its volume about

575 cm3. (By comparison, the world’s largest diamond, the Star of Africa,

has a volume of just over 30 cm3.)

a. If the ruby is attached to a vertically hanging spring with a spring con-

stant of 2.00 × 102 N/m so that the spring is stretched 15.8 cm what is

the gravitational force pulling the spring?

b. What is the mass of the jewel?

7. Mauna Kea on the island of Hawaii stands 4200 m above sea level. How-

ever, when measured from the island’s sea-submerged base, Mauna Kea

has a height of 10 200 m, making it the tallest single mountain in the

world. If you have a 4.20 × 103 m elastic cord with a spring constant of

3.20 × 10−2 N/m, what force would stretch the spring to 1.02 × 104 m?

8. Rising 348 m above the ground, La Gran Piedra in Cuba is the tallest

rock on Earth. Suppose an elastic band 2.00 × 102 m long hangs vertically

off the top of La Gran Piedra. If the band’s spring constant is 25.0 N/m,

how large must a mass be if, when it is attached to the band, it causes the

band to stretch all the way to the ground?

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Problem 12B 121

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Holt Physics

Problem 12BPERIOD OF A SIMPLE PENDULUM

Two friends in France use a pendulum hanging from the world’s highestrailroad bridge to exchange messages across a river. One friend attaches aletter to the end of the pendulum and releases it so that the pendulumswings across the river to the other friend. The bridge is 130.0 m abovethe river. How much time is needed for the letter to make one swingacross the river? Assume the river is 16.0 m wide.

S O L U T I O NGiven: L = 130.0 m g = 9.81 m/s2

Unknown: t = time required for pendulum to cross river = T/2 =?

Use the equation for the period of a simple pendulum. Then divide the period by

two to find the time of one swing across the river. The width of the river is not

needed to calculate the answer, but it must be small compared to the length of

the pendulum in order to use the equations for simple harmonic motion.

T = 2pL

g = 2p

9

1

.8

3

10.m

0 m

/s2 = 22.9 s

t = T

2 =

22

2

.9 s = 11.4 s

ADDITIONAL PRACTICE

1. An earthworm found in Africa was 6.7 m long. If this worm were a sim-

ple pendulum, what would its period be?

2. The shortest venomous snake, the spotted dwarf adder, has an average

length of 20.0 cm. Suppose this snake hangs by its tail from a branch and

holds a heavy prey with its jaws, simulating a pendulum with a length of

15.0 cm. How long will it take the snake to swing through one period?

3. If bamboo, which can grow 88 cm in a day, is grown for four days and then

used to make a simple pendulum, what will be the pendulum’s period?

4. A simple pendulum with a frequency of 6.4 × 10−2 Hz is as long as the

largest known specimen of Pacific giant seaweed. What is this length?

5. The deepest permafrost is found in Siberia, Russia. Suppose a shaft is

drilled to the bottom of the frozen layer, and a simple pendulum with a

length equal to the depth of the shaft oscillates within the shaft. In 1.00 h

the pendulum makes 48 oscillations. Find the depth of the permafrost.

6. Ganymede, the largest of Jupiter’s moons, is also the largest satellite in

the solar system. Find the acceleration of gravity on Ganymede if a sim-

ple pendulum with a length of 1.00 m has a period of 10.5 s.


NAME ______________________________________ DATE _______________ CLASS ____________________

P R O B L E M

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Holt Physics

Problem 12CPERIOD OF A MASS-SPRING SYSTEM

A large pearl was found in the Philippines in 1934. Suppose the pearl isplaced on a spring scale whose spring constant is 362 N/m. If the scale’s plat-form oscillates with a frequency of 1.20 Hz, what is the mass of the pearl?

S O L U T I O NGiven: k = 362 N/m f = 1.20 Hz

Unknown: m = ?

Use the equation for the period of a mass-spring system. Then express the period

in terms of frequency (T = 1/f ).

T = 2pm

k =

1

f

m = 4p

k2f 2 =

4p32(

6

1

2

.2

N

0

/

H

m

z)2 = 6.37 kg

ADDITIONAL PRACTICE

1. The hummingbird makes a humming sound with its wings, which beat

with a frequency of 90.0 Hz. Suppose a mass is attached to a spring with

a spring constant of 2.50 × 102 N/m. How large is the mass if its oscilla-

tion frequency is 3.00 × 10−2 times that of a hummingbird’s wings?

2. In 1986, a 35 × 103 kg watch was demonstrated in Canada. Suppose this

watch is placed on a huge trailer that rests on a lightweight platform, and

that oscillations equal to 0.71 Hz are induced. Find the trailer’s mass if the

platform acts like a spring scale with a spring constant equal to 1.0 × 106 N/m.

3. A double coconut can grow for 10 years and have a mass of 20.0 kg. If a

20.0 kg double coconut oscillates on a spring 42.7 times each minute,

what is the spring constant of the spring?

4. The monument commemorating the Battle of San Jacinto in Texas

stands almost 2.00 × 102 m and is topped by a 2.00 × 105 kg star. Imagine

that a 2.00 × 105 kg mass is placed on a spring platform. The platform re-

quires 0.80 s to oscillate from the top to the bottom positions. What is

the spring constant of the spring supporting the platform?

5. Suppose a 2662 kg giant seal is placed on a scale and produces a 20.0 cm

compression. If the seal and spring system are set into simple harmonic

motion, what is the period of the oscillations?

6. On average, a newborn human’s mass is just over 3.0 kg. However, in

1955, a 10.2 kg boy was born in Italy. Suppose this baby is placed in a crib

hanging from springs with a total spring constant of 2.60 × 102 N/m. If

the cradle is rocked with simple harmonic motion, what is its period?

Problem 12D 123

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P R O B L E M

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Holt Physics

Problem 12DWAVE SPEED

The world’s largest guitar, which was built by high school students in In-diana, has strings that are 9.0 m long. The fundamental vibration thatcan be induced on each string has a wavelength equal to twice the string’slength. If the wave speed in a string is 9.0 102 m/s, what is the frequencyof vibration?

S O L U T I O NGiven: f = 50.0 Hz L = 9.0 m

Unknown: v = ?

Use the equation for the speed of a wave. The wavelength is equal to twice the

length of the string (l = 2L).

v = fl = f(2L) = (50.0 Hz)[(2)(9.0 m)] = 9.0 × 102 m/s

ADDITIONAL PRACTICE

1. The speed of sound in sea water is about 1530 m/s. If a sound wave has a

frequency of 2.50 × 102 Hz, what is its wavelength in sea water?

2. Cicadas produce a sound that has a frequency of 123 Hz. What is the

wavelength of this sound in the air? The speed of sound in air is 334 m/s.

3. Human fingers are very sensitive, detecting vibrations with amplitudes as

low as 2.0 × 10–5 m. Consider a sound wave with a wavelength exactly

1000 times greater than the lowest amplitude detectable by fingers. What

is this wave’s frequency?

4. A nineteenth-century fisherman’s cottage in England is only 2.54 m long.

Suppose a fisherman whistles inside the cottage, producing a note that has

a wavelength that exactly matches the length of the house. What is the

whistle’s frequency? The speed of sound in air is 334 m/s.

5. The lowest vocal note in the classical repertoire is low D ( f = 73.4 Hz),

which occurs in an aria in Mozart’s opera Die Entführung aus dem Serail.

If low D has a wavelength of 4.50 m, what is the speed of sound in air?

6. The highest-pitched sound that a human ear can detect is about 21 kHz.

On the other hand, dolphins can hear ultrasound with frequencies up to

280 kHz. What is the speed of sound in water if the wavelength of ultra-

sound with a frequency of 2.80 × 105 Hz is 0.510 cm? How long would it

take this sound wave to travel to a dolphin 3.00 km away?


NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 13A

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INTENSITY OF SOUND WAVES

Kåre Walkert of Sweden reportedly snores loudly, with a record intensityof 4.5 × 10–8 W/m2. Suppose the intensity of Walkert’s snores are mea-sured 0.60 m from her mouth. What is the power associated with therecord snore?

S O L U T I O NGiven: Intensity = 4.5 × 10−8 W/m2

r = 0.60 m

Unknown: P = ?

Use the equation for the intensity of a spherical wave.

Intensity = 4p

P

r 2

P = 4pr2(Intensity) = 4p(0.60 m)2(4.5 × 10−8 W/m2)

P = 2.0 × 10−7 W

ADDITIONAL PRACTICE

1. Blue whales are the loudest creatures; they can emit sound waves with an

intensity of 3.0 × 10−3 W/m2. If this intensity is measured 4.0 m from its

source, what power is associated with the sound wave?

2. The whistling sound that is characteristic of the language known as “silbo,”

which is used on the Canary Island of Gomera, is detectable at 8.0 km. Use

the spherical wave approximation to find the power of a whistler’s sound.

Sound intensity at the hearing threshold is 1.0 × 10−12 W/m2.

3. Estimate how far away a cicada can be heard if the lowest audible inten-

sity of the sound it produces is 1.0 × 10−12 W/m2 and the power of a

cicada’s sound source is 2.0 × 10−6 W.

4. Howler monkeys, found in Central and South America, can emit a sound

that can be heard by a human several miles away. The power associated

with the sound is roughly 3.0 × 10−4 W. If the threshold of hearing of a

human is assumed to be 1.1 × 10−13 W/m2, how far away can a howler

monkey be heard.

5. In 1983, Roy Lomas became the world’s loudest whistler; the power of

his whistle was 1.0 × 10−4 W. What was the sound’s intensity at 2.5 m?

6. In 1988, Simon Robinson produced a sound having an intensity level of

2.5 × 10−6 W/m2 at a distance of 2.5 m. What power was associated with

Robinson’s scream?

Problem 13B 125

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P R O B L E M

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Holt Physics

Problem 13BHARMONICS

The tallest load-bearing columns are part of the Temple of Amun inEgypt, built in 1270 B.C. Find the height of these columns if a standingwave with a frequency of 47.8 Hz is generated in an open pipe that is astall as the columns. The sixth harmonic is generated. The speed of soundin air is 334 m/s.

S O L U T I O NGiven: f6 = 47.8 Hz v = 334 m/s

n = number of the harmonic = 6

Unknown: L = ?

When the pipe is open, the wavelength associated with the first harmonic (funda-

mental frequency) is twice the length of the pipe.

fn = n 2

v

L n = 1, 2, 3, . . .

L = n 2

v

fn = (6)

2

(

(

3

4

3

7

4

.8

m

H

/s

z

)

) = 21.0 m

ADDITIONAL PRACTICE

1. A 47.0 m alphorn was made in Idaho in 1989. An alphorn behaves like a

pipe with one end closed. If the frequency of the fifteenth harmonic is

26.7 Hz, how long is the alphorn? The speed of sound in air is 334 m/s.

2. A fully functional acoustic guitar over 8.0 m in length is on display in

Bristol, England. Suppose the speed of waves on the guitar’s strings is

5.00 × 102 m/s. If a third harmonic is generated on a string, so that the

sound produced in air has a wavelength of 3.47 m, what is the length of

the string? The speed of sound in air is 334 m/s.

3. The unsupported flagpole built for Canada’s Expo 86 has a height of

86 m. If a standing wave with a 19th harmonic is produced in an 86 m

open pipe, what is its frequency? The speed of sound in air is 334 m/s.

4. A power-plant chimney in Spain is 3.50 × 102 m high. If a standing wave

with a frequency of 35.5 Hz is generated in an open pipe with a length

equal to the chimney’s height and the 75th harmonic is present, what is

the speed of sound?

5. The world’s largest organ was completed in 1930 in Atlantic City, New

Jersey. Its shortest pipe is 4.7 mm long. If one end of this pipe is closed,

what is the number of harmonics created by an ultrasound with a wave-

length of 3.76 mm?


NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 14A

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ELECTROMAGNETIC WAVES

The atoms in an HCl molecule vibrate like two charged balls attached to theends of a spring. If the wavelength of the emitted electromagnetic wave is3.75 mm, what is the frequency of the vibrations?

S O L U T I O NGiven: l = 3.75 × 10−6 m

c = 3.00 × 108 m/s

Unknown: f = ?

Use the wave speed equation, and solve for l.

c = fl

f = lc

= 3

3

.

.

0

7

0

5

××

1

1

0

0

8

−6m

m

/s = 8.00 × 1013 Hz

ADDITIONAL PRACTICE

1. New-generation cordless phones use a 9.00 × 102 MHz frequency and can

be operated up to 60.0 m from their base. How many wavelengths of the

electromagnetic waves can fit between your ear and a base 60.0 m away?

2. The highest directly measured frequency is 5.20 × 1014 Hz, corresponding

to one of the transitions in iodine-127. How many wavelengths of elec-

tromagnetic waves with this frequency could fit across a dot on a book

page? Assume the dot is 2.00 × 10−4 m in diameter.

3. Commercial trucks cause about 18 000 lane-change and merging accidents

per year in the United States. To prevent many of them, a warning system

covering blind spots is being developed. The system uses electromagnetic

waves of frequency 2.40 × 1010 Hz. What is the wavelength of these waves?

4. A typical compact disc stores information in tiny pits on the disc’s surface.

A typical pit size is 1.2 mm. What is the frequency of electromagnetic waves

that have a wavelength equal to the typical CD pit size?

5. A new antiterrorist technique detects the differences in electromagnetic

waves emitted by humans and by weapons made of metal, plastic, or

ceramic. One possible range of wavelengths used with this technique is

from 2.0 mm to 5.0 mm. Calculate the associated range of frequencies.

6. The U.S. Army’s loudest loudspeaker is almost 17 m across and is trans-

ported on a special trailer. The sound is produced by an electromagnetic

coil that can generate a minimum frequency of 10.0 Hz. What is the wave-

length of these electromagnetic waves?

Problem 14B 127

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Holt Physics

Problem 14BCONCAVE MIRRORS

Lord Rosse, who lived in Ireland in the nineteenth century, built a reflect-ing telescope called the Leviathan. Lord Rosse used it for astronomicalobservations and discovered the spiral form of galaxies. Suppose theLeviathan’s mirror has a focal length of 2.50 m. Where would you place anobject in front of the mirror in order to form an image at a distance of3.75 m? What would the magnification be? If the image height were 6.0 cm, what would the object height be?

S O L U T I O NGiven: f = +2.50 m q = +3.75 m

h = 6.0 cm

The mirror is concave, so f is positive. The object is in front of

the mirror, so q is positive.

Unknown: p = ? M = ?

Diagram:

Choose the equation(s) or situation: Use the mirror equation for focal length

and the magnification formula.

p

1 +

1

q =

1

f M = −

p

q


p

1 =

1

f −

1

q


p

1 =

2.5

1

0 m −

3.7

1

5 m =

0

1

.4

m

00 −

0

1

.2

m

67 =

0

1

.1

m

33

p = 7.50 m

12

q = ?

3

3

2C F

1

p = 7.50 m

f = 2.50 m

1. DEFINE

2. PLAN

3. CALCULATE


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ADDITIONAL PRACTICE

1. In Alaska, the top of Mount McKinley has been seen from the top of

Mount Sanford, a distance of 370 km. An object is 3.70 × 102 km from a

giant concave mirror. If the focal length of the mirror is 2.50 × 102 km

what are the object distance and the magnification?

2. A human hair is about 80.0 mm thick. If one uses a concave mirror with a

focal length of 2.50 cm and obtains an image of −59.0 cm, how far has

the hair been placed from the mirror? What is the magnification of the

hair?

3. A mature blue whale may have a length of 28.0 m. How far from a con-

cave mirror with a focal length of 30.0 m must a 7.00-m-long baby blue

whale be placed to get a real image the size of a mature blue whale?

4. In 1950 in Seattle, Washington, there was a Christmas tree 67.4 m tall.

How far from a concave mirror having a radius of curvature equal to

12.0 m must a person 1.69 m tall stand to form a virtual image equal to

the height of the tree? Will the image be upright or inverted?

5. A stalagmite that is 32 m tall can be found in a cave in Slovakia. If a con-

cave mirror with a focal length of 120 m is placed 180 m from this stalag-

mite, how far from the mirror will the image form? What is the size of

the image? Is it upright or inverted? real or virtual?

6. The eye of the Atlantic giant squid has a diameter of 5.00 × 102 mm. If

the eye is viewed in a concave mirror with a radius of curvature equal to

the diameter of the eye and the eye is 1.000 × 103 mm from the mirror,

how far is the image from the mirror? What is the size of the image? Is

the image real or virtual?

7. Quick Bird is the first commercial satellite designed for forming high-

resolution images of objects on Earth. Suppose the satellite is 1.00 ×102 km above the ground and uses a concave mirror to form a primary

image of a 1.00 m object. If the image size is 4.00 mm and the image is

inverted, what is the mirror’s radius of curvature?

8. A stalactite with a length of 10.0 m was found in Brazil. If the stalactite is

placed 18.0 m in front of a concave mirror, a real image 24.0 m tall is

formed. Calculate the mirror’s radius of curvature.

Substitute the values for p and q to find the magnification of the image and h′ to

find the object height.

M = − 3

7

.

.

7

5

5

0

m

m =

h = − p

q

h′ = −

(7.50 m

3.7

)(

5

0

m

.060 m) =

The image appears between the focal point (2.50 m) and the center of curvature,

is smaller than the object, and is inverted (–1< M < 0). These results are con-

firmed by the ray diagram. The image is therefore real.

0.12 m

−0.500

4. EVALUATE

Problem 14C 129

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Holt Physics

Problem 14CCONVEX MIRRORS

The largest jellyfish ever caught had tentacles up to 36 m long, which isgreater than the length of a blue whale. Suppose the jellyfish is located infront of a convex spherical mirror 36.0 m away. If the mirror has a focallength of 12.0 m, how far from the mirror is the image? What is the imageheight of the jellyfish?

S O L U T I O NGiven: f = −12.0 m p = +36.0 m

h = 36 m

The mirror is convex, so f is negative. The object is in front of

the mirror, so p is positive.

Unknown: q = ? M = ?

Diagram:

Choose the equation(s) or situation: Use the mirror equation for focal length

and the magnification formula.

p

1 +

1

q =

1

f M = −

p

q


1

q =

1

f −

p

1


1

q = −

12.

1

0 m −

36.

1

0 m = −

0.

1

08

m

33 −

0.

1

02

m

78 −

0.

1

11

m

11

q =

Substitute the values for p and q to find the magnification of the image and h to

find the image height.

M = − −

3

9

6

.0

.0

01

m

m =

h′ = − q

p

h = −

(− 9.0

(

0

3

1

6.

m

0 m

)(3

)

6 m) = 9.001 m

0.250

−9.001 m

p = 36.0 m

2 23

31

F Cf =–12.0 m

q = ?

1

1. DEFINE

2. PLAN

3. CALCULATE


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ADDITIONAL PRACTICE

The image appears between the focal point (–12.0 m) and the mirror’s surface, as

confirmed by the ray diagram. The image is smaller than the object (M < 1) and

is upright (M > 0), as is also confirmed by the ray diagram.

1. The radius of Earth is 6.40 × 103 km. The moon is about 3.84 × 105 km

away from Earth and has a diameter of 3475 km. The Pacific Ocean sur-

face, which can be considered a convex mirror, forms a virtual image of

the moon. What is the diameter of that image?

2. A 10 g thread of wool was produced by Julitha Barber of Australia in

1989. Its length was 553 m. Suppose Barber is standing a distance equal

to the thread’s length from a convex mirror. If the mirror’s radius of cur-

vature is 1.20 × 102 m, what will the magnification of the image be?

3. Among the many discoveries made with the Hubble Space Telescope are

four new moons of Saturn, the largest being just about 70.0 km in diam-

eter. Suppose this moon is covered by a highly reflective coating, thus

forming a spherical convex mirror. Another moon happens to pass by at

a distance of 1.00 × 102 km. What is the image distance?

4. The largest scale model of the solar system was built in Peoria, Illinois. In

this model the sun has a diameter of 11.0 m. The real diameter of the sun

is 1.4 × 106 km. What is the scale to which the sun’s size has been reduced

in the model? If the model’s sun is a reflecting sphere, where in front of

the sphere is the object located?

5. Bob Henderson of Canada built a model railway to a scale of 1:1400.

How far from a convex mirror with a focal length of 20.0 mm should a

full-size engine be placed so that the size of its virtual image is the same

as that of the model engine?

6. The largest starfish ever discovered had a diameter of 1.38 m. Suppose

an object of this size is placed 6.00 m in front of a convex mirror. If the

image formed is just 0.900 cm in diameter (the size of the smallest

starfish), what is the radius of curvature of the mirror?

7. In 1995, a functioning replica of the 1936 Toyota Model AA sedan was

made in Japan. The model is a mere 4.78 mm in length. Suppose an ob-

ject measuring 12.8 cm is placed in front of a convex mirror with a focal

length of 64.0 cm. If the size of the image is the same as the size of the

model car, how far is the image from the mirror’s surface?

8. Some New Guinea butterflies have a wingspan of about 2.80 × 102 mm.

However, some butterflies which inhabit the Canary Islands have a

wingspan of only 2.00 mm. Suppose a butterfly from New Guinea is

placed in front a convex mirror. The image produced is the size of a but-

terfly from the Canary Islands. If the image is 50.0 cm from the mirror’s

surface, what is the focal length of the mirror?

Problem 15A 131

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Holt Physics

Problem 15A

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SNELL’S LAW

The smallest brilliant-cut diamond has a mass of about 15 µg and a heightof just 0.11 mm. Suppose a ray of light enters the diamond from the airand, upon contact with one of the gem’s facets, refracts at an angle of 22.2°.If the angle of incidence is 65.0°, what is the diamond’s index of refraction?

S O L U T I O NGiven: qi = 65.0° qr = 22.2° ni = 1.00

Unknown: nr = ?

Use the equation for Snell’s law.

ni(sin qi) = nr(sin qr)

nr = ni (

(

s

s

i

i

n

n

qq

r

i)

) = (1.00)

(

(

s

s

i

i

n

n

6

2

5

2

.

.

0

2

°°)

) = 2.40

ADDITIONAL PRACTICE

1. Extra dense flint glass has one of the highest indices of refraction of any

type of glass. Suppose a beam of light passes from air into a block of

extra dense flint glass. If the light has an angle of incidence of 72° and an

angle of refraction of 34°, what is the index of refraction of the glass?

2. The index of refraction of a clear oil is determined by passing a beam of light

through the oil and measuring the angles of incidence and refraction. If the

light in air approaches the oil’s surface at an angle of 47.9° to the normal

and moves into the oil at an angle of 29.0° to the normal, what is the oil’s

index of refraction? Assume the index of refraction for air is 1.00.

3. Someone on a glass-bottom boat shines a light through the glass into the

water below. A scuba diver beneath the boat sees the light at an angle of

17° with respect to the normal. If the glass’s index of refraction is 1.5 and

the water’s index of refraction is 1.33, what is the angle of incidence with

which the light passes from the glass into the water? What is the angle of

incidence with which the light passes from the air into the glass?

4. A beam of light is passed through a layer of ice into a fresh-water lake

below. The angle of incidence for the light in the ice is 55.0°, while the

angle of refraction for the light in the water is 53.8°. Calculate the index of

refraction of the ice, using 1.33 as the index of refraction of fresh water.

5. An arrangement of three glass blocks with indices of refraction of 1.5,

1.6, and 1.7 are sandwiched together. A beam of light enters the first

block from air at an angle of 48° with respect to the normal. What is the

angle of refraction after the light enters the third block?

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Holt Physics

Problem 15BLENSES

Suppose the smallest car that is officially allowed on United States roadsis placed upright in front of a converging lens. The lens, which has a focallength of 1.50 m, forms an image 75.0 cm tall and 2.00 m away. Calculatethe object distance, the magnification, and the object height.

S O L U T I O NGiven: q = +2.00 m f = +1.50 m h′ = −0.750 m

The image is behind the lens, so q is positive. The lens is con-

verging, so the focal length is positive (f > 0). The image is in-

verted, so h′ is negative.

Unknown: p = ? M = ? h = ?

Diagram:

Choose the equation(s) or situation: Use the thin-lens equation to calculate

the image distance. Then use the equation for magnification to calculate the

magnification and the object height.

1

f =

p

1 +

1

q M = −

p

q =

h

h

′


p

1 =

1

f −

1

q


p

1 =

1.5

1

0 m −

2.0

1

0 m =

0

1

.6

m

67 −

0

1

.5

m

00 =

0

1

.1

m

67 p =

M = − p

q = −

(

(

2

6

.

.

0

0

0

0

m

m

)

) =

h = M

h′ =

(−(−

0.

0

7

.

5

3

0

33

m

)

) =

Because −1 < M < 1, the image must form at a distance less than 2f but greater

than f , which is the case. At this position the image is real and inverted.

2.25 m

−0.333

6.00 m

1

32

1

FF

2F2F

32

p = ?

f = 1.50 m

fq = 2.00 m

h = ?

h' = –0.750 m

1. DEFINE

2. PLAN

3. CALCULATE

4. EVALUATE

132 Holt Physics Problem Workbook

Problem 15B 133

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ADDITIONAL PRACTICE

1. The National Museum of Photography, Film & Television, in England,

has a huge converging lens with a diameter of 1.37 m and a focal length

of 8.45 m. Suppose you use this lens as a magnifying glass. At what dis-

tance would a friend have to stand for the friend’s image to appear 25 m

in front of the lens? What is the image magnification?

2. The largest of seals is the elephant seal, while the smallest seal, the

Galápagos fur seal, is only 1.50 m in length. Suppose you use a diverg-

ing lens with a focal length of 8.58 m to observe an elephant seal. The

elephant seal’s image turns out to have the exact length of a Galápagos

fur seal and forms 6.00 m in front of the lens. How far away is the ele-

phant seal, and what is its length?

3. The common musk turtle, also called a “stinkpot,” has a length of

7.60 cm at maturity. Suppose a turtle with this length is placed in front

of a diverging lens that has a 14.0 cm focal length. If the turtle’s image

is 4.00 cm across, how far is the turtle from the lens? How far is the tur-

tle’s image from the lens?

4. The largest mammal on land, the elephant, can reach a height of 3.5 m.

The largest mammal in the sea, however, is much bigger. A blue whale,

which is also the largest animal ever to have lived on Earth, can be as

long as 28 m. If you use a diverging lens with a focal length of 10.0 m to

look at a 28.0-m-long blue whale, how far must you be from the whale

to see an image equal to an elephant’s height (3.50 m)?

5. The largest scorpions in the world live in India. The smallest scorpions

live on the shore of the Red Sea and are only about 1.40 cm in length.

Suppose a diverging lens with a focal length of 20.0 cm forms an image

that is 1.40 cm wide. If the image is 19.00 cm in front of the lens, what

is the object distance and size?

6. The ocean sunfish, Mola mola, produces up to 30 × 106 eggs at a time.

Each egg is about 1.3 mm in diameter. How far from a magnifying glass

with a focal length of 6.0 cm should an egg be placed to obtain an

image 5.2 mm in size? How far is it between the image and the lens?

7. In 1992, Thomas Bleich of Austin, Texas, produced a photograph nega-

tive of about 3500 attendants at a concert. The negative was more than

7 m long. Bleich used a panoramic camera with a lens that had a focal

length of 26.7 cm. Suppose this camera is used to take a picture of just

one concert attendant. If the attendant is 3.00 m away from the lens, how

far should the film be from the lens? What is the image magnification?

8. Komodo dragons, or monitors, are the largest lizards, having an average

length of 2.25 m. This is much shorter than the largest crocodiles. If a

crocodile is viewed through a diverging lens with a focal length of

5.68 m, its image is 2.25 m long. If the crocodile is 12.0 m from this

lens, what is the image distance? How long is the crocodile?


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9. The body of the rare thread snake is as thin as a match, and the longest

specimen ever found was only 108 mm long. If a thread snake of this

length is placed a distance equal to four times its length from a diverg-

ing lens and the lens has a focal length of 216 mm, how long is the

snake’s image? How far from the lens is the image?

10. Tests done by the staff of Popular Photography magazine revealed that

the zoom lenses available in stores have a focal length different from

what is written on them. Suppose one of these lenses, which is identi-

fied as having a focal length of 210 mm, yields an upright image of an

object located 117 mm away. If the image magnification is 2.4, what is

the true focal length of the lens?

11. In 1994, a model car was made at a scale of 1:64. This car traveled more

than 600 km in 24 h, setting a record. If this model car is placed under

an opaque projector, a real image will be projected. Suppose the image

on the screen has the same size as the actual, full-scale car. If the screen

is 12 m from the lens, what is the focal length of the lens? Is the image

upright or inverted?

12. The tallest man in history, Robert Wadlaw, was 2.72 m tall. The smallest

woman in history, Pauline Musters, had a height of 0.55 m. Suppose

Wadlaw is 5.0 m away from a converging lens. If his image is the same

size as Musters, what is the focal length of the lens?

13. Hummingbirds eggs, which have an average size of 10.0 mm, are the

smallest eggs laid by any bird. Suppose an egg is placed 12.0 cm from a

magnifying glass. A virtual image with a magnification of 3.0 is pro-

duced. What is the focal length of the lens?

14. In 1876, the Daily Banner, a newspaper printed in Roseberg, Oregon,

had pages that were 7.60 cm wide. What would be the width of this

newspaper’s image if the newspaper were placed 16.0 cm from a diverg-

ing lens with a focal length of 12.0 cm?

15. Estimates show that the largest dinosaurs were 48 m long. Suppose you

take a trip back in time with a camera that has a focal length of 110 mm.

Coming across a specimen of the largest dinosaur, you take its picture,

but to be safe and inconspicuous you take it from a distance of 120 m.

What length will the image have on the film?

16. The smallest spiders in the world are only about 0.50 mm across. On

the other hand, the goliath tarantula, of South America, can have a leg

span of about 280 mm. Suppose you use a diverging lens with a focal

length of 0.80 m to obtain an image that is 0.50 mm wide of an object

that is 280 mm wide. How far is the object from the lens? How far is the

image from the lens?

Problem 15C 135

NAME ______________________________________ DATE _______________ CLASS ____________________

P R O B L E MRutile, TiO2, has one of the highest indices of refraction: 2.80. Supposethe critical angle between rutile and an unknown liquid is 33.6°. What isthe liquid’s index of refraction?

S O L U T I O NGiven: qc = 33.6°

ni = 2.80

Unknown: nr = ?

Use the equation for critical angle.

sin qc = n

nr

i

nr = ni sin qc = (2.80)(sin 33.6°) = 1.54

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Holt Physics

Problem 15CCRITICAL ANGLE

ADDITIONAL PRACTICE

1. Light moves from glass into a substance of unknown refraction index. If the

critical angle for the glass is 46° and the index of refraction for the glass is

1.5, what is the index of refraction of the other substance?

2. The largest uncut diamond had a mass of more than 600 g. Eventually, the

diamond was cut into several pieces. Suppose one of those pieces is a cube

with sides 1.00 cm wide. If a beam of light were to pass from air into the

diamond with an angle of incidence equal to 75.0°, the angle of refraction

would be 23.3°. From this information, calculate the index of refraction

and the critical angle for diamond in air.

3. A British company makes optical fibers that are 13.6 km in length. If the

critical angle for the fibers in air is 42.1°, what is the index of refraction of

the fiber material?

4. In 1996, the Fiberoptic Link Around the Globe (FLAG) was started. It

initially involves placing a 27 000 km fiber optic cable at the bottom of

the Mediterranean Sea and the Indian Ocean. Suppose the index of re-

fraction of this fiber is 1.56 and the index of refraction of sea water 1.36,

what is critical angle for internal reflection in the fiber?

5. The world’s thinnest glass is 0.025 mm thick. If the index of refraction

for this glass is 1.52, what is the critical angle of ocean water? How far

will a ray of light travel in the glass if it undergoes one internal reflection

at the critical angle?


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Problem 16AINTERFERENCE AND DIFFRACTION

P R O B L E MTo help prevent cavities, scientists at the University of Rochester have de-veloped a method for melting tooth enamel without disturbing the innerlayers, or pulp, of the tooth. To accomplish this, short pulses from a laserare used. These laser pulses, which are in the microwave portion of theelectromagnetic spectrum, have a wavelength of 9.3 µm. Suppose thislaser is operated continuously with a double-slit arrangement. If the slitshave a separation of 45 µm, at what angle will the third-order maximumbe observed?

S O L U T I O NGiven: l = 9.3 mm = 9.3 × 10−6 m

d = 45 mm = 45 × 10−6 m

m = 3

Unknown: q = ?

Choose the equation(s) or situation: Because a maximum (bright) fringe is ob-

served, the equation for constructive interference should be used.

d(sin q) = ml


q = sin−1m

d

l


q = sin−13(

4

9

5

.3

××1

1

0

0−

−

6

6

m

m)

q =

The angle at which the third-order maximum appears is 38° from the central

maximum. Although the wavelength of the electromagnetic radiation is large

compared with that of visible light, the separation of the slits is much larger than

it would be in a visible-light double-slit setup. If the same slit separation that is

used with visible light (typically on the order of a few micrometers) were used

with microwave radiation, the interference pattern would not appear because

even for m = 1, sin q would be greater than one. This indicates that the condi-

tions for first-order constructive interference would not exist.

38°

1. DEFINE

2. PLAN

3. CALCULATE

4. EVALUATE

ADDITIONAL PRACTICE

1. Comet Hale-Bopp, which came close to Earth in 1997, has a complex

chemical composition. To understand it, scientists analyzed radiation

emitted from the comet’s nucleus. Carbon atoms in the comet emitted

Problem 16A 137

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radiation with a wavelength of 156.1 nm. Using a double-slit apparatus

with a slit separation of 1.20 × 103 nm to measure these wavelengths, at

what angle would a fifth-order maximum be observed?

2. A typical optic fiber has a thickness of only 6.00 × 103 nm. Consider a

beam from a standard He-Ne laser that has a wavelength equal to

633 nm. Suppose this beam is incident upon two parallel slits that are

separated by a distance equal to the width of a typical optic fiber. What is

the angle at which the first dark fringe would be observed?

3. The smallest printed and bound book, which contains the children’s

story “Old King Cole,” was published in 1985. The book’s width is about

0.80 mm. Imagine a double-slit apparatus with a separation equal to the

width of this book. What wavelength would produce a third-order mini-

mum at an angle of 1.6°?

4. The water in Earth’s atmosphere blocks most of the infrared waves com-

ing from space. In order to observe light of this wavelength, the Kuiper

Airborne Observatory has been developed. The observatory consists of

an optical telescope mounted inside a modified C-141 aircraft. The plane

flies at altitudes where the relative humidity is very low and where the in-

coming infrared radiation has not yet been significantly absorbed. Sup-

pose a double-slit arrangement with a 15.0 mm slit separation is used to

analyze infrared waves received by the telescope and that a second-order

maximum is observed at 19.5°. Determine the wave’s wavelength.

5. In 1995, Pan Xixing, of China, set a record for miniature writing by plac-

ing the text of a Chinese proverb on a human hair. If blue light with a

wavelength of 443 nm passes through two slits separated by a distance

equal to the width of an average character in that proverb, the fourth-

order minimum would be observed at an angle of 2.27°. Determine the

average width of each character.

6. A way to detect termites inside wooden structures has been developed at

the University of Minnesota. A detector perceives the high-frequency

sound waves produced by the termites’ chewing. These waves are then

converted into electromagnetic signals. Suppose these signals are at

“long” radio wavelengths and that a giant double-slit apparatus has been

built to observe interference of these radio waves. If the waves have a fre-

quency of 60.0 kHz, what would be the required slit separation for

observing the fourth-order maximum at 52.0°?

7. The Federal Communications Commission (FCC) assigns radio fre-

quencies to broadcasters to prevent stations close to each other from

transmitting at the same frequency. One reserved portion of the radio

spectrum is just beyond commercial FM frequencies and is used by

weather satellites. These broadcast at frequencies close to 137 MHz.

Suppose radiation with this frequency is incident on a double-slit

apparatus and that a second-order maximum is observed at 60.0°.

What is the slit separation? What is the highest order maximum that can

be observed for this radiation and with this apparatus?


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Holt Physics

Problem 16BDIFFRACTION GRATINGS

P R O B L E MTwo graduate students from the University of Nevada have developed Vene-tian blinds that open and close automatically by use of a solar sensor. Openblinds let in infrared radiation, which increases the temperature of theroom. Consider the blinds as a type of diffraction grating with a line separa-tion equal to 5.0 cm. Suppose the infrared waves pass through this gratingso that the third-order maximum is observed at an angle of 0.69°. What isthe wavelength of the infrared radiation?

S O L U T I O NGiven: q = 0.69°

d = 5.0 cm = 5.0 × 10−2 m

m = 3

Unknown: l = ?

Choose the equation(s) or situation: Use the equation for a diffraction grating.

d(sin q) = ml


l = d(si

m

n q)


l =

l =

The electromagnetic radiation that is diffracted slightly by the Venetian blinds is

in the infrared portion of the spectrum. To increase the angle at which the dif-

fraction fringes appear, it would be necessary to narrow the spacing between the

slats in the blinds.

2.0 × 10−4 m = 0.20 mm

(5.0 × 10−2 m)(sin 0.69°)

3

1. DEFINE

2. PLAN

3. CALCULATE

4. EVALUATE

ADDITIONAL PRACTICE

1. In 1996, a phone call was placed from the U.S. National Military Com-

mand Center to the U.S. Atlantic Command, 304 km away. The phone

signal, however, traveled 120 750 km because it was transmitted via a

Milstar communication satellite. The satellite uses superhigh-frequency

electromagnetic radiation to ensure reliable communication that is se-

cured against eavesdropping. Assume these waves are passed through a

diffraction grating with 1.00 × 102 lines/m. The first-order maximum

appears at an angle of 30.0°. Determine the wavelength and frequency of

the electromagnetic waves.

Problem 16B 139

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2. Micropipette tubes with an outer diameter of 0.02 mm are used in research

with living cells. Imagine a diffraction grating with a line separation of

0.020 mm. If this grating is used to analyze electromagnetic radiation, what

wavelength would produce a third-order maximum at an angle of 12°?

3. Most stars are believed to be very nearly spherical, but R Cassiopeiae is

much closer to having an oblong, or oval, shape. This remarkable fact was

discovered by photographing its image in the orange region of the visible

spectrum (at a wavelength of 714 nm). Suppose this radiation is passed

through a diffraction grating so that it produces a third-order maximum

at an angle of 12.0°. What is the line separation in the grating?

4. In 1995, a satellite called the X Ray Timing Explorer (XTE) was launched

by NASA. The satellite can analyze X rays coming from hot matter sur-

rounding massive and often compact objects, such as neutron stars and

black holes. Because X rays have such short wavelengths, the only diffrac-

tion gratings with sufficiently small slit separations are crystal lattices. In

crystals, the separation between atoms is small enough to diffract X rays.

Suppose X rays with a wavelength of 40.0 nm are incident on a crystal

lattice that has a “line” separation of 150.0 nm. At what angle will the

second-order maximum occur?

5. Electromagnetic radiation comes to Earth from hydroxyl radicals in giant

molecular clouds in space. The frequency of this radiation is 1612 MHz,

which is close to the microwave frequencies used for aircraft communi-

cations. Filtering out this background noise is an important task. Sup-

pose a technique similar to visible spectroscopy is performed with ambi-

ent microwave radiation using a microwave radio telescope. These waves

could be directed toward a large diffraction grating so that their wave-

lengths and intensities could be analyzed. If the grating’s line separation

is 45.0 cm, at what angle would the first-order maximum occur for mi-

crowaves with a frequency of 1612 MHz?

6. A new astronomical facility is set for completion on Mount Wilson, in

Georgia, by 1999. Named the Center for High Angular Resolution Astron-

omy (CHARA) Array, it comprises five telescopes whose individual im-

ages will be analyzed by computer. The resulting data will then be used to

form a single high-resolution image of distant galaxies. The technique is

similar to what is currently being done with radio telescopes. The CHARA

Array will be used to observe radiation in the infrared portion of the

spectrum with wavelengths as short as 2200 nm. Suppose 2200 nm light

passes through a diffraction grating with 64 × 103 lines/m and produces a

fringe at an angle of 34.0°. What order maximum will the fringe be?

7. A new technology to improve the image quality of large-screen televi-

sions has been developed recently. The entire screen would consist of tiny

cells, each equipped with a movable mirror. The mirrors would change

positions relative to the incoming signal, providing a brighter image. The

entire screen would look like a diffraction grating with 250 000 lines/m.

Imagine shining red light with a wavelength of 750 nm onto this grating.

What order maximum would be observed at an angle of 48.6°?


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Holt Physics

Problem 17ACOULOMB’S LAW

P R O B L E MSuppose you separate the electrons and protons in a gram of hydrogenand place the protons at Earth’s North Pole and the electrons at Earth’sSouth Pole. How much charge is at each pole if the magnitude of the elec-tric force compressing Earth is 5.17 × 105 N? Earth’s diameter is 1.27 × 107 m.

S O L U T I O NGiven: Felectric = 5.17 × 105 N

r = 1.27 × 107 m

kC = 8.99 × 109 N•m2/C2

Unknown: q = ?

Choose the equation(s) or situation: Rearrange the magnitude of the electric

force using Coulomb’s law.

q = Fele

kcC

tricr2

Substitute the values into the equation(s) and solve: .

q = q =

The electrons and the protons have opposite signs, so the electric force between

them is attractive. The large size of the force (equivalent to the weight of a

52 700 kg mass at Earth’s surface) indicates how strong the attraction between

opposite charges in atoms is.

9.63 × 104 C

(5.17 × 105 N)(1.27 × 107 m)2

8.99 × 109 N•m2/C2

1. DEFINE

2. PLAN

3. CALCULATE

4. EVALUATE

ADDITIONAL PRACTICE

1. The safe limit for beryllium in air is 2.0 × 10−6 g/m3, making beryllium

one of the most toxic elements. The charge on all electrons in the Be

contained in 1 m3 of air at the safe level is about 0.085 C. Suppose this

charge is placed 2.00 km from a second charge. Calculate the value of

the second charge if the magnitude of the electric force between the two

charges is 8.64 × 10−8 N.

2. Kalyan Ramji Sain, of India, had a mustache that measured 3.39 m from

end to end in 1993. Suppose two charges, q and 3q, are placed 3.39 m

apart. If the magnitude of the electric force between the charges is

2.4 × 10−6 N, what is the value of q?

Problem 17A 141

NAME ______________________________________ DATE _______________ CLASS ____________________

3. The remotest object visible to the unaided eye is the great galaxy Messier

31 in the constellation Andromeda. It is located 2.4 × 1022 m from Earth.

(By comparison, the sun is only about 1.5 × 1011 m away.) Suppose two

clouds containing equal numbers of electrons are separated by a distance

of 2.4 × 1022 m. If the magnitude of the electric force between the clouds

is 1.0 N, what is the charge of each cloud?

4. In 1990, a French team flew a kite that was 1034 m long. Imagine two

charges, +2.0 nC and −2.8 nC, at opposite ends of the kite. Calculate

the magnitude of the electric force between them. If the separation of

charges is doubled, what absolute value of equal and opposite charges

would exert the same electric force?

5. Betelgeuse, one of the brightest stars in the constellation of Orion, has a

diameter of 7.0 × 1011 m (500 times the diameter of the sun). Consider

two compact clouds with opposite charge equal to 1.0 × 105 C. If these

clouds are located 7.0 × 1011 m apart, what is the magnitude of the

electric force of attraction between them?

6. An Italian monk named Giovanni Battista Orsenigo was also a dentist.

From 1868 to 1903 he extracted exactly 2 000 744 teeth, which on average

amounts to about 156 teeth per day. Consider a group of protons equal

to the total number of teeth. If this group is divided in half, calculate the

charge of each half. Also calculate the magnitude of the electric force that

would result if the two groups of charges are placed 1.00 km apart.

7. The business district of London has about 4000 residents. However,

every business day about 320 000 people are there. Consider a group of

4.00 × 103 protons and a group of 3.20 × 105 electrons that are 1.00 km

apart. Calculate the magnitude of the electric force between them.

Calculate the magnitude of the electric force if each group contains

3.20 × 105 particles and if the separation distance remains the same.

8. In 1994, element 111 was discovered by an international team of physi-

cists. Its provisional name was unununium (Latin for “one-one-one”).

Find the distance between two equal and opposite charges, each having

a magnitude equal to the charge of 111 protons, if the magnitude of the

electric force between them is 2.0 × 10−28 N.

9. By 2005, the world’s tallest building will be the International Finance

Center in Taipei, Republic of China. Suppose a 1.00 C charge is placed

at both the base and the top of the International Finance Center. If the

magnitude of the electric force stretching the building is 4.48 × 104 N,

how tall is the International Finance Center?

10. A 44 000-piece jigsaw puzzle was assembled in France in 1992. Sup-

pose the puzzle were square in shape, and that a 5.00 nC charge is

placed at the upper right corner of the puzzle and a charge of

−2.50 nC is placed at the lower left corner. If the magnitude of the

electric force the two charges exert on each other were 1.18 × 10–11 N,

what would be the distance between the two charges? What would be

the length of the puzzle’s sides?

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Holt Physics

Problem 17BTHE SUPERPOSITION PRINCIPLE

P R O B L E MThe cinema screen installed at the Science Park, in Taejon, Korea, is 24.7 mhigh and 33.3 m wide. Consider the arrangement of charges shown below.If q1 = 2.00 nC, q2 = −3.00 nC, and q3 = 4.00 nC, find the magnitude and direction of the resultant electric force on q1.

REASONINGAccording to the superposition principle, the resultant force on the charge q1 is

the vector sum of the forces exerted by q2 and q3 on q1. First find the force ex-

erted on q1 by each charge, then use the Pythagorean theorem to find the magni-

tude of the resultant force on q1. Take the ratio of the resultant y component to

the resultant x component, and then take the arc tangent of that quantity to find

the direction of the resultant force on q1.

Given: q1 = 2.00 nC = 2.00 × 10−9 C

q2 = −3.00 nC = −3.00 × 10−9 C

q3 = 4.00 nC = 4.00 × 10−9 C

r1,2 = 24.7 m

r1,3 = 33.3 m

kC = 8.99 × 109 N•m2/C2

Unknown: F1,tot = ?

Diagram:

S O L U T I O N

q1 = 2.0 nC

q2 = −3.0 nC

q3 = 4.0 nC

r1,2 = 24.7 m

r1,3 = 33.3 m

F1,2

F1,3

F1,2 F1,2F1,3

F1,3

F1,tot

q1 q1

1. Calculate the magnitude of the forces with Coulomb’s law:

F2,1 = kC (

q

r2

2

,

q

1)12 = 8.99 × 109

N

C

•m2

2

F2,1 = 8.84 × 10−11 N

F3,1 = kC (

q

r3

3

,

q

1)12 = 8.99 × 109

N

C

•m2

2

F3,1 = 6.49 × 10−11 N

(4.00 × 10−9 C)(2.00 × 10−9 C)

(33.3 m)2

(3.00 × 10−9 C)(2.00 × 10−9 C)

(24.7 m)2

Problem 17B 143

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2. Determine the direction of the forces by analyzing the signs of the charges:

The force F2,1 is attractive because q1 and q2 have opposite signs. F2,1 is

directed along the positive y-axis, so its sign is positive.

The force F3,1 is repulsive because q1 and q3 have the same sign. F3,1 is

directed toward the negative x-axis, so its sign is negative.

3. Find the x and y components of each force:

For F2,1: Fx = F3,1 = −6.49 × 10−11 N; Fy = 0

For F3,1: Fy = F2,1 = 8.84 × 10−11 N; Fx = 0

4. Calculate the magnitude of the total force acting in both directions:

Fx,tot = Fx = −6.49 × 10−11 N

Fy,tot = Fy = 8.84 × 10−11 N

5. Use the Pythagorean theorem to find the magnitude of the resultant force:

F1,tot =√

(Fx,tot)2+ (Fy,tot)2 =√

(−6.49 × 10−11 N)2 + (8.84 × 10−11 N)2

F1,tot =

6. Use a suitable trigonometric function to find the direction of the resultant

force:

In this case, you can use the inverse tangent function.

tanq = F

F

x

y,

,

t

t

o

o

t

t = = −1.36

q = tan−1(−1.36) =

7. Evaluate your answer:

The resultant force makes an angle of 53.7° to the left and above the x-axis.

−53.7°

(8.84 × 10−11 N)(−6.49 × 10−11 N)

1.10 × 10−10 N

ADDITIONAL PRACTICE

1. In 1919 in Germany, a train of eight kites was flown 9740 m above the

ground. This distance is 892 m higher than Mount Everest. Consider the

arrangement of charges located at the various heights shown below. If

q1 = 2.80 mC, q2 = −6.40 mC, and q3 = 48.0 mC, find the magnitude and

direction of the resultant electric force acting on q1.

q2 = −6.40 mC

q3 = 48.0 mC

r1,2 = 892 m

r1,3 = 9740 m

q1 = 2.80 mC

2. In 1994, a group of British and Canadian athletes performed a rope slide

off the top of Mount Gibraltar, in Canada. The speed of the sliders at


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times exceeded 160 km/h. The total length of the slide was 1747 m. Sup-

pose several sliders are located on the rope as shown. Due to friction, they

acquire the electric charges shown. Find the magnitude and direction of the

resultant electric force acting on the athlete at the far right of the diagram.

3. In 1913, a special postage stamp was issued in China. It was 248 mm long

and 70.0 mm wide. Suppose equal charges of 1.0 nC are placed in the

corners of this stamp. Find the magnitude and direction of the resultant

electric force acting on the upper right corner (assume the widest part of

the stamp is aligned with the x-axis).

4. In 1993, a chocolate chip cookie was baked in Arcadia, California. It con-

tained about three million chips and was 10.7 m long and 8.7 m wide.

Suppose four charges are placed in the corners of that cookie as follows:

q1 = −12.0 nC at the lower left corner, q2 = 5.6 nC at the upper left corner,

q3 = 2.8 nC at the upper right corner, and q4 = 8.4 nC at the lower right

corner. Find the magnitude and direction of the resultant electric force

acting on q1.

5. In 1988, a giant firework was exploded at the Lake Toya festival, in Japan.

The shell had a mass of about 700 kg and produced a fireball 1.2 km in

diameter. Consider a circle with this diameter. Suppose four charges are

placed on the circle’s perimeter so that the lines between them form a

square with sides parallel to the x- or y-axes. The charges have the follow-

ing strengths and locations: q1 = 16.0 mC at the upper left “corner,”

q2 = 2.4 mC at the upper right corner, q3 = −3.2 mC at the lower right

corner, and q4 = −4.0 mC at the lower left corner. Find the magnitude

and direction of the resultant electric force acting on q1. (Hint: Find the

distances between the charges first.)

6. American athlete Jesse Castenada walked 228.930 km in 24 h in 1976,

setting a new record. Consider an equilateral triangle with a perimeter

equal to the distance Castenada walked. Suppose the charges are placed

at the following vertices of the triangle: q1 = 8.8 nC at the bottom left

vertex, q2 = −2.4 nC at the bottom right vertex, and q3 = 4.0 nC at the

top vertex. Find the magnitude and direction of the resultant electric force

acting on q1.

q1 = 2.0 nC

q3 = 4.0 nC

q4 = 5.5 nC

r1,2 = 5.00 × 102 m

r1,4 = 1747 m

q2 = 3.0 nCr1,3 = 1.00 × 103 m

Problem 17C 145

NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 17CEQUILIBRIUM

P R O B L E MIn 1955, a water bore that was 2231 m deep was drilled in Montana.Consider two charges, q2 = 1.60 mC and q1, separated by a distance equalto the depth of the well. If a third charge, q3 = 1.998 µC, is placed 888 mfrom q2 and is between q2 and q1, this third charge will be in equilibrium.What is the value of q1?

S O L U T I O N1. DEFINE

2. PLAN

Given: q2 = 1.60 mC = 1.60 × 10−3 C

q3 = 1.998 mC = 1.998 × 10−6 C

r3,2 = 888 m

r3,1 = 2231 m − 888 m = 1342 m

r2,1 = 2231 m

kC = 8.99 × 109 N•m2/C2

Unknown: q1 = ?

Diagram:

3. CALCULATE

4. EVALUATE

q2 = 1.60 mCq3 = 1.998 mC

r3,1 = 1342 m

r2,1 = 2231 m

r3,2 = 888 m

q1

Choose the equation(s) or situation: The force exerted on q3 by q2 will be op-

posite the force exerted on q3 by q1. The resultant force on q3 must be zero in

order for the charge to be in equilibrium. This indicates that F3,1 and F3,2 must

be equal to each other.

F3,1 = kC(q

r3

3

,

q

1)12 and F3,2 = kC(q

r3

3

,

q

2)22

F3,1 = F3,2

kC(q

r3

3

,

q

1)12 = kC(q

r3

3

,

q

2)22

Rearrange the equation(s) to isolate the unknown(s): q3 and kC cancel.

q1 = q2r

r3

3

,

,

1

2

2


q1 = (1.60 × 10−3 C)1838482m

m

2

= 3.65 × 10−3 C

q1 =

Because q1 is a little more than twice as large as q2, the third charge (q3) must be

farther from q1 for the forces on q3 to balance.

3.65 mC

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ADDITIONAL PRACTICE

1. Hans Langseth’s beard measured 5.33 m in 1927. Consider two charges,

q1 = 2.50 nC and an unspecified charge, q2, are separated 5.33m. A

third charge of 1.0 nC is placed 1.90 m away from q1. If the net electric

force on this third charge is zero, what is q2?

2. The extinct volcano Olympus Mons, on Mars, is the largest mountain

in the solar system. It is 6.00 102 km across and 24 km high. Suppose

a charge of 75 mC is placed 6.0 102 km from a unspecified charge. If

a third charge of 0.10 mC is placed 24 km from the first charge and the

net electric force on this third charge is zero, how large is the unspeci-

fied charge?

3. Earth’s mass is about 6.0 × 1024 kg, while the moon’s mass is

7.3 × 1022 kg. What equal charges must be placed on Earth and the

moon to make the net force between them zero?

4. In 1974, an emerald with a mass of 17.23 kg was found in Brazil. Sup-

pose you want to hang this emerald on a string that is 80.0 cm long and

has a breaking strength of 167.6 N. To hang the jewel safely, you remove

a certain charge from the emerald and place it at the pivot point of the

string. What is the minimum possible value of this charge?

5. Little Pumpkin, a miniature horse owned by J. C. Williams, Jr., of South

Carolina, had a mass of about 9.00 kg. Consider Little Pumpkin on a

twin-pan balance. If the mass on the other pan is 8.00 kg and r equals

1.00 m, what equal and opposite charges must be placed as shown in the

diagram below to maintain equilibrium?

6. The largest bell that is in use today is in Mandalay, Myanmar, formerly

called Burma. Its mass is about 92 × 103 kg. Suppose the bell is sup-

ported in equilibrium as shown in the figure below. Calculate the value

for the charge q.

m+q

−q

r

m = 92 × 103 kg

l1 = 1.0 m+q

−q

r = 2.5 ml2 = 8.0 m

Problem 17C 147

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7. In more than 30 years, Albert Klein, of Calfornia, drove 2.5 106 km

in one automobile. Consider two charges, q1 2.0 C and q2 6.0 C,

separated by Klein’s total driving distance. A third charge, q3 4.0 C, is

placed on the line connecting q1 and q2. How far from q1 should q3 be

placed for q3 to be in equilibrium?

8. A 55 mC charge and a 137 mC charge are separated by 87 m. Where

must a 14 mC charge be placed between these other two charges in

order for the net electric force on it to be zero?

9. In 1992, a Singapore company made a rope that is 56 cm in diameter

and has an estimated breaking strength of 1.00 × 108 N. Suppose this

rope is used to connect two strongly charged asteroids in space. If the

charges of the asteroids are q1 = 1.80 × 104 C and q2 = 6.25 × 104 C,

what is the minimum length that the rope can have and still remain in-

tact? Neglect the effects of gravity.

10. The CN Tower, in Toronto, Canada, is 553 m tall. Suppose two balls,

each with a mass of 5.00 kg and a charge of 40.0 mC, are placed at the

top and bottom of the tower, respectively. The ball at the top is then

dropped. At what height is the acceleration on the ball zero?

11. Mycoplasma is the smallest living organism known. Its mass has an esti-

mated value of 1.0 × 10−16 g. Suppose two specimens of this organism

are placed 1.0 m apart and one electron is placed on each. If the

medium through which the Mycoplasma move exerts a resistive force

on the organisms, how large must that force be to balance the force of

electrostatic repulsion?

12. The parasitic wasp Carapractus cinctus has a mass of 5.0 × 10−6 kg,

which makes it one of the smallest insects in the world. If two such

wasps are given equal and opposite charges with an absolute value of

2.0 × 10−15 C and are placed 1.00 m from each other on a horizontal

smooth surface, what extra horizontal force must be applied to each

wasp to keep it from sliding? Take into account both gravitational and

electric forces between the wasps.

13. In 1995, a single diamond was sold for more than $16 million. It was not

the largest diamond in the world, but its mass was an impressive 20.0 g.

Consider such a diamond resting on a horizontal surface. It is known

that if the diamond is given a charge of 2.0 mC and a charge of at least

–8.0 mC is placed on that surface at a distance of 1.7 m from it, then the

diamond will barely keep from sliding. Calculate the coefficient of static

friction between the diamond and the surface.


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Holt Physics

Problem 17DELECTRIC FIELD STRENGTH

P R O B L E MThe Seto-Ohashi bridge, linking the two Japanese islands of Honshu andShikoku, is the longest “rail and road” bridge, with an overall length of12.3 km. Suppose two equal charges are placed at the opposite ends of thebridge. If the resultant electric field strength due to these charges at thepoint exactly 12.3 km above one of the bridge’s ends is 3.99 10–2 N/Cand is directed at 75.3° above the positive x-axis, what is the magnitude ofeach charge?

REASONINGAccording to the superposition principle, the resultant electric field strength at the

point above the bridge is the vector sum of the electric field strengths produced by

q1 and q2 . First, find the components of the electric field strengths produced by

each charge, then combine components in the x and y directions to find the elec-

tric field strength components of the resultant vector. Equate this to the compo-

nents in the x and y directions of the electric field vector. Finally, rearrange the

equation to solve for the charge.

Given: Etot = 3.99 × 10−4 N/C

q = 75.3°

r1 = 12.3 km = 1.23 × 104 m

kC = 8.99 × 109 N•m2/C2

S O L U T I O N

q2 12.3 km q1

Φ

E1 E 2

E total = 3.99 × 10–2 N/C

q = 75.3°

r1 =12.3 km

r2

Problem 17D 149

NAME ______________________________________ DATE _______________ CLASS ____________________

Unknown: q1 = ? q2 = ?

The distance r2 must be calculated from the information in the diagram. Because r2

forms the hypotenuse of a right triangle whose sides equal r1, it follows that

r2 =√

(r1)2+ (r1)2 =√

2(r1)2 = 1.74 × 104 m

The angle that r2 makes with the coordinate system equals the inverse tangent of

the ratio of the vertical to the horizontal components. Because these components

are equal,

tan f = 1.00, or f = 45.0°

1. Find the x and y components of each electric field strength vector:

At this point, the direction of each component must be taken into account.

For E1: Ex,1 = 0

Ey,1 = E1 = (

k

rC

1

q

)21

For E2: Ex,2 = E2 cos (45.0°) =

Ey,2 = E2 sin (45.0°) =

2. Calculate the magnitude of the total electric field strength in both the

x and y directions:

Ex,tot = Ex,1 + Ex,2 = = Etot cos (75.3°)

Ey,tot = Ey,1 + Ey,2 = (

k

rC

1

q

)12 + = Etot sin (75.3°)

5. Rearrange the equation(s) to isolate the unknown(s):

q2 =

q2 = q1 = 4.82 10–4 C

(3.99 × 10−2 N/C)cos(75.3°)√

2(1.74 104 m)2

8.99 109 N•m2/C2

Etot cos (75.3°)√

2(r2)2

kC

kCq2√2(r2)2

kCq2√2(r2)2

kCq2√2(r2)2

kCq2√2(r2)2

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ADDITIONAL PRACTICE

1. The world’s largest tires have a mass of almost 6000 kg and a diameter of

3.72 m each. Consider an equilateral triangle with sides that are 3.72 m

long each. If equal positive charges are placed at the points on either end

of the triangle’s base, what is the direction of the resultant electric field

strength vector at the top vertex? If the magnitude of the electric field

strength at the top vertex equals 0.145 N/C, what are the two quantities

of charge at the base of the triangle?

2. The largest fountain is found at Fountain Hills, Arizona. Under ideal

conditions, the 8000 kg column of water can reach as high as 190 m.

Suppose a 12 nC charge is placed on the ground and another charge of

unknown quantity is located 190 m above the first charge. At a point on

the ground 120 m from the first charge, the horizontal component of the

resultant electric field strength is found to be Ex = 1.60 × 10−2 N/C.

Using this information, calculate the unknown quantity of charge.

3. Pontiac Silverdome Stadium, in Detroit, Michigan, is the largest air-

supported building in the world. Suppose a charge of 18.0 mC is placed

at one end of the stadium and a charge of −12.0 mC is placed at the other

end of the stadium. If the electric field halfway between the charges is

22.3 N/C, directed toward the –12.0 mC charge, what is the length of the

stadium?

4. In 1897, a Ferris wheel with a diameter of 86.5 m was built in London.

The wheel held 10 first-class and 30 second-class cabins, and each cabin

was capable of carrying 30 people. Consider two cabins positioned

exactly opposite each other. Suppose one cabin has an unbalanced charge

of 4.8 nC and the other cabin has a charge of 16 nC. At what distance

from the 4.8 nC charge along the diameter of the wheel would the

strength of the resultant electric field be zero?

5. Suppose three charges of 3.6 mC each are placed at three corners of the

Imperial Palace in Beijing, China, which has a length of 960 m and a width

of 750 m. What is the strength of the electric field at the fourth corner?

6. The world’s largest windows, which are in the Palace of Industry and

Technology in Paris, France, have a maximum width of 218 m and a

maximum height of 50.0 m. Consider a rectangle with these dimensions.

If charges are placed at its corners, as shown in the figure below, what

is the electric field strength at the center of the rectangle? The value of

q is 6.4 nC.

q

3q

q

2q

E50.0 m

218 m

θ

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Problem 18A 151

NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 18AELECTRICAL POTENTIAL ENERGY

P R O B L E MIn 1988, a sand castle 8.37 km long was built in Britain. Suppose a chargeof 5.60 10−3 C is placed at one end of the sand castle, while an unspecifiedcharge is placed at the opposite end. If the electrical potential energy associ-ated with these charges is –50.5 J, what is the quantity of the second charge?

S O L U T I O NGiven: PEelectric = −50.5 J

q2 = 5.60 × 10−3 C

r = 8.37 km

= 8.37 × 103 m

kC = 8.99 × 109 N•m2/C2

Unknown: q1 = ?

Use the equation for the electrical potential energy associated with a pair of

charges.

q1 = r P

k

E

c

e

qle

2

ctric

q1 =

q1 = 8.40 × 10−3 C

(8.37 × 103 m)(−50.5 J)(8.99 × 109 N•m2/C2)(5.60 × 10−3 C)

ADDITIONAL PRACTICE

1. In 1995, a NASA test vehicle called Pathfinder flew 15.4 km above Earth’s

surface, setting a new altitude record for solar-powered unmanned aircraft.

Suppose the electrical potential energy of two equal charges separated by a

distance of 15.4 km is 0.868 J. What is the magnitude of each charge?

2. Each side of the Pentagon, the main building occupied by the United

States Department of Defense, is about 281 m long. Suppose a charge of

2.40 × 10−7 C is separated from a different charge by a distance of 281 m.

If the electrical potential energy of the charges is −2.0 × 10−5 J, what is

the magnitude of the second charge?

3. The service stairway for the Niesenbahn funicular railway, in Switzerland,

has 11 674 steps and is 2365 m high. If a certain charge is moved up that

stairway, it gains 4.80 × 10−4 J of electrical potential energy in Earth’s

electric field, which is directed downward and has a strength of

1.50 × 102 N/C. Find the magnitude of the charge.

4. In 1987, all 46 members of the Illawarra Mini Bike Training Club in Aus-

tralia simultaneously rode on the same motorcycle. Suppose the club

members rode a distance d. If two 44 mC charges are separated by a


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distance equal to d so that their potential energy is 1.083 × 10−2 J, what is

the value of d?

5. One of the world’s earliest lighthouses was the Pharos, in Alexandria,

Egypt. Built around 220 B.C. and destroyed by an earthquake about 600

years ago, the Pharos was taller than any lighthouse existing today. Sup-

pose two charges, −16.0 mC and 24.0 mC, are separated by a distance

equal to the height of the Pharos. If the work required to separate the

charges by a vast (infinite) distance is 2.83 × 104 J, how far apart are the

charges initially?

6. The main span of the Humber Estuary Bridge in Great Britain is 1410 m

long, causing its towers to be almost 4 cm out of parallel to compensate

for the curvature of Earth. Suppose a uniform electric field with a

strength of 380 N/C is set up from one end of the main span to the other

and an electron moves along the entire length of this field. What is the

change in electrical potential energy? Is the change positive or negative?

7. In 1987, a chimney 275 m tall was razed in South Africa by a demolition

team consisting of experts from the United States and Great Britain.

Suppose there was an object with a 12.5 nC charge at the top of the

chimney that fell to the ground after the demolition. What was the

change in the electrical potential energy if Earth’s electric field strength

has a magnitude of 1.50 × 102 N/C and a downward direction?

8. The Idaho National Engineering Laboratory announced in 1995 a new

plan for a mass-transit project. Called the CyberTran system, this project

will consist of low-mass cars running on elevated tracks. Each car will

carry up to 32 passengers and will be powered by two electric motors

that together deliver 150 kW of power. The projected speed of these cars

is 250 km/h.

a. Suppose there is no dissipation of energy through friction between

the car and the track. How much work would have to be done to

raise the speed of a fully-loaded car from rest to 2.50 × 102 km/h?

Assume that the car has a mass of 5.00 × 102 kg and that the pas-

sengers have a total mass of 2.000 × 103 kg. (Hint: Recall the

work–kinetic energy theorem from Section 5E.)

b. For the situation described in part (a), how much is the car dis-

placed? (Hint: Use the definition of power and the equation for

displacement from Section 2C to find the solution.)

c. Suppose two equal charges are separated by the distance calculated

in part (b) so that the electrical potential energy of the two-charge

system equals the energy conveyed by work to the car in part (a).

What is the magnitude of each charge?

Problem 18B 153

NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 18BELECTRIC POTENTIAL

P R O B L E MThe longest railway platform is in Kharagpur, India. Suppose you place twocharges, −8.1 × 10−8 C and −2.4 × 10−7 C, at opposite ends of this platform.The electric potential at a point 6.60 × 102 m from the greater charge is−7.5 V. What is the distance between the charges?

S O L U T I O N

Given: q1 = −8.1 × 10−8 C

q2 = −2.4 × 10−7 C

r2 = 6.60 × 102 m

V = −7.5 V

kC = 8.99 × 109 N• m2/C2

Unknown: d = ?

Use the equation for the electric potential near a point charge.

r1 = d − r2

V = kC q

r = kC

q

r1

1 + q

r2

2 = kCd q

−1

r2 +

q

r2

2k

V

C −

q

r2

2 = d

q

−1

r2

d = + r2

d =

− + 660 m

d = + 660 m

d = −4

−.7

8.

×1 ×

101−01

−

0

8

C

C

/m + 660 m = 830 m

−8.1 × 10−8 C[−8.3 × 10−10 C/m + 3.6 × 10−10 C/m]

(−2.4 × 10−8 C)

660 m

−7.5 V(8.99 × 109 N•m2/C2)

q1

k

V

c −

q

r2

2

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1. The Sydney Harbour Bridge, in Australia, is the world’s widest long-span

bridge. Suppose two charges, −12.0 nC and −68.0 nC, are separated by

the width of the bridge. The electric potential along a line between the

charges at a distance of 16.0 m from the −12.0 nC charge is −25.3 V. How

far apart are the charges?

ADDITIONAL PRACTICE


NAME ______________________________________ DATE _______________ CLASS ____________________

2. The Hughes H4 Hercules, nicknamed the Spruce Goose, has a wingspan

of 97.5 m, which is greater than the wingspan of any other plane. Sup-

pose two charges, 18.0 nC and 92.0 nC, are placed at the tips of the

wings. If an electric potential of 53.3 V is measured at a certain point

along the wings, how far is that point from the 92.0 nC charge?

3. A Canadian company has developed a scanning device that can detect

drugs, explosives, and even cash that are being smuggled across state

borders. The scanner uses accelerated protons to generate gamma rays,

which can easily penetrate through most substances that are less than a

few centimeters thick. At the heart of the scanner is a compact power

supply that can produce an electric potential as large as 1.0 × 106 V. Find

the value of a point charge q that would create an electric potential of

1.0 × 106 V at a distance of 12 cm.

4. Gravitational potential and electric potential are described by similar

mathematical equations. Suppose Earth has a uniform distribution of pos-

itive charge on its surface. A test particle with a mass of 1.0 kg and a charge

of 1.0 C is placed at some distance from Earth. What must the total charge

on Earth’s surface be for the test particle to experience equal gravitational

and electric potentials? (Hint: Obtain the equation for gravitational poten-

tial by comparing the gravitational force equation from Section 7I to

Coulomb’s law for electrostatic force. Assume that the entire charge on

Earth’s surface can be treated as a point charge at Earth’s center.)

5. Overall, the matter in the sun is electrically neutral. However, the tem-

peratures within the sun are too high for electrons to remain with posi-

tively charged nuclei for more than a fraction of a second. Suppose the

positive and negative charges in the sun could be separated into two

clouds with a separation of 1.5 × 108 km, which is equal to the average

distance between Earth and the sun.

a. Calculate the charge in each cloud. Assume that the sun, which has

a mass of 1.97 × 1030 kg, consists entirely of hydrogen. The mass of

one hydrogen atom is 1.67 × 10−27 kg.

b. Calculate the electric potential for the two clouds at a distance of

1.1 × 108 km (the distance between the sun and Venus) from the

proton cloud.

6. The freight station at Hong Kong’s waterfront is the largest multilevel in-

dustrial building in the world. The station is 292 m long and 276 m wide,

and inside there are more than 26 km of roadways. Consider a rectangle

that is 292 m long and 276 m wide. If charges +1.0q, −3.0q, +2.5q, and

+4.0q are placed at the vertices of this rectangle, what is the electric

potential at the rectangle’s center? The value of q is 64 nC.

7. The largest premechanical earthwork was built about 700 years ago in

Africa along the boundaries of the Benin Empire. The total length of the

earthwork is estimated to be 16 000 km. Consider an equilateral triangle

in space formed by three equally charged asteroids with charges of

7.2 × 10−2 C. If the length of each side of the triangle is 1.6 × 104 km,

what is the electric potential at the midpoint of any one of the sides?

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Problem 18B 155

NAME ______________________________________ DATE _______________ CLASS ____________________

8. The Royal Dragon restaurant in Bangkok, Thailand, can seat over 5000

customers, making it the largest restaurant in the world. The area of the

restaurant is equal to that of a square with sides measuring 184 m each. If

three charges, each equal to 25.0 nC, are placed at three vertices of this

square, what is the electric potential at the fourth vertex?

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Holt Physics

Problem 18CCAPACITANCE

P R O B L E MConsider a parallel-plate capacitor the size of Nauru, an island with anarea of just 21 km2. If 110 V is applied across the capacitor to store 55 J ofelectrical potential energy, what is the capacitance of this capacitor? Ifthe area of its plates is the same as the area of Nauru, what is the plateseparation?

S O L U T I O NGiven: A = 21 km2 = 21 × 106 m2

PEelectric = 55 J

∆V = 110 V

e0 = 8.85 × 10−12 C2/N•m2

Unknown: C = ? d = ?

To determine capacitance and the plate separation, use the definition for capaci-

tance energy and the equation for a parallel-plate capacitor.

PEelectric = 12

C(∆V )2

C = 2

(

P

∆E

Vele

)c2tric =

(

2

1

(

1

5

0

5

V

J

)

)2 =

C = e0 A

d

d = e0 C

A

d = 8.85 × 10−12 N

C

•m

2

2d = 2.0 × 10−2 m = 2.0 cm

(21 × 106 m2)(9.1 × 10−3 F)

9.1 × 10−3 F

ADDITIONAL PRACTICE

1. To improve the short-range acceleration of an electric car, a capacitor

may be used. Charge is stored on the capacitor’s surface between a

porous composite electrode and electrolytic fluid. Such a capacitor can

provide a potential difference of nearly 3.00 × 102 V. If the energy

stored in this capacitor is 17.1 kJ, what is the capacitance?

2. A huge capacitor bank powers the large Nova laser at Lawrence Liver-

more National Lab, in California. Each capacitor can store 1450 J of

electrical potential energy when a potential difference of 1.0 × 104 V is

applied across its plates. What is the capacitance of this capacitor?

Problem 18C 157

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3. Air becomes a conductor if the electric field across it exceeds

3.0 × 106 V/m. What is the maximum charge that can be accumulated

on an air-filled capacitor with a 0.2 mm plate separation and a plate

area equal to the area of the world’s largest painting (6.7 × 103 m2)?

4. The world’s largest hamburger, which was made in Wyoming, had a

radius of about 3.1 m. Suppose you build an air-filled parallel-plate

capacitor with an area equal to that of the hamburger and a plate sepa-

ration of 1.0 mm. The largest electric field that air can sustain before its

insulating properties break down and it begins conducting electricity is

3.0 MV/m. What is the maximum charge that you will be able to store

in the capacitor?

5. In 1987, an ultraviolet flash of light was produced at the Los Alamos

National Lab. For 1.0 ps, the power of the flash was 5.0 × 1015 W. If all

of this energy was provided by a battery of charged capacitors with a

total capacitance of 0.22 F, what was the potential difference across

these capacitors?

6. The International Exposition Center in Cleveland, Ohio, covers

2.32 × 105 m2. If a capacitor with this same area and a plate separa-

tion of 1.5 cm is charged to 0.64 mC, what is the energy stored in

the capacitor?

7. In 1990, a pizza with a radius of 18.0 m was made in South Africa. Sup-

pose you make an air-filled capacitor with parallel plates whose area is

equal to that of the pizza. If a potential difference of 575 V is applied

across this capacitor, it will store just 3.31 J of electrical potential en-

ergy. What are the capacitance and plate separation of this capacitor?

8. When the keys on a computer keyboard are pressed, the plate separation

of small parallel-plate capacitors mounted under the keys changes from

about 5.00 mm to 0.30 mm. This causes a change in capacitance, which

triggers the electronic circuitry. If the area of the plates is 1.20 cm2, find

the change in the capacitance. Is it positive or negative?

9. Tristan da Cunha, a remote island inhabited by a few hundred people,

has an area of 98 km2. Suppose a 0.20 F parallel-plate capacitor with a

plate area equal to 98 km2 is built. What is the plate separation?

10. The largest apple pie ever made was rectangular in shape and had an

area of 7.0 m × 12.0 m. Consider a parallel-plate capacitor with this

area. Assume that the capacitor is filled with air and the distance be-

tween the plates is 1.0 mm.

a. What is the capacitance of the capacitor?

b. What potential difference would have to be placed across the ca-

pacitor for it to store 1.0 J of electrical potential energy?

11. The largest lasagna in the world was made in California in 1993. It had

an area of approximately 44 m2. Imagine a parallel-plate capacitor with

this area that is filled with air. If a potential difference of 30.0 V is

placed across the capacitor, the stored charge is 2.5 mC.

a. Calculate the capacitance of the capacitor.

b. Calculate the distance that the plates are separated.

c. Calculate the electrical potential energy that is stored in the

capacitor.


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Problem 19A 159

NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 19ARELATING CURRENT AND CHARGE

P R O B L E MAmtrak introduced an electric train in 2000 that runs between New Yorkand Boston. With a travel time of 3.00 h, the train is not superfast, but it is-comfortable, very safe, and environmentally friendly. Find the chargepassing through the engine during the trip if the feeding current is 1.80 ×102 A.

S O L U T I O NGiven: ∆t = 3.00 h = 1.08 × 104 s

I = 1.80 × 102 A

Unknown: ∆Q = ?

Use the equation for electric current.

I = ∆∆Q

t

∆Q = I∆t = (1.80 × 102 A)(1.08 × 104 s) = 1.94 × 106 C

ADDITIONAL PRACTICE

1. An 8.00 × 106 kg electromagnet built in Switzerland draws a current of

3.00 × 102 A. How much charge passes through the magnet in 2.4 min?

2. On July 16, 2186, a total solar eclipse lasting 7 min, 29 s, will be observed

over the mid-Atlantic Ocean. During the eclipse, an observer may need a

flashlight to see. If the flashlight draws a current of 0.22 A, how much

charge will pass through the light bulb during the eclipse?

3. The National Institute of Standards and Technology has built a clock that

is off by only 3.3 ms a year. Consider a current of 0.88 A in a wire. How

many electrons pass through a cross-section of the wire in 3.3 × 10−6 s?

4. In England, a miniature electric bicycle has been constructed. The Ni-Cd

batteries help the rider pedal for up to 3.00 h. If a charge of 1.51 × 104 C

passes through the motor during those 3.00 h, what is the current?

5. Because high-temperature superconducting cables cannot sustain high

currents, their applicability is limited. However, a prototype of a high-

temperature superconductor that can transfer a charge of 1.8 × 105 C in

6.0 min has been constructed. To what current does this correspond?

6. In 1994, a prototype of what was claimed to be the first “practical” elec-

tric car was introduced. Recharging the batteries is not very practical,

though. Calculate the average time needed to recharge the car’s batteries

if a 13.6 A current carries 4.40 × 105 C to the batteries.


NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 19BRESISTANCE

P R O B L E MA medical belt pack with a portable laser for in-the-field medical purposeshas been constructed. The laser draws a current of 2.5 A and the circuitryresistance is 0.6 Ω. What is the potential difference across the laser?

S O L U T I O NGiven: I = 2.5 A R = 0.6 Ω

Unknown: ∆V = ?

Use the definition of resistance.

∆V = IR = (2.5 A)(0.6 Ω) = 1.5 V

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ADDITIONAL PRACTICE

1. Electric eels, found in South America, can provide a potential difference

of 440 V that draws a current of 0.80 A through the eel’s prey. Calculate

the resistance of the circuit (the eel and prey).

2. It is claimed that a certain camcorder battery can provide a potential dif-

ference of 9.60 V and a current of 1.50 A. What is the resistance through

which the battery must be discharged?

3. A prototype electric car is powered by a 312 V battery pack. What is the

resistance of the motor circuit when 2.8 × 105 C passes through the cir-

cuit in 1.00 h?

4. In 1992, engineers built a 2.5 mm long electric motor that can be driven

by a very low emf. What is the potential difference if it draws a 3.8 A cur-

rent through a 0.64 Ω resistor?

5. A team from Texas A&M University has built an electric sports car with a

maximum motor current of 2.4 × 103 A. Determine the potential differ-

ence that provides this current if the circuit resistance is 0.30 Ω.

6. Stanford University scientists have constructed the Orbiting Picosatellite

Automated Launcher (OPAL). OPAL can launch disposable “picosatel-

lites” the size of hockey pucks. Each picosatellite will be powered by a

3.0 V battery for about an hour. If the satellite’s circuitry were to have a

resistance of 16 Ω, what current would be drawn by the satellite?

7. For years, California has been striving for all zero-emission vehicles on

its roads. In 1995, a street bus with a range of 120 km was built. This bus

is powered by batteries delivering 6.00 × 102 V. If the circuit resistance is

4.4 Ω, what is the current in the bus’s circuit?

Problem 19C 161

NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 19CELECTRIC POWER

P R O B L E MIn 1994, a group of students at Lawrence Technological University, inSouthfield, Michigan, built a car that combines a conventional diesel en-gine and an electric direct-current motor. The power delivered by themotor is 32 kW. If the resistance of the car’s circuitry is 8.0 Ω, find thecurrent drawn by the motor.

S O L U T I O NGiven: P = 32 kW = 3.2 × 104 W

R = 8.0 Ω

Unknown: I = ?

Because power and resistance are known, use the second form of the power equa-

tion to solve for I.

P = I 2R

I = R

P = (3.2

(8

×.0

1 Ω0

4

)W) = 63 A

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ADDITIONAL PRACTICE

1. A flying source of light is being developed that will consist of a metal-

halide lamp lifted by a helium-filled balloon. The maximum power rat-

ing for the lamp available for the device is 12 kW. If the lamp’s resistance

is 2.5 × 102 Ω, what is the current in the lamp?

2. The first American hybrid electric bus operated in New York in 1905.

The gasoline-fueled generator delivered 33.6 kW to power the bus. If the

generator supplied an emf of 4.40 × 102 V, how large was the current?

3. A compact generator has been designed that can jump-start a car, though

the generator’s mass is only about 10 kg. Find the maximum current that

the generator can provide at 12.0 V if its maximum power is 850 W.

4. Fuel cells combine gaseous hydrogen and oxygen to effectively and

cleanly produce energy. Recently, German engineers produced a fuel cell

that can generate 4.2 × 1010 J of electricity in 1.1 × 103 h. What potential

difference would this fuel cell place across a 40.0 Ω resistor?

5. Omega, a laser built at the University of Rochester, New York, generated

6.0 × 1013 W for 1.0 ns in 1995. If this power is provided by 8.0 × 106 V

placed across the circuit, what is the circuit’s resistance?

6. In 1995, Los Alamos National Lab developed a model electric power

plant that used geothermal energy. Find the plant’s projected power

output if the plant produces a current of 6.40 × 103 A at 4.70 × 103 V.


NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 19DCOST OF ELECTRICAL ENERGY

P R O B L E MIn 1995, the fortune of Sir Muda Hassanal Bolkiah, Sultan of Brunei, wasestimated at $37 billion. Suppose this money is used to pay for the energyused by an ordinary 1.0 kW microwave oven at a rate of $0.086/kW•h.How long can the microwave oven be powered?

S O L U T I O NGiven: Cost of energy = $0.086/kW•h P = 1.0 kW

Purchase power = $37 × 109

Unknown: ∆t = ?

First, calculate the number of kilowatt-hours that can be purchased by dividing

the total amount of money ($) by the cost of energy ($/kW•h). Then divide the

energy used (kW•h) by the power (kW) to find the time.

Energy = ($37 × 109)1$k

0

W

.08

•

6

h = 4.3 × 1011 kW•h

∆t = Ene

P

rgy =

(4.3 ×(1

1

.0

01

k

1

W

kW

)

•h) = 4.3 × 1011 h = 4.9 My

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ADDITIONAL PRACTICE

1. The ten reactors of the nuclear power station in Fukushima, Japan, pro-

duce 8.8 × 109 W of electric power. If you have $1.0 × 106, how many hours

of the station’s energy output can you buy? Assume a price of $0.081/kW•h.

2. A power plant in Hawaii produces electricity by using the difference in

temperature between surface water and deep-ocean water. The plant pro-

duces 104 kW. Suppose this energy is sold at a rate of $0.120/kW•h. For

how long would $18 000 worth of this energy last?

3. For basketball fans, a flexible light source that can be attached around any

basketball hoop rim has been developed. This source reportedly lasts for

1.0 × 104 h. How much power will this light source provide over its lifetime

if its overall cost of operation is $23 and the energy cost is $0.086/kW•h?

4. Fluorescent lamps with resistances that can be adjusted from 80.0 Ω to

400.0 Ω are being produced. If such a lamp is connected to a 110 V emf

source, how much will it cost to operate the lamp at its maximum rated

power for 24 h? The cost of energy is $0.086/kW•h.

5. A solar-cell installation that can convert 15.5 percent of the sun’s energy

into electricity has been built. The device delivers 1.0 kW of power. If it

produces energy at a cost of $0.080/kW•h, how much energy must the

sun provide to produce $1000.00 worth of energy?

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Problem 20A 163

NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 20ARESISTORS IN SERIES

P R O B L E MA particular electronic-code lock provides over 500 billion combinations.Moreover, it can sustain an electric shock of 1.25 × 105 V. Suppose this potential difference is applied across a series connection of the followingresistors: 11.0 kΩ, 34.0 kΩ, and 215 kΩ. What is the equivalent resistancefor the circuit? What current would pass through the resistors?

S O L U T I O NGiven: R1 = 11.0 kΩ = 11.0 × 103 Ω

R2 = 34.0 kΩ = 34.0 × 103 ΩR3 = 215 kΩ = 215 × 103 Ω∆V = 1.25 × 105 V

Unknown: Req = ? I = ?

Choose the equation(s) or situation: Because the resistors are in series, the

equivalent resistance can be calculated from the equation for resistors in series.

Req = R1 + R2 + R3

The equation relating potential difference, current, and resistance can be used to

calculate the current.

∆V = IReq


I = R

∆

e

V

q


Req = (11.0 × 103 Ω + 34.0 × 103 Ω + 215 × 103 Ω)

Req =

I = (

(

2

1

.

.

6

2

0

5

××

1

1

0

05

5

ΩV)

) =

For resistors connected in series, the equivalent resistance should be greater than

the largest resistance in the circuit.

2.60 × 105 Ω > 2.15 × 105 Ω

0.481 A

2.60 × 105 Ω

1. DEFINE

2. PLAN

3. CALCULATE

4. EVALUATE

ADDITIONAL PRACTICE

1. The Large Optics Diamond Turning Machine at the Lawrence Livermore

National Lab in California can cut a human hair lengthwise 3000 times!

That would produce pieces with a cross-sectional area of 10−7 mm2. A

1.0 m long silver wire with this cross-sectional area would have a resistance


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of 160 kΩ. Consider three pieces of silver wire connected in series. If their

lengths are 2.0 m, 3.0 m, and 7.5 m, and the resistance of each wire is pro-

portional to its length, what is the equivalent resistance of the connection?

2. Most of the 43 × 103 kg of gold that sank with the British ship HMS Lau-

rentis in 1917 has been recovered. If this gold were drawn into a wire

long enough to wrap around Earth’s equator five times, its electrical re-

sistance would be about 5.0 × 108 Ω. Consider three resistors with resis-

tances that are exactly 1/3, 2/7, and 1/5 the resistance of the gold wire.

What equivalent resistance would be produced by connecting all three

resistors in series?

3. A 3 mm thick steel wire that stretches for 5531 km has a resistance of

about 82 kΩ. If you connect in series three resistors with the values

16 kΩ, 22 kΩ, and 32 kΩ, what value must the fourth resistor have for

the equivalent resistance to equal 82 kΩ?

4. The largest operating wind generator in the world has a rotor diameter of

almost 100 m. This generator can deliver 3.2 MW of power. Suppose you

connect a 3.0 kΩ resistor, a 4.0 kΩ resistor, and a 5.0 kΩ resistor in series.

What potential difference must be applied across these resistors in order

to dissipate power equal to 1.00 percent of the power provided by the

generator? What is the current through the resistors? (Hint: Recall the re-

lation between potential difference, resistance, and power.)

5. The resistance of loudspeakers varies with the frequency of the sound

they produce. For example, one type of speaker has a minimum resis-

tance of 4.5 Ω at low frequencies and 4.0 Ω at ultra-low frequencies, and

it has a peak resistance of 16 Ω at a high frequency. Consider a set of re-

sistors with resistances equal to 4.5 Ω, 4.0 Ω, and 16.0 Ω. What values of

the equivalent resistance can be obtained by connecting any two of them

in different series connections?

6. Standard household potential difference in the United States is 1.20 × 102 V.

However, many electric companies allow the residential potential difference

to increase up to 138 V at night. Suppose a 2.20 × 102 Ω resistor is con-

nected to a constant potential difference. A second resistor is provided in an

alternate circuit so that when the potential difference rises to 138 V, the sec-

ond resistor connects in series with the first resistor. This changes the resis-

tance so that the current in the circuit is unchanged. What is the value of

the second resistor?

7. The towers of the Golden Gate Bridge, in San Francisco, California, are

about 227 m tall. The supporting cables of the bridge are about 90 cm

thick. A steel cable with a length of 227 m and a thickness of 90 cm

would have a resistance of 3.6 × 10−5 Ω. If a 3.6 × 10−5 Ω resistor is con-

nected in series with an 8.4 × 10−6 Ω resistor, what power would be dissi-

pated in the resistors by a 280 A current? (Hint: Recall the relation be-

tween current, resistance, and power.)

Problem 20B 165

NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 20BRESISTORS IN PARALLEL

P R O B L E MA light bulb in a camper’s flashlight is labeled 2.4 V, 0.70 A. Find theequivalent resistance and the current if three of these light bulbs are connected in parallel to a standard C size 1.5 V battery.

S O L U T I O NGiven: ∆V1 = 2.4 V I1 = 0.70 A

∆V2 = 2.4 V I2 = 0.70 A

∆V3 = 2.4 V I3 = 0.70 A

∆V = 1.5 V

Unknown: Req = ? I = ?

Choose the equation(s) or situation: Because the resistors (bulbs) are in parallel,

the equivalent resistance can be calculated from the equation for resistors in parallel.

R

1

eq =

R

1

1 +

R

1

2 +

R

1

3

To calculate the individual resistances, use the definition of resistance.

∆Vn = InRn

The following form of the equation can be used to calculate the current.

∆V = IReq


Rn = ∆

I

V

n

n I = R

∆

e

V

q


R1 = ∆

I

V

1

1 = (

(

0

2

.

.

7

4

0

V

A

)

) = 3.4 Ω

R2 = ∆

I

V

2

2 = (

(

0

2

.

.

7

4

0

V

A

)

) = 3.4 Ω

R3 = ∆

I

V

3

3 = (

(

0

2

.

.

7

4

0

V

A

)

) = 3.4 Ω

R

1

eq =

R

1

1 +

R

1

2 +

R

1

3 =

(3.4

3

Ω) =

0

1

.8

Ω8

Req =

I = (

(

1

1

.

.

1

5

ΩV)

) =

For resistors connected in parallel, the equivalent resistance should be less than

the smallest resistance in the circuit.

1.1 Ω < 3.4 Ω

1.4 V

1.1 Ω

1. DEFINE

2. PLAN

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3. CALCULATE

4. EVALUATE


NAME ______________________________________ DATE _______________ CLASS ____________________

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ADDITIONAL PRACTICE

1. A certain full-range loudspeaker has a maximum resistance of 32 Ω at

45 Hz, a resistance of 5.0 Ω for most audible frequencies, and a resistance

of only 1.8 Ω at 20 kHz. Consider three resistors with resistances of

1.8 Ω, 5.0 Ω, and 32 Ω. Find the equivalent resistance if they are

connected in parallel.

2. The Large Electron Positron ring, near Geneva, Switzerland, is one of the

biggest scientific instruments on Earth. The circumference of the ring is

27 km. A copper wire with this length and a cross-sectional area of 1 mm2

will have a resistance of about 450 Ω. Consider a parallel connection of

three resistors with resistances equal to 1.0, 2.0, and 0.50 times the resis-

tance of the copper wire, respectively. What is the equivalent resistance?

3. Cars on the Katoomba Scenic Railway are pulled along by winding ca-

bles, and at one point, they move along a 310 m span that makes an angle

of 51° with the horizontal. A 310 m steel cable that is 4 cm thick would

have an estimated resistance of 2.48 × 10−2 Ω. An equivalent resistance

of 6.00 × 10−3 Ω can be obtained if two resistors, one having the same

resistance as the steel cable, are connected in parallel. Find the resistance

of the second resistor.

4. In 1992 in Atlanta, 1 724 000 United States quarters were placed side by

side in a straight line. Suppose these quarters were stacked to form a

cylindrical tower. If the influence of the air gaps between coins is negligi-

ble, the resistance of the tower can be estimated easily. Find the resistance

if the parallel connection of four resistors that have resistances equal to

exactly 1, 3, 7, and 11 times the tower’s resistance yields an equivalent

resistance of 6.38 × 10−2 Ω.

5. The largest piece of gold ever found had a mass of about 70 kg. If you

were to draw this mass of gold out into a thin wire with a cross-sectional

area of 2.0 mm2, the wire would have a length of 1813 km. The wire

would also have a resistance per unit length of 1.22 × 10−2 Ω/m.

a. What is the resistance of the wire?

b. Suppose the wire were cut into pieces having resistance of exactly

1/2, 1/4, 1/5, and 1/20 of the wire’s resistance, respectively. If these

pieces are reconnected in parallel, what is the equivalent resistance

of the four pieces?

6. A powerful cordless drill uses a 14.4 V battery to deliver 225 W of power.

Treating the drill as a resistor, find its resistance. If a single 14.4 V battery

is connected to four “drill” resistors that are connected in parallel, what

are the equivalent resistance and the battery current?

7. The total length of the telephone wires in the Pentagon is 3.22 × 105 km.

Suppose these wires have a resistance of 1.0 × 10−2 Ω/m. If all the wires

are cut into 1.00 × 103 km pieces and all pieces are connected in parallel

to a AA battery (∆V = 1.50 V), what would the current through the wires

be? Assume that a AA battery can sustain this current.

Problem 20C 167

NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 20CEQUIVALENT RESISTANCE

P R O B L E MA certain amplifier can drive five channels with a load of 8.0 Ω each.Consider five 8.0 Ω resistors connected as shown. What is the equivalentresistance?

R1 = 8.0 ΩR3 = 8.0 Ω

R2 = 8.0 Ω

R5 = 8.0 Ω

R4 = 8.0 Ω

REASONING

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Divide the circuit into groups of series and parallel resistors. This way, the meth-

ods presented in determining equivalent resistance for resistors in series and par-

allel can be used to calculate the equivalent resistance for each group.

S O L U T I O N1. Redraw the circuit as a group of resistors along one side of the circuit.

Bends in a wire do not affect the circuit and do not need to be represented in a

schematic diagram. Redraw the circuit without corners, keeping the arrange-

ment of the circuit elements the same and disregarding the emf source.

2. Identify components in series, and calculate their equivalent resistance.

At this stage, there are no resistors in series.

3. Identify components in parallel, and calculate their equivalent resistance.

Resistors in group (a) are in parallel. For group (a):

Re

1

q,a =

R

1

1 +

R

1

2 =

(8.0

2

Ω) =

0

1

.2

Ω5

Req,a = 4.0 Ω

R1 = 8.0 Ω

Req

R5 = 8.0 Ω

R4 = 8.0 ΩR3 = 8.0 Ω

R2 = 8.0 Ω

Req,a

Req,b

(a)

(b)

(c)


NAME ______________________________________ DATE _______________ CLASS ____________________

4. Repeat steps 2 and 3 until the resistors in the circuit are reduced to a

single equivalent resistance.

Resistors in group (b) are in series.

Req,b = Req,a + R3 + R4 = 4.0 Ω + 8.0 Ω + 8.0 Ω = 20.0 Ω

Resistors in group (c) are in parallel.

R

1

eq =

Re

1

q,b +

R

1

5 =

(20.

1

0 Ω) +

(8.0

1

Ω) =

0.

1

05

Ω00

+ 0

1

.1

Ω2

Req = 0

1

.1

Ω7

−1

= 5.9 Ω

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ADDITIONAL PRACTICE

1. In 1993, the gold reserves in the United States were about 8.490 × 106 kg.

If all that gold were made into a thick wire with a cross-sectional area

of 1 cm2, its total resistance would be about 6.60 × 102 Ω. If the same

operation were applied to the gold reserves of Germany, France, and

Switzerland, the resistances would be 2.40 × 102 Ω, 2.00 × 102 Ω, and

2.00 × 102 Ω, respectively. Now consider all four resistors connected as

shown in the circuit diagram below. Find the equivalent resistance.

2. In 1920, there was an electric car that could travel at about 40 km/h and

that had about a 45 km range. The car was powered by a 24 V battery.

Suppose this battery is connected to a combination of resistors, as shown

in the circuit diagram below. What is the battery current?

R1 = 2.0 Ω R2 = 4.0 Ω

R3 = 6.0 Ω

R4 = 3.0 Ω

∆V = 24 V

R1 = 6.60 × 102 Ω R2 = 2.40 × 102 ΩR4 = 2.00 × 102 Ω

R3 = 2.00 × 102 Ω

Problem 20C 169

NAME ______________________________________ DATE _______________ CLASS ____________________

3. By adding water to the Enviro-Gen portable power pack, the device can

generate 12 V for up to 40.0 h. If this device powers a combination of

small appliances with the resistances shown in the circuit diagram below,

what will be the net current for the circuit?

4. In 1995, the most powerful wind generator in the United States was the

Z-40, which has a rotor diameter of 40 m. The machine is capable of

producing 5.00 × 102 A at 1.00 × 103 V, assuming 100 percent efficiency.

Suppose a direct current of 5.00 × 102 A is produced when a potential

difference of 1.00 × 103 V is placed across a circuit of resistors, as shown

in the diagram below. What is the equivalent resistance of the circuit?

What is the power dissipated in the circuit?

5. The longest-lasting battery in the world is at Oxford University, in

England. It was built in 1840 and was still working in 1977, producing

a 1.0 × 10−8 A current. The battery provided a potential difference of

2.00 × 103 V. If the battery is connected to a group of resistors, as shown

in the circuit diagram below, find the value of the equivalent resistance

and the value of r.

R1 = r R2 = 3r

R3 = 2r R4 = 4r

∆V = 2.00 × 103 V

R1 = 1.5 Ω

R2 = 3.0 Ω

R3 = 1.0 Ω

∆V = 1.00 × 103 V

R1 = 2.5 Ω R2 = 3.5 Ω

R3 = 3.0 Ω

R4 = 4.0 Ω

R5 = 1.0 Ω

∆V = 12 V

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NAME ______________________________________ DATE _______________ CLASS ____________________

6. During World War II, a high-powered searchlight was produced that had

a power rating of 6.0 × 105 W. Assuming a potential difference of 220 V

across the searchlight, find the resistance of the light bulb in that search-

light. Find the equivalent resistance for several of these light bulbs

connected as shown in the circuit diagram below. What is the total power

dissipated in the circuit?R1 R2

R3

R4 R5

R6

∆V = 220 V

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Problem 20D 171

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NAME ______________________________________ DATE _______________ CLASS ____________________

CURRENT IN AND POTENTIAL DIFFERENCE ACROSS A RESISTOR

P R O B L E M

Holt Physics

Problem 20D

REASONINGFirst find the equivalent resistance of the circuit. From this, determine the total

circuit current. Then rebuild the circuit in steps, calculating the current and po-

tential difference for the equivalent resistance of each group until the current in

and potential difference across the specified 8.0 Ω resistor are known.

S O L U T I O N1. Determine the equivalent resistance of the circuit.

The equivalent resistance, which was calculated in the sample problem of the

previous section, is 5.7 Ω.

2. Calculate the total current in the circuit.

Substitute the potential difference and equivalent resistance in ∆V = IR, and

rearrange the equation to find the current delivered by the battery.

I = R

∆

e

V

q =

(

(

1

5

2

.9

.0

ΩV

)

) = 2.0 A

3. Determine a path from the equivalent resistance found in step 1 to the

specified resistor.

Review the path taken to find the equivalent resistance in the diagram below,

and work backward through this path. The equivalent resistance for the entire

circuit is the same as the equivalent resistance for group (c). The top resistors

in group (c), in turn, form the equivalent resistance for group (b), and the

rightmost resistor in group (b) is the specified 8.0 Ω resistor.

R1 = 8.0 Ω

Req

R5 = 8.0 Ω

R4 = 8.0 ΩR3 = 8.0 Ω

R2 = 8.0 Ω

Req,a

Req,b

(a)

(b)

(c)

For the circuit from the previous section’s sample problem, determinethe current in and potential difference across the 8.0 Ω resistor (R4) inthe figure below.

R1 = 8.0 Ω

R2 = 8.0 Ω

R3 = 8.0 Ω R4 = 8.0 Ω

R5 = 8.0 Ω

∆V = 12.0 V

NAME ______________________________________ DATE _______________ CLASS ____________________


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4. Follow the path determined in step 3, and calculate the current in and po-

tential difference across each equivalent resistance. Repeat this process

until the desired values are found.

Regroup, evaluate, and calculate.

Replace the circuit’s equivalent resistance with group (c), as shown in the fig-

ure. The resistors in group (c) are in parallel, so the potential difference across

each resistor is equal to the potential difference across the equivalent resistance,

which is 12.0 V. The current in the equivalent resistance in group (b) can now

be calculated using ∆V = IR.

Given: ∆V = 12.0 V Req,b = 20.0 Ω

Unknown: Ib = ?

Ib = R

∆

e

V

q,b =

(

(

2

1

0

2

.

.

0

0

ΩV)

) = 0.600 A

Regroup, evaluate, and calculate.

Replace the 20.0 Ω resistor with group (b). The resistors R3, Req,b , and R4 in

group (b) are in series, so the current in each resistor is the same as the cur-

rent in the equivalent resistance, which equals 0.600 A.

Ib =

The potential difference across the 8.0 Ω resistor at the right can be calculated

using ∆V = IR.

Given: Ib = 0.600 A R4 = 8.0 Ω

Unknown: ∆V = ?

∆V = IR = (0.600 A)(8.0 Ω) =

The current through the specified resistor is 0.600 A, and the potential differ-

ence across it is 4.8 V.

4.8 V

0.600 A

ADDITIONAL PRACTICE

1. Recall from the previous section the high-powered searchlight with the

power rating of 6.0 × 105 W. For a potential difference of 220 V placed

across the light bulb of this searchlight, you found a value for the bulb’s

resistance. You also determined the equivalent resistance for the circuit

shown in the figure below.

a. Calculate the potential difference across and current in the search-

light bulb labeled R3.

R1 R2

R3

R4 R5

R6

∆V = 220 V

Problem 20D 173

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NAME ______________________________________ DATE _______________ CLASS ____________________

b. Calculate the potential difference across and current in the search-


c. Calculate the potential difference across and current in the search-


2. Recall the portable power pack that can provide 12 V for 40.0 h. The

device powers a combination of small appliances with the resistances

shown in the circuit diagram below. In the previous section, you calcu-

lated the equivalent resistance and net current for this circuit.

a. Calculate the potential difference across and current in the 1.0 Ωappliance.

b. Calculate the potential difference across and current in the 2.5 Ωappliance.

c. Calculate the potential difference across and current in the 4.0 Ωappliance.

d. Calculate the potential difference across and current in the 3.0 Ωappliance.

3. Recall the longest-lasting battery in the world, which was constructed at

Oxford University in 1840. In 1977, the terminal voltage of the battery

was 2.00 × 103 V. Suppose the battery is placed in the circuit shown in the

diagram below. Determine the equivalent resistance of the circuit, and

then find the following:

a. the potential difference and current in the 5.0 Ω resistor (R4).

b. the potential difference and current in the 2.0 Ω resistor (R3).

c. the potential difference and current in the 7.0 Ω resistor (R5).

d. the potential difference and current in the 3.0 × 101 Ω resistor (R7).

R2 = 3.0 Ω R3 = 2.0 Ω

R4 = 5.0 Ω

R1 = 15 Ω

R5 = 7.0 Ω R6 = 3.0 Ω

R7 = 3.0 × 101 Ω

∆V = 2.00 × 103 V

R1 = 2.5 Ω R2 = 3.5 Ω

R3 = 3.0 Ω

R4 = 4.0 Ω

R5 = 1.0 Ω

∆V = 12 V


NAME ______________________________________ DATE _______________ CLASS ____________________

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Holt Physics

Problem 21AMAGNITUDE OF A MAGNETIC FIELD

P R O B L E MA proposed shock absorber uses the properties of tiny magnetizable par-ticles suspended in oil. In the presence of a magnetic field, the particlesform chains, making the liquid nearly solid. Suppose this liquid is acti-vated by a magnetic field with a magnitude of 0.080 T. What force will aparticle with a charge of 2.0 × 10−11 C experience if it moves perpendicu-lar to the magnetic field with a speed of 4.8 cm/s?

S O L U T I O NGiven: q = 2.0 × 10−11 C

v = 4.8 cm/s = 4.8 × 10−2 m/s

B = 0.080 T = 8.0 × 10−2 T = 8.0 × 10−2 N/A•m

Unknown: Fmagnetic = ?

Use the equation for the magnitude of a magnetic field.

B = Fma

q

g

v

netic

Fmagnetic = qvB

Fmagnetic = (2.0 × 10−11 C)( 4.8 × 10−2 m/s)(8.0 × 10−2 N/A•m)

Fmagnetic = 7.7 × 10−14 N

ADDITIONAL PRACTICE

1. In 1995, construction began on the world’s most powerful electromag-

net, which will be capable of producing a magnetic field with a strength

of 45 T. If an electron enters this field at a right angle with a speed of

7.5 × 106 m/s, what will be the magnetic force acting on the electron?

2. Transrapid, the world’s first train using magnetic field levitation, is de-

signed to glide 1.0 cm above the track at speeds of up to 450 km/h. Sup-

pose a charged particle with a charge of 12 × 10−9 C and a speed of

450 km/h moves at right angles to a 2.4 T magnetic field. What is the

magnetic force acting on the particle?

3. Europe’s fastest train can move at speeds of up to 350 km/h. What is the

magnetic force on a particle with a 36 nC charge that travels 350 km/h at

a 30.0° angle with respect to Earth’s magnetic field? Assume the strength

of Earth’s magnetic field is 7.0 × 10−5 T. (Hint: Recall that a magnetic field

affects only the velocity component perpendicular to the field.)

4. The U.S. Navy plans to use electromagnetic catapults to launch planes

from aircraft carriers. These catapults will be able to accelerate a 45 ×103 kg plane to a launch speed of 260 km/h. If an electron travels at a

Problem 21A 175

NAME ______________________________________ DATE _______________ CLASS ____________________

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.speed of 2.60 × 102 km/h perpendicular to a magnetic field, so that the

force acting on the electron is 3.0 × 10−17 N, what is the magnetic field

strength?

5. A solar-powered vehicle developed at Massachusetts Institute of Technol-

ogy can reach an average speed of 60.0 km/h. Due to the presence of

Earth’s magnetic field, a magnetic force acts on the electrons that move

through the car’s circuitry. Suppose an electron moves at 60.0 km/h per-

pendicular to Earth’s magnetic field. If a 2.0 × 10−22 N force acts on the

electron, what is the magnetic field strength at that location?

6. By 1905, a locomotive powered by an electric motor (which was in turn

powered by an internal-combustion engine) cruised between the East

Coast and West Coast. Suppose a particle with a charge of 88 × 10−9 C

moves at the same speed as the locomotive in a 0.32 T magnetic field. If

the magnetic force on the particle is 1.25 × 10−6 N, what is the particle’s

(and locomotive’s) speed?

7. NASA plans a shuttle launch system, called MagLifter, that would speed

up the spacecraft using superconducting magnets. A space shuttle would

be accelerated along a track 4.0 km long. At the end of the track, the speed

of the shuttle would be great enough to place it into orbit around Earth.

a. At what speed is an electron moving in a powerful 6.4 T magnetic

field if it experiences a force of 2.76 × 10−16 N?

b. If this is the shuttle’s final speed at the end of the track, how long

does it take the shuttle to glide along the track, assuming that the

shuttle undergoes constant acceleration from rest?

8. The mass spectrometer is a device that was first used to separate the differ-

ent isotopes of an element. In a mass spectrometer, ionized atoms with the

same speed enter a region in which a strong magnetic field is maintained.

The field, which is perpendicular to the plane in which the atoms move,

exerts a force that keeps the atoms in a circular path. Because all of the

atoms have the same charge and speed, the magnetic force exerted on them

is the same. However, the atoms have slightly different masses, and there-

fore the centripetal acceleration on each varies slightly. The centripetal

force depends on the speed of the atoms and the radius of the circular path

the atoms follow. Isotopes with larger masses have trajectories with larger

radii. As a result, the more-massive isotopes are detected farther from the

center of the apparatus than isotopes with smaller masses.

a. Suppose the magnetic field in a mass spectrometer has a field

strength of 0.600 T and that two lithium atoms, each with a single

positive charge of 1.60 × 10−19 C and a speed of 2.00 × 105 m/s, enter

that field. What is the magnetic force exerted on these two atoms?

b. If the masses of the atoms are 9.98 × 10−27 kg and 11.6 × 10−27 kg,

respectively, what will be the difference in the radius for each

atom’s path? (Hint: The magnetic force equals the force required to

keep the atoms in a circular path.)

Holt Physics

Problem 21BFORCE ON A CURRENT-CARRYING CONDUCTOR

P R O B L E M

S O L U T I O NGiven: B = 1.0 × 103 T

Fmagnetic = 1.6 × 102 N

l = 25 cm = 0.25 m

Unknown: I = ?

Use the equation for force on a current-carrying conductor perpendicular to a

magnetic field.

I = Fma

B

g

l

netic

I = = 0.64 A1.6 × 102 N

(1.0 × 103 T)(0.25 m)


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ADDITIONAL PRACTICE

1. In 1964, a magnet at the Francis Bitter National Magnet Laboratorycreated a magnetic field with a magnitude of 22.5 T. Ten megawatts ofpower was required to generate this field. If a wire that is 12 cm longand that carries a current of 8.4 × 10−2 A is placed in this field at a rightangle to the field, what is the force acting on it?

2. One of the subway platforms in downtown Chicago is 1066 m long.Suppose the current in one of the service wires below the platform is0.80 A. What is the magnitude of Earth’s magnetic field at that locationif the field exerts a force of 6.3 × 10−2 N on the wire? Assume the cur-rent is perpendicular to the field.

3. The distance between the pylons of the power line across the AmeralikFjord in Greenland is 5.376 km. If this length of wire carries a currentof 12 A and experiences a force of 3.1 N because of Earth’s magneticfield, what is the magnitude of the field in that region? Assume the wiremakes an angle of 38° with the field. (Hint: Only the component of themagnetic field that is perpendicular to the current contributes to themagnetic force.)

4. In February 1996, NASA extended a 21.0 km conducting tether from thespace shuttle Columbia in order to see how much power could be gener-ated by interacting with Earth’s magnetic field. Suppose the

In March 1967, a 1000 T magnetic field was obtained in a lab for a frac-tion of a second. Maximum sustained fields presently do not exceed 50 T.If a 25 cm long wire is perpendicular to a 1.0 × 103 T magnetic field and the magnetic force on the wire is 1.6 × 102 N, what is the current in the wire?

Problem 21B 177

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.magnetic field at the shuttle’s location has a magnitude of 6.40 × 10−7 T.What current must be induced in the tether, located perpendicular tothe field, to create a magnetic force of 1.80 × 10−2 N?

5. Magnetic fields are created by the currents in home appliances. Sup-

pose the magnetic field in the vicinity of an electric blow-dryer has a

magnitude of about 2.5 × 10−4 T. If a wire 4.5 cm long in such a blow-

dryer experiences a magnetic force of 3.6 × 10−7 N, what current must

exist in the wire? Assume the wire is perpendicular to the field.

6. At Sandia National Laboratories, in Albuquerque, New Mexico, an inter-

esting project has been developed. Engineers have proposed a train that

will roll on existing rails and use existing wheels but be propelled by mag-

netic forces. Suppose such a train is pushed by a 5.0 × 105 N force and re-

sults from the interaction between a current in many wires and a magnetic

field with a magnitude of 3.8 T that is oriented perpendicular to the wires.

Find the total length of the wires if they carry a 2.00 × 102 A current.

7. The largest electric sign used for advertising was set on the Eiffel

Tower, in Paris. It was lighted by 250 000 bulbs, which required an im-

mense number of electric cables. Find the total length of those cables if

a single straight cable of identical length experiences a force of 16.1 N

in Earth’s magnetic field. Assume the magnitude of the magnetic field

is 6.4 × 10−5 T and the wire, carrying a current of 2.8 A, is placed

perpendicular to the field.

8. Cows have four stomachs, so if a cow swallows a nail, it is hard to re-

move it. To extract the nail, a veterinarian gives the cow a strong mag-

net to swallow. Suppose that such a “cow magnet” has a magnetic field

magnitude of 0.040 T. If a 55 cm straight length of wire located near

the magnet carries a current of 0.10 A and is positioned so that the

angle between the magnetic field lines and the current is 45°, what is

the magnetic force that the current will experience? (Hint: Only the

component of the magnetic field that is perpendicular to the current

contributes to the magnetic force.)

9. For many years, the strongest magnetic field ever produced had a mag-

nitude of 38 T. Suppose a straight wire with a length of 2.0 m is located

perpendicular to this field. What current would have to pass through

the wire in order for the magnetic force to equal the weight of a gradu-

ate student with a mass of 75 kg?

10. The longest straight span of railroad tracks stretches 478 km in south-

western Australia. Suppose an electric current is sent through one of

the rails and that a force of 0.40 N is exerted by Earth’s magnetic field.

If the magnetic field has a strength of 7.50 × 10−5 T at that location and

is perpendicular to the rail, how large is the current?


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Holt Physics

Problem 22AINDUCED EMF AND CURRENT

P R O B L E MIn 1994, a unicycle with a wheel diameter of 2.5 cm was ridden 3.6 m in LasVegas, Nevada. Suppose the wheel has 200 turns of thin wire wrappedaround its rim, creating loops with the same diameter as the wheel. An emfof 9.6 mV is induced when the wheel is perpendicular to a magnetic fieldthat steadily decreases from 0.68 T to 0.24 T. For how long is the emfinduced?

S O L U T I O NGiven: N = 200 turns

D = 2.5 cm = 2.5 × 10−2 m

Bi = 0.68 T

Bf = 0.24 T

emf = 9.6 mV = 9.6 × 10−3 V

q = 0.0°

Unknown: ∆t = ?

Choose the equation(s) or situation: Use Faraday’s law of magnetic induction to

find the induced emf in the coil. Only the magnetic field strength changes with time.

emf = −N ∆[AB

∆(c

t

osq)] = −NA(cos q)

∆∆

B

t

Use the equation for the area of a circle to calculate the area (A).

A = pr 2 = p D

2

2


∆t = −NA (cos q)e

∆m

B

f

Substitute values into the equation(s) and solve:

A = p 2.5 × 1

2

0−2 m2

= 4.9 × 10−4 m2

∆t = −(200)(4.9 × 10−4 m2)[cos(0.0°)](0

(

.

9

2

.

4

6

T

×−10

0−.368

V

T

)

)

∆t =

The induced emf is directed through the coiled wire so that the magnetic field

produced opposes the decrease in the applied magnetic field. The rate of this

change is indicated by the positive value of time.

4.5 s

1. DEFINE

2. PLAN

3. CALCULATE

4. EVALUATE

Problem 22A 179

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.ADDITIONAL PRACTICE

1. The Pentagon covers an area of 6.04 × 105 m2, making it one of the

world’s largest office buildings. Suppose a huge loop of wire is placed on

the ground so that it covers the same area as the Pentagon. If the loop is

pulled at opposite ends so that its area decreases, an emf will be induced

because of the component of Earth’s magnetic field that is perpendicular

to the ground. If the field component has a strength of 6.0 × 10−5 T and

the average induced emf in the loop is 0.80 V, how much time passes be-

fore the loop’s area is reduced by half?

2. A Japanese-built Ferris wheel has a diameter of 100.0 m and can carry

almost 500 people at a time. Suppose a huge magnet is used to create

a field with an average strength of 0.800 T perpendicular to the wheel.

The magnet is then pulled away so that the field steadily decreases to

zero over time. If the wheel is a single conducting circular loop and the

induced emf is 46.7 V, find the time needed for the magnetic field to

decrease to zero.

3. In 1979, a potential difference of about 32.0 MV was measured in a lab in

Tennessee. The maximum magnetic fields, obtained for very brief periods

of time, had magnitudes around 1.00 × 103 T. Suppose a coil with exactly

50 turns of wire and a cross-sectional area of 4.00 × 10−2 m2 is placed per-

pendicular to one of these extremely large magnetic fields, which quickly

drops to zero. If the induced emf is 32.0 MV, in how much time does the

magnetic field strength decrease from 1.00 × 103 T to 0.0 T?

4. The world’s largest retractable roof is on the SkyDome in Toronto,

Canada. Its area is 3.2 × 104 m2, and it takes 20 min for the roof to fully

retract. If you have a coil with exactly 300 turns of wire that changes its

area from 0.0 m2 to 3.2 × 104 m2 in 20.0 min, what will be the emf in-

duced in the coil? Assume that a uniform perpendicular magnetic field

with a strength of 4.0 × 10−2 T passes through the coil.

5. At Massachusetts Institute of Technology, there is a specially shielded

room in which extremely weak magnetic fields can be measured. As of

1994, the smallest field measured had a magnitude of 8.0 × 10−15 T.

Suppose a loop having an unknown number of turns and an area equal

to 1.00 m2 is placed perpendicular to this field and the magnitude of the

field strength is increased tenfold in 3.0 × 10−2 s. If the emf induced is

equal to −1.92 × 10−11 V, how many turns are in the loop?

6. An electromagnet that has a mass of almost 8.0 × 106 kg was built at the

CERN particle physics research facility in Switzerland. As part of the de-

tector in one of the world’s largest particle accelerators, this magnet cre-

ates a fairly large magnetic field with a magnitude of 0.50 T. Consider a

coil of wire that has 880 equal turns. Suppose this loop is placed perpen-

dicular to the magnetic field, which is gradually decreased to zero in 12 s.

If an emf of 147 V is induced, what is the area of the coil?

Holt Physics

Problem 22BINDUCTION IN GENERATORS

P R O B L E MIn Virginia, a reversible turbine was built to serve as both a power genera-tor and a pump. When operating as a generator, the turbine can deliver al-most 5 × 108 W of power. The normal angular speed of the turbine rotor is 26.9 rad/s. Suppose the generator produces a maximum emf of 12 kV atthis angular speed. If the generator’s magnetic field strength is 0.12 T, howmany turns of wire, each with an area of 40.0 m2, are used in the generator?

S O L U T I O NGiven: w = 26.9 rad/s

B = 0.12 T

A = 40.0 m2

maximum emf = 12 kV = 12 × 103 V

Unknown: N = ?

Choose the equation(s) or situation: Use the maximum emf equation for a

generator.

maximum emf = NBAw


N = maxim

AB

u

wm emf


N = =

Although the magnetic field of the generator is fairly large and the area of the

loops is very large, several turns of wire are needed to produce a large maximum

emf. This gives some indication that at least one—and often several—of the

physical attributes of a generator must be large to induce a sizable emf.

93 turns(12 × 103 V)

(40.0 m2)(0.12 T)(26.9 rad/s)

1. DEFINE

2. PLAN


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3. CALCULATE

4. EVALUATE

ADDITIONAL PRACTICE

1. A gas turbine with rotors that are 5.0 cm in diameter was built in 1989.

The turbine’s rotors spin with a frequency of 833 Hz. Suppose a coil of

wire has a cross-sectional area equal to that of the turbine rotor. This coil

turns with the same frequency as the turbine rotor and is perpendicular

to a 8.0 × 10−2 T magnetic field. If the maximum emf induced in the coil

is 330 V, how many turns of wire are there in the coil?

Problem 22B 181

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.2. A company in Michigan makes solar-powered lawn mowers. The batter-

ies can run a 3 kW motor for over an hour, and the blades spin at

335 rad/s, which is about 30 percent faster than the blades on a gasoline-

powered mower. Suppose the kinetic energy of rotation of the blades is

used to generate electricity. The generator is built so that the coil turns

at 335 rad/s, creating a maximum emf of 214 V. The coil is placed in a

8.00 × 10−2 T magnetic field. Each of the turns of wire on the rotating

coil has an area of 0.400 m2. Use this information to calculate the num-

ber of turns of wire.

3. Tom Archer designed a self-propelled vertical Catherine wheel (its rim is

used to mount fireworks) that was 19.3 m in diameter. In 1994, the wheel

was successfully demonstrated. It made a few turns at an average angular

speed of 0.52 rad/s. If a similar wheel with exactly 40 turns of wire

wrapped around the rim is placed in a uniform magnetic field and ro-

tated about an axis that is along the wheel’s diameter, an emf will be gen-

erated. Suppose the maximum induced emf is 2.5 V. If the angular speed

of the wheel’s rotation is 0.52 rad/s, what is the magnitude of the mag-

netic field strength?

4. The Garuda, an airplane propeller designed in Germany in 1919, was

6.90 m across and had an angular speed of 57.1 rad/s during flight. Con-

sider a generator producing a maximum emf of 8.00 × 103 V. If the rotor

has 236 square turns, each with 6.90 m sides, and if the angular speed of

rotation equals 57.1 rad/s, what is the magnitude of the magnetic field in

which the rotor must be turned?

5. In Japan, a 5 mm long working model of a car has been built. The motor,

less than 1 mm long, uses a coil with 1000 turns of wire and is powered

by a 3.0 V emf source. Consider a mini-generator that uses a coil with ex-

actly 1000 turns of wire, each with an area of 8.0 cm2. If the coil is placed

in a 2.4 × 10−3 T magnetic field, what is the angular speed in rad/s

needed to produce a maximum emf of 3.0 V?

6. In 1995, a turbine was built that had a rotor shaft suspended magnetically,

almost fully eliminating friction. This allows the extremely high angular

speeds needed to create a large emf to be achieved. Suppose the turbine

turns a coil that contains exactly 640 turns of wire, each with an area of

0.127 m2. This generator produces a maximum emf of 24.6 kV while ro-

tating in a 8.00 × 10−2 T magnetic field. What is the angular speed in rad/s

of the coil?

7. At the University of Virginia is a centrifuge whose rotor is magnetically

supported in a vacuum; this allows for extremely low retarding forces.

The frequency of rotation of this centrifuge is 1.0 × 103 Hz. Consider an

electrical generator with this same frequency. The coil is placed in a

magnetic field that has a magnitude of 0.22 T. If the coil of the generator

has 250 circular loops, each with a 12 cm radius, what is the maximum

emf that can be induced?

Holt Physics

Problem 22CRMS CURRENTS AND POTENTIAL DIFFERENCES

P R O B L E MIn 1945, turbo-electric trains in the United States were capable of speedsexceeding 160 km/h. Steam turbines powered the electric generators,which in turn powered the driving wheels. Each generator producedenough power to supply an rms potential difference of 6.0 × 103 V acrossan 18 Ω resistor. What was the maximum potential difference across theresistor? What was the maximum current in the resistor? What was therms current in the resistor? What was the generator’s power output?

S O L U T I O NGiven: ∆Vrms = 6.0 × 103 V R = 18 Ω

Unknown: ∆Vmax = ? Imax = ? Irms = ? P = ?

Choose the equation(s) or situation: Use the equation relating maximum and

rms potential differences to calculate the maximum potential difference. Use the

definition of resistance to calculate the maximum current, then use the equation

relating maximum and rms currents to calculate rms current. Power can be cal-

culated from the product of rms current and rms potential difference.

∆Vmax =√

2(∆Vrms)

Imax = ∆V

Rmax

Irms = I√ma

2x

P = ∆VrmsIrms


∆Vmax = (1.41)(6.0 × 103 V)

=

Imax = (8.5

(1

×8

1

Ω03

)

V)

=

Irms = (0.707)(4.7 × 102 A)

=

P = (6.0 × 103 V)(3.3 × 102 A)

=

To determine whether severe rounding errors occurred through the various cal-

culations, obtain the product of the maximum current and maximum potential

difference. The product of ∆Vmax and Imax should equal 2P, which for this prob-

lem equals 4.0 × 106 W.

2.0 × 106 W

3.3 × 102 A

4.7 × 102 A

8.5 × 103 V

1. DEFINE

2. PLAN


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3. CALCULATE

4. EVALUATE

Problem 22C 183

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1. In 1963, the longest single-span “rope way” for cable cars opened in Cali-

fornia. The rope way stretched about 4 km from the Coachella Valley to

Mount San Jacinto. Suppose the rope way, which is actually a steel cable,

becomes icy. To de-ice the cable, you can connect its two ends to a

120 V (rms) generator. If the resistance of the cable is 6.0 × 10−2 Ω,

a. what will the rms current in the cable be?

b. what will the maximum current in the cable be?

c. what power will be dissipated by the cable, thus melting the ice?

2. In 1970, a powerful sound system was set up on the Ontario Motor

Speedway in California to make announcements to more than 200 000

people over the noise of 50 racing cars. The acoustic power of that system

was 30.8 kW. If the system was driven by a generator that provided an

rms potential difference of 120.0 V and only 10.0 percent of the supplied

power was transformed into acoustic power, what was the maximum

current in the sound system?

3. Modern power plants typically have outputs of over 10 × 106 kW. But in

1905, the Ontario Power Station, built on the Niagara River, produced

only 1.325 × 105 kW. Consider a single generator producing this power

when it is connected to a single load (resistor). If the generated rms po-

tential difference is 5.4 × 104 V, what is the maximum current and the

value of the resistor?

4. Stability of potential difference is a major concern for all high-emf

sources. In 1996, James Cross of the University of Waterloo in Canada,

constructed a compact power supply that produces a stable potential dif-

ference of 1.024 × 106 V. It can provide 2.9 × 10−2 A at this potential dif-

ference. If these values are the rms quantities for an alternating current

source, what are the maximum potential difference and current?

5. Certain species of catfish found in Africa have “power plants” similar to

those of electric eels. Though the electricity generated is not as powerful

as that of some eels, the electric catfish can discharge 0.80 A with a po-

tential difference of 320 V. Consider an ac generator in a circuit with a

load. If its maximum values for potential difference and current are the

same as the potential difference and current for the catfish, what are the

rms values for the potential difference and current? What is the resis-

tance of the load?

6. A wind generator installed on the island of Oahu, Hawaii, has a rotor

that is about 100 m in diameter. When the wind is strong enough, the

generator can produce a maximum current of 75 A in a 480 Ω load. Find

the rms potential difference across the load.

7. The world’s first commercial tidal power plant, built in France, has a power

output of only 6.2 × 104 kW, produced by 24 generators. Find the power

produced by each generator. If one of these generators is connected to a

120 kΩ resistor, find the rms and maximum currents in it.

Holt Physics

Problem 22DTRANSFORMERS

P R O B L E MThe span between the pylons for the power line serving the BuksefjordenHydro Power Station, in Greenland, is about 5 km. These power linescarry electricity at a high potential difference, which is then stepped-down to the standard European household potential difference of 220 V.If the transformer that does this has a primary with 1.5 × 105 turns and asecondary with 250 turns, what is the potential difference across the pri-mary? (Note: Reduction of potential difference usually takes place overseveral steps, not one.)

S O L U T I O N

Given: N1 = 1.5 × 105 turns N2 = 250 turns

∆V2 = 220 V

Unknown: ∆V1 = ?

Choose the equation(s) or situation: Use the transformer equation.

∆∆

V

V2

1 =

N

N2

1


∆V1 = ∆V2 N

N

2

1


∆V1 = (220 V)(1.

(

5

2

×50

10

tu

5

r

t

n

u

s

r

)

ns)

∆V1 =

The primary should have 130 kV. The step-down factor for the transformer is

600:1.

1.3 × 105 V

1. DEFINE

2. PLAN


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3. CALCULATE

4. EVALUATE

ADDITIONAL PRACTICE

1. The most powerful electromagnet in the world uses 24 MW of power.

Consider a transformer that transmits 24 MW of power. The primary

has 5600 turns, and the secondary has 240 turns. If the secondary poten-

tial difference is 4.1 kV, what is the primary potential difference?

2. In 1990, New Jersey was the most densely populated state in the United

States, with 403 people per square kilometer. Consider a transformer with

74 turns in the primary and 403 turns in the secondary. If a 650 V potential

difference exists between the terminals of the secondary, what is the poten-

tial difference between the terminals of the primary?

Problem 22D 185

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.3. In 1992, a battery whose longest dimension was 70 nm was made at the

University of California at Irvine. It produced a 2.0 × 10−2 V potential

difference for almost an hour. Suppose an ac generator producing this

potential difference is connected to a transformer that contains exactly

400 turns in the primary and exactly 3600 turns in the secondary. If this

emf is created between the terminals of the primary of a step-up trans-

former, what is the potential difference between the terminals of the sec-

ondary? If the 2.0 × 10−2 V is applied to the secondary instead, what is

the potential difference created between the terminals of the primary?

4. The hydraulic turbines installed at the third power plant at Grand

Coulee, in Washington, are almost 10 m in diameter. Each turbine is

large enough to produce a current of 1.0 × 103 A at 765 kV. If a trans-

former steps this potential difference down to 540 kV and the primary

contains 2.8 × 103 turns, how many turns must be in the secondary?

5. In 1965, the biggest power failure to date left about 30 million people in

the dark for several hours. About 200 000 km2, including Ontario,

Canada, and several northeastern states in the United States were af-

fected. Such failures can usually be avoided because all major electric

grids are interconnected. Transformers are needed in these connections.

For example, a transformer can increase the potential difference from

230 kV to 345 kV, which is the typical potential difference for transmis-

sion lines in the United States. If the primary in such a step-up trans-

former has 1.2 × 104 turns, how many turns are in the secondary?

6. The sunroof on some cars doubles as a solar battery. In strong sunlight, it

produces about 20.0 W of power.

a. If this power is transmitted by the primary of a transformer with

120 V across it, what is the current in the primary?

b. What is the potential difference across the secondary if the primary

contains only 36 percent of the number of turns in the secondary?

7. Electric cars, though still few in number, are appearing on the roads. By

using a 220 V potential difference to recharge their batteries instead of

the standard 120 V, the cars could be recharged in 3 to 6 h. One process,

developed at the Electric Power Research Institute in California, suggests

using 220 V outlets with a 30.0 A charging current. If a transformer is

used to increase the potential difference from 120 V to 220 V, what is the

current in the primary? How many turns does the primary have if the

secondary has 660 turns?


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Holt Physics

Problem 23AQUANTUM ENERGY

P R O B L E MFree-electron lasers can be used to produce a beam of light with variablewavelength. Because the laser can produce light with wavelengths as longas infrared waves or as short as X rays, its potential applications are muchgreater than for a laser that can produce light of only one wavelength.If such a laser produces photons of energies ranging from 1.034 eV to620.6 eV, what are the minimum and the maximum wavelengths corre-sponding to these photons?

S O L U T I O NGiven: E1 = 1.034 eV

= (1.034 eV)1.60 × 10−19 e

J

V = 1.65 × 10−19 J

E2 = 620.6 eV

= (620.6 eV)1.60 × 10−19 e

J

V = 9.93 × 10−17 J

h = 6.63 × 10−34 J•s

c = 3.00 × 108 m/s

Unknown: lmin = ? lmax = ?

Use the equation for the energy of a quantum of light. Use the relationship

between the frequency and wavelength of electromagnetic waves.

E = hf

f = lc

Substitute for f in the first equation, and rearrange to solve for wavelength.

E = h

lc

l = h

E

c

Substitute values into the equation.

lmax =

lmax = 1.21 × 10−6 m

lmax =

lmin =

lmin = 2.00 × 10−9 m

lmin = 2.00 nm

(6.63 × 10−34 J•s)(3.00 × 108 m/s)

(9.93 × 10−17 J)

1210 nm

(6.63 × 10−34 J•s)(3.00 × 108 m/s)

(1.65 × 10−19 J)

Problem 23A 187

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1. In 1974, IBM researchers announced that X rays with energies of

1.29 × 10−15 J had been guided through a “light pipe” similar to optic

fibers used for visible and near-infrared light. Calculate the wavelength

of one of these X-ray photons.

2. Some strains of Mycoplasma are the smallest living organisms. The wave-

length of a photon with 6.6 × 10−19 J of energy is equal to the length of

one Mycoplasma. What is that wavelength?

3. Of the various types of light emitted by objects in space, the radio signals

emitted by cold hydrogen atoms in regions of space that are located be-

tween stars are among the most common and important. These signals

occur when the “spin” angular momentum of an electron in a hydrogen

atom changes orientation with respect to the “spin” angular momentum of

the atom’s proton. The energy of this transition is equal to a fraction of an

electron-volt, and the photon emitted has a very low frequency. Given that

the energy of these radio signals is 5.92 × 10−6 eV, calculate the wavelength

of the photons.

4. The camera with the fastest shutter speed in the world was built for re-

search with high-power lasers and can expose individual frames of film

with extremely high frequency. If the frequency is the same as that of a

photon with 2.18 × 10−23 J of energy, calculate its magnitude.

5. Wireless “cable” television transmits images using radio-band photons

with energies of around 1.85 × 10−23 J. Find the frequency of these

photons.

6. In physics, the basic units of measurement are based on fundamental phys-

ical phenomena. For example, one second is defined by a certain transition

in a cesium atom that has a frequency of exactly 9 192 631 770 s−1. Find

the energy in electron-volts of a photon that has this frequency. Use

the unrounded values for Planck’s constant (h = 6.626 0755 × 10−34 J•s)

and for the conversion factor between joules and electron volts (1 eV =1.602 117 33 × 10−19 J).

7. Consider an electromagnetic wave that has a wavelength equal to 92 cm,

a length that corresponds to the longest ear of corn grown to date. What

is the frequency corresponding to this wavelength? What is its photon

energy? Express the answer in joules and in electron-volts.

8. The slowest machine in the world, built for testing stress corrosion, can

be controlled to operate at speeds as low as 1.80 × 10−17 m/s. Find the

distance traveled at this speed in 1.00 year. Calculate the energy of the

photon with a wavelength equal to this distance.

Holt Physics

Problem 23BPHOTOELECTRIC EFFECT

P R O B L E MUltraviolet radiation, which is part of the solar spectrum, causes a photo-electric effect in certain materials. If the kinetic energy of the photoelec-trons from an aluminum sample is 5.6 × 10−19 J and the work function ofaluminum is about 4.1 eV, what is the frequency of the photons that pro-duce the photoelectrons?

S O L U T I O NGiven: KEmax = 5.6 × 10−19 J = 3.5 eV

hft = 4.1 eV

h = 6.63 × 10−34 J•s = 4.14 × 10−15 eV•s

c = 3.00 × 108 m/s

Unknown: f = ?

Use the equation for the maximum kinetic energy of an ejected photoelectron to

calculate the frequency of the photon.

KEmax = h

lc − hft

f = KEma

hx + hft

f =

f = 1.8 × 1015 Hz

[3.5 eV + 4.1 eV]4.14 × 10−15 eV/s


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1. The melting point of mercury is about −39°C, which makes it convenient

for many applications, such as thermometers and thermostats. Mercury

also has some unusual applications, such as in “liquid mirrors.” By spin-

ning a pool of mercury, a perfect parabolic surface can be obtained for

use as a concave mirror. If the surface of mercury is exposed to light, a

photoelectric effect can be observed. If the work function of mercury is

4.5 eV, what is the frequency of photons that produce photoelectrons

with kinetic energies of 3.8 eV?

2. The largest all-metal telescope mirror, which was used in Lord Rosse’s

telescope, the Leviathan, was produced in 1845 from a copper-tin alloy.

The work function of the surface of that mirror can be estimated as

4.3 eV. Calculate the frequency of the photons that would produce

photoelectrons having a kinetic energy of 3.2 eV.

Problem 23B 189

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.3. Values for the work function have been experimentally determined for

most nonradioactive, elemental metals. The smallest work function,

which is 2.14 eV, belongs to the element cesium. The largest work func-

tion, which is 5.9 eV, belongs to the element selenium.

a. What is the wavelength of the photon that will just have the thresh-

old energy for cesium?

b. What is the wavelength of the photon that will just have the thresh-

old energy for selenium?

4. Carbon is a nonmetal, yet it is a conductor of electricity, and it exhibits

photoelectric properties. Calculate the work function and the threshold

frequency for carbon if photons with a wavelength of 2.00 × 102 nm pro-

duce photoelectrons moving at a speed of 6.50 × 105 m/s.

5. Two giant water jugs made in 1902 for the Maharaja of Jaipur, India, are

the largest single-piece silver items on Earth. Each jug has a capacity of

about 8 m3 and a mass of more than 240 kg. If the surface of one of these

jugs is exposed to UV light that has a frequency of 2.2 × 1015 Hz, a pho-

toelectric effect is observed. If the photoelectrons have 4.4 eV of kinetic

energy, find the work function and the threshold frequency of silver.

6. Rhenium is one of the rarest elements on Earth. Estimates indicate that on

average there is less than 1 mg of rhenium in a kilogram of Earth’s crust

and about 4 ng of rhenium in a liter of sea water. Rhenium also has one of

the highest work functions of any metal. Suppose that a rhenium sample is

exposed to light with a wavelength of 2.00 × 102 nm and that photoelec-

trons with kinetic energies of 0.46 eV are emitted. Using this information,

calculate the work function and threshold frequency for rhenium.

7. Sodium-vapor lamps, which are widely used in streetlights, produce yel-

low light with a principal wavelength of 589 nm. Would this sodium light

cause a photoelectric effect on the surface of solid sodium (hft = 2.3 eV)?

If the answer is yes, what is the energy of those photoelectrons? If the an-

swer is no, how much energy does the photon lack?

8. Lithium’s unusual electric properties make it an ideal material for high-

capacity batteries. The purity of a thin lithium foil, used in a lithium-

polymer “sandwich” to create an efficient battery for solar-powered cars,

is extremely important. One way to assess a metal’s purity is by means of

the photoelectric effect. If the work function of lithium is 2.3 eV, what is

the kinetic energy of the photoelectrons produced by violet light with a

wavelength of 410 nm?

9. Lead and zinc are vital elements in the construction of electric batteries.

For example, the largest battery in the world, in use in California, is a

lead-acid battery, while the most durable battery in the world, working

continuously since 1840, is a zinc-sulfur battery. Zinc and lead have simi-

lar work functions: 4.3 eV and 4.1 eV, respectively. Suppose certain pho-

tons have just enough energy to cause a photoelectric effect in zinc. If the

same photons were to strike the surface of lead, what would be the speed

of the photoelectrons?

Holt Physics

Problem 23CDE BROGLIE WAVES

P R O B L E MIn 1974, the most massive elementary particle known was the y ¢ particle,which has a mass of about four times the mass of a proton. If the deBroglie wavelength is 3.615 × 10−11 m when it has a speed of 2.80 × 103 m/s,what is the particle’s mass?

S O L U T I O NGiven: l = 3.615 × 10−11 m v = 2.80 × 103 m/s

h = 6.63 × 10−34 J•s

Unknown: m = ?

Use the equation for the de Broglie wavelength.

m = lh

v = = 6.55 × 10−27 Kg

(6.63 × 10−34 J•s)(3.615 × 10−11 m)(2.80 × 103 m/s)


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ADDITIONAL PRACTICE

1. The world’s smallest watch, made in Switzerland, has a fifteen-jewel

mechanism and is less than 5 mm wide. When this watch has a speed of

3.2 m/s, its de Broglie wavelength is 3.0 × 10−32 m. What is the mass of the

watch?

2. Discovered in 1983, Z° was still the most massive particle known in 1995.

If the de Broglie wavelength of the Z° particle is 6.4 × 10−11 m when the

particle has a speed of 64 m/s, what is the particle’s mass?

3. Although beryllium, Be, is toxic, the Be2+ ion is harmless. When a Be2+

ion is accelerated through a potential difference of 240 V, the ion’s de

Broglie wavelength is 4.4 × 10−13 m. What is the mass of the Be2+ ion?

4. In 1972, a powder with a particle size of 2.5 nm was produced. At what

speed should a neutron move to have a de Broglie wavelength of 2.5 nm?

5. The graviton is a hypothetical particle that is believed to be responsible

for gravitational interactions. Although its existence has not been proven,

cosmological observations and theories indicate that its mass, which is

theoretically zero, has an upper limit of 7.65 × 10−70 kg. What speed must

a graviton have for its de Broglie wavelength to be 5.0 × 1032 m? (Gravi-

tons are predicted to have a speed equal to that of light.)

6. The average mass of the bee hummingbird is about 1.6 g. What is the de

Broglie wavelength of this variety of hummingbird if it is flying at 3.8 m/s?

7. In 1990, Dale Lyons ran 42 195 m, in 3 h, 47 min, while carrying a spoon

with an egg in it. What was Lyons’ average speed during the run? If the

egg’s mass was 0.080 kg, what was its de Broglie wavelength?

Problem 25A 191

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.Holt Physics

Problem 25ABINDING ENERGY

P R O B L E MEach egg of Caraphractus cinctus, a parasitic wasp, has a mass of about 2.0 × 10−10 kg. Consider the formation of 16

8 O from H atoms and neu-trons. How many nuclei of 16

8 O must be formed to produce a mass defectequal to the mass of one Caraphractus cinctus egg? What is the total bind-ing energy of these 16

8 O nuclei? The atomic mass of 168 O is 15.994 915 u.

S O L U T I O NGiven: megg = 2.0 × 10−10 kg

element formed = 168O

Z = 8 N = 16 − 8 = 8

atomic mass of 168O = 15.994 915 u

atomic mass of H = 1.007 825 u

mn = 1.008 665 u

Unknown: ∆m = ?

n = number of 168O nuclei formed = ?

total Ebind = ?

Choose the equation(s) or situation: First find the mass defect using the equa-

tion for mass defect.

∆m = Z(atomic mass of H) + Nmn − atomic mass

To determine the number of oxygen-16 nuclei that must be formed to produce a

total mass defect equal to the mass of a Caraphractus cinctus egg, calculate the

ratio of the mass of one egg to the mass defect.

the number of 168O nuclei formed = n =

m

∆e

mgg

Finally, to find the total binding energy of all 168O nuclei, use the equation for the

binding energy of a nucleus and multiply it by n.

total Ebind = nEbind = n∆mc2

total Ebind = n∆m (931.50 MeV/u)


∆m = 8(1.007 825 u) + 8(1.008 665 u) − 15.994 915 u

∆m = 8.062 600 u + 8.069 320 u − 15.994 915 u

∆m = 0.137 005 u

n = (2

(

.

0

0

.1

×37

10

0

−

0

1

5

0

u

k

)

g) 1.66 ×

1

1

u

0−27 kg =

total Ebind = (8.8 × 1017)(0.137 005 u)(931.50 MeV/u)

total Ebind = 1.1 × 1020 MeV

8.8 × 1017 nuclei

1. DEFINE

2. PLAN

3. CALCULATE


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ADDITIONAL PRACTICE

1. In 1993, the United States had more than 100 operational nuclear reactors

producing about 30 percent of the world’s nuclear energy, or 610 TW•h.

a. Find the mass defect corresponding to a binding energy equal to

that energy output.

b. How many 21H nuclei would be needed to provide this mass defect?

c. How many 5626Fe nuclei would be needed to provide this mass defect?

d. How many 22688Ra nuclei would be needed to provide this mass defect?

2. In 1976, Montreal hosted the Summer Olympics. To complete the new

velodrome, the 4.1 × 107 kg roof had to be raised 10.0 cm to be placed in

the exact position.

a. Find the energy needed to raise the roof.

b. Find the mass of 5626Fe that is formed when an amount of energy equal

to that calculated in part (a) is obtained from binding H atoms and

neutrons in iron-56 nuclei.

3. Nuclear-energy production worldwide was 2.0 × 103 TW•h in 1993. What

mass of 23592U releases an equivalent amount of energy in the form of binding

energy?

4. In 1993, the United States burned about 2.00 × 108 kg of coal to produce

about 2.1 × 1019 J of energy. Suppose that instead of burning coal, you

obtain energy by forming coal (126C) out of H atoms and neutrons. What

amount of coal must be formed to provide 2.1 × 1019 J of energy? As-

sume 100 percent efficiency.

5. The sun radiates energy at a rate of 3.9 × 1026 J/s. Suppose that all the

sun’s energy occurs because of the formation of 42He from H atoms and

neutrons. Find the number of reactions that take place each second.

6. Sulzer Brothers, a Swiss company, made powerful diesel engines for the

container ships built for American President Lines. The power of each

12-cylinder engine is about 42 MW. Suppose the turbines use the forma-

tion of 147N for the energy-releasing process. What mass of nitrogen

would have to be formed to provide enough energy for 24 h of continu-

ous work? Assume the turbines are 100 percent efficient.

7. A hundred years ago, the most powerful hydroelectric plant in the world

produced 3.84 × 107 W of electric power. Find the total mass of 126C

atoms that must be formed each second from H atoms and neutrons to

produce the same power output.

For every nucleus of 168O formed, the mass defect is 0.137 005 u, and the mass of

the nucleus formed is 15.994 915 u. When ∆m has a total value of 2.0 × 10−10 kg,

8.8 × 1017 nuclei, or 2.3 × 10−8 kg of 168O will have formed.

4. EVALUATE

Problem 25B 193

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.Holt Physics

Problem 25BNUCLEAR DECAY

P R O B L E MThe most stable radioactive nuclide known is tellurium-128. It was discovered in 1924, and its radioactivity was proven in 1968. This isotopeundergoes two-step beta decay. Write the equations that correspond tothis reaction.

S O L U T I O NGiven: 128

52Te → −10e + X + v

X → −10e + Z + v

Unknown: the daughter elements X and Z

The mass numbers and atomic numbers on both sides of the expression must be

the same so that both charge and mass are conserved during the course of this

particular decay reaction.

Mass number of X = 128 − 0 = 128

Atomic number of X = 52 − (−1) = 53

The periodic table shows that the nucleus with an atomic number of 53 is iodine, I.

Thus, the first step of the process is as follows:

12852Te → −1

0e + 12853I + v

A similar approach for the second beta decay reaction gives the following equa-

tion. Again, the emission of an electron does not change the mass number of the

nucleus. It does, however, change the atomic number by 1.

Mass number of Z = 128 − 0 = 128

Atomic number of Z = 53 − (−1) = 54

The periodic table shows that the nucleus with an atomic number of 54 is xenon,

Xe. Thus the next step of the process is as follows:

12853I → −1

0e + 12854Xe + v

The complete two-step reaction is described by the two balanced equations below.

12852Te → −1

0e + 12853I + v

12853I → −1

0e + 12854Xe + v

ADDITIONAL PRACTICE

1. Standard nuclear fission reactors use 23592U for fuel. However, the supply

of this uranium isotope is limited. Its concentration in natural uranium-

238 is low, and the cost of enrichment is high. A good alternative is the

breeder reactor in which the following reaction sequence occurs: 23892U

captures a neutron, and the resulting isotope emits a −10e particle to form

23993Np. This nuclide emits a second −1

0e particle to form 23994Pu, which is

fissionable and can be used as an energy-producing material. Write bal-

anced equations for each of the reactions described.

2. Radon has the highest density of any gas. Under normal conditions

radon’s density is about 10 kg/m3. One of radon’s isotopes undergoes

two alpha decays and then one beta decay (b −) to form 21283Bi. Write the

equations that correspond to these reaction steps.

3. Every element in the periodic table has isotopes, and cesium has the

most: as of 1995, 37 isotopes of cesium had been identified. One of ce-

sium’s most stable isotopes undergoes beta decay (b −) to form 13556Ba.

Write the equation describing this beta-decay reaction.

4. Fission is the process by which a heavy nucleus decomposes into two

lighter nuclei and releases energy. Uranium-235 undergoes fission when

it captures a neutron. Several neutrons are produced in addition to the

two light daughter nuclei. Complete the following equations, which de-

scribe two types of uranium-235 fission reactions.

23592U + 10n → 144

56Ba + 8936Kr + ____

23592U + 10n → 140

54Xe + ____ + 210n

5. The maximum safe amount of radioactive thorium-228 in the air is

2.4 × 10–19 kg/m3, which is equivalent to about half a kilogram distrib-

uted over the entire atmosphere. One reason for this substance’s high

toxicity is that it undergoes alpha decay in which gamma rays are pro-

duced as well. Write the equation corresponding to this reaction.

6. The 1930s were notable years for nuclear physics. In 1931, Robert Van de

Graaff built an electrostatic generator that was capable of creating the

high potential differences needed to accelerate charged particles. In 1932,

Ernest O. Lawrence and M. Stanley Livingston built the first cyclotron. In

the same year, Ernest Cockcroft and John Walton observed one of the

first artificial nuclear reactions. Complete the following equation for the

nuclear reaction observed by Cockcroft and Walton.

11p + 73Li → ____ + 42He

7. Among the naturally occurring elements, astatine is the least abundant,

with less than 0.2 g present in Earth’s entire crust. The isotope 21785At

accounts for only about 5 × 10−9 g of all astatine. However, this highly

radioactive isotope contributes nothing to the natural abundance of

astatine because when it is created, it immediately undergoes alpha

decay. Write the equation for this decay reaction.


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Problem 25C 195

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ADDITIONAL PRACTICE

Holt Physics

Problem 25CMEASURING NUCLEAR DECAY

P R O B L E MTellurium-128, the most stable of all radioactive nuclides, has a half-lifeof about 1.5 × 1024 years. Determine the decay constant of 128

52 Te. Howlong would it take for 75 percent of a sample of this isotope to decay?

S O L U T I O NGiven: T1/2 = 1.5 × 1024 year

N = 1.00 − 0.75 = 0.25 = 25 percent remaining

Unknown: l = ? t = ?

Choose the equation (s) or situation: To calculate the decay constant, use the

equation for determining a half-life.

T1/2 = 0.6

l93

After 75 percent of the original sample decays, 25 percent, or one-fourth, of the

parent nuclei remain. The rest of the nuclei have decayed into daughter nuclei. By

definition, after the elapsed time of one half-life, half of the parent nuclei remain.

After two half-lives, one fourth (or half of one-half) of the parent nuclei remain.

The time for 75 percent of the original sample to decay is therefore two half-lives.

t = 2(T1/2)


l = 0

T

.6

1

9

/2

3


l = (1.5 ×

0

1

.6

0

9234 year) =

t = 2(1.5 × 1024 year) =

Because the half-life of tellurium-128 is on the order of 1024 years, the decay

constant is on the order of the reciprocal of the half-life, or 10−24 year. This

corresponds to one decay event per year for a mole (127.6 g) of pure

tellurium-128.

3.0 × 1024 year

4.6 × 10−24 year−1

1. DEFINE

2. PLAN

3. CALCULATE

4. EVALUATE

1. In 1995, an Ethiopian athlete, Halie Gebrselassie, ran 10 km in 26 min,

43.53 s. Lead-214 has a half-life similar to Gebrselassie’s run time. Find

the decay constant for 21482Pb. What percent of a sample of this isotope

would decay in a time interval equal to five times Gebrselassie’s run time?

2. Thorium-228, the most toxic of radioactive substances, has a half-life of

1.91 years. How long would it take for a sample of this isotope to de-

crease its toxicity by 93.75 percent?

3. In 1874, a huge cloud of locusts covered Nebraska, occupying an area of

almost 5.00 × 105 km2. The number of insects in that cloud was esti-

mated as 1013. Consider a sample of 14456Ba, which has a half-life of 11.9 s,

containing 1.00 × 1013 atoms. Approximately how long would it take for

1.25 × 1012 atoms to remain?

4. The oldest living tree in the world is a bristlecone pine in California

named Methuselah. Its estimated age is 4800 years. Suppose a sample of226

88Ra began to decay at the time the pine began to grow. What percent of

the sample would remain now? The half-life for 22688Ra is 1600 years.

5. The oldest fish on record is an eel that lived to the age of 88 years in a

museum aquarium in Sweden. After that period of time, only about 1

1

6 of

a sample of 21482Pb would be left. Find the decay constant for 210

82Pb.

6. In 1994, Peter Hogg sailed across the Pacific Ocean on his trimaran Aotea

in 34 days, 6 h, and 26 min. During that time period, only about 5

1

12 of a

sample of radon-222 would not decay. Estimate the decay constant of222

86Rn in (days)−1.

7. Lithium-5 is the least stable isotope known. Its mean lifetime is

4.4 × 10−22 s. Use the definition of a radionuclide’s mean lifetime as the

reciprocal of the decay constant (T = 1/l) to calculate the half-life of

lithium-5.


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Chapter 1The Science of Physics

II

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1. distance = 4.35 light yearsdistance = 4.35 light years ×

9.4

1

6

l

1

ig

×ht

1

y

0

e

1

a

5

r

m = 4.12 × 1016 m

a. distance = 4.12 × 1016 m × 1

10

M6 m

m =

b. distance = 4.12 × 1016 m × 10

1−p12

m

m = 4.12 × 1028 pm

4.12 × 1010 Mm

Additional Practice 1A

Givens Solutions

2. energy = 1.2 × 1044 Ja. energy = 1.2 × 1044 J ×

1

1

0

k3

J

J =

b. energy = 1.2 × 1044 J × 1

1

0−n9

J

J = 1.2 × 1053 nJ

1.2 × 1041 kJ

6. m = 1.90 × 105 kg m = 1.90 × 105 kg × 1.78 ×

1

1

e

0

V−36 kg = 1.07 × 1041 eV

a. m = 1.07 × 1041 eV × 1

10

M6 e

e

V

V =

b. m = 1.07 × 1041 eV × 1

1

01

T2

e

e

V

V = 1.07 × 1029 TeV

1.07 × 1035 MeV

4. distance = 152 100 000 kma. distance = 152 100 000 km ×

1

1

00

k

0

m

m ×

10

1−y2

m4 m =

b. distance = 152 100 000 km × 1

1

00

k

0

m

m ×

1

1

02

Y4

m

m = 1.521 × 10−13 Ym

1.521 × 1035 ym

5. energy = 2.1 × 1015 W •ha. energy = 2.1 × 1015 W •h ×

1

1

J

W

/s ×

36

1

0

h

0 s =

b. energy = 7.6 × 1018 J × 1

1

0

G9

J

J = 7.6 × 109 GJ

7.6 × 1018 J

3. m = 1.0 × 10−16 ga. m = 1.0 × 10−16 g ×

1

1

01

P5

g

g =

b. m = 1.0 × 10−16 g × 10

1−1

fg5 g

=

c. m = 1.0 × 10−16 g × 10

1−a1

g8 g

= 1.0 × 102 ag

0.10 fg

1.0 × 10−31 Pg

Section Two — Problem Workbook Solutions II Ch. 1–1

II

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7. m = (200)(2 × 1030 kg) =4 × 1032 kg a. m = 4 × 1032 kg × × =

b. m = 4 × 1032 kg × 1

1

0

k

3

g

g ×

1

1

01

E8

g

g = 4 × 1017 Eg

4 × 1038 mg103mg

1 g103 g1 kg

Givens Solutions

8. area = 166 241 700 km2

depth = 3940 m

V = volume = area × depth

V = (166 241 700 km2)(3940 m) × 11

00

k

0

m

m

2

V = 6.55 × 1017 m3

a. V = 6.55 × 1017 m3 × 10

1

6

m

cm3

3

=

b. V = 6.55 × 1017 m3 × 10

1

9

m

m3

m3

= 6.55 × 1026 mm3

6.55 × 1023 cm3

Holt Physics Solution ManualII Ch. 1–2


Chapter 2Motion In One Dimension

II

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1. ∆x = 443 m

vavg = 0.60 m/s

∆t = v

∆

av

x

g =

0

4

.6

4

0

3

m

m

/s = 740 s = 12 min, 20 s


Givens Solutions

2. vavg = 72 km/h

∆x = 1.5 km∆t =

v

∆

av

x

g = = 75 s

1.5 km

72 k

h

m36

1

0

h

0 s

3. ∆x = 5.50 × 102 m

vavg = 1.00 × 102 km/h

vavg = 85.0 km/h

a. ∆t = v

∆

av

x

g = =

b. ∆x = ∆vavg∆t

∆x = (85.0 km/h)36

1

0

h

0 s

1

1

0

k

3

m

m(19.8 s) = 468 m

19.8 s5.50 × 102 m

1.00 × 102 k

h

m 36

1

0

h

0 s 11

00

k

0

m

m

4. ∆x1 = 1.5 km

v1 = 85 km/h

∆x1 = 0.80 km

v2 = 67 km/h

a. ∆ttot = ∆t1 + ∆t2 = ∆v

x

1

1 + ∆v

x

2

2

∆ttot = + = 64 s + 43 s =

b. vavg = = = = 77 km/h2.3 km

(107 s)36

1

0

h

0 s

1.5 km + 0.80 km(64 s + 43 s)

3

1

60

h

0

∆x1 + ∆x2∆t1 + ∆t2

107 s0.80 km

67 k

h

m36

1

0

h

0 s

1.5 km

85 k

h

m36

1

0

h

0 s

5. r = 7.1 × 104 km

∆t = 9 h, 50 min

∆x = 2πr

vavg = = =

vavg =4.5 × 108 m

(590 min)16

m

0

i

s

n

4.5 × 108 m

(540 min + 50 min)16

m

0

i

s

n

2π(7.1 × 107 m)

(9 h)60

1

m

h

in + 50 min1

6

m

0

i

s

n

∆x∆t

vavg =

Thus the average speed = 1.3 × 104 m/s.

On the other hand, the average velocity for this point is zero, because the point’s dis-placement is zero.

1.3 × 104 m/s


Givens Solutions

6. ∆x = –1.73 km

∆t = 25 s

7. vavg,1 = 18.0 km/h

∆t1 = 2.50 s

∆t2 = 12.0 s

a. ∆x1 = vavg,1∆t1 = (18.0 km/h)36

1

0

h

0 s 110k

3

m

m(2.50s) = 12.5 m

∆x2 = –∆x1 = –12.5 m

vavg,2 = ∆∆

x

t2

2 = –

1

1

2

2

.

.

0

5

s

m =

b. vavg,tot = ∆∆x

t1

1

++

∆∆

x

t2

2 = 12.

2

5

.5

m

0

+s +

(−1

1

2

2

.0

.5

s

m) = =

c. total distance traveled = ∆x1 – ∆x2 = 12.5 m – (–12.5 m) = 25.0 m

total time of travel = ∆t1 + ∆t2 = 2.50 s + 12.0 s = 14.5 s

average speed = to

t

t

o

a

t

l

a

d

l

i

t

s

i

t

m

an

e

ce =

2

1

5

4

.

.

0

5

m

s = 1.72 m/s

0.0 m/s0.0 m14.5 s

–1.04 m/s

8. ∆x = 2.00 × 102 km

∆t = 5 h, 40 min, 37 s

vavg = (1.05)vavg

∆x = 12

∆x

a. vavg = ∆∆

x

t = =

2.0

2

0

0

×43

1

7

05

s

m

vavg =

b. ∆t = = = 9.73 × 103 s

2.00 ×2

105 m

(1.05)9.79 m

s

∆xvavg

9.79 m/s = 35.2 km/h

2.00 × 105 m

5 h 360

h

0 s + 40 min

m

60

in

s + 37 s

vavg = = –1.73

2

×5

1

s

03 m = –69 m/s = –250 km/h

∆x∆t

∆t = (9.73 × 103 s)36

1

0

h

0 s = 2.70 h

(0.70 h)60

1

m

h

in = 42 min

∆t = 2 h, 42 min

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1. vi = 0 km/h = 0 m/s

aavg = 1.8 m/s2

∆t = 1.00 min

vf = aavg ∆t + vi = (1.80 m/s2)(1.00 min)16

m

0

i

s

n + 0 m/s =

vf = 108 m/s = (108 m/s) 36

1

0

h

0 s110k

3m

m = 389 km/h

108 m/s

Additional Practice 2B

Givens Solutions

2. ∆t = 2.0 min

aavg = 0.19 m/s2

vi = 0 m/s

vf = aavg ∆t + vi = (0.19 m/s2) (2.0 min)16

m

0

i

s

n + 0 m/s = 23 m/s

3. ∆t = 45.0 s

aavg = 2.29 m/s2

vi = 0 m/s

vf = aavg ∆t + vi = (2.29 m/s2)(45.0 s) + 0 m/s = 103 m/s

5. ∆x = (15 hops) 110

h

.0

o

m

p

= 1.50 × 102 m

∆t = 60.0 s

∆t stop = 0.25 s

vf = 0 m/s

vi = vavg = +2.50 m/s

a. vavg = ∆∆

x

t =

1.50

60

×.0

10

s

2 m =

b. aavg = v

∆f

t

−

sto

v

p

i = 0 m/s

0

−.2

2

5

.5

s

0 m/s =

−0

2

.2

.5

5

0

m

m

/

/

s

s = −1.0 × 101 m/s2

+2.50 m/s

6. ∆x = 1.00 × 102 m, backward= −1.00 × 102 m

∆t = 13.6 s

∆t = 2.00 s

vi = 0 m/s

vf = vavg

vavg = ∆∆

x

t =

−1.00

13

×.6

1

s

02 m = −7.35 m/s

aavg = vf

∆−t

vi = = 3.68 m/s2−7.35 m/s − 0 m/s

2.00 s

7. ∆x = 150 m

vi = 0 m/s

vf = 6.0 m/s

vavg = 3.0 m/s

a. ∆t = v

∆

av

x

g =

3

1

.

5

0

0

m

m

/s =

b. aavg = vf

∆−t

vi = 6.0

5

m

.0

/s

×−1

0

01

m/s = 0.12 m/s2

5.0 × 101 s

4. ∆x = 29 752 m

∆t = 2.00 h

vi = 3.00 m/s

vf = 4.13 m/s

∆t = 30.0 s

a. vavg = ∆∆

x

t = =

b. aavg = ∆∆

v

t = =

1.

3

1

0

3

.0

m

s

/s = 3.77 × 10−2 m/s24.13 m/s − 3.00 m/s

30.0 s

4.13 m/s29 752 m

(2.00 h)36

1

0

h

0 s


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8. vi = +245 km/h

aavg = −3.0 m/s2

vf = vi −(0.200) vi

Givens Solutions

9. ∆x = 3.00 km

∆t = 217.347 s

aavg = −1.72 m/s2

vf = 0 m/s

vi = vavg = ∆∆

x

t = = 13.8 m/s

tstop = vf

aa

−

vg

vi = = = 8.02 s−13.8 m/s−1.72 m/s2

0 m/s − 13.8 m/s

−1.72 m/s2

3.00 × 103 m

217.347 s

10. ∆x = +5.00 × 102 m

∆t = 35.76 s

vi = 0 m/s

∆t = 4.00 s

vmax = vavg + (0.100) vavg

vf = vmax = (1.100)vavg = (1.100)∆∆

x

t = (1.100)5.0

3

0

5

×.7

1

6

0

s

2 m = +15.4 m/s

aavg = ∆∆t

v

=

vf

∆−t

vi = = + 3.85 m/s215.4 m/s − 0 m/s

4.00 s

1. ∆x = 115 m

vi = 4.20 m/s

vf = 5.00 m/s

∆t = vi

2∆+

x

vf =

4.20

(

m

2)

/

(

s

1

+15

5.

m

00

)

m/s =

(2

9

)

.

(

2

1

0

1

m

5

/

m

s

) = 25.0 s

Additional Practice 2C

2. ∆x = 180.0 km

vi = 3.00 km/s

vf = 0 km/s

∆t = vi

2∆+

x

vf = =

3

3

.

6

0

0

0

.0

km

km

/s = 1.2 × 102 s

(2)(180.0 km)3.00 km/s + 0 km/s

3. vi = 0 km/h

vf = 1.09 × 103 km/h

∆x = 20.0 km

∆x = 5.00 km

vi = 1.09 × 103 km/h

vf = 0 km/h

a. ∆t = vi

2

+∆x

vf =

∆t = =

b. ∆t = vi

2

+∆x

vf =

∆t = = 33.0 s10.0 × 103 m

(1.09 × 103 km/h)36

1

0

h

0 s11

00

k

0

m

m

(2)(5.00 × 103 m)

(1.09 × 103 km/h + 0 km/h)36

1

0

h

0 s11

00

k

0

m

m

132 s40.0 × 103 m

(1.09 × 103 km/h)36

1

0

h

0 s11

00

k

0

m

m

(2)(20.0 × 103 m)

(1.09 × 103 km/h + 0 km/h)36

1

0

h

0 s11

00

k

0

m

m

vi = 245 k

h

m36

1

0

h

0 s 110k

3

m

m = +68.1 m/s

vf = (1.000 − 0.200) vi = (0.800)(68.1 m/s) = +54.5 m/s

∆t = vf

aa

−

vg

vi = 54.5 m

−3

/s

.0

−m

6

/

8

s

.21 m/s

= = 4.5 s−13.6 m/s− 3.0 m/s2


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4. vi = vavg = 518 km/h

vf = (0.600) vavg

∆t = 2.00 min

vavg = 518 k

h

m60

1

m

h

in11

0

k

3

m

m = 8.63 × 103 m/min

∆x = 12

(vi + vf)∆t = 12

[vavg + (0.600) vavg]∆t = 12

(1.600)(8.63 × 103 m/min)(2.00 min)

∆x = 13.8 × 103 m = 13.8 km

Givens Solutions

5. ∆t = 30.0 s

vi = 30.0 km/h

vf = 42.0 km/h

∆x = 12

(vi + vf)∆t = 12

(30.0 km/h + 42.0 km/h) 36

1

0

h

0 s(30.0 s)

∆x = 12

72.0 k

h

m36

1

0

h

0 s(30.0 s)

∆x = 3.00 × 10−1 km = 3.00 × 102 m

1. vi = 186 km/h

vf = 0 km/h = 0 m/s

a = −1.5 m/s2

∆t = vf −

a

vi = = −−

5

1

1

.5

.7

m

m

/s

/2s

= 34 s

0 m/s − (186 km/h) 36

1

0

h

0 s 11

0

k

3

m

m

−1.5 m/s2

Additional Practice 2D

6. vf = 96 km/h

vi = 0 km/h

∆t = 3.07 s

∆x = 12

(vi + vf)∆t = 12

(0 km/h + 96 km/h) 36

1

0

h

0 s11

0

k

3

m

m(3.07 s)

∆x = 12

96 × 103 m

h(8.53 + × 10−4 h) = 41 m

7. ∆x = 290.0 m

∆t = 10.0 s

vf = 0 km/h = 0 m/s

vi = 2

∆∆t

x − vf =

(2)(

1

2

0

9

.

0

0

.0

s

m) − 0 m/s =

(Speed was in excess of 209 km/h.)

58.0 m/s = 209 km/h

8. ∆x = 5.7 × 103 km

∆t = 86 h

vf = vi + (0.10) vi

vf + vi = 2

∆∆t

x

vi (1.00 + 0.10) + vi = 2

∆∆t

x

vi = (2)

(

(

2

5

.

.

1

7

0

×)(

1

8

0

6

3

h

k

)

m) = 63 km/h

9. vi = 2.60 m/s

vf = 2.20 m/s

∆t = 9.00 min

∆x = 12

(vi + vf)∆t = 12

(2.60 m/s + 2.20 m/s)(9.00 min)m

60

in

s = 1

2(4.80 m/s)(5.40 × 102 s)

∆x = 1.30 × 103 m = 1.30 km

2. vi = −15.0 m/s

vf = 0 m/s

a = +2.5 m/s2

vi = 0 m/s

vf = +15.0 m/s

a = +2.5 m/s

For stopping:

∆t1 = vf

a

− vi = 0 m/s

2

−.5

(−m

1

/

5

s

.2

0 m/s) =

1

2

5

.5

.0

m

m

/s

/2s

= 6.0 s

For moving forward:

∆t2 = vf

a

− vi = = 1

2

5

.5

.0

m

m

/s

/2s

= 6.0 s

∆t tot = ∆t1 + ∆t2 = 6.0 s + 6.0 s = 12.0 s

15.0 m/s − 0.0 m/s

2.5 m/s2


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3. vi = 24.0 km/h

vf = 8.0 km/h

a = −0.20 m/s2

∆t = vf −

a

vi

∆t =

∆t = = 22 s

−16.0 k

h

m36

1

0

h

0 s11

0

k

3

m

m

−0.20 m/s2

(8.0 km/h − 24.0 km/h) 36

1

0

h

0 s 11

0

k

3

m

m

−0.20 m/s2

4. v1 = 65.0 km/h

vi,2 = 0 km/h

a2 = 4.00 × 10−2 m/s2

∆x = 2072 m

For cage 1:

∆x = v1∆t1

∆t1 = ∆v1

x = =

For cage 2:

∆x = vi,2∆t2 + 12

a2∆t22

Because vi,2 = 0 km/h,

∆t2 = 2

a

∆2

x =

4.0

(2

0)

×(2

1

00

7−2

2 m

m)

/s2 =

Cage 1 reaches the bottom of the shaft in nearly a third of the time required for cage 2.

322 s

115 s2072 m

(65.0 km/h)36

1

0

h

0 s 110k

3

m

m

5. ∆x = 2.00 × 102 m

v = 105.4 km/h

vi,car = 0 m/s

a. ∆t = = =

b. ∆x = vi,car ∆t + 12

acar∆t2

acar = 2

∆∆t2

x =

(2)(2

(

.

6

0

.

0

83

×s

1

)

02

2 m) = 8.57 m/s2

6.83 s2.00 × 102 m

105.4 k

h

m 36

1

0

h

0 s 110k

3

m

m

∆xv

Givens Solutions

7. vi = 3.17 × 102 km/h

vf = 2.00 × 102 km/h

∆t = 8.0 s

a = vf

∆−t

vi =

a = =

∆x = vi∆t + 12

a∆t2 = (3.17 × 102 km/h)36

1

0

h

0 s11

0

k

3

m

m(8.0 s) + 1

2(−4.1 m/s2)(8.0 s)2

∆x = (7.0 × 102 m) + (−130 m) = +570 m

−4.1 m/s2

(−117 km/h)36

1

0

h

0 s11

0

k

3

m

m

8.0 s

(2.00 × 102 km/h − 3.17 × 102 km/h)36

1

0

h

0 s11

0

k

3

m

m

8.0 s

6. vi = 6.0 m/s

a = 1.4 m/s2

∆t = 3.0 s

∆x = vi∆t + 12

a∆t2 = (6.0 m/s)(3.0 s) + 12

(1.4 m/s2)(3.0 s)2 = 18 m + 6.3 m = 24 m


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10. vi = 24.0 m/s

a = −0.850 m/s2

∆t = 28.0 s

vf = vi + a∆t = 24.0 m/s + (− 0.850 m/s2)(28.0 s) = 24.0 m/s − 23.8 m/s = +0.2 m/s

Givens Solutions

11. a = +2.67 m/s2

∆t = 15.0 s

∆x = +6.00 × 102m

vi ∆t = ∆x − 12

a∆t2

vi = ∆∆

x

t − 1

2a∆t =

6.00

15

×.0

10

s

2 m − 1

2(2.67 m/s2)(15.0 s) = 40.0 m/s − 20.0 m/s = +20.0 m/s

12. a = 7.20 m/s2

∆t = 25.0 s

vf = 3.00 × 102 ms

vi = vf − a∆t

vi = (3.00 × 102 m/s) − (7.20 m/s2)(25.0 s) = (3.00 × 102 m/s) − (1.80 × 102 m/s)

vi = 1.20 × 102 m/s

13. vi = 0 m/s

∆x = 1.00 × 102 m

∆t = 12.11 s

∆x = vi∆t + 12

a∆t2

Because vi = 0 m/s,

a = 2

∆∆t2

x =

(2)(

(

1

1

.0

2

0

.1

×1

1

s)

02

2 m) = 1.36 m/s2

8. vi = 0 m/s

vf = 3.06 m/s

a = 0.800 m/s2

∆t2 = 5.00 s

∆t1 = vf

a

− vi = 3.0

0

6

.

m

80

/

0

s

m

−/

0

s2m/s

= 3.82

∆x1 = vi∆t1 + 12

a∆t12 = (0 m/s) (3.82 s) + 1

2(0.800 m/s2) (3.82 s)2 = 5.84 m

∆x2 = vf∆t2 = (3.06 m/s)(5.00 s) = 15.3 m

∆xtot = ∆x1 + ∆x2 = 5.84 m + 15.3 m = 21.1 m

9. vf = 3.50 × 102 km/h

vi = 0 km/h = 0 m/s

a = 4.00 m/s2

∆t = (vf

a

− vi) = =

∆x = vi∆t + 12

a∆t2 = (0 m/s)(24.3 s) + 12

(4.00 m/s2)(24.3 s)2

∆x = 1.18 × 103 m = 1.18 km

24.3 s

(3.50 × 102 km/h − 0 km/h) 36

1

0

h

0 s 11

0

k

3

m

m

(4.00 m/s2)

14. vi = 3.00 m/s

∆x = 1.00 × 102 m

∆t = 12.11 s

a = 2(∆x

∆−t 2

vi∆t) =

a =

a = (

(

2

1

)

2

(

.

6

1

4

1

m

s)2)

= 0.87 m/s2

(2)(1.00 × 102 m − 36.3 m)

(12.11 s)2

(2)[1.00 × 102 m − (3.00 m/s)(12.11 s)]

(12.11 s)2

15. vf = 30.0 m/s

vi = 18.0 m/s

∆t = 8.0 s

a = vf

∆−t

vi = = 12

8

.0

.0

m

s

/s = 1.5 m/s230.0 m/s − 18.0 m/s

8.0 s


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1. vi = 0 km/h

vf = 965 km/h

a = 4.0 m/s2∆x =

vf2

2

−a

vi2

=

∆x = 7.19

8

×.0

1

m

04

/

m

s2

2/s2

= 9.0 × 103 m = 9.0 km

(965 km/h) 2 − (0 km/h)236

1

0

h

0 s

2

110k

3

m

m

2

(2)(4.0 m/s2)

2. vi = (0.20) vmax

vmax = 2.30 × 103 km/h

vf = 0 km/h

a = −5.80 m/s2

∆x = vf

2

2

−a

vi2

=

∆x = −1.6

−3

11

×.6

1

m

04

/

m

s2

2/s2

= 1.41 × 103 m = 1.41 km

(0 km/h)2 − (0.20)2 (2.30 × 103 km/h)2 36

1

0

h

0 s

2

110k

3

m

m

2

(2)(−5.80 m/s2)

3. vf = 9.70 × 102 km/h

vi = (0.500)vf

a = 4.8 m/s2

∆x = vf

2

2

−a

vi2

=

∆x =

∆x =

∆x = = 5.7 × 103 m = 5.7 km5.45 × 104m2/s2

9.6 m/s2

(7.06 × 105 km2/h2)36

1

0

h

0 s

2

110k

3

m

m

2

(2)(4.8 m/s2)

(9.41 × 105 km2/h)2 −2.35 × 105 km2/h2)36

1

0

h

0 s

2

110k

3

m

m

2

(2)(4.8 m/s2)

(9.70 × 102 km/h)2 −(0.50)2 (9.70 × 102 km/h)236

1

0

h

0 s

2

110k

3

m

m

2

(2)(4.8 m/s2)

Additional Practice 2E

Givens Solutions

4. vi = 8.0 m/s

∆x = 40.0 m

a = 2.00 m/s2

vf =√

2a∆x + vi2 =

√(2)(2.0m/s2)(40.m)+ (8.0m/s)2 =

√1.60 × 102m2/s2 + 64m2/s2

vf =√

224m2/s2 = ± 15 m/s = 15 m/s

5. ∆x = +9.60 km

a = −2.0 m/s2

vf = 0 m/s

vi = vf2− 2a∆x = (0 m/s)2 − (2)(−2.0m/s2)(9.60× 103 m)

vi =√

3.84 × 104m2/s2 = ±196 m/s = +196 m/s

7. ∆x = 44.8 km

∆t = 60.0 min

a = −2.0 m/s2

∆x = 20.0 m

vi = 12.4 m/s

a. vavg = ∆∆

x

t = =

b. vf =√

2a∆x+ vi2 =

√(2)(−2.0m/s2)(20.0m)+ (12.4 m/s)2 =

√(−80.0 m2/s2)+ 154 m2/s2

vf =√

74 m2/s2 = ±8.6 m/s = 8.6 m/s

12.4 m/s44.8 × 103 m

(60.0 min)(60 s/min)

6. a = +0.35 m/s2

vi = 0 m/s

∆x = 64 m

vf =√

2a∆x + vi2 =

√(2)(0.35 m/s2)(64m)+ (0m/s)2

vf =√

45 m2s2 = ±6.7 m/s = +6.7 m/s


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3. ∆y = −443 m + 221 m = −222 m

a = −9.81 m/s2

vi = 0 m/s

vf =√

2a∆y − vi2 =

√(2)(−9.81 m/s2)(−222m)− (0m/s)2 =

√4360 m2/s2

vf = ±66.0 m/s = −66.0 m/s

Givens Solutions

4. ∆y = +64 m

a = −9.81 m/s2

∆t = 3.0 s

∆y = vi∆t + 12

a∆t2

vi = = = 64 m

3.

+0

4

s

4 m

vi = 1

3

0

.

8

0

m

s = 36 m/s initial speed of arrow = 36 m/s

3.0 s

∆t

5. ∆y = −111 m

∆t = 3.80 s

a = −9.81 m/s2

∆y = vi ∆t + 12

a∆t2

vi = = =

vi = −

3

4

.

0

8

.

0

2

s

m = −10.6 m/s

−111 m + 70.8 m

3.80 s

3.80 s

∆t

9. ∆x = 4.0 × 102 m

∆t = 11.55

vi = 0 km/h

vf = 2.50 × 102 km/h

1. ∆y = −343 m

a = −9.81 m/s2

vi = 0 m/s

vf =√

2a∆y + vi2 =

√(2)(−9.81 m/s2)(−343m)+ (0m/s)2 =

√6730 m2/s2

vf = ±82.0 m/s = −82.0 m/s

Additional Practice 2F

2. ∆y = +4.88 m

vi = +9.98 m/s

a = −9.81 m/s2

vf =√

2a∆y + vi2 =

√(2)(−9.81 m/s2)(4.88m)+ (9.98 m/s)2 =

√−95.7 m2/s2 + 99.6m2/s2

vf =√

3.90 m2/s2 = ±1.97 m/s = ±1.97 m/s

a = vf

2

2∆−

x

vi2

=

a = 4.8

8

2

.0

××1

1

0

0

3

2m

m

2/s2

= 6.0 m/s2

(2.50 × 102 km/h)2 − (0 km/h)236

1

0

h

0 s

2

110

k

3

m

m

2

(2)(4.0 × 102 m)

8. ∆x = 2.00 × 102 m

a = 1.20 m/s2

vf = 25.0 m/s

vi =√

vf2− 2a∆x =√

(25.0m/s)2 − (2)(1.20m/s2)(2.00× 102 m)

vi =√

625m2/s2 − 4.80× 102m2/s2

vi =√

145m2/s2 = ±12.0 m/s = 12.0 m/s

10. vi = 25.0 km/h

vf = 0 km/h

∆x = 16.0 m

a = vf

2

2∆−

x

vi2

=

a = = −1.51 m/s2−4.82 m2/s2

32.0 m

(0 km/h)2 − (25.0 km/h)236

1

0

h

0 s

2

110

k

3

m

m

2

(2)(16.0 m)

64 m − 12

(−9.81 m/s2)(3.0 s)2

∆y − 12

a∆t2 −111 m − 12

(−9.81 m/s2)(3.80 s)2

∆y − 12

a∆t2


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9. ∆ymax = +21 cm

a = −9.81 m/s2

vf = 0 m/s

∆y = +7.0 cm

vi =√

vf2− 2a∆ymax =√

(0 m/s)2 − (2)(−9.81m/s2)(2.1 × 10−1m) =√

4.1m2/s2

vi = +2.0 m/s

For the flea to jump +7.0 cm = +7.0 × 10−2 m = ∆y ,

∆y = vi∆t + 12

a∆t2 or 12

a∆t2 + vi∆t − ∆y = 0

Solving for ∆t by means of the quadratic equation,

∆t =

∆t =

∆t = =

∆t = = 0.37 s or 0.04 s

To choose the correct value for ∆t, insert ∆t, a, and vi into the equation for vf .

vf = a∆t + vi = (−9.81 m/s2)(0.37 s) + 2.0 m/s

vf = (−3.6 m/s) + 2.0 m/s = −1.6 m/s

vf = a∆t + vi = (−9.81 m/s2)(0.04 s) + 2.0 m/s

vf = (−0.4 m/s) + 2.0 m/s = +1.6 m/s

Because vf is still directed upward, the shorter time interval is correct. Therefore,

∆t = 0.04 s

2.0 m/s ± 1.6 m/s

9.81 m/s2

2.0 m/s ±√

2.6m2/s29.81 m/s2

−2.0 m/s ±√

4.0m2/s2 − 1.4 m2/s2−9.81 m/s2

−2.0 m/s ±√

(2.0 m/s)2 − (2)(−9.81 m/s2)(−7.0× 10−2m)−9.81 m/s2

−vi ± (vi)2− 42

a(−∆y)

22

a

6. ∆y = −228 m

a = −9.81 m/s2

vi = 0 m/s

When vi = 0 m/s,

∆t = 2

a

∆y = =

In the presence of air resistance, the sandwich would require more time to fall be-cause the downward acceleration would be reduced.

6.82 s(2)(−228 m)−9.81 m/s2

Givens Solutions

7. vi = 12.0 m/s, upward =+12.0 m/s

vf = 3.0 m/s, upward =+3.0 m/s

a = −9.81 m/s2

yi = 1.50 m

∆y = vf

2

2

−a

vi2

= =

∆y = = 6.88 m

height of nest from ground = h

∆y = h − yi h = ∆y + yi = 6.88 m + 1.50 m = 8.38 m

−135 m2/s2

−19.6 m/s2

9.0 m3/s2 − 144 m2/s2

(2)(−9.81 m/s2)

(3.0 m/s)2 − (12.0 m/s)2

(2)(−9.81 m/s2)

8. ∆y = +43 m

a = −9.81 m/s2

vf = 0 m/s

Because it takes as long for the ice cream to fall from the top of the flagpole to theground as it does for the ice cream to travel up to the top of the flagpole, the free-fallcase will be calculated.

Thus, vi = 0 m/s, ∆y = −43 m, and ∆y = 12

a∆t2.

∆t = 2

a

∆y = = 3.0 s

(2)(−43 m)−9.81 m/s2


Chapter 3Two-Dimensional Motion and Vectors

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1. ∆tx = 7.95 s

∆y = 161 m

d = 226 m

d2 = ∆x2 + ∆y2

∆x =√

d2− ∆y2 =√

(226 m)2 − (161m)2 =√

5.11 × 104 m2− 2.59× 104 m2

∆x =√

2.52 × 104 m2 = 159 m

∆x =

v = ∆∆

t

x

x =

1

7

5

.9

9

5

m

s = 20.0 m/s

159 m


Givens Solutions

2. d1 = 5.0 km

θ1 = 11.5°

d2 = 1.0 km

q2 = −90.0°

∆xtot = d1(cos q1) + d2(cos q2) = (5.0 km)(cos 11.5°) + (1.0 km)[cos(−90.0°)]

∆xtot = 4.9 km

∆ytot = d1(sin q1) + d2(sin q2) = (5.0 km)(sin 11.5°) + (1.0 km)[sin(−90.0°)] = 1.0 km − 1.0 km

∆ytot = 0.0 km

d =√

(∆xtot)2 + (∆ytot)2 =√

(4.9 km)2 + (0.0km)2

d =

q = tan−1 ∆∆x

yt

t

o

o

t

t = tan−1 04.

.

0

9

k

k

m

m = 0.0°, or due east

4.9 km

3. ∆x = 5 jumps

1 jump = 8.0 m

d = 68 m

d2 = ∆x2 + ∆y2

∆y =√

d2− ∆x2 =√

(68m)2 − [(5)(8.0m)]2 =√

4.6× 103 m2− 1.6 × 103 m2

∆y =√

3.0× 103 m = 55 m

number of jumps northward = 8.0

5

m

5

/

m

jump = 6.9 jumps =

q = tan−1 ∆∆

x

y = tan−1 (5)

5

(

5

8.

m

0 m) = 36° west of north

7 jumps

4. ∆x = 25.2 km

∆y = 21.3 km

d =√

∆x2+ ∆y2 =√

(25.2km)2 + (21.3 km)2

d =√

635km2+ 454 km2 =√

1089 km2

d =

q = tan−1 ∆∆

x

y = tan−1 221

5

.

.

3

2

k

k

m

m

q = 42.6° south of east

33.00 km


Givens Solutions

5. ∆y = −483 m

∆x = 225 mq = tan−1

∆∆

x

y = tan−1 2−2458m

3 = −65.0° =

d =√

∆x2+ ∆y2 =√

(225 m)2 + (−483m)2

d =√

5.06 × 104 m2+ 2.33× 105 m2 =√

2.84 × 105 m2

d = 533 m

65.0° below the waters surface

6. v = 15.0 m/s

∆tx = 8.0 s

d = 180.0 m

d2 = ∆x2 + ∆y2 = (v∆tx)2 + (v∆ty)2

d2 = v2(∆tx2 + ∆ty

2)

∆ty =d

v2

− ∆tx2 = 115

8

.0

0

. 0

m m

/s

2

− (8.0s)2 =√

144s2 − 64s2 =√

8.0× 101 s2

∆ty = 8.9 s

7. v = 8.00 km/h

∆tx = 15.0 min

∆ty = 22.0 min

d =√

∆x2+ ∆y2 =√

(v∆tx)2 + (v∆ty)2

= v√

∆tx2+ ∆ty2

d = (8.00 km/h)60

1

m

h

in

√(15.0min)2 + (22.0 min)2

d = (8.00 km/h)60

1

m

h

in

√225min2+ 484 min2

d = 86.000

m

k

i

m

n

√709min2 =

q = tan−1∆∆

x

y = tan−1

v

v

∆∆

t

t

x

y = tan−1 ∆∆

t

t

x

y = tan−1 212

5

.

.

0

0

m

m

i

i

n

n

q = 55.7° north of east

3.55 km

1. d = (5)(33.0 cm)

∆y = 88.0 cmq = sin−1

∆d

y = sin−1(5)

8

(

8

3

.

3

0

.0

cm

cm) =

∆x = d(cos q) = (5)(33.0 cm)(cos 32.2°) = 1.40 × 102 cm to the west

32.2° north of west

2. q = 60.0°

d = 10.0 m

∆x = d(cos q) = (10.0 m)(cos 60.0°) =

∆y = d(sin q) = (10.0 m)(sin 60.0°) = 8.66 m

5.00 m

3. d = 10.3 m

∆y = −6.10 m

Finding the angle between d and the x-axis yields,

q1 = sin−1 ∆d

y = sin−1 −160..130m

m = −36.3°

The angle between d and the negative y-axis is therefore,

q = −90.0 − (−36.3°) = −53.7°

q =

d2 + ∆x2 + ∆y2

∆x =√

d2− ∆y2 =√

(10.3m)2 − (−6.10 m)2 =√

106m2− 37.2m2 =√

69 m2

∆x = ±8.3 m

53.7° on either side of the negative y-axis


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Givens Solutions

4. d = (8)(4.5 m)

q = 35°∆x = d(cos q) = (8)(4.5 m)(cos 35°) =

∆y = d(sin q) = (8)(4.5 m)(sin 35°) = 21 m

29 m

5. v = 347 km/h

q = 15.0°

vx = v(cos q) = (347 km/h)(cos 15.0°) =

vy = v(sin q) = (347 km/h)(sin 15.0°) = 89.8 km/h

335 km/h

6. v = 372 km/h

∆t = 8.7 s

q = 60.0°

d = v∆t = (372 km/h)36

1

0

h

0 s(103 m/km)(8.7 s) = 9.0 × 102 m

∆x = d(cos q) = (9.0 × 102 m)(cos 60.0°) =

∆y = d(sin q) = (9.0 × 102 m)(sin 60.0°) = 780 m north

450 m east

Additional Practice 3C, p. 21

7. d = 14 890 km

q = 25.0°

∆t = 18.5 h

vavg = ∆d

t =

1.48

1

9

8

×.4

1

5

0

h

4 km =

vx = vavg(cos q) = (805 km/h)(cos 25.0°) =

vy = vavg(sin q) = (805 km/h)(sin 25.0°) = 340 km/h south

730 km/h east

805 km/h

8. vi = 6.0 × 102 km/h

vf = 2.3 × 103 km/h

∆t = 120 s

q = 35° with respect to horizontal

a = ∆∆

v

t =

vf

∆−t

vi

a =

a =

a = 3.9 m/s2

ax = a(cos q) = (3.9 m/s2)(cos 35°) =

ay = a(sin q) = (3.9 m/s2)(sin 35°) = 2.2 m/s2 vertically

3.2 m/s2 horizontally

(1.7 × 103 km/h) 36

1

0

h

0 s(103 m/km)

1.2 × 102 s

(2.3 × 103 km/h − 6.0 × 102 km/h)36

1

0

h

0 s(103 m/km)

1.2 × 102 s

1. ∆x1 = 250.0 m

d2 = 125.0 m

q2 = 120.0°

∆x2 = d2(cos q2) = (125.0 m)(cos 120.0°) = −62.50 m

∆y2 = d2(sin q2) = (125.0 m)(sin 120.0°) = 108.3 m

∆xtot = ∆x1 + ∆x2 = 250.0 m − 62.50 m = 187.5 m

∆ytot = ∆y 1 + ∆y2 = 0 m + 108.3 m = 108.3 m

d =√

(∆xtot)2 + (∆ytot)2 =√

(187.5 m)2 + (108.3m)2

d =√

3.516× 104 m2+ 1.173 × 104 m2 =√

4.689× 104 m2

d =

q = tan−1∆∆x

yt

t

o

o

t

t = tan−1110

8

8

7

.

.

3

5

m

m = 30.01° north of east

216.5 m


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2. v = 3.53 × 103 km/h

∆t1 = 20.0 s

∆t2 = 10.0 s

q1 = 15.0°

q2 = 35.0°

∆x1 = v∆t1(cos q1)

∆x1 = (3.53 × 103 km/h)36

1

0

h

0 s(103 m/km)(20.0 s)(cos 15.0°) = 1.89 × 104 m

∆y1 = v∆t1(sin q1)

∆y1 = (3.53 × 103 km/h)36

1

0

h

0 s(103 m/km)(20.0 s)(sin 15.0°) = 5.08 × 103 m

∆x2 = v∆t2(cos q2)

∆x2 = (3.53 × 103 km/h)36

1

0

h

0 s(103 m/km)(10.0 s)(cos 35.0°) = 8.03 × 103 m

∆y2 = v∆t2(sin q2)

∆y2 = (3.53 × 103 km/h)36

1

0

h

0 s(103 m/km)(10.0 s)(sin 35.0°) = 5.62 × 103 m

∆ytot = ∆y1 + ∆y2 = 5.08 × 103 m + 5.62 × 103 m =

∆xtot = ∆x1 + ∆x2 = 1.89 × 104 m + 8.03 × 103 m =

d =√

(∆xtot)2 + (∆ytot)2 =√

(2.69× 104 m)2 + (1.07× 104 m)2

d =√

7.24 × 108 m2+ 1.11× 108 m2 =√

8.35 × 108 m

d =

q = tan−1∆∆x

yt

t

o

o

t

t = tan−112.

.

0

6

7

9

××

1

1

0

0

4

4m

m

q = 21.7° above the horizontal

2.89 × 104 m

2.69 × 104 m

1.07 × 104 m

Givens Solutions

3. ∆x1 + ∆x2 = 2.00 × 102 m

∆y1 + ∆y2 = 0

q1 = 30.0°

q2 = −45.0°

v = 11.6 km/h

∆y1 = d1(sin q1) = −∆y2 = −d2(sin q2)

d1 = −d2 ssi

i

n

n

qq

2

1 = −d2si

s

n

i

(

n

−3

4

0

5

.

.

0

0

°°)

= 1.41d2

∆x1 = d1(cos q1) = (1.41d2)(cos 30.0°) = 1.22d2

∆x2 = d2(cos q2) = d2[cos(−45.0°)] = 0.707d2

∆x1 + ∆x2 = d2(1.22 + 0.707) = 1.93d2 = 2.00 × 102 m

d2 =

d1 = (1.41)d2 = (1.41)(104 m) =

v = 11.6 km/h = (11.6 km/h)36

1

0

h

0 s(103 m/km) = 3.22 m/s

∆t1 = d

v1 = 31

.2

4

2

7

m

m

/s = 45.7 s

∆t2 = d

v2 = 31

.2

0

2

4

m

m

/s = 32.3 s

∆ttot = ∆t1 + ∆t2 = 45.7 s + 32.3 s = 78.0 s

147 m

104 m


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5. v = 57.2 km/h

∆t1 = 2.50 h

∆t2 = 1.50 h

θ2 = 30.0°

d1 = v∆t1 = (57.2 km/h)(2.50 h) = 143 km

d2 = v∆t2 = (57.2 km/h)(1.50 h) = 85.8 km

∆tot = d1 + d2(cos q2) = 143 km + (85.8 km)(cos 30.0°) = 143 km + 74.3 km = 217 km

∆ytot = d2(sin q2) = (85.8 km)(sin 30.0°) = 42.9 km

d =√

(∆xtot)2 + (∆ytot)2 =√

(217 km)2 + (42.9 km)

d =√

4.71 × 104 km2+ 1.84× 103 km2 =√

4.89 × 104 km2

d =

q = tan−1∆∆x

yt

t

o

o

t

t = tan−1422

1

.

7

9

k

k

m

m = 11.2° north of east

221 km

1. vx = 9.37 m/s

∆y = −2.00 m

g = 9.81 m/s2

∆t = 2

−∆g

y =

∆vx

x

∆x = vx 2

−∆g

y = (9.37 m/s) (2

−)

9

(.

−82

1

. 0

m0

/m

s2) = 5.98 m

The river is 5.98 m wide.

4. v = 925 km/h

∆t1 = 1.50 h

∆t2 = 2.00 h

q2 = 135°

d1 = v∆t1 = (925 km/h)(103 m/km)(1.50 h) = 1.39 × 106 m

d2 = v∆t2 = (925 km/h)(103 m/km)(2.00 h) = 1.85 × 106 m

∆x1 = d1 = 1.39 × 106 m

∆y1 = 0 m

∆x2 = d2(cos q2) = (1.85 × 106 m)(cos 135°) = −1.31 × 106 m

∆y2 = d2(sin q2) = (1.85 × 106 m)(sin 135°) = 1.31 × 106 m

∆xtot = ∆x1 + ∆x2 = 1.39 × 106 m + (− 1.31 × 106 m) = 0.08 × 106 m

∆ytot = ∆y1 + ∆y2 = 0 m + 1.31 × 106 m = 1.31 × 106 m

d =√

(∆xtot)2 + (∆ytot)2 =√

(0.08× 106 m)2 + (1.31 × 106 m)2

d =√

6× 109 m2+ 1.72× 1012m2 =√

1.73 × 1012m2

d =

q = tan−1∆∆x

yt

t

o

o

t

t = tan−110.

.

3

0

1

8

××

1

1

0

0

6

6

m

m = 86.5° = 90.0° − 3.5°

q = 3.5° east of north

1.32 × 106 m = 1.32 × 103 km

Givens Solutions


2. ∆x = 7.32 km

∆y = −8848 m

g = 9.81 m/s2

∆t = 2

−∆g

y =

∆vx

x

vx = 2

−∆g

y ∆x =

(2−)9

(−.8

8

18

m

48

/s

m

2

) (7.32 × 103 m) =

No. The arrow must have a horizontal speed of 172 m/s, which is much greater than100 m/s.

172 m/s


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4. vx = 372 km/h

∆x = 40.0 m

g = 9.81 m/s2

∆t = ∆vx

x

∆y = − 1

2g ∆t2 =

−2

g

v

∆

x

x2

2

=

∆y = −0.735 m

The ramp is 0.735 m above the ground.

−(9.81 m/s2)(40.0 m)2

(2)(372 km/h)36

1

0

h

0 s110k

3

m

m

2

5. ∆x = 25 m

vx = 15 m/s

g = 9.81 m/s2

h = 25 m

∆t = ∆vx

x

∆y = − 1

2g∆t2 =

−2

g

v

∆

x

x2

2

=

∆y = h − h′ = −14 m

h′ = h − ∆y = 25 m − (−14 m)

= 39 m

−(9.81 m/s2)(25 m)2

(2)(15 m/s)2

6. l = 420 m

∆y =

∆x = l

g = 9.81 m/s2

−l2

7. ∆y = −2.45 m

v = 12.0 m/s

g = 9.81 m/s2

vy2 = −2g∆y

v2 = vx2 + vy

2 = vx2 − 2g∆y

vx =√

v2+ 2g∆y =√

12.0 m/s2+ (2)(9.81m/s2)(−2.45 m)

vx =√

144m2/s2 − 48.1m2/s2

=√

96 m2/s2

vx = 9.8 m/s

3. ∆x = 471 m

vi = 80.0 m/s

g = 9.81 m/s2

∆t = ∆vx

x

∆y = − 1

2 g ∆t2 =

−2

g

v

∆

x2x2

= = −1.70 × 102 m

The cliff is 1.70 × 102 m high.

−(9.81 m/s2)(471 m)2

(2)(80.0 m/s)2

Givens Solutions

∆t = 2

−∆g

y =

∆vx

x

vx = 2

−∆g

y ∆x =

(

−29

)(

.8

−121m

0/

m

s2

) (420 m) = 64 m/s


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Givens Solutions

1. ∆x = 201.24 m

q = 35.0°

g = 9.81 m/s2

∆y = vi (sin q) ∆t − 1

2g∆t2 = vi (sin q) −

1

2g∆t = 0

∆x = vi(cos q)∆t

∆t = vi(c

∆o

x

s q)

vi(sin q) = 1

2g vi(c

∆o

x

s q)

vi = 2(sin g

q∆)(

x

cosq) =

vi = 45.8 m/s

(9.81 m/s2)(201.24 m)(2)(sin 35.0°)(cos 35.0°)

2. ∆x = 9.50 × 102 m

q = 45.0°

g = 9.81 m/s2

Using the derivation shown in problem 1,

vi = 2(sin qg∆

)(x

cosq) =

vi = 96.5 m/s

At the top of the arrow’s flight:

v = vx = vi(cos q ) = (96.5 m/s)(cos 45.0°) = 68.2 m/s

(9.81 m/s2)(9.50 × 102 m)(2)(sin 45.0°)(cos 45.0°)

3. ∆x = 27.5 m

q = 50.0°

g = 9.81 m/s2


vi = 2(sin qg∆

)(x

cosq) =

vi = 16.6 m/s

(9.81 m/s2)(27.5 m)2(sin 50.0°)(cos 50.0°)

8. ∆y = −1.95 m

vx = 3.0 m/s

g = 9.81 m/s2

vy2 = −2g ∆y

v =√

vx2+ vy2 =√

vx2− 2g∆y

v =√

(3.0 m/s)2 − (2)(9.81m/s2)(−1.95 m)

v =√

9.0m2/s2 + 38.3m2/s2 =√

47.3 m2/s2 =

q = tan−1 v

v

x

y = tan−1 = tan−1 q = 64° below the horizontal

√−(2)(9.81 m/s2)(–1.95m)

3.0 m/s

√−2g∆y

vx

6.88 m/s


4. ∆x = 44.0 m

q = 45.0°

g = 9.81 m/s2


a. vi = 2(sin qg∆

)(x

cosq) =

vi = 20.8 m/s

(9.81 m/s2)(44.0 m)

(2)(sin 45.0°)(cos 45.0°)


b. At maximum height, vy, f = 0 m/s

vy, f2 = vy, i

2 − 2g∆y = 0

∆ymax = = vi

2(s

2

in

g

q)2

= = 11.0 m

c. ∆ymax = v

2i

g

2

= (2

(2

)(

0

9

.8

.8

m

1 m

/s)

/

2

s2 = 22.1 m

The brick’s maximum height is 22.1 m.

The brick’s maximum height is 11.0 m.

(20.8 m/s)2(sin 45.0°)2

(2)(9.81 m/s2)

vy, i2

2g

Givens Solutions

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5. ∆x = 76.5 m

q = 12.0°

g = 9.81 m/s2

At maximum height, vy, f = 0 m/s.

vy, f2 = vy, i

2 − 2g∆y = 0

∆ymax = v

2y,

gi2

= vi

2(s

2

i

g

n q)2

Using the derivation for vi2 from problem 1,

∆ymax = 2(sin

g

q∆)(

x

cos q)

(sin

2g

q)2

= ∆4

x

(c

(s

o

i

s

n

qq)

) =

∆x(t

4

an q)

∆ymax = (76.5 m)(

4

tan 12.0°) = 4.07 m

6. vrunner = 5.82 m/s

vi,ball = 2vrunner

In x-direction,

vi,ball(cos q) = 2vrunner(cos q) = vrunner

2(cos q) = 1

q = cos−112

= 60°

7. vi = 8.42 m/s

q = 55.2°

∆t = 1.40 s

g = 9.81 m/s2

For first half of jump,

∆t1 = 1.4

2

0 s = 0.700 s

∆y = vi(sin q)∆t1 − 12

g∆t12 = (8.42 m/s)(sin 55.2°)(0.700 s) − 1

2(9.81 m/s2)(0.700 s)2

∆y = 4.84 m − 2.40 m = 2.44 m

∆x = vi(cos q)∆t

∆x = (8.42 m/s) (cos 55.2°)(1.40 s) = 6.73 m

The fence is 2.44 m high.

8. vi = 2.2 m/s

q = 21°

∆t = 0.16 s

g = 9.81 m/s2

∆x = vi(cos q)∆t = (2.2 m/s) (cos 21°)(0.16 s) =

Maximum height is reached in a time interval of ∆2

t

∆ymax = vi (sin q)∆2

t − 1

2g

∆2

t

2

∆ymax = (2.2 m/s)(sin 21°) 0.1

2

6 s − 1

2 (9.81 m/s2)0.1

2

6 s

2

∆ymax = 6.3 × 10−2 m − 3.1 × 10−2 m = 3.2 × 10−2 m = 3.2 cm

The flea’s maximum height is 3.2 cm.

0.33 m


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Givens Solutions

1. vse = 126 km/h north

vgs = 40.0 km/h east

vge = vgs2 + vse2 = (40.0km/h)2 + (126km/h)2

vge = 1.60 × 103 km2/h2+ 1.59× 104km2/h2

vge = 1.75 × 104 km/h =

q = tan−1 v

v

g

se

s = tan−1 410

2

.

6

0

k

k

m

m

/

/

h

h = 72.4° north of east

132 km/h


2. vwe = −3.00 × 102 km/h

vpw = 4.50 × 102 km/h

∆x = 250 km

vpe = vpw + vwe = 4.50 × 102 km/h − 3.00 × 102 km/h = 1.50 × 102 km/h

∆t = v

∆

p

x

e =

1.50

2

×50

10

k2m

km/h = 1.7 h

3. vtw = 9.0 m/s north

vwb = 3.0 m/s east

∆t = 1.0 min

vtb = vtw + vwb

vtb =√

vtw2+ vwb2 =

√(9.0 m/s)2 + (3.0m/s)2 =

√81 m2/s2 + 9.0 m2/s2

vtb =√

9.0× 101 m2/s2vtb = 9.5 m/s

∆x = vtb∆t = (9.5 m/s)(1.0 min) 16

m

0

i

s

n =

q = tan−1v

vw

tw

b = tan−139.

.

0

0

m

m

/

/

s

s = 18° east of north

570 m

4. vsw = 40.0 km/h forward

vfw = 16.0 km/h forward

∆x = 60.0 m

vsf = vsw − vfw = 40.0 km/h − 16.0 km/h = 24.0 km/h toward fish

∆t = ∆vs

x

f = = 9.00 s

60.0 m

(24.0 km/h)36

1

0

h

0 s110k

3

m

m

5. v1E = 90.0 km/h

v2E = −90.0 km/h

∆t = 40.0 s

v12 = v1E − v2E

v12 = 90.0 km/h − (−90.0 km/h) = 1.80 × 102 km/h

∆x = v12∆t = (1.80 × 102 km/h) 36

1

0

h

0 s 110k

3

m

m(40.0 s) = 2.00 × 103 m = 2.00 km

The two geese are initially 2.00 km apart

6. vme = 18.0 km/h forward

Vre = 0.333 Vme= 6.00 km/h forward

∆x = 12.0 m

vmr = vme − vre

vmr = 18.0 km/h − 6.0 km/h = 12.0 km/h

∆t = v

∆

m

x

r =

(12

1

.

2

0

.0

km

m

/h) 36

1

0

h

0 s 110k

3m

m

∆t = 3.60 s


Chapter 4Two-Dimensional Motion and Vectors

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1. mw = 75 kg

mp = 275 kg

g = 9.81 m/s2

The normal force exerted by the platform on the weight lifter’s feet is equal to andopposite of the combined weight of the weightlifter and the pumpkin.

Fnet = Fn − mwg − mpg = 0

Fn = (mw + mp)g = (75 kg + 275 kg) (9.81 m/s2)

Fn = (3.50 × 102 kg)(9.81 m/s2) = 3.43 × 103 N

Fn = 3.43 × 103 N upward against feet


Givens Solutions

2. mb = 253 kg

mw = 133 kg

g = 9.81 m/s2

Fnet = Fn,1 + Fn,2 − mbg − mwg = 0

The weight of the weightlifter and barbell is distributed equally on both feet, so thenormal force on the first foot (Fn,1) equals the normal force on the second foot (Fn,2).

2Fn,1 = (mb + mw)g = 2Fn,2

Fn,1 = Fn,2 = (mb +

2

mb)g =

Fn,1 = Fn,2 = = 1.89 × 103 N

Fn,1 = Fn,2 = 1.89 × 103 N upward on each foot

(386 kg)(9.81 m/s2)

2

(253 kg + 33 kg)9.81 m

s2

2

3. Fdown = 1.70 N

Fnet = 4.90 N

Fnet2 = Fforward

2 + Fdown2

Fforward =√

Fnet2 − Fdown

2 =√

(4.90N)2 − (1.70 N)2Fforward =

√21.1 N2 = 4.59 N

4. m = 3.10 × 102 kg

g = 9.81 m/s2

q1 = 30.0°

q2 = − 30.0°

Fx,net = ΣFx = FT,1(sin q1) + FT,2(sin q2) = 0

Fy,net = ΣFy = FT,1(cos q1) + FT,2(cos q2) + Fg = 0

FT,1(sin 30.0°) = −FT,2[sin (−30.0°)]

FT,1 = FT,2

FT,1(cos q1) + FT,1(cos q2) = −Fg = mg

FT,1(cos 30.0°) + FT,1[cos (−30.0°)] = (3.10 × 102 kg)(9.81 m/s2)

FT,1 =

FT,1 = FT,2 =

As the angles q1 and q2 become larger, cos q1 and cos q2 become smaller. Therefore,FT,1 and FT,2 must become larger in magnitude.

1.76 × 103 N

(3.10 × 102 kg)(9.81 m/s2)(2)(cos 30.0°)[cos(−30.0°)]


Givens Solutions

5. m = 155 kg

FT,1 = 2FT,2

g = 9.81 m/s2

q1 = 90° − q2

Fx,net = FT,1(cos q1) − FT,2(cos q2) = 0

Fy,net = FT,1(sin q1) + FT,2(sin q2) − mg = 0

FT,1[(cos q1) − 1

2(cos q2)] = 0

2 (cos q1) = cos q2 = cos(90° − q1) = sin q1

2 = tan q1

q1 = tan−1(2) = 63°

q2 = 90° − 63° = 27°

FT,1(sin q1) + FT

2,1(sin q2) = mg

FT,1 =

FT,1 = = (155

0

k

.8

g

9

)(

+9.

0

8

.

1

2

m

3

/s2) =

(155 kg

1

)

.1

(9

2

.81 m)

FT,1 =

FT,2 = 6.80 × 102 N

1.36 × 1.36 × 103 N

(155 kg)(9.81 m/s2)(sin 63°) + (sin

2

27°)

mg(sin θ1) +

1

2 (sin θ2)

1. vi = 173 km/h

vf = 0 km/h

∆x = 0.660 m

m = 70.0 kg

g = 9.81 m/s2

a = vf

2

2∆−

x

vi2

=

a = −1.75 × 103 m/s2

F = ma = (70.0 kg)(−1.75 × 103 m/s2) =

Fg = mg = (70.0 kg)(9.81 m/s2) =

The force of deceleration is nearly 178 times as large as David Purley’s weight.

6.87 × 102 N

−1.22° × 105 N

[(0 km/h)2 − (173 km/h)2](103 m/km)2(1 h/3600 s)2

(2)(0.660 m)

2. m = 2.232 × 106 kg

g = 9.81 m/s2

anet = 0 m/s2

a. Fnet = manet = Fup − mg

Fup = manet + mg = m(anet + g) = (2.232 × 106 kg)(0 m/s2 + 9.81 m/s2)

Fup = = mg

b. Fdown = mg(sin q)

anet = F

mnet =

Fup −m

Fdown = mg − m

m

g(sin q)

anet = g(1 − sin q) = (9.81 m/s2)[1.00 − (sin 30.0°)] = 9.81

2

m/s2

= 4.90 m/s2

anet = 4.90 m/s2 up the incline

2.19 × 107 N

3. m = 40.00 mg= 4.00 × 10−5 kg

g = 9.807 m/s2

anet = (400.0)g

Fnet = Fbeetle − Fg = manet = m(400.0) g

Fbeetle = Fnet + Fg = m(400.0 + 1)g = m(401)g

Fbeetle = (4.000 × 10−5 kg)(9.807 m/s2)(401) =

Fnet = Fbeetle − Fg = m(400.0) g = (4.000 × 10−5 kg)(9.807 m/s2)(400.0)

Fnet =

The effect of gravity is negligible.

1.569 × 10−1 N

1.573 × 10−1 N


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Givens Solutions

4. ma = 54.0 kg

mw = 157.5 kg

anet = 1.00 m/s2

g = 9.81 m/s2

The net forces on the lifted weight is

Fw,net = mwanet = F ′ − mwg

where F ′ is the force exerted by the athlete on the weight.

The net force on the athlete is

Fa,net = Fn,1 + Fn,2 − F ′ − mag = 0

where Fn,1 and Fn,2 are the normal forces exerted by the ground on each of the ath-lete’s feet, and −F ′ is the force exerted by the lifted weight on the athlete.

The normal force on each foot is the same, so

Fn,1 = Fn,2 = Fn and

F ′ = 2Fn − mag

Using the expression for F ′ in the equation for Fw,net yields the following:

mwanet = (2Fn − mag) − ma g

2Fn = mw(anet + g) + mag

Fn = mw(anet +

2

g) + mag =

Fn =

Fn = = 223

2

2 N = 1116 N

Fn,1 − Fn,2 = Fn = 1116 N upward

1702 N + 5.30 × 102 N

2

(157.5 kg)(10.81 m/s2) + (54.0 kg)(9.81 m/s2)

2

(157.5 kg)(1.00 m/s2 + 9.81 m/s2) + (54.0 kg)

2

5. m = 2.20 × 102 kg

anet = 75.0 m/s2

g = 9.81 m/s2

Fnet = manet = Favg − mg

Favg = m(anet + g) = (2.20 × 102 kg)(75.0 m/s2 + 9.81 m/s2)

Favg = (2.20 × 102 kg)(84.8 m/s2) = 1.87 × 104 N

Favg = 1.87 × 104 N upward

6. m = 2.00 × 104 kg

∆t = 2.5

vi = 0 m/s

vf = 1.0 m/s

g = 9.81 m/s2

anet = vf

∆−t

vi = (1.0 m/

2

s

.

−5

0

s

.0 m/s) = 0.40 m/s2

Fnet = manet = FT − mg

FT = manet + mg = m(anet + g)

FT = (2.00 × 104 kg)(0.40 m/s2 + 9.81 m/s2)

FT = (2.00 × 104 kg)(10.21 m/s2) = 2.04 × 105 N

FT = 2.04 × 105 N

7. m = 2.65 kg

q1 = q2 = 45.0°

anet = 2.55 m/s2

g = 9.81 m/s2

Fx,net = FT,1(cos q1) − FT,2(cos q2) = 0

FT,1(cos 45.0°) = FT,2(cos 45.0°)

FT,1 = FT,2

Fy,net = manet = FT,1(sin q1) + FT,2(sin q2) − mg

FT = FT,1 = FT,2

q = q1 = q2

FT(sin q) + FT(sin q) = m(anet + g)

2FT(sin q) = m(anet + g)


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FT = m

2

(a

(sn

ie

nt +

q)

g) =

FT = = 23.2 N

FT,1 = 23.2 N

FT,2 = 23.2 N

(2.65 kg)(12.36 m/s2)

(2)(sin 45.0°)

(2.65 kg)(2.55 m/s2 = 9.81 m/s2)

(2)(sin 45.0°)

Givens Solutions

8. m = 20.0 kg

∆x = 1.55 m

vi = 0 m/s

vf = 0.550 m/s

anet = vf

2

2∆−

x

vi2

= = 9.76 × 10−2 m/s2

Fnet = manet = (20.0 kg)(9.76 × 10−2 m/s2) = 1.95 N

(0.550 m/s)2 − (0.00 m/s)2

(2)(1.55 m)

9. mmax = 70.0 kg

m = 45.0 kg

g = 9.81 m/s2

Fmax = mmaxg = FT

Fmax = (70.0 kg)(9.81 m/s2) = 687 N

Fnet = manet = FT − mg = Fmax − mg

anet = Fm

max − g =

4

6

5

8

.

7

0

N

kg − 9.81 m/s2 = 15.3 m/s2 − 9.81 m/s2 = 5.5 m/s2

anet = 5.5 m/s2 upward

10. m = 3.18 × 105 kg

Fapplied = 81.0 × 103 N

Ffriction = 62.0 × 103 N

Fnet = Fapplied − Ffriction = (81.0 × 103 − 62.0 × 103 N)

Fnet = 19.0 × 103 N

anet = F

mnet = 31.

9

1

.

8

0

××

1

1

0

05

3

k

N

g = 5.97 × 10−2 m/s2

11. m = 3.00 × 103 kg

Fapplied = 4.00 × 103 N

q = 20.0°

Fopposing = (0.120) mg

g = 9.81 m/s2

Fnet = manet = Fapplied(cos q) − Fopposing

anet =

anet =

anet = = 3

2

.0

.3

0

××

1

1

0

0

2

3N

kg

anet = 7.7 × 10−2 m/s2

3.76 × 103 N − 3.53 × 103 N

3.00 × 103 kg

(4.00 × 103 N)(cos 20.0°) − (0.120)(3.00 × 103 kg)(9.81 m/s2)

3.00 × 103 kg

Fapplied(cos q) − (0.120) mg

m

12. mc = 1.600 × 103 kg

mw = 1.200 × 103 kg

vi = 0 m/s

g = 9.81 m/s2

∆y = 25.0 m

For the counterweight: The tension in the cable is FT.

Fnet = FT − mwg = mwanet

For the car:

Fnet = mcg − FT = mcanet

Adding the two equations yields the following:

mcg − mwg = (mw + mc)anet

anet = (m

mc

c

−+

m

mw

w

)g =

anet = = 1.40 m/s2(4.00 × 102 kg)(9.81 m/s2)

2.800 × 103 kg

(1.600 × 103 kg − 1.200 × 103 kg)(9.81 m/s2)

1.600 × 103 kg + 1.200 × 103 kg


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13. m = 409 kg

d = 6.00 m

q = 30.0°

g = 9.81 m/s2

Fapplied = 2080 N

vi = 0 m/s

a. Fnet = Fapplied − mg(sin q) = 2080 N − (409 kg)(9.81 m/s2)(sin 30.0°)

Fnet = 2080 N − 2010 N = 70 N

Fnet =

b. anet = F

mnet =

4

7

0

0

9

N

kg = 0.2 m/s2

anet =

c. d = vi∆t + 1

2anet ∆t2 = (0 m/s)∆t +

1

2(0.2 m/s2)∆t2

∆t = (

(

20

)

.(

2

6 .

m

00

/sm2)

) = 8 s

0.2 m/s2 at 30.0° above the horizontal

70 N at 30.0° above the horizontal

vf =√

2anet∆y+ vi2 =

√(2)(1.40 m/s2)(25.0m)+ (0m/s)2

vf = 8.37 m/s

Givens Solutions

14. amax = 0.25 m/s2

Fmax = 57 N

Fapp = 24 N

a. m = F

am

m

a

a

x

x = 0.2

5

5

7

m

N

/s2 =

b. Fnet = Fmax − Fapp = 57 N − 24 N = 33 N

anet = F

mnet =

2.3

3

×3

1

N

02 kg = 0.14 m/s2

2.3 × 102 kg

15. m = 2.55 × 103 kg

FT = 7.56 × 103 N

qT = −72.3°

Fbuoyant = 3.10 × 104 N

Fwind = −920 N

g = 9.81 m/s2

∆y = −45.0 m

vi = 0 m/s

a. Fx,net = ΣFx = max,net = FT(cos qT) + Fwind

Fx,net = (7.56 × 103 N)[cos(−72.3°)] − 920 N = 2.30 × 103 N − 920 N = 1.38 × 103 N

Fy,net = ΣFy = may,net = FT(sin qT) + Fbuoyant + Fg = FT(sin qT) + Fbuoyant − mg

Fy,net = (7.56 × 103 N)[sin(−72.3°)] = 3.10 × 104 N − (2.55 × 103 kg)(9.81 m/s2)

Fy,net = −7.20 × 103 N + 3.10 × 104 N − 2.50 × 104 = −1.2 × 103 N

Fnet =√

(Fx,net)2+ (Fy,net)2 =√

(1.38× 103 N)2 + (−1.2× 103 N)2

Fnet =√

1.90 × 106 N2+ 1.4 × 106 N2

Fnet =√

3.3× 106 N2 = 1.8 × 103 N

q = tan−1FFx

y,

,

n

n

e

e

t

t = tan−1−1.

1

3

.

8

2

××

1

1

0

03

3

N

N

q = −41°

Fnet =

b. anet = F

mnet =

2

1

.5

.8

5

××

1

1

0

0

3

3N

kg

anet =

c. Because vi = 0

∆y = 1

2 ay,net ∆t2

∆x = 1

2 ax,net ∆t2

∆x = a

ax

y,

,

n

n

e

e

t

t ∆y = ∆y = ta

∆n

y

q

∆x = ta

−n

4

(

5

−.0

41

m

°) = 52 m

anet(cos q)anet(sin q)

0.71 m/s2

1.8 × 103 N at 41° below the horizontal


1. m = 11.0 kg

mk = 0.39

g = 9.81 m/s2


Givens Solutions

2. m = 2.20 × 105 kg

ms = 0.220

g = 9.81 m/s2

Fs,max = msFn = msmg

Fs,max = (0.220)(2.20 × 105 kg)(9.91 m/s2) = 4.75 × 105 N

3. m = 25.0 kg

Fapplied = 59.0 N

q = 38.0°

ms = 0.599

g = 9.81 m/s2

Fs,max = msFm

Fn = mg(cos q) + Fapplied

Fs,max = ms[mg(cos q) = Fapplied] = (0.599)[(25.0 kg)(9.81 m/s2)(cos 38.0° + 59.0 N]

Fs,max = (0.599)(193 N + 59 N) = (0.599)(252 N) =

Alternatively,

Fnet = mg(sin q) − Fs,max = 0

Fs,max = mg(sin q) = (25.0 kg)(9.81 m/s2)(sin 38.0°) = 151 N

151 N

Fk = mkFn = mkmg

Fk = (0.39) (11.0 kg)(9.81 m/s2) = 42.1 N

4. q = 38.0°

g = 9.81 m/s2

Fnet = mg(sin q) − Fk = 0

Fk = mkFn = mkmg(cos q)

mkmg(cos q) = mg(sin q)

mk = c

si

o

n

s

qq

= tan q = tan 38.0°

mk = 0.781

5. q = 5.2°

g = 9.81 m/s2




mk = c

si

o

n

s

qq

= tan q = tan 5.2°

mk = 0.091

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ed.

10. m = 3.00 × 103 kg

q = 31.0°

g = 9.81 m/s2




mk = c

si

o

n

s

qq

= tan q = tan 31.0°

mk =

Fk = mkmg(cos q) = (0.601)(3.00 × 103 kg)(9.81 m/s2)(cos 31.0°)

Fk =

Alternatively,

Fk = mg(sin q) = (3.00 × 103 kg)(9.81 m/s2)(sin 31.0°) = 1.52 × 104 N

1.52 × 104 N

0.601

6. m = 281.5 kg

q = 30.0°

Fnet = 3mg(sin q) − ms(3mg)(cos q) − Fapplied = 0

Fapplied = mg

ms = = =

ms = (3

1

)

.

(

5

c

0

o

−s 3

1

0

.0

.0

0

°) =

(3)(c

0

o

.

s

5

3

0

0.0°)

ms = 0.19

(3)(sin 30.0°) − 1.00

(3)(cos 30.0°)

3(sin q) − 1.00

3(cos q)

3mg(sin q) − mg

3mg(cos q)

Givens Solutions

7. m = 1.90 × 105 kg

ms = 0.460

g = 9.81 m/s2

Fnet = Fapplied − Fk = 0

Fk = mkFn = mkmg

Fapplied = mkmg = (0.460)(1.90 × 105 kg)(9.81 m/s2)

Fapplied = 8.57 × 105 N

8. Fapplied = 6.0 × 103 N

mk = 0.77

g = 9.81 m/s2

Fnet = Fapplied − Fk = 0

Fk = mkFn

Fn = Fap

mp

k

lied = 6.0

0

×.7

1

7

03 N =

Fn = mg

m = F

gn =

7

9

.8

.8

×1

1

m

0

/

3

s

N2 = 8.0 × 102 kg

7.8 × 103 N

9. Fapplied = 1.13 × 108 N

ms = 0.741

Fnet = Fapplied − Fs,max = 0

Fs,max = msFn = msmg

m = Fa

mpp

sglied =

(0.

1

7

.

4

1

1

3

)(

×9

1

.8

0

1

8

m

N

/s2 = 1.55 × 102 kg


II

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1. Fapplied = 130 N

anet = 1.00 m/s2

mk = 0.158

g = 9.81 m/s2


Fk = mkFn = mkmg

manet + mkmg = Fapplied

m(anet + mkg) = Fapplied

m = an

F

e

a

t

p

+pli

med

kg =

m = = = 51 kg130 N2.55 m/s2

130 N1.00 m/s2 + 1.55 m/s2

130 N1.00 m/s2 + (0.158)(9.81 m/s2)

2. Fnet = −2.00 × 104 N

q = 10.0°

mk = 0.797

g = 9.81 m/s2

Fnet = manet = mg(sin q) − Fk


m[g(sin q) − mkg(cos q)] = Fnet

m = g[sin q −

Fn

me

k

t

(cos q)] =

m = = (9.8

−1

2.

m

00

/s

×2)

1

(−04

0.

N

611)

m =

Fn = mg(cos q) = (3.34 × 103 kg)(9.81 m/s2)(cos 10.0°) = 3.23 × 104 N

3.34 × 103 kg

−2.00 × 104 N(9.81 m/s2)(0.174 − 0.785)

−2.00 × 104 N(9.81 m/s2)[(sin 10.0°) − (0.797)(cos 10.0°)]

3. Fnet = 6.99 × 103 N

q = 45.0°

mk = 0.597



m[g(sin q) − mkg(cos q)] = Fnet

m = g[sin q −

Fn

me

k

t

(cos q)] =

m = = (9.8

6

1

.9

m

9

/

×s2

1

)

0

(

3

0.

N

285)

m =

Fn = mg(cos q) = (2.50 × 103 kg)(9.81 m/s2)(cos 45.0°) = 1.73 × 104 N

2.50 × 103 kg

6.99 × 103 N(9.81 m/s2)(0.707 − 0.422)

6.99 × 103 N(9.81 m/s2)[(sin 45.0°) − (0.597)(cos 45.0°)]

4. m = 9.50 kg

q = 30.0 °

Fapplied = 80.0 N

anet = 1.64 m/s2

g = 9.81 m/s2

Fnet = manet = Fapplied − Fk − mg(sin q)


mkmg(cos q) = Fapplied − manet − mg(sin q)

mk =

mk = 80.0 N − (9.50 kg)[1.64 m/s2 + (9.81 m/s2)(sin 30.0°)]

(9.50 kg)(9.81 m/s2)(cos 30.0°)

Fapplied − m[anet + g (sin q)]

mg(cos q)


Givens Solutions


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6. q = 38.0°

mk = 0.100

g = 9.81 m/s2



manet = mg[sin q − mk(cos q)]

anet = g[sin q − mk(cos q)] = (9.81 m/s2)[(sin 38.0°) − (0.100)(cos 38.0°)]

anet = (9.81 m/s2)(0.616 − 7.88 × 10−2) = (9.81 m/s2)(0.537)

anet =

Acceleration is independent of the rider’s and sled’s masses. (Masses cancel.)

5.27 m/s2

Givens Solutions

mk = =

mk = =

mk = 0.222

17.9 N(9.50 kg)(9.81 m/s2)(cos 30.0°)

80.0 N − 62.1 N(9.50 kg)(9.81 m/s2)(cos 30.0°)

80.0 N − (9.50 kg)(6.54 m/s2)(9.50 kg)(9.81 m/s2)(cos 30.0°)

80.0 N − (9.50 kg)[1.64 m/s2 + 4.90 m/s2)

(9.50 kg)(9.81 m/s2)(cos 30.0°)

5. m = 1.89 × 105 kg

Fapplied = 7.6 × 105 N

anet = 0.11 m/s2


Fk = Fapplied − manet = 7.6 × 105 N − (1.89 × 105)(0.11 m/s2) = 7.6 × 105 N − 2.1 × 104 N

Fk = 7.4 × 105 N

7. ∆t = 6.60 s

q = 34.0°

mk = 0.198

g = 9.81 m/s2

vi = 0 m/s



manet = mg[sin q − mk(cos q)]

anet = g[sin q − mk(cos q)] = (9.81 m/s2)[(sin 34.0°) − (0.198)(cos 34.0°)]

anet = (9.81 m/s2)(0.559 − 0.164) = (9.81 m/s2)(0.395)

anet =

vf = vi + anet∆t = 0 m/s + (3.87 m/s2)(6.60 s)

vf = 25.5 m/s2 = 92.0 km/h

3.87 m/s2


Chapter 5Work and Energy

II

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1. W = 1.15 × 103 J

m = 60.0 kg

g = 9.81 m/s2

q = 0°

W = Fd(cos q) = mgd(cos q)

d = mg(

W

cos q) =

d = 195 m

1.15 × 105 J(60.0 kg)(9.81 m/s2)(cos 0°)


Givens Solutions

2. m = 1.45 × 106 kg

g = 9.81 m/s2

q = 0°

W = 1.00 × 102 MJ

F = (2.00 × 10−2) mg

W = Fd(cos q)

d = F(c

W

os q) =

d = 352 m

1.00 × 108 J(2.00 × 10−2)(1.45 × 106 kg)(9.81 m/s2)(cos 0.00°)

3. m = 1.7 g

W = 0.15 J

anet = 1.2 m/s2

q = 0°

g = 9.81 m/s2

4. m = 5.40 × 102 kg

W = 5.30 × 104 J

g = 9.81 m/s2

q = 30.0°

q ′ = 0°

W = Fd(cos q ′) = Fd

F = mg(sin q)

W = mg(sin q)d

d = mg(

W

sin q) =

d = 20.0 m

5.30 × 104 J(5.40 × 102 kg)(9.81 m/s3)(sin 30.0°)

Fnet = manet = F − mg

F = manet + mg

W = Fd(cos q) = m(anet + g)d(cos q)

d = m(anet +

W

g)(cos q) =

d =

d = 8.0 m

0.15 J(1.7 × 10−3 kg)(11.0 m/s2)

0.15 J(1.7 × 10−3 kg)(1.2 m/s2 + 9.81 m/s2)(cos 0°)


II

5. d = 5.45 m

W = 4.60 × 104 J

q = 0°

Fnet = Flift − Fg = 0

F = Flift = Fg

W = Fd(cos q) = Fgd(cos q)

Fg = d(c

W

os q) =

(5.

4

4

.

5

60

m

×)(

1

c

0

o

4

s

J

0°) = 8.44 103 N

Givens Solutions

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6. d = 52.0 m

m = 40.0 kg

W = 2.04 × 104 J

q = 0°

F = d(c

W

os q) =

(5

2

2

.

.

0

0

4

m

×)

1

(c

0

o

4

s

J

0°) = 392 N

7. d = 646 m

W = 2.15 × 105 J

q = 0°

F = d(c

W

os q) =

(64

2

6

.1

m

5 ×)(

1

co

0

s

5

0

J

°) = 333 N

8. m = 1.02 × 103 kg

d = 18.0 m

angle of incline = q = 10.0°

q ′ = 0°

g = 9.81 m/s2

mk = 0.13

Fnet = Fg − Fk = mg (sin q) − mkmg(cos q)

Wnet = Fnetd(cos q ′) = mgd(cos q ′)[(sin q) − mk(cos q)]

Wnet = (1.02 × 103 kg)(9.81 m/s2)(18.0 m)(cos 0°)[(sin 10.0°) − (0.13)(cos 10.0°)]

Wnet = (1.02 × 103 kg)(9.81 m/s2)(18.0 m)(0.174 − 0.128)

Wnet = (1.02 × 103 kg)(9.81 m/s2)(18.0 m)(0.046)

Wnet = 8.3 × 103 J

9. d = 881.0 m

Fapplied = 40.00 N

q = 45.00°

Fk = 28.00 N

q ′ = 0°

Wnet = Fnetd(cos q ′)

Fnet = Fapplied(cos q) − Fk

Wnet = [Fapplied(cos q) − Fk]d(cos q ′)

Wnet = [40.00 N(cos 45.00°) − 28.00° N](881.0 m)(cos q)

Wnet = (28.28 N − 28.00 N)(881.0 m) = (0.28 N)(881.0 m)

Wnet = 246.7 J

10. m = 9.7 × 103 kg

q = 45°

F = F1 = F2 = 1.2 × 103 N

d = 12 m

Wnet = Fnetd(cos q) = (F1 + F2)d(cos q) = 2Fd(cos q)

Wnet = (2)(1.2 × 103 N)(12 m)(cos 45°) = 2.0 × 104 J

11. m = 1.24 × 103 kg

F1 = 8.00 × 103 N east

F2 = 5.00 × 103 N 30.0°

south of east

d = 20.0 m south

Only F2 contributes to the work done in moving the flag south.

q = 90.0° − 30.0° = 60.0°

Wnet = Fnetd(cos q) = F2d(cos q) = (5.00 × 103 N)(20.0 m)(cos 60.0°)

Wnet = 5.00 × 104 J


II

1. ∆x = 1.00 × 102 m

∆t = 9.85 s

KE = 3.40 × 103 J

v = ∆∆t

x

KE = 12

mv2 = 12

m ∆∆

x

t

2

m = 2K

∆E

x

∆2t2

= = 66.0 kg(2)(3.40 × 103 J)(9.85 s)2

(1.00 × 102 m)2


Givens SolutionsC

opyr

ight

©by

Hol

t, R

ineh

art a

nd W

inst

on.A

ll rig

hts

rese

rved

.

2. v = 4.00 × 102 km/h

KE = 2.10 × 107 Jm =

2

v

K2E

= = 3.40 × 103 kg(2)(2.10 × 107 J)

(4.00 × 102 km/h)2 (103 m km)2 (1 h/3600 s)2

3. v = 50.3 km/h

KE = 6.54 × 103 Jm =

2

v

K2E

= = 67.0 kg(2)(6.54 × 103 J)

(50.3 km/h)2(103 m/km)2 (1 h/3600 s)

4. v = 318 km/h

KE = 3.80 MJm =

2

v

K2E

= = 974 kg(2)(3.80 × 106 J)

(318 km/h)2 (103 m/km)2 (1 h/3600 s)

5. m = 51.0 kg

KE = 9.96 × 104 Jv =

2m

KE = (2)(9

5

.9

1

6

.0

×kg104

J) = 62.5 m/s = 225 km/h

6. ∆x = 93.625 km

∆t = 24.00 h

m = 55 kg

a. vavg = ∆∆

x

t =

(24

9

.

.

0

3

0

62

h

5

)(

×3

1

6

0

0

4

0

m

s/h) =

b. KE = 12

mv2 = 12

(55 kg)(1.084 m/s)2 = 32 J

1.084 m/s

7. m = 3.38 × 1031 kg

KE = 1.10 × 1042 Jv =

2m

KE = (23

)

.(

3

18

.1×0 1

×0

1310

4

k

2

g

J) = 2.55 × 105 m/s = 255 km/s

8. m = 680 kg

v = 56.0 km/h

KELB = 3.40 × 103 J

a. KE = 12

mv2 = 12

(680 kg)[(56.0 km/h)(103 m/km)(1 h/3600 s)]2 = 8.23 × 104 J

b. K

K

E

E

L

p

B

b = 3

8

.

.

4

2

0

××

1

1

0

0

4

3J

J =

2

1

4

9. v = 11.2 km/s

m = 2.3 × 105 kg

KE = 12

mv2 = 12

(2.3 × 105 kg)(11.2 × 103 m/s)2 = 1.4 × 1013 J


II


Givens Solutions

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serv

ed.

1. d = 227 m

m = 655 g

g = 9.81 m/s2

Fresistance = (0.0220)mg

q = 0°

KEi = 0 J

Wnet = ∆KE = KEf − KEi = KEf

Wnet = Fnetd(cos q)

Fnet = Fg − Fresistance = mg − (0.0220)mg = mg(1 − 0.0220)

KEf = mg(1 − 0.0220)d(cos q) = (655 × 10−3 kg)(9.81 m/s2)(1 − 0.0220)(227 m)(cos 0°)

KEf = (0.655 kg)(9.81 m/s2)(0.9780)(227 m)

KEf = 1.43 × 103 J

2. vi = 12.92 m/s

Wnet = −2830 J

m = 55.0 kg

Wnet is the work done by friction.

Wnet = ∆KE = KEf − KEi = KEf − 12

mvi2

KEf = Wnet + 12

mvi2 = − 2830 J + 1

2(55.0 kg)(12.92 m/s)2 = −2830 J + 4590 J

KEf = 1.76 × 103 J

3. m = 25.0 g

hi = 553 m

hf = 353 m

vi = 0 m/s

vf = 30.0 m/s

g = 9.81 m/s2

q = 0°

Wnet = ∆KE = KEf − KEi = 12

mvf2 − 1

2mvi

2

Wnet = Fnetd(cos q)

Fnet = Fg − Fr = mg − Fr

d = hi − hf

Wnet = (mg − Fr)(hi − hf)(cos q)

mg − Fr =12

m (vf2 − vi

2)

(hi – hf)(cos q)

Fr = mg − 2(h

(

i

v

−f2

h

−

f)(

v

ci2

o

)

s q) = (25.0 × 10−3 kg)9.81 m/s2 −

Fr = (25.0 × 10−3 kg)9.81 m/s2 − (

9

2

.

)

0

(

0

2.

×00

10

×

2

1

m

02

2/

m

s2

)

Fr = (25.0 × 10−3 kg)(9.81 m/s2 − 2.25 m/s2) = (25.0 × 10−3 kg)(7.56 m/s2)

Fr = 0.189 N

(30.0 m/s)2 − (0 m/s)2

(2)(553 m − 353 m)(cos 0°)

4. vi = 404 km/h

Wnet = −3.00 MJ

m = 1.00 × 103 kg


mvf2 − 1

2mvi

2

12

mvf2 = 1

2mvi

2 + Wnet

vf = vi2+ 2W

mnet = [(404km/h)(103m/km)(1h/3600s)]2 +(2)1

(

.−00

3.0

×0 1×

031k

0

g6

J)

vf =√

1.26 × 104 m2/s2 − 6.00× 103 m2/s2 =√

6.6× 103 m2/s2

vf = 81 m/s = 290 km/h

5. m = 45.0 g

hi = 8848.0 m

hf = 8806.0 m

vi = 0 m/s

vf = 27.0 m/s

g = 9.81 m/s2

q = 0°


mvf2 − 1

2mvi

2

Wnet = Fnetd(cos q)

Fnet = Fg − Fr = mg − Fr

d = hi − hf

Wnet = mg(hi − hf)(cos q) − Fr(hi − hf)(cos q)

− Fr(hi − hf)(cos q) = Fr(hi − hf)(cos 180° + q) = Wr

12

m(vf2 − vi

2) = mg(hi − hf)(cos q) + Wr


II

6. vf = 35.0 m/s

vi = 25.0 m/s

Wnet = 21 kJ


mvf2 − 1

2mvi

2

m = v

2

f2

W

−n

vet

i2 = =

m = 6.0

4

×2

1

×0

120

m

3 J2/s2 = 7.0 × 101 kg

42 × 103 J1220 m2/s2 − 625 m2/s2

(2)(21 × 103 J)(35.0 m/s)2 − (25.0 m/s)2

Givens Solutions

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ed.

7. vi = 104.5 km/h

vf = 12

vi

mk = 0.120

g = 9.81 m/s2

q = 180°


mvf2 − 1

2mvi

2

Wnet = Wkd(cos q) = Fkd(cos q) = mkmgd(cos q)

12

m(vf2 − vi

2) = mkmgd(cos q)

d = 2m

v

k

f

g

2

(

−co

v

si2

q) =

d = =

d = 268 m

−(3)130

.6

4

0

.5

0 m/s

2

−(8)(0.120)(9.81 m/s2)

130

.6

4

0

.5

0 m/s

2

14

− 1−(2)(0.120)(9.81 m/s2)

(2)(0.120)(9.81 m/s2)(cos 180°)


Wr = m[12

(vf2 − vi

2) − g(hi − hf)(cos q)] = (45.0 × 10−3 kg)12

(27.0 m/s)2 − 12

(0 m/s)2

−(9.81 m/s2)(8848.0 m − 8806.0 m)(cos 0°)Wr = (45.0 × 10−3 kg)[364 m2/s2 − (9.81 m/s2)(42.0 m)]

Wr = (45.0 × 10−3 kg)(364 m2/s2 − 412 m2/s2)

Wr = (45.0 × 10−3 kg)(−48 m2/s2) = −2.16 J

1. h = 6.13/2 m = 3.07 m

PEg = 4.80 kJ

g = 9.81 m/s2

m = P

g

E

h

g = (9.81

4.

m

80

/s

×2)

1

(

0

3

3

.0

J

7 m) = 1.59 × 102 kg

2. h = 1.70 m

PEg = 3.04 × 103 J

g = 9.81 m/s2

m = P

g

E

h

g = (9.81

3.

m

04

/s

×2)

1

(

0

1

3

.7

J

0 m) = 182 kg

3. PEg = 1.48 × 107 J

h = (0.100)(180 km)

g = 9.81 m/s2

m = P

g

E

h

g = = 83.8 kg1.48 × 107 J

(9.81 m/s2)(0.100)(180 × 103 m)

4. m = 3.6 × 104 kg

PEg = 8.88 × 108 J

g = 9.81 m/s2

h = P

m

E

g

g = = 2.5 × 103 m = 2.5 km8.88 × 108 J

(3.6 × 104 kg)(9.81 m/s2)

[(104.5 km/h)(103 m/km)(1 h/3600 s)]2[12

2

− (1)2]


II

5. P

m

Eg = 20.482 m2/s2

g = 9.81 m/s2

P

m

Eg = gh = 20.482 m2/s2

h = 20.482

g

m2/s2

= 20

9

.

.

4

8

8

1

2

m

m

/

2

s

/2

s2

= 2.09 m

Givens Solutions

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6. k = 3.0 × 104 N/m

PEelastic = 1.4 × 102 Jx = ±

2PE

kelastic = ± (32.

)

0(1

×.4

10

×41N

0/

2

mJ) = +9.7 × 10−2 m = 9.7 cm

7. m = 51 kg

g = 9.81 m/s2

h = 321 m − 179 m = 142 m

k = 32 N/m

x = 179 m − 104 m = 75 m

PEtot = PEg + PEelastic

Set PEg = 0 J at the river level.

PEg = mgh = (51 kg)(9.81 m/s2)(142 m) = 7.1 × 104 J

PEelastic = 12

kx2 = 12

(32 N/m)(75 m)2 = 9.0 × 104 J

PEtot = (7.1 × 104 J) + (9.0 × 104 J) = 1.6 × 105 J

8. h2 = 4080 m

h1 = 1860 m

m = 905 kg

g = 9.81 m/s2

∆PEg = PEg,2 − PEg,1 = mg(h2 − h1) = (905 kg)(9.81 m/s2)(4080 m − 1860 m)

∆PEg = (905 kg)(9.81 m/s2)(2220 m) = 1.97 × 107 J

9. m = 286 kg

k = 9.50 × 103 N/m

g = 9.81 m/s2

x = 59.0 cm

h1 = 1.70 m

h2 = h1 − x

a. PEelastic = 12

kx2 = 12

(9.50 × 103 N/m)(0.590 m)2 =

b. PEg,1 = mgh1 = (286 kg)(9.81 m/s2)(1.70 m) =

c. h2 = 1.70 m − 0.590 m = 1.11 m

PEg,2 = mgh2 = (286 kg)(9.81 m/s2)(1.11 m) =

d. ∆PEg = PEg,2 − PEg,1 = (3.11 × 103 J) − (4.77 × 103 J) =

The answer in part (d) is approximately equal in magnitude to that in (a); theslight difference arises from rounding. The increase in elastic potential energycorresponds to a decrease in gravitational potential energy; hence the differencein signs for the two answers.

−1.66 × 103 J

3.11 × 103 J

4.77 × 103 J

1.65 × 103 J


II

10. ∆x = 9.50 × 102 m

q = 45.0°

m = 65.0 g

g = 9.81 m/s2

x = 55.0 cm

a. vx = vi(cos q) = ∆∆

x

t

∆t = vi(c

∆o

x

s q)

vertical speed of the arrow for the first half of the flight = vi(sin q) = g∆2

t

vi(sin q) = 2vi(

g

c

∆o

x

s q)

vi = 2(sin g

q∆)(

x

cosq) = = 96.5 m/s

KEi = 12

mvi2 = 1

2(65.0 × 10−3 kg)(96.5 m/s)2 =

b. From the conservation of energy,

PEelastic = KEi

12

kx2 = KEi

k = 2K

x2

Ei = (55.

(

0

2)

×(3

1

0

0

3−2

J)

m)2 =

c. KEi = PEg,max + KEf

KEf = 12

mvx2 = 1

2m[(vi(cos q)]2 = 1

2(65.0 × 10−3 kg)(96.5 m/s)2(cos 45.0°)2 = 151 J

PEg,max = KEi − KEf = 303 J − 151 J = 152 J

hmax = PE

m

g,m

gax =

h = 238 m

152 J(65.0 × 10−3 kg)(9.81 m/s2)

2.00 × 103 N/m

303 J

(9.81 m/s2)(9.50 × 102 m)(2)(sin 45.0°)(cos 45.0°)

Givens Solutions

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ed. 1. m = 118 kg

hi = 5.00 m

g = 9.81 m/s2

vi = 0 m/s

KEf = 4.61 kJ


mghi + 12

mvi2 = mghf + KEf

mghf = mghi + 12

mvi2 − KEf

hf = hi + v

2i

g

2

− K

m

E

g

f = 5.00 m + (2)

(

(

0

9.

m

81

/s

m

)2

/s2) −

(118

4

k

.6

g

1

)(

×9.

1

8

0

1

3

m

J

/s2)

hf = 5.00 m − 3.98 m = 1.02 m above the ground


2. vf = 42.7 m/s

hf = 50.0 m

vi = 0 m

g = 9.81 m/s2


mghi + 12

mvi2 = mghf + 1

2mvf

2

hi = hf + vf

2

2

−g

vi2

= 50.0 m + = 50.0 m + 92.9 m

hi =

The mass of the nut is not needed for the calculation.

143 m

(42.7 m/s)2 − (0 m/s)2

(2)(9.81 m/s2)


II

3. hi = 3150 m

vf = 60.0 m/s

KEi = 0 J

g = 9.81 m/s2


mghi = mghf + 12

mvf2

hf = hi − v

2

f

g

2

= 3150 m − (2

(

)

6

(

0

9

.

.

0

81

m

m

/s

/

)

s

2

2) = 3150 m − 183 m

hf = 2970 m

Givens Solutions

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4. hi = 1.20 × 102 m

hf = 30.0 m

m = 72.0 kg

g = 9.81 m/s2

KEi = 0 J

5. hf = 250.0 m

∆ME = −2.55 × 105 J

m = 250.0 kg

g = 9.81 m/s2

∆ME = PEf − KEi = mghf − 12

mvi2

vi = 2ghf− 2∆m

ME = (2)(9.81 m/s2)(250.0 m)−(2)(−2

2

5.

0

5.

5

0×k g

105

J)

vi =√

4.90 × 103 m2/s2 + 2.04× 103 m2/s2 =√

6.94 × 103 m2/s2

vi = 83.3 m/s = 3.00 × 102 km/h

6. hi = 12.3 km

m = 120.0 g

g = 9.81 m/s2

KEi = 0 J

∆h = hi − hf = 3.2 km


PEi − PEf = KEf

KEf = PEi − PEf = mghi − mghf = mg∆h

KEf = mg∆h = (0.1200 kg)(9.81 m/s2)(3.2 × 103 m) =

PEf = mghf = mg(hi − ∆h) = (0.1200 kg)(9.81 m/s2)(12.3 × 103 m − 3.2 × 103 m)

PEf = (0.1200 kg)(9.81 m/s2)(9.1 × 103 m) =

Alternatively,

PEf = PEi − KEf = mghi − KEf

PEf = (0.1200 kg)(9.81 m/s2)(12.3 × 103 m) − 3.8 × 103 J = 1.45 × 104 J − 3.8 × 103 J

PEf = 1.07 × 104 J

1.1 × 104 J

3.8 × 103 J

7. h = 68.6 m

v = 35.6 m/s

g = 9.81 m/s2

PEf = 0 J

KEi = 0 J

MEi = PEi = mgh

MEf = KEf = 12

mv2

percent of energy dissipated = (MEi −

M

M

E

E

i

f)(100) = (100)

percent of energy dissipated = (100)

percent of energy dissipated = (100)

percent of energy dissipated = (673 J −

6

6

7

3

3

4

J

J)(100) =

(39

6

J

7

)

3

(1

J

00) = 5.8 percent

(9.81 m/s2)(68.6 m) − 12

(35.6 m/s)2

(9.81 m/s2)(68.6 m)

gh − 12

v2

gh

mgh − 12

mv2)

mgh


PEi − PEf = KEf

KEf = ∆PE = mg(hi − hf)

KEf = (72.0 kg)(9.81 m/s2)(1.20 × 102 m − 30.0 m) = (72.0 kg)(9.81 m/s2)(9.0 × 101 m)

KEf =

vf = 2Km

Ef = (2)(6

7

.

24

.0×k 1

g04

J)

vf = 42 m/s

6.4 × 104 J

II

1. P = 56 MW

∆t = 1.0 hW = P∆t = (56 × 106 W)(1.0 h)(3600 s/h) = 2.0 × 1011 J


Givens SolutionsC

opyr

ight

©by

Hol

t, R

ineh

art a

nd W

inst

on.A

ll rig

hts

rese

rved

.


2. ∆t = 62.25 min

P = 585.0 W

W = P∆t = (585.0 W)(62.25 min)(60 s/min) = 2.185 × 106 J

3. h = 106 m

m = 14.0 kg

g = 9.81 m/s2

P = 3.00 × 102 W

q = 0°

W = Fgd(cos q) = Fgd = mgh

∆t = W

P =

m

P

gh = = 48.5 s

(14.0 kg)(9.81 m/s2)(106 m)

3.00 × 102 W

4. P = 2984 W

W = 3.60 × 104 J∆t =

W

P =

3.6

2

0

98

×4

1

W

04 J = 12.1 s

5. ∆t = 3.0 min

W = 54 kJP =

∆W

t = = 3.0 × 102 W

54 × 103 J(3.0 min)(60 s/min)

6. ∆t = 16.7 s

h = 18.4 m

m = 72.0 kg

g = 9.81 m/s2

q = 0°

W = Fgd (cos q) = mgh

P = ∆W

t =

m

∆g

t

h =

P = 778 W

(72.0 kg)(9.81 m/s2)(18.4 m)

16.7 s

Section Two—Problem Workbook Solutions II Ch. 6–1

II

HR

W m

ater

ial c

opyr

ight

ed u

nder

not

ice

appe

arin

g ea

rlier

in th

is b

ook.

6ChapterMomentum and Collisions


Givens Solutions

1. v = 40.3 km/h

p = 6.60 × 102 kg •m/sm =

v

p = = 59.0 kg

6.60 × 102 kg •m/s(40.3 × 103 m/h)(1 h/3600 s)

2. mh = 53 kg

v = 60.0 m/s to the east

ptot = 7.20 × 103 kg •m/s tothe east

ptot = mhv + mpv

mp = ptot −

v

mhv =

mp = = = 67 kg4.0 × 103 kg •m/s

60.0 m/s

7.20 × 103 kg •m/s − 3.2 × 103 kg •m/s

60.0 m/s

7.20 × 103 kg •m/s − (53 kg)(60.0 m/s)

60.0 m/s

3. m1 = 1.80 × 102 kg

m2 = 7.0 × 101 kg

ptot = 2.08 × 104 kg•m/s tothe west

= −2.08 × 104 kg•m/s

v = m1

p

+tot

m2 = =

v = −83.2 m/s = 83.2 m/s to the west

−2.08 × 104 kg•m/s

2.50 × 102 kg

−2.08 × 104 kg•m/s1.80 × 102 kg + 7.0 × 101 kg

4. m = 83.6 kg

p = 6.63 × 105 kg•m/s

v = m

p =

6.63 ×83

1

.

0

6

5

k

k

g

g•m/s = 7.93 × 103 m/s = 7.93 km/s

1. m = 9.0 × 104 kg

vi = 0 m/s

vf = 12 cm/s upward

F = 6.0 × 103 N

∆t = =

∆t = = 1.8 s(9.0 × 104 kg)(0.12 m/s)

6.0 × 103 N

(9.0 × 104 kg)(0.12 m/s) − (9.0 × 104 kg)(0 m/s)

6.0 × 103 N

mvf − mviF


5. m = 6.9 × 107 kg

v = 33 km/h

6. h = 22.13 m

m = 2.00 g

g = 9.81 m/s2

p = mv = (6.9 × 107 kg)(33 × 103 m/h)(1 h/3600 s) = 6.3 × 108 kg •m/s

mgh = 12

mvf2

vf =√

2ghp = mvf = m

√2gh = (2.00 × 10−3 kg)

√(2)(9.81 m/s2)(22.13 m)

p = 4.17 × 10−2 kg •m/s downward


II

HR

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ater

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rlier

in th

is b

ook.

2. m = 1.00 × 106 kg

vi = 0 m/s

vf = 0.20 m/s

F = 12.5 kN

∆p = mvf − mvi = (1.00 × 106 kg)(0.20 m/s) − (1.00 × 106 kg)(0 m/s)

∆p =

∆t = ∆F

p = = 16 s

2.0 × 105 kg •m/s

12.5 × 103 N

2.0 × 105 kg •m/s

Givens Solutions

3. h = 12.0 cm

F = 330 N, upward

m = 65 kg

g = 9.81 m/s2

The speed of the pogo stick before and after it presses against the ground can be de-termined from the conservation of energy.

PEg = KE

mgh = 12

mv2

v = ± √

2gh

For the pogo stick’s downward motion,

vi = − √

2gh

For the pogo stick’s upward motion,

vf = + √

2gh

∆p = mvf − mvi = m√

2gh − m − √

2gh∆p = 2m

√2gh

∆t = ∆F

p = 2m

√

F

2gh =

∆t = 0.60 s

(2)(65 kg)√

(2)(9.81 m/s2)(0.120 m)330 N

4. m = 6.0 × 103 kg

F = 8.0 kN to the east

∆t = 8.0 s

vi = 0 m/s

vf = =

vf = 11 m/s, east

(8.0 × 103 N)(8.0 s) + (6.0 × 103 kg)(0 m/s)

6.0 × 103 kg

F∆t + mvim

5. vi = 125.5 km/h

m = 2.00 × 102 kg

F = −3.60 × 102 N

∆t = 10.0 s

vf =

vf =

vf = = =

or vf = (16.8 × 10−3 km/s)(3600 s/h) = 60.5 km/h

16.8 m/s3.37 103 kg •m/s

2.00 102 kg

−3.60 × 103 N •s + 6.97 × 103 kg •m/s

2.00 × 102 kg

(−3.60 × 102 N)(10.0 s) + (2.00 × 102 kg)(125.5 × 103 m/h)(1 h/3600 s)

2.00 × 102 kg

F∆t + mvim

6. m = 45 kg

F = 1.6 × 103 N

∆t = 0.68 s

vi = 0 m/s

vf = F∆t

m

+ mvi =

vf = = 24 m/s(1.6 × 103 N)(0.68 s)

45 kg

(1.6 × 103 N)(0.68 s) + (45 kg)(0 m/s)

45 kg


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7. m = 4.85 × 105 kg

vi = 20.0 m/s northwest

vf = 25.0 m/s northwest

∆t = 5.00 s

F = =

F = =

F = 4.8 × 105 N northwest

2.4 × 106 kg•m/s

5.00 s

1.21 × 107 kg•m/s − 9.70 × 106 kg•m/s

5.00 s

(4.85 × 105 kg)(25.0 m/s) − (4.85 × 105 kg)(20.0 m/s)

5.00 s

mvf − mvi∆t

Givens Solutions

8. vf = 12.5 m/s upward

m = 70.0 kg

∆t = 4.00 s

vi = 0 m/s

F = = = 219 N

F = 219 N upward

(70.0 kg)(12.5 m/s) − (70.0 kg)(0 m/s)

4.00 s

mvf − mvi∆t

9. m = 12.0 kg

h = 40.0 m

∆t = 0.250 s

vf = 0 m/s

g = 9.81 m/s2

From conservation of energy, vi = −√

2gh

∆p = mvf − mvi = mvf − m– √

2gh∆p = (12.0 kg)(0 m/s) + (12.0 kg)

√(2)(9.81 m/s2)(40.0m) = 336 kg •m/s

F = ∆∆t

p = = 1340 N = 1340 N upward

336 kg •m/s

0.250 s

1. F = 2.85 × 106 N backward= −2.85 × 106 N

m = 2.0 × 107 kg

vi = 3.0 m/s forward = +3.0 m/s

vf = 0 m/s

∆t = 21 s

∆p = F∆t = (−2.85 × 106 N)(21 s)

∆p =

∆x = 12

(vi + vf)∆t = 12

(3.0 m/s + 0 m/s)(21 s) = 32 m forward

−6.0 × 107 kg•m/s forward or 6.0 × 107 kg•m/s backward


2. m = 6.5 × 104 kg

F = −1.7 × 106 N

vi = 1.0 km/s

∆t = 30.0 s

∆p = F∆t = (−1.7 × 106 N)(30.0 s) =

vf = ∆p +

m

mvi =

vf = = = 220 m/s

∆x = 12

(vi + vf)∆t = 12

(1.0 × 103 m/s + 220 m/s)(30.0 s) = 12

(1.2 × 103 m/s)(30.0 s)

∆x = 1.8 × 104 m = 18 km

1.4 × 107 kg•m/s

6.5 × 104 kg

−5.1 × 107 kg•m/s + 6.5 × 107 kg•m/s

6.5 × 104 kg

−5.1 × 107 kg•m/s + (6.5 × 104 kg)(1.0 × 103 m/s)

6.5 × 104 kg

−5.1 × 107 kg•m/s

Givens Solutions


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3. m = 2.03 × 104 kg

vi = 5.00 m/s to the east = 5.00 m/s

∆t = 20.3 s

F = 1.20 × 103 N to the west

∆p = F∆t = (−1.20 × 103 N)(20.3 s) =

vf = ∆p +

m

mvi =

vf = =

vf = 3.73 m/s

∆x = 12

(vi + vf)∆t = 12

[5.00 m/s + (3.73 m/s)](20.3 s) = 12

(8.73 m/s)(20.3 s)

∆x = 88.6 m = 88.6 m to the east

7.58 × 104 kg•m/s

2.03 × 104 kg

−2.44 × 104 kg•m/s + 1.02 × 105 kg•m/s

2.03 × 104 kg

−2.44 × 104 kg•m/s + (2.03 × 104 kg)(5.00 m/s)

2.3 × 104 kg

2.44 × 104 kg•m/s to the west

4. m = 113 g

vi = 2.00 m/s to the right

vf = 0 m/s

∆t = 0.80 s

F = mvf

∆−t

mvi = =

F = −0.28 N =

∆x = 12

(vi + vf)∆t = 12

(2.00 m/s + 0 m/s)(0.80 s)

∆x = 0.80 m to the right

0.28 N to the left

−(0.113 kg)(2.00 m/s)

0.80 s

(0.113 kg)(0 m/s) − (0.113 kg)(2.00 m/s)

0.80 s

6. h = 68.6 m

m = 1.00 × 103 kg

F = −2.24 × 104 N

g = 9.81 m/s2

vf = 0 m/s

From conservation of energy,

vi =√

2gh

∆p = mvf − mvi = mvf − m√

2gh

∆t = ∆F

p = mvf −

F

m√

2gh =

∆t = =

∆x = 12

(vi + vf)∆t = 12

(√

2gh + vf)∆t

∆x = 12

√

(2)(9.81 m/s2)(68.6m) + 0 m/s(1.64 s) = 30.1 m

1.64 s−(1.00 × 103 kg)

√(2)(9.81 m/s2)(68.6m)

−2.24 × 104 N

(1.00 × 103 kg)(0 m/s) − (1.00 × 103 kg)√

(2)(9.81. m/s2)(68.6m)−2.24 × 104 N

5. m = 4.90 × 106 kg

vi = 0.200 m/s

vf = 0 m/s

∆t = 10.0 s

F = ∆∆

p

t =

mvf

∆−t

mvi =

F = –9.80 104 N

F =

∆x = 12

(vi + vf)∆t = 12

(0.200 m/s + 0 m/s)(10.0 s)

∆x = 1.00 m

9.80 × 104 N opposite the palace’s direction of motion

–(4.90 106 kg)(0.200 m/s)

10.0 s

(4.90 × 106 kg)(0 m/s) − (4.90 × 106 kg)(0.200 m/s)

10.0 s


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7. m = 100.0 kg

vi = 4.5 × 102 m/s

vf = 0 m/s

F = −188 N

∆t = mvf

F

− mvi =

∆t = =

∆x = 12

(vi + vf)∆t = 12

(4.5 × 102 m/s + 0 m/s)(240 s) = 5.4 × 104 m = 54 km

The tunnel is long.54 km

240 s−(100.0 kg)(4.5 × 102 m/s)

−188 N

(100.0 kg)(0 m/s) − (100.0 kg)(4.5 × 102 m/s)

−188 N

Givens Solutions

1. m1 = 3.3 × 103 kg

v1,i = 0 m/s

v2,i = 0 m/s

v2,f = 2.5 m/s to the right = 2.5 m/s

v1,f = 0.050 m/s to the left = –0.050 m/s

Because the initial momentum is zero, the final momentum is also zero, and so

m2 = −m

v2

1v

,f

1,f = = 66 kg−(3.3 × 103 kg)(−0.050 m/s)

2.5 m/s

2. m1 = 1.25 × 103 kg

v1,i = 0 m/s

v2,i = 0 m/s

v2,f = 1.40 m/s backward = –1.40 m/s

∆t1 = 4.0 s

∆x1 = 24 cm forward= 24 cm

Because the initial momentum is zero, the final momentum is also zero, and so

v1,f = ∆∆

x

t1

1 = 0

4

.2

.0

4

s

m = 0.060 m/s forward

m2 = −m

v2

1v

,f

1,f = = 54 kg−(1.25 × 103 kg)(0.060 m/s)

−1.40 m/s

3. m1 = 114 kg

v2, f = 5.32 m/s backward = −5.32 m/s

v1, f = 3.41 m/s forward = +3.41 m/s

m2 = 60.0 kg

m1vi + m2vi = m1v1, f + m2v2, f

vi = m1v

m1,

1

f

++

m

m

2

2v2, f =

vi = = 7.0 ×

1

1

7

0

4

1

k

k

g

g•m/s = 0.40 m/s

vi = 0.40 m/s forward

389 kg•m/s − 319 kg•m/s

174 kg

(114 kg)(3.41 m/s) + (60.0 kg)(−5.32 m/s)

114 kg + 60.0 kg

4. m1 = 5.4 kg

v1, f = 7.4 m/s forward = +7.4 m/s

v2, f = 1.4 m/s backward = −1.4 m/s

m2 = 50.0 kg

m1vi + m2vi = m1v1, f + m2v2, f

vi = =

vi = = −3.0 ×

55

1

.

0

4

1

k

k

g

g•m/s = –0.54 m/s

vi = 0.54 m/s backward

4.0 × 101 kg•m/s − 7.0 × 101 kg•m/s

55.4 kg

(5.4 kg)(7.4 m/s) + (50.0 kg)(−1.4 m/s)

5.4 kg + 50.0 kg

m1v1, f + m2v2, fm1 + m2


5. m1 = 3.4 × 102 kg

v2, f = 9.0 km/h northwest= −9.0 km/h

v1, f = 28 km/h southeast= +28 km/h

m2 = 2.6 × 102 kg

m1vi + m2vi = m1v1, f + m2v2, f

vi = =

vi = =

vi = 12 km/h to the southeast

7.2 × 103 kg•km/h

6.0 × 102 kg

9.5 × 103 kg•km/h − 2.3 × 103 kg•km/h

6.0 × 102 kg

(3.4 × 102 kg)(28 km/h) + (2.6 × 102 kg)(−9.0 km/h)

3.4 × 102 kg + 2.6 × 102 kg

m1v1, f + m2v2, fm1 + m2

Givens Solutions

6. mi = 3.6 kg

m2 = 3.0 kg

v1,i = 0 m/s

v2,i = 0 m/s

v2,f = 2.0 m/s to the left= –2.0 m/s

Because the initial momentum is zero, the final momentum must also equal zero.

mi v1,f = −m2v2,f

v1,f = −m

m2v

1

2,f = = 1.7 m/s = 1.7 m/s to the right−(3.0 kg)(−2.0 m/s)

3.6 kg

7. m1 = 449 kg

v1,i = 0 m/s

v2,i = 0 m/s

v2,f = 4.0 m/s backward= –4.0 m/s

m2 = 60.0 kg

∆t = 3.0 s

Because the initial momentum is zero, the final momentum must also equal zero.

v1,f = −m

m2

1

v2,f = = 0.53 m/s = 0.53 m/s forward

∆x = v1,f∆t = (0.53 m/s)(3.0 s) = 1.6 m forward

−(60.0 kg)(−4.0 m/s)

449 kg


1. m1 = 155 kg

v1,i = 6.0 m/s forward

v2,i = 0 m/s

vf = 2.2 m/s forward

m2 = =

m2 = =

m2 = 270 kg

590 kg•m/s

2.2 m/s

930 kg•m/s –340 kg•m/s

2.2 m/s

(155 kg)(6.0 m/s) − (155 kg)(2.2 m/s)

2.2 m/s − 0 m/s

m1v1,i − m1vfvf − v2,i


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2. v1,i = 10.8 m/s

v2,i = 0 m/s

vf = 10.1 m/s

m1 = 63.0 kg

m2 = =

m2 = = = 4.4 kg44 kg •m/s10.1 m/s

6.80 × 102 kg •m/s − 6.36 × 102 kg •m/s

10.1 m/s

(63.0 kg)(10.8 m/s) − (63.0 kg)(10.1 m/s)

10.1 m/s − 0 m/s


3. v1, i = 4.48 m/s to the right

v2, i = 0 m/s

vf = 4.00 m/s to the right

m2 = 54 kg

m1 = = (54 k

0

g

.

)

4

(

8

4

m

.00

/s

m/s)

m1 = 450 kg

(54 kg)(4.00 m/s) − (54 kg)(0 m/s)

4.48 m/s − 4.00 m/s


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6. m1 = 9.50 kg

v1, i = 24.0 km/h to the north

m2 = 32.0 kg

vf = 11.0 km/h to the north

v2, i =

v2, i =

v2, i = =

= 228

32

k

.

g

0

•k

k

m

g

/h

v2, i = 7.12 km/h to the north

456 kg•km/h − 228 kg•km/h

32.0 kg

(41.5 kg)(11.0 km/h) − 228 kg•km/h

32.0 kg

(9.5 kg + 32.0 kg)(11.0 km/h) − (9.50 kg)(24.0 km/h)

32.0 kg

(m1 + m2)vf − m1v,1m2

7. m1 = m2

v1,i = 89 km/h

v2,i = 69 km/h

Because m1 = m2, vf = v1,i +

2

v2,i = = 158 k

2

m/h = 79 km/h

vf = 79 km/h

89 km/h + 69 km/h

2

8. m1 = 3.0 × 103 kg

m2 = 2.5 × 102 kg

v2,i = 3.0 m/s down = –3.0 m/s

v1,i = 1.0 m/s up = +1.0 m/s

vf = =

vf = =

vf = 0.69 m/s = 0.69 m/s upward

2.2 103 kg•m/s

3.2 103 kg

3.0 103 kg•m/s –7.5 102 kg•m/s

3.2 103 kg

(3.0 × 103 kg)(1.0 m/s) + (2.5 × 102 kg)(−3.0 m/s)

(3.0 × 103 kg) + (2.5 × 102 kg)

m1v1,i + m2v2,im1 + m2

9. m1 = (2.267 × 103 kg) +(5.00 × 102 kg) = 2.767 ×103 kg

m2 = (1.800 × 103 kg) +(5.00 × 102 kg) = 2.300 ×103 kg

v1,i = 2.00 m/s to the left = –2.00 m/s

v2,i = 1.40 m/s to the right= +1.40 m/s

vf = =

vf = = –23

5

1

0

0

67

kg

k

•

g

m/s = –0.456 m/s

vf = 0.456 m/s to the left

–5.53 103 kg •m/s + 3220 kg •m/s

5.067 × 103 kg

(2.767 × 103 kg)(−2.00 m/s) + (2.300 × 103 kg)(1.40 m/s)

2.767 × 103 kg + 2.300 × 103 kg

m1v1,i + m2v2,im1 + m2

4. m1 = 28 × 103 kg

m2 = 12 × 103 kg

v1,i = 0 m/s

vf = 3.0 m/s forward

v2,i =

v2,i =

v2,i =

v2,i = 1.0 × 101 m/s forward

(4.0 104 kg)(3.0 m/s)

12 103 kg

(28 × 103 kg + 12 × 103 kg)(3.0 m/s) − (28 × 103 kg)(0 m/s)

12 × 103 kg

(m1 + m2)vf − m1v1,im2

5. m1 = 227 kg

m2 = 267 kg

v1,i = 4.00 m/s to the left= –4.00 m/s

vf = 0 m/s

v2,i =

v2,i = = 3.40 m/s

v2,i = 3.40 m/s to the right

(227 kg + 267 kg)(0 m/s) − (227 kg)(–4.00 m/s)

267 kg

(m1 + m2)vf − m1v1,im2

Givens Solutions


Givens Solutions

1. m1 = 2.0 g

v1,i = 2.0 m/s forward = +2.0 m/s

m2 = 0.20 g

v2,i = 8.0 m/s backward = −8.0 m/s forward

vf =

vf =

vf =

vf = 2.4

2

.2

10–

1

3

0

k–3

g•

k

m

g

/s = 1.1 m/s forward

KEi = 12

m1v1,i2 + 1

2m2v2,i

2

KEi = 2

1(2.0 × 10−3 kg)(2.0 m/s)2 + 2

1(0.20 × 10−3 kg)(−8.0 m/s)2

KEi = 4.0 × 10−3 J + 6.4 × 10−3 J = 1.04 × 10−3 J

KEf = 2

1(m1 + m2)vf2 =

2

1(2.0 × 10−3 kg + 0.20 × 10−3 kg)(1.1 m/s)2

KEf = 2

1(2.2 × 10−3 kg)(1.1 m/s)2

∆KE = KEf − KEi = 1.3 × 10−3 J − 1.04 × 10−2 J = −9.1 × 10−3 J

fraction of total KE dissipated = ∆K

K

E

E

i = = 0.88

9.1 × 10−3 J1.04 × 10−2 J

4.0 10–3 kg•m/s − 1.6 10–3 kg•m/s

2.2 10–3 kg

(2.0 10–3 kg)(2.0 m/s) + (0.20 10–3 kg)(−8.0 m/s)

2.0 10–3 kg + 0.20 10–3 kg

m1v1,i + m2v2,im1 + m2


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2. m1 = 313 kg

v1,i = 6.00 m/s away fromshore

v2,i = 0 m/s

vf = 2.50 m/s away fromshore

m2 = =

m2 = = = 4.4 × 102 kg

KEi = 2

1m1v1,i2 +

2

1m2v2,i2

KEi = 2

1(313 kg)(6.00 m/s)2 + 2

1(4.40 × 102 kg)(0 m/s)2 = 5630 J

KEf = 2

1(m1 + m2)vf2

KEf = 2

1(313 kg + 4.40 × 102 kg)(2.50 m/s)2 = 2

1(753 kg)(2.50 m/s)2 = 2350 J

∆KE = KEf − KEi = 2350 J − 5630 J = −3280 J

1.10 × 103 kg •m/s

2.50 m/s

1880 kg •m/s − 782 kg •m/s

2.50 m/s

(313 kg)(6.00 m/s) − (313 kg)(2.50 m/s)

2.50 m/s − 0 m/s


3. m1 = m2 = 111 kg

v1, i = 9.00 m/s to the right= +9.00 m/s

v2, i = 5.00 m/s to the left= −5.00 m/s

vf = m1v

m1,

1

i ++

m

m2

2

v2, i =

vf = = 44

2

4

2

k

2

g

k

•m

g

/s = 2.00 m/s to the right

KEi = 2

1m1v1,i2 +

2

1m2v2,i2 =

2

1(111 kg)(9.00 m/s)2 + 2

1(111 kg)(–5.00 m/s)2

KEi = 4.50 × 103 J + 1.39 × 103 J = 5.89 × 103 J

KEf = 12

(m1 + m2)vf2 =

2

1(111 kg 111 kg)(2.00 m/s)2 = 2

1(222 kg)(2.00 m/s)

KEf = 444 J

∆KE = KEf − KEi = 444 J − 5.89 × 103 J = −5450 J

999 kg•m/s − 555 kg•m/s

222 kg

(111 kg)(9.00 m/s) + (111 kg)(−5.00 m/s)

111 kg + 111 kg


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Givens Solutions

5. m1 = 4.00 × 105 kg

v1, i = 32.0 km/h

m2 = 1.60 × 105 kg

v2, i = 45.0 km/h

vf = m1v

m1,

1

i ++

m

m2

2v2,i =

vf = =

vf = 35.7 km/h

KEi = 12

m1v1,i2 + 1

2m2v2,i

2

KEi = 12

(4.00 × 105 kg)(32.0 × 103 m/h)(1 h/3600 s)2 + 12

(1.60 × 105 kg)

(45.0 × 103 m/h)2(1 h/3600 s)2

KEi = 1.58 × 107 J + 1.25 × 107 J = 2.83 × 107 J

KEf = 12

(m1 + m2)vf2

KEf = 12

(4.00 × 105 kg + 1.60 × 105 kg)(35.7 × 103 m/h)2(1 h/3600 s)2

= 12

(5.60 × 105 kg)(35.7 × 103 m/h)2(1 h/3600 s)2

KEf = 2.75 × 107 J

∆KE = KEf − KEi = 2.75 × 107 J − 2.83 × 107 J = −8 × 105 J

2.00 × 107 kg•km/h

5.60 × 105 kg

1.28 × 107 kg•km/h + 7.20 × 106 kg•km/h

5.60 × 105 kg

(4.00 × 105 kg)(32.0 km/h) + (1.60 × 105 kg)(45.0 km/h)

4.00 × 105 kg + 1.60 × 105 kg

4. m1 = m2 = 60.0 kg + 50.0 kg= 110.0 kg

v1, i = 106.0 km/h to the east= +106.0 km/h

v2, i = 75.0 km/h to the west= −75.0 km/h

vf = =

vf = = 3.41 ×

22

1

0

0

.

3

0

k

k

g

g

•km/h

vf = 15.5 km/h to the east

KEi = 12

m1v1, i2 + 1

2m2v2, i

2

KEi = 12

(110.0 kg)(106.0 × 103 m/h)2(1 h/3600 s)2 + 12

(110.0 kg)(−75.0 × 103 m/h)2

(1 h/3600 s)2

KEi = 4.768 × 104 J + 2.39 × 104 J = 7.16 × 104 J

KEf = 12

(m1 + m2)vf2

KEf = 12

(110.0 kg + 110.0 kg)(15.5 × 103 m/h)2(1 h/3600 s)2

= 12

(220.0 kg)(15.5 × 103 m/h)2(1 h/3600 s)2

KEf = 2.04 × 103 J

∆KE = KEf − KEi = 2.04 × 103 J − 7.16 × 104 J = −6.96 × 104 J

1.166 × 104 kg•km/h − 8.25 × 103 kg•km/h

220.0 kg

(110.0 kg)(106.0 km/h) + (110.0 kg)(−75.0 km/h)

110.0 kg + 110.0 kg

m1v1,i + m2v2,im1 + m2

Givens Solutions


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6. m1 = 21.3 kg

v1,i = 0 m/s

m2 = 1.80 × 10−1 kg

vf = 6.00 × 10−2 m/s

v2,i = (m1 + m

m

2)

2

vf –m1v1,i

v2,i =

v2,i = 7.17 m/s

KEi = 12

m1v1,i2 + 1

2m2v2,i

2

KEi = 12

(21.3 kg)(0 m/s)2 + 12

(1.80 × 10−1 kg)(7.17 m/s)2

KEi = 0 J 4.63 J = 4.63 J

KEf = 12

(m1 + m2)vf2

KEf = 12

(21.3 kg 0.180 kg)(6.00 × 10−2 m/s)2 = 12

(21.5 kg)(6.00 × 10−2 m/s)2

KEf = 3.87 × 10−2 J

∆KE = KEf − KEi = 3.87 × 10−2 J − 4.63 J = −4.59 J

(21.5 kg)(6.00 10−2 m/s)

1.80 10–1 kg

(21.3 kg + 1.80 × 10−1 kg)(6.00 × 10−2 m/s) – (21.3 kg)(0 m/s)

1.80 × 10−1 kg

7. m1 = 122 g

m2 = 96.0 g

v2,i = 0 m/s

Because v2,i = 0 m/s, m1v1,i = (m1 + m2)vf

v1,i =

fraction of KE dissipated = ∆K

K

E

E

i =

KEf

K

−Ei

KEi =

fraction of KE dissipated =


m1vf2 + m2vf

2 − (m1vf )2 + 2m1m2vf2 + (m2vf )2

m1

(m1 + m2)vf2 − m1(m1 +

m

m

1

2)vf2

m1(m1 +m

m

1

2)vf2

12

m1 + m2vf2 − 1

2m1v1,i

2

12

m1v1,i2

(m1 + m2)vfm1

(m1vf )2 + 2m1m2vf

2 + (m2vf )2

m1


fraction of KE dissipated = =

fraction of KE dissipated = = =

The fraction of kinetic energy dissipated can be determined without the initial veloc-ity because this value cancels, as shown above. The initial velocity is needed to findthe decrease in kinetic energy.

−0.440−171.5 g

3.90 × 102 g

−96.0 g − 75.5 g122 g + 192 g + 75.5 g

−96.0 g − (9

1

6

2

.0

2

g

g

)2

122 g + (2)(96.0 g) + (9

1

6

2

.0

2

g

g

)2

−m2 − m

m2

1

2

m1 + 2m2 + m

m2

1

2

vf2m1 + m2 − m1 − 2m2 −

m

m2

1

2

vf2m1 + 2m2 +

m

m2

1

2


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Additional Practice 6G

Givens Solutions

2. m1 = 18.40 kg

m2 = 56.20 kg

v2, i = 5.000 m/s to the left = −5.000 m/s

v2, f = 6.600 × 10−2 m/s to the left

= − 6.600 × 10−2 m/s

v1, f = 10.07 m/s to the left= −10.07 m/s

Momentum conservation

m1v1, i + m2v2, i = m1v1, f + m2v2, f

v1, i =

v1, i =

v1, i = = 92

1

.

8

0

.4

k

0

g•

k

m

g

/s = 5.00 m/s

v1, i =

Conservation of kinetic energy (check)

12

m1v1, i2 + 1

2m2v2, i

2 = 12

m1v1, f2 + 1

2m2v2, f

2

12

(18.40 kg)(5.00 m/s)2 + 12

(56.20)(−5.000 m/s)2 = 12

(18.40 kg)(−10.07 ms)2

+ 12

(56.20 kg)(−6.600 × 10−2 m/s)2

2.30 × 102 J + 702.5 J = 932.9 J + 0.1224 J

932 J = 933 J

The slight difference arises from rounding.


−185.3 kg•m/s − 3.709 kg•m/s + 281.0 kg•m/s

18.40 kg

(18.40 kg)(−10.07 m/s) + (56.20 kg)(−6.600 × 10−2 m/s) − (56.20 kg)(−5.000 m/s)

18.40 kg

m1v1, f + m2v2, f − m2v2, im1

1. m2 = 0.500 m1

v1, i = 3.680 × 103 km/h

v1, f = −4.40 × 102 km/h

v2, f = 5.740 × 103 km/h


m1v1, i + m2v2, i = m1v1, f + m2v2, f

v2, i = =

v2, i = (2.00)v1, f + v2, f − (2.00)v1, i = (2.00)(−4.40 × 102 km/h) + 5.740 × 103 km/h − (2.00)(3.680 × 103 km/h) = −8.80 × 102 km/h + 5.740 × 103 km/h − 7.36 × 103 km/h

v2, i =


12

m1v1, i2 + 1

2m2v2, i

2 = 12

m1v1, f2 + 1

2m2v2, f

2

12

m1v1, i2 + 1

2(0.500)m1v2, i

2 = 12

m1v1, f2 + 1

2(0.500)m1v2, f

2

v1, i2 + (0.500)v2, i

2 = v1, f2 + (0.500)v2, f

2

(3.680 × 103 km/h)2 + (0.500)(−2.50 × 103 km/h)2 = (−4.40 × 102 km/h)2 + (0.500)(5.740 × 103 km/h)2

1.354 × 107 km2/h2 + 3.12 × 106 km2/h2 = 1.94 × 105 km2/h2 + 1.647 × 107 km2/h2

1.666 × 107 km2/h2 = 1.666 × 107 km2/h2

−2.50 × 103 km/h

m1v1, f + (0.500)m1v2, f − m1v1, i(0.500)m1

m1v1, f + m2v2, f − m1v1, im2

Givens Solutions


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3. m1 = m2

v1, i = 5.0 m/s to the right

= +5.0 m/s

v1, f = 2.0 m/s to the left

= −2.0 m/s

v2, f = 5.0 m/s to the right

= +5.0 m/s


m1v1, i + m2v2, i = m1v1, f + m2v2, f

v2, i = = v1, f + v2, f − v1, i

v2, i = −2.0 m/s + 5.0 m/s − 5.0 m/s = −2.0 m/s

v2, i =


12

miv1, i2 + 1

2m2v2, i

2 = 12

m1v1, f2 + 1

2m2v2, f

2

v1,i2 + v2,i

2 = v1, f2 + v2, f

2

(5.0 m/s)2 + (−2.0 m/s)2 = (−2.0 m/s)2 + (5.0 m/s)2

25 m2/s2 + 4.0 m2/s2 = 4.0 m2/s2 + 25 m2/s2

29 m2/s2 = 29 m2/s2

2.0 m/s to the left

m1v1, f + m2v2, f − m1v1, im2

4. m1 = 45.0 g

v1, i = 273 km/h to the right

= +273 km/h

v2, i = 0 km/h

v1, f = 91 km/h to the left

= −91 km/h

v2, f = 182 km/h to the right

= +182 km/h


m1v1, i + m2v2, i = m1v1, f + m2v2, f

m2 = m1v

v1

2

,

,

f

i −−

v

m

2

1

,

v

f

1, i =

m2 = =

m2 =


12

m1v1,i2 + 1

2m2v2,i

2 = 12

m1v1, f2 + 1

2m2v2, f

12

(45.0 g)(273 103 m/h)2(1 h/3600 s)2 + 12

(90.1 g)(0 m/s)2

= 12

(45.0 g)(−91 × 103 m/h)2(1 h/3600 s)2 + 12

(90.1 g)(182 × 103 m/h)2 (1 h/3600 s)2

129 J + 0 J = 14 J + 115 J

129 J = 129 J

90.1 g

−16.4 103 g •km/h

−182 km/h

−4.1 103 g •km/h − 12.3 103 g •km/h

−182 km/h

(45.0 g)(−91 km/h) − (45.0 g)(273 km/h)

0 km/h − 182 km/h

5. v1,i = 185 km/h to the right 185 km/h

v2,i = 0 km/h

vi,f = 80.0 km/h to the left = −80.0 km/h

m1 = 5.70 10–2 kg


m1v1, i + m2v2,i = m1v1,f + m2v2,f

m

m1

2v1,i −

m

m1

2v1,f = v2,f –v2,i

m

m1

2 [185 km/h − (−80.0 km/h)] = v2,f − 0 km/h

v2,f = m

m1

2 (265 km/h) to the right

Conservation of kinetic energy

KEi = 12

m1v1,i2 + 1

2m2v2,i

2 = 12

m1v1,i2

KEf = 12

m1v1, f2 + 1

2m2v2,f

2

12

m1v1,i2 = 1

2m1v1,f

2 + 12

m2v2,f2

m

m1

2(v1,i)

2 = m

m1

2 (v1,f)

2 + v2,f2

m

m1

2(185 km/h)2 =

m

m1

2 (−80.0 km/h)2 + v2,f

2

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Givens Solutions

m

m1

2(3.42× 104 km2/h2) −

m

m1

2(6.40× 103 km2/h2) = v2,f

v2,f = m

m1

2(2.78× 104km2/h2) =

m

m1

2 167 km/h

Equating the two results for v2,f yields the ratio of m1 to m2.

m

m1

2 (265 km/h) =

m

m1

2 (167 km/h)

265 km/h = m

m2

1 (167 km/h)

m

m2

1 = 216

6

5

7

k

k

m

m

/

/

h

h

2= 2.52

m2 = (2.52) m1 (2.52)(5.70 10–2 kg)

m2 = 0.144 kg

6. m1 = 4.00 × 105 kg

m2 = 1.60 × 105 kg

v1, i = 32.0 km/h to the right

v2, i = 36.0 km/h to the right

v1, f = 35.5 km/h to the right


m1v1, i + m2v2, i = m1v1, f + m2v2, f

v2, f =

v2, f =

v2, f =

v2, f = 4.4

1

×.6

1

0

0

×

6

1

k

0

g5•k

k

m

g

/h

v2, f =


12

m1v1, i2 + 1

2m2v2, i

2 = 12

m1v1, f2 + 1

2m2v2, f

2

12

(4.00 × 105 kg)(32.0 × 103 m/h)2(1 h/3600 s)2 + 12

(1.60 × 105 kg)(36.0 × 103 m/h)2

(1 h/3600 s)2 = 12

(4.00 × 105 kg)(35.5 × 103 m/h)2(1 h/3600 s)2 + 12

(1.60 × 105 kg)

(28 103 m/h)2(1 h/3600 s)2

1.58 × 107 J + 8.00 × 106 J = 1.94 × 107 J + 4.8 × 106 J

2.38 × 107 J = 2.42 × 107 J


28 km/h to the right

1.28 × 107 kg•km/h + 5.76 × 106 kg•km/h − 1.42 × 107 kg•km/h

1.60 × 105 kg

(4.00 × 105 kg)(32.0 km/h) + (1.60 × 105 kg)(36.0 km/h) − (4.00 × 105 kg)(35.5 km/h)

1.60 × 105 kg

m1v1, i + m2v2, i − m1v1, fm2

7. m1 = 5.50 × 105 kg

m2 = 2.30 × 105 kg

v1, i = 5.00 m/s to the right= +5.00 m/s

v2, i = 5.00 m/s to the left= −5.00 m/s

v2, f = 9.10 m/s to the right= +9.10 m/s


m1v1,i + m2v2,i = m1v1,f + m2v2,f

v1,f =

v1,f =

v1,f = = −0.89 m/s right

v1,f =


12

m1v1, i2 + 1

2m2v2, i

2 = 12

m1v1, f2 + 1

2m2v2, f

2

12

(5.50 × 105 kg)(5.00 m/s)2 + 12

(2.30 × 105 kg)(−5.00 m/s)2 = 12

(5.50 × 105 kg)(−0.89 m/s)2 + 1

2(2.30 × 105 kg)(9.10 × m/s)2

6.88 × 106 J + 2.88 × 106 J = 2.2 × 105 J + 9.52 × 106 J

9.76 × 106 J = 9.74 × 106 J


0.89 m/s left

2.75 106 kg•m/s – 1.15 106 kg•m/s –2.09 × 106 kg•m/s

5.50 × 105 kg

(5.50 × 105 kg)(5.00 m/s) + (2.30 × 105 kg)(−5.00 m/s) − (9.10 m/s)

5.50 × 105 kg

m1v1,i + m2v2,i − m2v2,fm1

Givens Solutions


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Chapter 7Rotational Motion and the Law of Gravity

II

1. r = 10.0 km

∆q = +15.0 rad∆s = r∆q = (10.0 km)(15.0 rad) =

The particle moves in the positive, or , direction around the neutron star’s “north” pole.

counterclockwise

1.50 × 102 km


Givens Solutions

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2. ∆q = 3(2p rad)

r = 6560 km

∆s = r∆q = (6560 km)[(3)(2p rad)] = 1.24 × 105 km

3. r = 1.40 ×

2

105 km

= 7.00 × 104 km

∆q = 1.72 rad

rE = 6.37 × 103 km

a. ∆s = r∆q = (7.00 × 104 km)(1.72 rad) =

b. ∆qE = ∆rE

s = = 3.00 rev, or 3.00 orbits

(1.20 × 105 km)(1 rev/2p rad)

6.37 × 103 km

1.20 × 105 km

4. ∆q = 225 rad

∆s = 1.50 × 106 kmr =

∆∆

qs

= 1.50

22

×5

1

r

0

a

6

d

km = 6.67 × 103 km

5. r = 5.8 × 107 km

∆s = 1.5 × 108 km∆q =

∆r

s =

1

5

.

.

5

8

××

1

1

0

0

8

7k

k

m

m = 2.6 rad

6. ∆s = −1.79 × 104 km

r = 6.37 × 103 km

∆q = ∆r

s =

−6

1

.3

.7

7

9

××

1

1

0

03

4

k

k

m

m = −2.81 rad


1. r = 1.82 m

wavg = 1.00 × 10−1 rad/s

∆t = 60.0 s

∆q = wavg∆t = (1.00 × 10−1 rad/s)(60.0 s) =

∆s = r∆q = (1.82 m)(6.00 rad) = 10.9 m

6.00 rad

2. ∆t = 120 s

wavg = 0.40 rad/s∆q = wavg∆t = (0.40 rad/s)(120 s) = 48 rad

3. r = 30.0 m

∆s = 5.0 × 102 m

∆t = 120 s

wavg = ∆∆qt

= r

∆∆s

t =

(3

5

0

.

.

0

0

×m

1

)(

0

1

2

2

m

0 s) = 0.14 rad/s


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4. ∆q = 16 rev

∆t = 4.5 minwavg =

∆∆qt

= = 0.37 s(16 rev)(2p rad/rev)(4.5 min)(60 s/min)

Givens Solutions

5. wavg = 2p rad/24 h

∆q = 0.262 rad∆t =

w∆

a

q

vg = = 1.00 h

0.262 rad

22

p4

r

h

ad

6. r = 2.00 m

∆s = 1.70 × 102 km

wavg = 5.90 rad/s

∆t = w∆

a

q

vg =

rw∆

a

s

vg = = 1.44 × 104 s = 4.00 h

1.70 × 105 m(2.00 m)(5.90 rad/s)

1. aavg = 2.0 rad/s2

w1 = 0 rad/s

w2 = 9.4 rad/s

∆t = w2

a−

avg

w1

∆t =

∆t = 4.7 s

9.4 rad/s − 0.0 rad/s

2.0 rad/s2

2. ∆tJ = 9.83 h

aavg = −3.0 × 10−8 rad/s2

w2 = 0 rad/s

w1 = ∆∆

t

q

J =

(9.83 h

2p)(3

ra

6

d

00 s/h) = 1.78 × 10−4 rad/s

∆t = w2

a−

avg

w1 =

∆t = 5.9 × 103 s

0.00 rad/s − 1.78 × 10−4 rad/s

−3.0 × 10−8 rad/s2

3. w1 = 2.00 rad/s

w2 = 3.15 rad/s

∆t = 3.6 s

aavg = w2

∆−t

w1 = = 1.1

3

5

.6

ra

s

d/s

aavg = 0.32 rad/s2

3.15 rad/s − 2.00 rad/s

3.6 s


4. w1 = 8.0 rad/s

w2 = 3w1 = 24 rad/s

∆t = 25 s

aavg = w2

∆−t

w1 = 24 rad/s

2

−5

8

s

.0 rad/s =

16

2

r

5

ad

s

/s

aavg = 0.64 rad/s2

5. ∆t1 = 365 days

∆q1 = 2p rad

aavg = 6.05 × 10−13 rad/s2

∆t2 = 12.0 days

w1 = ∆∆qt2

1 = = 1.99 × 10−7 rad/s

w2 = w1 + aavg∆t2 = 1.99 × 10−7 rad/s + (6.05 × 10−13 rad/s2)(12.0 days)(24 h/day)(3600 s/h)

w2 = 1.99 × 10−7 rad/s + 6.27 × 10−7 rad/s = 8.26 × 10−7 rad/s

2p rad(365 days)(24 h/day)(3600 s/h)

6. w1 = 0 rad/s

aavg = 0.800 rad/s2

∆t = 8.40 s

aavg = w2

∆−t

w1

w2 = w1 + aavg∆t

w2 = 0 rad/s + (0.800 rad/s2)(8.40 s)

w2 = 6.72 rad/s


II


Givens SolutionsC

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©by

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nd W

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hts

rese

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.

1. wi = 5.0 rad/s

a = 0.60 rad/s2

∆t = 0.50 min

wf = wi + a∆t

wf = 5.0 rad/s + (0.60 rad/s2)(0.50 min)(60.0 s/min)

wf = 5.0 rad/s + 18 rad/s

wf = 23 rad/s

2. a = 1.0 × 10−10 rad/s2

∆t = 12 h

wi = 27

2

.

p3

r

d

a

a

d

ys

wi = 27

2

.

p3

r

d

a

a

d

ys124

da

h

y36

1

0

h

0 s = 2.66 × 10−6 rad/s

wf = wi + a∆t = 2.66 × 10−6 rad/s + (1.0 × 10−10 rad/s2)(12 h)(3600 s/h)

wf = 2.66 × 10−6 rad/s + 4.3 × 10−6 rad/s = 7.0 × 10−6 rad/s

3. r = 2

4

p3

r

m

ad

wi = 0 rad/s

∆s = 160 m

a = 5.00 × 10−2 rad/s2

∆q = ∆r

s

wf2 = wi

2 + 2a∆q = wi2 +

2ar

∆s

wf = wi2 +

2 ar∆s = (0 rad/s)2 +

wf = 1.5 rad/s

(2)(5.00 × 10−2 rad/s2)(160 m)

24p3r

m

ad

4. ∆s = 52.5 m

a = −3.2 × 10−5 rad/s2

wf = 0.080 rad/s

r = 8.0 cm

∆q = ∆r

s

wf2 = wi

2 + 2a∆q = wi2 +

2ar

∆s

wi = wf2 −

2ar∆s = (0.080 rad/s)2 −

wi =√

6.4× 10−3rad2/s2 + 4.2 × 10−2rad2/s2 =√

4.8× 10−2rad2/s2

wi = 0.22 rad/s

(2)(−3.2 × 10−5 rad/s2)(52.5 m)

8.0 × 10−2 m

5. r = 3.0 m

wi = 0.820 rad/s

wf = 0.360 rad/s

∆s = 20.0 m

∆q = ∆r

s

a = wf

2

2

∆−qwi

2

= =

a = =

a = −4.1 × 10−2 rad/s2

−0.542 rad2/s2

(2)230..00m

m

0.130 rad2/s2 − 0.672 rad2/s2

(2)230..00m

m

(0.360 rad/s)2 − (0.820 rad/s)2

(2)230..00m

m

wf2 − wi

2

2

∆r

s


II

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6. r = 1.0 km

wi = 5.0 × 10−3 rad/s

∆t = 14.0 min

∆q = 2p rad

∆q = wi∆t + 12

a∆t2

a = 2(∆q

∆−t2

wi∆t) =

a = = = 6.0 × 10−6 rad/s2(2)(2.1 rad)[(14.0 min)(60 s/min)]2

(2)(6.3 rad − 4.2 rad)[(14.0 min)(60 s/min)]2

(2)[2p rad − (5.0 × 10−3 rad/s)(14.0 min)(60 s/min)]

[(14.0 min)(60 s/min)]2

7. wi = 7.20 × 10−2 rad/s

∆q = 12.6 rad

∆t = 4 min, 22 s

∆t = (4 min)(60 s/min) + 22 s = 262 s

∆q = wi∆t + 12

a∆t2

a = 2(∆q

∆−t2

wi∆t) =

a = = (2)(

(

−2

6

6

.

2

3

s

r

)

a2

d/s) = −1.8 × 10−4 rad/s2(2)(12.6 rad − 18.9 rad)

(262 s)2

(2)[12.6 rad − (7.20 × 10−2 rad/s)(262 s)]

(262 s)2

8. wi = 27.0 rad/s

wf = 32.0 rad/s

∆t = 6.83 s

aavg = wf

∆−t

wi = = 5.

6

0

.8

r

3

ad

s

/s

aavg = 0.73 rad/s2

32.0 rad/s − 27.0 rad/s

6.83 s

9. a = 2.68 × 10−5 rad/s2

∆t = 120.0 s

wi = 2p

1

r

2

ad

∆q = wi∆t + 12

a∆t2

∆q = 21

p2

r

h

ad(1 h/3600 s)(120.0 s) + 1

2(2.68 × 10−5 rad/s2)(120.0 s)2

∆q = 1.7 × 10−2 rad + 1.93 × 10−1 rad = 0.210 rad

10. wi = 6.0 × 10−3 rad/s

wf = 3wi = 18 × 10−3 rad/s

a = 2.5 × 10−4 rad/s2

∆q = wf

2

2

−a

wi2

∆q = =

∆q = = 0.56 rad2.8 × 10−4 rad2/s2

5.0 × 10−4 rad/s2

3.2 × 10−4 rad2/s2 − 3.6 × 10−5 rad2/s2

(2)(2.5 × 10−4 rad/s2)

(18 × 10−3 rad/s)2 − (6.0 × 10−3 rad/s)2

(2)(2.5 × 10−4 rad/s2)

11. wi = 9.0 × 10−7 rad/s

wf = 5.0 × 10−6 rad/s

a = 7.5 × 10−10 rad/s2

∆t = wf

a− wi

∆t = =

∆t = 5.5 × 103 s = 1.5 h

4.1 × 10−6 rad/s7.5 × 10−10 rad/s2

5.0 × 10−6 rad/s − 9.0 × 10−7 rad/s

7.5 × 10−10 rad/s2


II

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12. r = 7.1 m

∆s = 500.0 m

wi = 0.40 rad/s

a = 4.0 × 10−3 rad/s2

∆q = ∆r

s = wi∆t + 1

2a∆t2

12

a∆t2 + wi∆t − ∆r

s = 0

Using the quadratic equation:

∆t =

∆t =

∆t = =

Choose the positive value:

∆t = = 0.4

ra

5

d

r

/

a

s

d2

/s = 1.1 × 102 s

−0.40 rad/s + 0.85 rad/s

4.0 × 10−3 rad/s2

– 0.40 rad/s ±√

0.72 rad2/s24.0 × 10−3 rad/s2

−0.40 rad/s ±√

0.16 rad2/s2 + 0.56rad2/s24.0 × 10−3 rad/s2

−0.40 rad/s ± (0.40rad/s)2 + (4)12

(4.0× 10−3rad/s2)50

7

0

.1

.0m

m

(2)1

2(4.0 × 10−3 rad/s2)

− wi ± wi2 − 41

2a

−∆rs

21

2a


1. w = 4.44 rad/s

vt = 4.44 m/sr =

wvt =

4

4

.4

.4

4

4

r

m

ad

/

/

s

s = 1.00 m

2. vt = 16.0 m/s

w = 1.82 × 10−5 rad/sr =

wvt =

1.82

1

×6.

1

0

0

m−5

/

r

s

ad/s =

circumference = 2pr = (2p)(879 km) = 5.52 × 103 km

8.79 × 105 m = 879 km

3. w = 5.24 × 103 rad/s

vt = 131 m/s

r = wvt =

5.24

1

×31

10

m3/

r

s

ad/s = 2.50 × 10−2 m = 2.50 cm

4. vt = 29.7 km/s

r = 1.50 × 108 kmw =

v

rt =

1.5

2

0

9.

×7

1

k

0

m8/

k

s

m = 1.98 × 10−7 rad/s

5. r = 19.0

2

mm = 9.50 mm

w = 25.6 rad/s

vt = rw = (9.50 × 10−3 m)(25.6 rad/s) = 0.243 m/s


II

Givens Solutions

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1. r = 32 m

at = 0.20 m/s2a =

a

rt =

0.2

3

0

2

m

m

/s2

a = 6.2 × 10−3 rad/s2


2. r = 8.0 m

at = −1.44 m/s2a =

a

rt =

−1.

8

4

.

4

0

m

m

/s2

= −0.18 rad/s2

3. ∆w = −2.4 × 10−2 rad/s

∆t = 6.0 s

at = − 0.16 m/s2

a = ∆∆wt =

−2.4 ×6

1

.0

0−

s

2 rad/s = −4.0 × 10−3 rad/s2

r = a

at =

−4.0

−0

×.1

1

6

0−m3

/

r

s

a

2

d/s2 = 4.0 × 101 m

4. ∆q ′ = 14 628 turns

∆t′ = 1.000 h

at = 33.0 m/s2

wi = 0 rad/s

∆q = 2p rad

r = a

at

a = wf

2

2∆−

qwi

2

wf =

r = =

r = 2

− (0 rad/s)2 = 0.636 m(14 628 turns)(2p rad/turn)

(1.000 h)(3600 s/h)

2at∆q

∆∆qt′

′

2

− wi2

∆∆qt′

′

2

− wi2

2∆q

∆q ′∆t ′

5. r = 56.24 m

wi = 6.00 rad/s

wf = 6.30 rad/s

∆t = 0.60 s

at = ra

a = wf

∆−t

wi

at = rwf

∆−t

wi = (56.24 m) =

at = 28 m/s2

(56.24 m)(0.30 rad/s)

0.60 s

6.30 rad/s − 6.00 rad/s

0.60 s

at

(2)(33.0 m/s2)(2p rad)

6. r = 1.3 m

∆q = 2p rad

∆t = 1.8 s

wi = 0 rad/s

at = ra

α = 2(∆q

∆−t2wi∆t)

at = r2(∆q∆−t2wi∆t) = (1.3 m)

at = 5.0 m/s2

(2)[2p rad − (0 rad/s)(1.8 s)]

(1.8 s)2


II

Givens SolutionsC

opyr

ight

©by

Hol

t, R

ineh

art a

nd W

inst

on.A

ll rig

hts

rese

rved

.

1. vt = 0.17 m/s

ac = 0.29 m/s2r =

v

at

c

2

= (0

0

.

.

1

2

7

9

m

m

/

/

s

s

)2

2

= 0.10 m

Additional Practice 7G

2. w = 2p rad/day

ac = 2.65 × 10−7 m/s2

ac = rw2

r = wac

2 =

r = 50.1 m

2.65 × 10−7 m/s2

[(2p rad/day)(1 day/24 h)(1 h/3600 s)]2

3. r = 58.4

2

cm = 29.2 cm

ac = 8.50 × 10−2 m/s2

vt =√

rac =√

(29.2× 10−2m)(8.50 × 10−2m/s2

vt = 0.158 m/s

4. r = 12

2

cm = 6.0 cm

ac = 0.28 m/s2

vt =√

rac =√

(6.0 × 10−2m)(0.28 m/s2) = 0.13 m/s

5. r = 20.0 m

∆t = 16.0 s

∆q = 2p rad

ac = rw2 = r ∆∆qt

2

ac = (20.0

(

m

16

)

.

(

0

2

s

p)2

rad)2

= 3.08 m/s2

6. ∆t = 1.000 h

∆s = 47.112 km

r = 6.37 × 103 kmac =

v

rt2

= = r

∆∆

s

t

2

2

ac = = 2.69 × 10−5 m/s2(47 112 m)2

(6.37 × 106 m)[(1.000 h)(3600 s/h)]2

∆∆

s

t

2

r

Additional Practice 7H

1. m1 = 235 kg

m2 = 72 kg

r = 25.0 m

Fc = 1850 N

mtot = m1 + m2 = 235 kg + 72 kg = 307 kg

Fc = mtot ac = mtot v

rt2

vt = m

rF

toc

t = (25.03

m

0)

7

( 1

k8

g

50N) = 12.3 m/s

2. m = 30.0 g

r = 2.4 m

FT = 0.393 N

g = 9.81 m/s2

FT = Fg + Fc = mg + mv

rt2

vt = r(FT

m− mg) =

vt = =

vt = 2.8 m/s

(2.4 m)(0.099 N)

30.0 × 10−3 kg

(2.4 m)(0.393 N − 0.294 N)

30.0 × 10−3 kg

(2.4 m)[0.393 N − (30.0 × 10−3 kg)(9.81 m/s2)]

30.0 × 10−3 kg


2. m1 = 3.08 × 104 kg

r = 1.27 × 107 m

Fg = 2.88 × 10−16 N

G = 6.673 × 10−11 N•m2

kg2

m2 = =

m2 = 2.26 × 104 kg

(2.88 × 10−16 N)(1.27 × 107 m)2

6.673 × 10−11 N

k

•

g

m2

2

(3.08 × 104 kg)

Fgr2

Gm1

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3. vt = 8.1 m/s

r = 4.23 m

m1 = 25 g

g = 9.81 m/s2

Fg = Fc

m1g = m2

r

vt2

m2 = m

v1

t2

gr

m2 = = 1.6 × 10−2 kg(25 × 10−3 kg)(9.81 m/s2)(4.23 m)

(8.1 m/s)2

4. vt = 75.57 km/h

m = 92.0 kg

Fc = 12.8 N

Fc = m v

rt2

r = m

F

v

c

t2

=

r = 3.17 × 103 m = 3.17 km

(92.0 kg)[(75.57 km/h)(103 m/km)(1 h/3600 s)]2

12.8 N

5. m = 75.0 kg

r = 446 m

vt = 12 m/s

g = 9.81 m/s2

Fc = m

r

vt2

= =

FT = Fc + mg = 24 N + (75.0 kg)(9.81 m/s2)

FT = 24 N + 736 N = 7.60 × 102 N

24 N(75.0 kg)(12 m/s)2

446 m

1. r = 6.3 km

Fg = 2.5 × 10−2 N

m1 = 3.0 kg

G = 6.673 × 10−11 N•m2

kg2

m2 = =

m2 = 5.0 × 1015 kg

(2.5 × 10−2 N)(6.3 × 103 m)2

6.673 × 10−11 N

k

•

g

m2

2

(3.0 kg)

Fgr2

Gm1

3. m1 = 5.81 × 104 kg

r = 25.0 m

Fg = 5.00 × 10−7 N

G = 6.673 × 10−11 N•m2

kg2

m2 = =

m2 = 80.6 kg

(5.00 × 10−7 N)(25.0 m)2

6.673 ×10−11 N

k

•

g

m2

2

(5.81 × 104 kg)

Fgr2

Gm1

Givens Solutions

Additional Practice 7I


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Givens Solutions

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7. m1 = 318mE

m2 = 50.0 kg

VJ = 1323VE

mE = 5.98 × 1024 kg

rE = 6.37 × 106 m

G = 6.673 × 10−11 N•m2

kg2

If VJ = 1323 VE, then rJ =√3 1323rE.

Fg = Gm

rJ

12

m2 =

Fg = 1.30 × 103 N

(6.673 × 10−11 N •m2/kg2)(318)(5.98 × 1024 kg)(50.0 kg)

[(√3 1323)(6.37 × 106 m)]2

4. m1 = 621 g

m2 = 65.0 kg

Fg = 1.0 × 10−12 N

G = 6.673 × 10−11 N•m2

kg2

r = Gm

F1

g

m2

r = = 52 m(6.673 × 10−11 N •m2/kg2)(0.621 kg)(65.0 kg)

1.0 × 10−12 N

5. m1 = m2 = 1.0 × 108 kg

Fg = 1.0 × 10−3 N

G = 6.673 × 10−11 N•m2

kg2

r = Gm

F1

g

m2r = r = 2.6 × 104 m = 26 km

(6.673 × 10−11 N •m2/kg2)(1.0 × 108 kg)2

1.0 × 10−3 N

6. ms = 25 × 109 kg

m1 = m2 = 12

ms

r = 1.0 × 103 km

G = 6.673 × 10−11 N•m2

kg2

Fg = Gm

r12

m2 = = 1.0 × 10−2 N6.673 × 10−11

N

k

•

g

m2

2

12

(25 × 109 kg)2

(1.0 × 106 m)2

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1. m = 3.00 × 105 kg

q = 90.0° − 45.0° = 45.0°

t = 3.20 × 107 N •m

g = 9.81 m/s2

t = Fd(sin q) = mgl (sin q)

l = mg(s

tin q)

l =

l = 15.4 m

3.20 × 107 N •m(3.00 × 105 kg)(9.81 m/s2)(sin 45.0°)


Givens Solutions

2. tnet = 9.4 kN •m

m1 = 80.0 kg

m2 = 120.0 kg

g = 9.81 m/s2

tnet = t1 + t2 = F1d1(sin q1) + F2d2(sin q2)

q1 = q2 = 90°, so

tnet = F1d1 + F2d2 = m1g + m2gl

l =

l =9.4 × 103 N •m

+ (120.0 kg)(9.81 m/s2)

=

l = = 6.0 m9.4 × 103 N •m1.57 × 103 N

9.4 × 103 N•m392 N + 1.18 × 103 N(80.0 kg)(9.81 m/s2)

2

tnetm

21g + m2g

l2

3. tnet = 56.0 N •m

m1 = 3.9 kg

m2 = 9.1 kg

d1 = 1.000 m − 0.700 m =0.300 m

g = 9.81 m/s2


q1 = q2 = 90°, so

tnet = F1d1 + F2d2 = m1gd1 + m2g(1.000 m − x)

x = 1.000 m − tnet

m

−

2

m

g1gd1

x = 1.000 m −

x = = = 1.000 m − 0.50 m

x = 0.50 m = 5.0 × 101 cm

45 N•m(9.1 kg)(9.81 m/s2)

56.0 N•m − 11 N•m(9.1 kg)(9.81 m/s2)

56.0 N •m − (3.9 kg)(9.81 m/s2)(0.300 m)

(9.1 kg)(9.81 m/s2)

8ChapterRotational Equilibrium and Dynamics

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4. t = −1.3 × 104 N •m

l = 6.0 m

d = 1.0 m

q = 90.0° − 30.0° = 60.0°

t = Fd(sin q) = −Fg(l − d)(sin q)

Fg = = =

Fg = 3.0 × 103 N

1.3 × 104 N •m(5.0 m)(sin 60.0°)

−(−1.3 × 104 N •m)(6.0 m − 1.0 m)(sin 60.0°)

−t(l − d)(sin q)

Givens Solutions

5. R = 76

2

m = 38 m

q = 60.0°

t = −1.45 × 106 N •m

t = Fd(sin q) = −FgR(sin q)

Fg = R(s

−in

tq)

=

Fg = 4.4 × 104 N

−(−1.45 × 106 N •m)

(38 m)(sin 60.0°)

6. m1 = 102 kg

m2 = 109 kg

l = 3.00 m

l 1 = 0.80 m

l 2 = 1.80 m

g = 9.81 m/s2


q1 = q2 = 90°, so

tnet = F1d1 + F2d2 = m1g − l 1 + m2g − l 2tnet = (102 kg)(9.81 m/s2)3.0

2

0 m − 0.80 m + (109 kg)(9.81 m/s2)3.0

2

0 m − 1.80 m

tnet = (102 kg)(9.81 m/s2)(1.50 m − 0.80 m) + (109 kg)(9.81 m/s2)(1.50 m − 1.80 m)

tnet = (102 kg)(9.81 m/s2)(0.70 m) + (109 kg)(9.81 m/s2)(−0.30 m)

tnet = 7.0 × 102 N •m − 3.2 × 102 N •m

tnet = 3.8 × 102 N •m

l2

l2

7. m = 5.00 × 102 kg

d1 = 5.00 m

t = 6.25 × 105 N •m

g = 9.81 m/s2

q1 = 90.0° − 10.0° = 80.0°

d2 = 4.00 m

q2 = 90°

a. t = Fd(sin q) = mgd1(sin q1)

t = (5.00 × 102 kg)(9.81 m/s2)(5.00 m)(sin 80.0°)

t =

b. tnet = Fd2(sin q2) − t = Fd2(sin q2) − mgd1(sin q1)

F =

F = =

F = 1.62 × 105 N

6.49 × 105 N •m

4.00 m

6.25 × 105 N •m + 2.42 × 104 N •m

4.00 m (sin 90°)

tnet + mgd1(sin q1)

d2(sin q2)

2.42 × 104 N •m


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1. t1 = 2.00 × 105 N •m

t2 = 1.20 × 105 N •m

h = 24 m

Apply the second condition of equilibrium, choosing the base of the cactus as thepivot point.

tnet = t1 − t2 − Fd(sin q) = 0

Fd(sin q) = t1 − t2

For F to be minimum, d and sin q must be maximum. This occurs when the force isperpendicular to the cactus (q = 90°) and is applied to the top of the cactus (d = h =24 m).

Fmin = t1 −

h

t2 =

Fmin = = 3.3 × 103 N applied to the top of the cactus8.0 × 104 N•m

24 m

2.00 × 105 N •m − 1.20 × 105 N •m

24 m

Givens Solutions

2. m1 = 40.0 kg

m2 = 5.4 kg

d1 = 70.0 cm

d2 = 100.0 cm − 70.0 cm = 30.0 cm

g = 9.81 m/s2

Apply the first condition of equilibrium.

Fn − m1g − m2g − Fapplied = 0

Fn = m1g + m2g + Fapplied = (40.0 kg)(9.81 m/s2) + (5.4 kg)(9.81 m/s2) + Fapplied

Fn = 392 N + 53 N + Fapplied = 455 N + Fapplied

Apply the second condition of equilibrium, using the fulcrum as the location for the axis of rotation.

Fappliedd2 + m2gd2 − m1gd1 = 0

Fapplied = m1gd1

d

−

2

m2gd2 =

Fapplied = =

Fapplied =

Substitute the value for Fapplied into the first-condition equation to solve for Fn.

Fn = 455 N + 863 N = 1318 N

863 N

259 N•m0.300 m

275 N•m − 16 N•m

0.300 m

(40.0 kg)(9.81 m/s2)(0.700 m) − (5.4 kg)(9.81 m/s2)(0.300 m)

0.300 m

3. m = 134 kg

d1 = 2.00 m

d2 = 7.00 m − 2.00 m = 5.00 m

q = 60.0°

g = 9.81 m/s2

Apply the first condition of equilibrium in the x and y directions.

Fx = Fapplied (cos q) − Ff = 0

Fy = Fn − Fapplied (sin q) − mg = 0

To solve for Fapplied, apply the second condition of equilibrium, using the fulcrum asthe pivot point.

Fapplied (sin q)d2 − mgd1 = 0

Fapplied = d2

m

(s

g

in

d1

q) =

Fapplied =

Substitute the value for Fapplied into the first-condition equations to solve for Fn andFf .

Fn = Fapplied(sin q) + mg = (607 N)(sin 60.0°) + (134 kg)(9.81 m/s2)

Fn = 526 N + 1.31 × 103 N =

Ff = Fapplied(cos q) = (607 N)(cos 60.0°) = 304 N

1.84 × 103 N

607 N

(134 kg)(9.81 m/s2)(2.00 m)

(5.00 m)(sin 60.0°)



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4. m = 8.8 × 103 kg

d1 = 3.0 m

d2 = 15 m − 3.0 m = 12 m

q = 20.0°

g = 9.81 m/s2

Apply the first condition of equilibrium in the x and y directions.

Fx = Ffulcrum,x − F (sin q) = 0

FyFfulcrum,y − F (cos q) − mg = 0

To solve for F, apply the second condition of equilibrium •, using the fulcrum as thepivot point.

Fd2 − mg d1 (cos q) = 0

F = mg d1

d

(

2

cos q) =

F =

Substitute the value for F into the first-condition equations to solve for the compo-nents of Ffulcrum.

Ffulcrum,x = F (sin q) = (2.0 × 104 N)(sin 20.0°)

Ffulcrum,x =

Ffulcrum,y = F (cos q) + mg = (2.0 × 104 N)(cos 20.0°) + (8.8 × 104 kg)(9.81 m/s2)

Ffulcrum,y = 1.9 × 104 N + 8.6 × 105 N = 8.8 × 105 N

6.8 × 103 N

2.0 × 104 N

(8.8 × 103 kg)(9.81 m/s2)(3.0 m)(cos 20.0°)

12 m

Givens Solutions

6. m1 = 3.6 × 102 kg

m2 = 6.0 × 102 kg

l = 15 m

l 1 = 5.0 m

g = 9.81 m/s2

Apply the second condition of equilibrium, using the pool’s edge as the pivot point.

Assume the total mass of the board is concentrated at its center.

m1 g d − m2 g − l 1 = 0

d = =

d = = =

d = 4.2 m from the pool’s edge

(6.0 × 102 kg)(2.5 m)

3.6 × 102 kg

(6.0 × 102 kg)(7.5 m − 5.0 m)

3.6 × 102 kg

(6.0 × 102 kg)15

2

m − 5.0 m

3.6 × 102 kg

m2l2

− l 1

m1

m2 g l2

− l 1

m1 g

l2

5. m1 = 64 kg

m2 = 27 kg

d1 = d2 = 3.0

2

0 m = 1.50 m

Fn = 1.50 × 103 N

g = 9.81 m/s2

Apply the first condition of equilibrium to solve for Fapplied.

Fn − m1 g − m2 g − Fapplied = 0

Fapplied = Fn − m1 g − m2 g = 1.50 × 103 N − (64 kg)(9.81 m/s2) − (27 kg)(9.81 m/s2)

Fapplied = 1.50 × 103 N − 6.3 × 102 N − 2.6 × 103 N = 6.1 × 102 N

To solve for the lever arm for Fapplied, apply the second condition of equilibrium,using the fulcrum as the pivot point.

Fapplied d + m2 g d2 − m1 g d1 = 0

d = m1 g d

F1

ap

−

pl

m

ied

2 g d2 =

d = =

d = 0.89 m from the fulcrum, on the same side as the less massive seal

5.4 × 102 N •m

6.1 × 102 N

9.4 × 102 N •m − 4.0 × 102 N •m

6.1 × 102 N

(64 kg)(9.81 m/s2)(1.50 m) − (27 kg)(9.81 m/s2)(1.50 m)

6.1 × 102 N


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7. m = 449 kg

l = 5.0 m

F1 = 2.70 × 103 N

g = 9.81 m/s2

Apply the first condition of equilibrium to solve for F2.

F1 + F2 − mg = 0

F2 = mg − F1

F2 = (449 kg)(9.81 m/s2) − 2.70 × 103 N = 4.40 × 103 N − 2.70 × 103 N = 1.70 × 103 N

Apply the second condition of equilibrium, using the left end of the platform as thepivot point.

F2 l − m g d = 0

d = =

d = 1.9 m from the platform’s left end

(1.70 × 103 N)(5.0 m)(449 kg)(9.81 m/s2)

F2 lm g

Givens Solutions

8. m1 = 414 kg

l = 5.00 m

m2 = 40.0 kg

F1 = 50.0 N

g = 9.81 m/s2

Apply the first condition of equilibrium to solve for F2.

F1 + F2 − m1 g − m2 g = 0

F2 = m1 g + m2 g − F1 = (m1 + m2) g − F1

F2 = (414 kg + 40.0 kg)(9.81 m/s2) − 50.0 N = (454 kg)(9.81 m/s2) − 50.0 N = 4.45 ×103 N − 50.0 N

F2 = 4.40 × 103 N

Apply the second condition of equilibrium, using the supported end (F1) of the stickas the rotation axis.

F2 d − m1 g − m2 g l = 0

d = =

d = =

d = 2.75 m from the supported end

(247 kg)(9.81 m/s2)(5.00 m)

4.40 × 103 N

(207 kg + 40.0 kg)(9.81 m/s2)(5.00 m)

4.40 × 103 N

414

2

kg + 40.0 kg(9.81 m/s2)(5.0 m)

4.40 × 103 N

m

21 + m2 g l

F2

l2

1. R = 50.0 m

M = 1.20 × 106 kg

t = 1.0 × 109 N •m

a = tI

= M

tR2 =

a = 0.33 rad/s2

1.0 × 109 N •m(1.20 × 106 kg)(50.02)


2. M = 22 kg

R = 0.36 m

t = 5.7 N •m

a = tI

= M

tR2 =

a = 2.0 rad/s2

5.7 N•m(22 kg)(0.36 m)2


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3. M = 24 kg

l = 2.74 m

F = 1.8 N

The force is applied perpendicular to the lever arm, which is half the pencil’s length.

Therefore,

t = F d (sin q) = F

a = tI

= =

a = 0.16 rad/s2

(1.8 N)2.7

2

4 m

1

1

2 (24 kg)(2.74 m)2

F l2

1

1

2 M l 2

l2

Givens Solutions

4. M = 4.07 × 105 kg

R = 5.0 m

t = 5.0 × 104 N •m

a = tI

=

a =

a = 9.8 × 10−3 rad/s2

(5.0 × 104 N •m)12

(4.07 × 105 kg)(5.0 m)2

t12

MR2

5. R = 2.00 m

F = 208 N

a = 3.20 × 10−2 rad/s2

The force is applied perpendicular to the lever arm, which is the ball’s radius.

Therefore,

t = F d (sin q) = F R

T = at

= F

aR =

I = 1.30 × 104 kg •m2

(208 N)(2.00 m)3.20 × 10−2 rad/s2

6. r = 8.0 m

t = 7.3 × 103 N•m

a = 0.60 rad/s2

I = at

= mr2

I = =

m = r

I2 = = 1.9 × 102 kg

1.2 × 104 kg•m2

(8.0 m)2

1.2 × 104 kg •m27.3 × 103 N•m

0.60 rad/s2

7. vt,i = 2.0 km/s

l = 15.0 cm

∆t = 80.0 s

t = −0.20 N•m

vt,f = 0 m/s

I = at

= = −0.20 N•mI =

=

I = 6.0 × 10−4 kg •m2

0 m/s − 2.0 × 103 m/s

0.15

2

0 m(80.0 s)

vt,f − vt,il

d

2 ∆t

t

wf

∆−t

wi

t

−0.20 N•m

(0

−.0

2

7

.0

5

×m

1

)

0

(

3

80

m

.0

/s

s)


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8. R = 1.7

2

0 m = 0.85 m

t = 125 N •m

∆t = 2.0 s

wi = 0 rad/s

wf = 12 rad/s

I = at

= MR2

I = at

= = =

I =

M = R

I2 = = 29 kg

21 kg•m2

(0.85 m)2

21 kg •m2

125 N•m6.0 rad/s2

125 N •m

12 rad/

2

s

.0

−s

0 rad/s

t

wf

∆−t

wi

Givens Solutions

9. R = 3.00 m

M = 17 × 103 kg

wi = 0 rad/s

wf = 3.46 rad/s

∆t = 12 s

t = I a = 12

MR2wf

∆−t

wit = = 2.2 × 104 N•m

(17 × 103 kg)(3.00 m)2(3.46 rad/s − 0 rad/s)

(2)(12 s)

10. R = 4.0 m

M = 1.0 × 108 kg

wi = 0 rad/s

wf = 0.080 rad/s

∆t = 60.0 s

t = Ia = 12

MR2wf

∆−t

wit = = 1.1 × 106 N •m

(1.0 × 108 kg)(4.0 m)2 (0.080 rad/s − 0 rad/s)

(2)(60.0 s)

11. I = 2.40 × 103 kg•m2

∆q = 2(2p rad) = 4p rad

∆t = 6.00 s

wi = 0 rad/s

∆q = wi ∆t + 12

a ∆t2

Because wi = 0,

∆q = 12

a∆t2

a = 2

∆∆t2q

t = Ia = 2

∆I

t

∆2q

=

t = 1.68 × 103 N •m

(2)(2.40 × 103 kg•m2)(4p rad)

(6.00 s)2

12. m = 7.0 × 103 kg

r = 18.3 m

at = 25 m/s2

t = Ia = (mr2)a

rt = mrat

t = (7.0 × 103 kg)(18.3 m)(25 m/s2)

t = 3.2 × 106 N •m


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1. ri = 4.95 × 107 km

vi = 2.54 × 105 Km/h

vf = 1.81 × 105 km/h

Li = Lf

Iiwi = If wf

mri2

v

ri

i = mrf2

v

rf

fri vi = rf vf

rf = ri

v

v

f

i =

rf = 6.95 × 107 km

(4.95 × 107 km)(2.54 × 105 km/h)

1.81 × 105 km/h


Givens Solutions

2. vi = 399 km/h

vf = 456 km/h

R = 0.20 m

∆q = 20 rev

Li = Lf

Ii wi = If wf

m ri2

v

ri

i = mrf2

v

rf

fri vi = rf vf

rf = ri − ∆s = ri − R ∆q

ri vi = (ri − R ∆q) vf

ri (vf − vi) = (R ∆q) vf

ri = v

vf

f

R

−∆v

q

i =

=

ri = 2.0 × 102 m

(456 km/h)(0.20 m)(20 rev)(2p rad/rev)

57 km/h

(456 km/h)(0.20 m)(20 rev)(2p rad/rev)

456 km/h − 399 km/h

3. M = 25.0 kg

R = 15.0 cm

wi = 4.70 × 10−3 rad/s

wf = 4.74 × 10−3 rad/s

Li = Lf

Ii wi = If wf

If = =25

MR2 wi

If = = 0.223 kg•m2

Ii = 25

MR2 = 25

(25.0 kg)(0.150 m)2 = 0.225 kg•m2

∆I = If − Ii = 0.223 kg•m2 − 0.225 kg•m2 = −0.002 kg•m2

The moment of inertia decreases by 0.002 kg•m2.

(2)(25.0 kg)(0.150 m)2(4.70 × 10−3 rad/s)

(5)(4.74 × 10−3 rad/s)

wf

Ii wiwf


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4. vi = 395 km/h

ri = 1.20 × 102 m

∆∆

r

t = 0.79 m/s

∆t = 33 s

rf = ri − ∆∆

r

t∆t

Li = Lf

Ii wi = Lf wf

mri2

v

ri

i = mrf2

v

rf

fri vi = rf vf = ri −

∆∆

r

t ∆tvf

vf = =

vf = =

vf = 5.0 × 102 km/h

(1.20 × 102 m)(395 km/h)

94 m

(1.20 × 102 m)(395 km/h)

1.20 × 102 m − 26 m

(1.20 × 102 m)(395 km/h)1.20 × 102 − (0.79 m/s)(33 s)

ri vi

[ri − ∆∆

r

t ∆t]

Givens Solutions

5. ri = 10.

2

0 m = 5.00 m

rf = 4.0

2

0 m = 2.00 m

wi = 1.26 rad/s

Li = Lf

Iiwi = Ifwf

mri2wi = mrf

2wf

wf = ri

r

2

f

w2

i =

wf = 7.88 rad/s

(5.00 m)2(1.26 rad/s)

(2.00 m)2

6. R = 3.00 m

M = 1.68 × 104 kg

ri = 2.50 m

rf = 3.00 m

m = 2.00 × 102 kg

wi = 3.46 rad/s

Li = Lf

Iiwi = Ifwf

2

1MR2 + mri2wi = 1

2MR2 + mrf

2wf

wf =

wf =2

1(1.68 × 104 kg)(3.00 m)2 + (2.00 × 102 kg)(2.50 m)2(3.46 rad/s)

21(1.68 × 104 kg)(3.00 m)2 + (2.00 × 102 kg)(3.00 m)2

wf =

wf =

wf = 3.43 rad/s

∆w = wf − wi = 3.43 rad/s − 3.46 rad/s = −0.03 rad/s

The angular speed decreases by 0.03 rad/s.

(7.68 × 104 kg •m2)(3.46 rad/s)

7.74 × 104 kg •m2

(7.56 × 104 kg •m2 + 1.25 × 103 kg •m2)(3.46 rad/s)

7.56 × 104 kg •m2 + 1.80 × 103 kg •m2

2

1MR2 + mri2wi

12

MR2 + mrf2


1. m = 407 kg

h = 57.0 m

vf = 12.4 m/s

wf = 28.0 rad/s

g = 9.81 m/s2

MEi = MEf

mgh = 12

mvf2 + 1

2 I wf

2

12

I wf2 = mgh − 1

2 mvf

2

I = 2mgh

w−

f2mvf

2

= m(2g

wh

f2− vf

2)

I =

I = =

I = 5.0 × 102 kg •m2

(407 kg)(9.7 × 102 m2/s2)

(28.0 rad/s)2(407 kg)(1.12 × 103 m2/s2 − 154 m2/s2)

(28.0 rad/s)2

(407 kg)[(2)(9.81 m/s2)(57.0 m) − (12.4 m/s)2]

(28.0 rad/s)2

Givens Solutions

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Additonal Practice 8E

2. h = 5.0 m

g = 9.81 m/s2

MEi = MEf

mgh = 12

mvf2 +

2

1Iwf2

mgh = 12

mvf2 + 1

2(mr2)

v

r

f2

2

mgh = mvf

212

+ 12

= mvf2

vf =√

gh =√

(9.81m/s2)(5.0 m)

vf = the mass is not required7.0 m/s

3. h = 1.2 m

g = 9.81 m/s2

MEi = MEf

12

mvi2 + 1

2Iwi

2 = mgh

2

1mvi2 + 1

21

2mr2

v

ri2

2

= mgh

mvi2 1

2 + 1

4 = 3

4 mvi

2 = mgh

vi = 4

3

gh = = 4.0 m/s

(4)(9.81 m/s2)(1.2 m)

3

4. vf = 12.0 m/s

I = 0.80mr2

g = 9.81 m/s2

MEi = MEf

mgh = 2

1mvf2 +

2

1Iwf2 =

2

1mvf2 +

2

1(0.80 mr2)v

rf

2

mgh = mvf2

2

1 + 0.

2

80 = 0.90 mvf

2

h = 0.90

g

vf2

= (0.9

9

0

.

)

8

(

1

12

m

.0

/s

m2

/s)2

h = 13 m

II

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5. vi = 5.4 m/s

g = 9.81 m/s2

q = 30.0°

MEi = MEf

2

1mvi2 +

2

1Iwi2 = mgh = mgd(sin q)

mgd(sin q) = 2

1mvi2 +

2

125

mr2 v

ri2

2

mgd(sin q) = mvi

2 2

1 + 15

= 1

7

0 mvi

2

d = 10 g

7

(

v

sii

n

2

q) = = 4.2 m

(7)(5.4 m/s)2

(10)(9.81 m/s2)(sin 30.0°)

Givens Solutions

6. r = 2.0 m

wf = 5.0 rad/s

g = 9.81 m/s2

m = 4.8 × 103 kg

MEi = MEf

mgh = 2

1mvf2 +

2

1Iwf2

mgh = 2

1mr2wf2 +

2

125

mr2 wf2

mgh = mr2wf2

2

1 + 15

= 1

7

0 mr2 wf

2

h = =

h =

KEtrans = 2

1mvf2 =

2

1mr2wf2

KEtrans = 2

1(4.8 × 103 kg)(2.0 m)2(5.0 rad/s)2

KEtrans = 2.4 × 105 J

7.1 m

(7)(2.0 m)2(5.0 rad/s)2

(10)(9.81 m/s2)

1

7

0r2wf

2

g

7. m = 5.55 kg

h = 1.40 m

g = 9.81 m/s2

MEi = MEf

mgh = 2

1mvf2 +

2

1Iwf2

mgh = 2

1mvf2 +

2

125

mr2 v

r

f2

2

mgh = mvf

2 2

1 + 15

= 1

7

0 mvf

2

vf = 107

gh = = 4.43 m/s

KErot = 2

1Iwf2 =

5

1mvf2

KErot = = 21.8 J(5.55 kg)(4.43 m/s)2

5

(10)(9.81 m/s2)(1.40 m)

7


Chapter 9Fluid Mechanics

II

1. mp = 1158 kg

V = 3.40 m3

ρ = 1.00 × 103 kg/m3

g = 9.81 m/s2

FB = Fg

ρVg = (mp + mr)g

mr = ρV − mp = (1.00 × 103 kg/m3)(3.40 m3) − 1158 kg = 3.40 × 103 kg − 1158 kg

mr = 2.24 × 103 kg


Givens Solutions

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2. V = 4.14 × 10−2 m3

apparent weight =3.115 × 103 N

ρsw = 1.025 × 103 kg/m3

g = 9.81 m/s2

FB = Fg − apparent weight

ρswVg = mg − apparent weight

m = ρswV + apparen

g

t weight = (1.025 × 103 kg/m3)(4.14 × 10−2 m3) +

3.1

9

1

.8

5

1

×m

1

/

0

s

3

2N

m = 42.4 kg + 318 kg = 3.60 × 102 kg

3. l = 3.00 m

A = 0.500 m2

rfw = 1.000 × 103 kg/m3

rsw = 1.025 × 103 kg/m3

Fnet,1 = Fnet,2 = 0

FB,1 − Fg,1 = FB,2 − Fg,2

rfwVg − mg = rswVg − (m + mballast)g

mballastg = (rsw − rfw)Vg

mballast = (rsw − rfw)Al

m0 = (1.025 × 103 kg/m3 − 1.000 × 103 kg/m3)(0.500 m2)(3.00 m)

m0 = (25 kg/m3)(0.500 m2)(3.00 m) = 38 kg

4. A = 3.10 × 104 km2

h = 0.84 km

r = 1.025 × 103 kg/m3

g = 9.81 m/s2

FB = rVg = rAhg

FB = (1.025 × 103 kg/m3)(3.10 × 1010 m2)(840 m)(9.81 m/s2) = 2.6 × 1017 N

5. m = 4.80 × 102 kg

g = 9.81 m/s2

apparent weight = 4.07 × 103 N

FB = Fg − apparent weight = mg − apparent weight

FB = (4.80 × 102 kg)(9.81 m/s2) − 4.07 × 103 N = 4.71 × 103 N − 4.07 × 103 N

FB = 640 N


II

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6. h = 167 m

H = 1.50 km

rsw = 1.025 × 103 kg/m3

Fg,i = FB

riVig = rswVswg

ri(h + H)Ag = rswHAg

ri = h

rs

+wH

H

ri = = = 921 kg/m3(1.025 × 103 kg/m3)(1.50 × 103 m)

1670 m

(1.025 × 103 kg/m3)(1.50 × 103 m)

167 m + 1.50 × 103 m

7. l = 1.70 × 102 m

r = 13.

2

9 m = 6.95 m

msw = 2.65 × 107 kg

a = 2.00 m/s2

g = 9.81 m/s2

Fnet = Fg − FB

msuba = msubg − mswg

ρsubVa = ρsubVg − mswg

ρsub(g − a)V = mswg

ρsub = (g

m

−sw

a

g

)V =

(g −m

a)s

(w

πg

r2 l )

ρsub =

ρsub =

ρsub = 1.29 × 103 kg/m3

(2.65 × 107 kg)(9.81 m/s2)

(9.81 m/s2)(π)(6.95 m)2(1.70 × 102 m)

(2.65 × 107 kg)(9.81 m/s2)(9.81 m/s2 − 2.00 m/s2)(π)(6.95 m)2(1.70 × 102 m)

8. V = 6.00 m3

∆ apparent weight = 800 N

ρwater = 1.00 × 103 kg/m3

g = 9.81 m/s2

Fg,1 = Fg,2

FB,1 + apparent weight in water = FB,2 + apparent weight in PEG solution

ρwaterVg + apparent weight in water − apparent weight in PEG solution = ρsolnVg

ρsoln =

ρsoln =

ρsoln = =

ρsoln = 1.01 × 103 kg/m3

5.97 × 104 N(6.00 m3)(9.81 m/s2)

5.89 × 104 N + 800 N(6.00 m3)(9.81 m/s2)

(1.00 × 103 kg/m3)(6.00 m3)(9.81 m/s2) + 800 N

(6.00 m3)(9.81 m/s2)

ρwaterVg + ∆ apparent weight

Vg


1. P = 1.01 × 105 Pa

A = 3.3 m2

F = PA (1.01 × 105 Pa)(3.3 m2) = 3.3 × 105 N


II

2. P = 4.0 × 1011 Pa

r = 50.0 m

F = PA = P(πr2)

F = (4.0 × 1011 Pa)(π)(50.0 m)2

F = 3.1 × 1015 N

Givens Solutions

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3. m1 = 181 kg

m2 = 16.0 kg

A1 = 1.8 m2

g = 9.81 m/s2

P1 = P2

A

F1

1 =

A

F2

2

A2 = F2

F

A

1

1 = 3m

m

2

1

g

g

A1 = 3m

m

2

1

A1

A2 = = 0.48 m2(3)(16.0 kg)(1.8 m2)

181 kg

4. m = 4.0 × 107 kg

F2 = 1.2 × 104 N

A2 = 5.0 m2

g = 9.81 m/s2

P1 = P2

A

F1

1 =

A

F2

2

A1 = A

F2F

2

1 = A

F2m

2

g

A1 = = 1.6 × 105 m2(5.0 m2)(4.0 × 107 kg)(9.81 m/s2)

1.2 × 104 N

5. P = 2.0 × 1016 Pa

F = 1.02 × 1031 N

A = P

F =

1

2

.

.

0

0

2

××1

1

0

01

3

6

1

P

N

a =

A = 4πr2

r = = r = 6.4 × 106 m

5.1 × 1014 m2

4π

A4π

5.1 × 1014 m2

6. F = 4.6 × 106 N

r = 38

2

cm = 19 cm

P = A

F

Assuming the squid’s eye is a sphere, its total surface area is 4πr2. The outer half ofthe eye has an area of

A = 2πr2

P = 2π

F

r2 = (2

4

π.6

)(

×0.

1

1

0

9

6

m

N

)2 = 2.0 × 107 Pa

7. A = 26.3 m2

F = 1.58 × 107 N

Po = 1.01 × 105 Pa

P = A

F =

1.5

2

8

6.

×3

1

m

07

2N

=

Pgauge = P − Po = 6.01 × 105 Pa − 1.01 × 105 Pa = 5.00 × 105 Pa

6.01 × 105 Pa

II

1. h = (0.800)(16.8 m)

P = 2.22 × 105 Pa

Po = 1.01 × 105 Pa

g = 9.81 m/s2

P = Po + ρgh

ρ = P

g

−h

Po = =

ρ = 918 kg/m3

1.21 × 105 Pa(9.81 m/s2)(0.800)(16.8 m)

2.22 × 105 Pa − 1.01 × 105 Pa(9.81 m/s2)(0.800)(16.8 m)


Givens Solutions

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2. h = −950 m

P = 8.88 × 104 Pa

Po = 1.01 × 105 Pa

g = 9.81 m/s2

P = Po + ρgh

ρ = = =

ρ = 1.3 kg/m3

−1.2 × 104 Pa(9.81 m/s2)(−950 m)

8.88 × 104 Pa − 1.01 × 105 Pa

(9.81 m/s2)(−950 m)

P − Pog

1

h

3. P = 13.6P0

P0 = 1.01 × 105 Pa

r = 1.025 × 103 kg/m3

g = 9.81 m/s2

P = P0 + rgh

h = 13.6P

r0

g

− P0 = 12

r.6

g

P0

h = = 127 m(12.6)(1.01 × 105 Pa)

(1.025 × 103 kg/m3)(9.81 m/s2)

4. P = 4.90 × 106 Pa

P0 = 1.01 × 105 Pa

r = 1.025 × 103 kg/m3

g = 9.81 m/s2

P = P0 + rgh

h = P

r−

g

P0 =

h = = 477 m4.80 × 106 Pa

(1.025 × 103 kg/m3)(9.81 m/s2)

4.90 × 106 Pa − 1.01 × 105 Pa(1.025 × 103 kg/m3)(9.81 m/s2)

6. h = 10 916 m

P0 = 1.01 × 105 Pa

r = 1.025 × 103 kg/m3

g = 9.81 m/s2

P = P0 + rgh = 1.01 × 105 Pa + (1.025 × 103 kg/m3)(9.81 m/s2)(10 916 m)

P = 1.01 × 105 Pa + 1.10 × 108 Pa = 1.10 × 108 Pa

5. h = 245 m

Po = 1.01 × 105 Pa

ρ = 1.025 × 103 kg/m3

g = 9.81 m/s2

P = Po + ρgh

P = 1.01 × 105 Pa + (1.025 × 103 kg/m3)(9.81 m/s2)(245 m)

P = 1.01 × 105 Pa + 2.46 × 106 Pa

P = 2.56 × 106 Pa



II

1. P1 = (1 + 0.12)P2 = 1.12 P2

v1 = 0.60 m/s

v2 = 4.80 m/s

ρ = 1.00 × 103 kg/m3

h1 = h2

P1 + 12

ρv12 + ρgh1 = P2 + 1

2ρv2

2 + ρgh2

h1 = h2, and P2 = 1

P

.11

2, so the equation simplifies to

P1 + 12

ρv12 =

1

P

.11

2 + 1

2ρv2

2

P11 − 1.

1

12 = 1

2ρ(v2

2 − v12)

P1 = =

P1 =

P1 = = 1.06 × 105 Pa(1.00 × 103 kg/m3)(22.6 m2/s2)

(2)(0.107)

(1.00 × 103 kg/m3)(23.0 m2/s2 − 0.36 m2/s2)

(2)(1 − 0.893)

(1.00 × 103 kg/m3)[(4.80 m/s)2 − (0.60 m/s)2]

(2)1 − 1.

1

12

ρ(v22 − v1

2)

(2)1 − 1.

1

12


Givens SolutionsC

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nd W

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on.A

ll rig

hts

rese

rved

.

2. r1 = 4.1

2

0 m = 2.05 m

v1 = 3.0 m/s

r2 = 2.7

2

0 m = 1.35 m

P2 = 82 kPa

ρ = 1.00 × 103 kg/m3

h1 = h2

A1v1 = A2v2

v2 = A

A1v

2

1 = ππr1

r

2

2

v2

1 = r1

r2

2

2v1

v2 = = 6.9 m/s

P1 + 12

rv12 + ρgh1 = P2 + 1

2rv2

2 + ρgh2

h1 = h2, so the equation simplifies to

P1 + 2

1rv12 = P2 +

2

1rv22

P1 = P2 + 2

1r(v22 − v1

2)

P1 = 82 × 103 Pa + 2

1(1.00 × 103 kg/m3)[(6.9 m/s)2 − (3.0 m/s)2]

P1 = 82 × 103 Pa + 2

1(1.00 × 103 kg/m3)(48 m2/s2 − 9.0 m2s2)

P1 = 82 × 103 Pa + 2

1(1.00 × 103 kg/m3)(39 m2/s2)

P1 = 82 × 103 Pa + 2.0 × 104 Pa = 10.2 × 104 Pa = 102 kPa

(2.05 m)2(3.0 m/s)

(1.35 m)2

3. h2 − h1 = 12

h

∆x = 19.7 m

To find the horizontal speed of the cider, recall that for a projectile with no initialvertical speed,

∆x = v∆t

∆y = − 12

g ∆t2 = − 12

h

∆t = h

g

v = ∆∆

x

t = =

g∆hx2

P1 + 1

2 ρv1

2 + ρgh1 = P2 + 12

ρv22 + ρgh2

∆x

h

g


II

Assuming the vat is open to the atmosphere, P1 = P2.

Also assume v2 ≈ 0. Therefore, the equation simplifies to

12

ρv12 + ρgh1 = ρgh2

12

v12 = 1

2

g∆hx2

2

= g(h2 − h1) = g12

h

g ∆

h

x2

= gh

h2 = ∆x2

h = ∆x = 19.7 m

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4. v1 = 59 m/s

g = 9.81 m/s2P1 +

2

1rv12 + rgh1 = P2 +

2

1rv22 + rgh2

Assume v2 ≈ 0 and P1 = P2 = P0.

2

1rv12 + rgh1 = rgh2

h2 − h1 = v

21

g

2

= (2)

(

(

5

9

9

.8

m

1

/

m

s)

/

2

s2) = 1.8 × 102 m

5. h2 − h1 = 66.0 m

g = 9.81 m/s2P1 +

2

1rv12 + rgh1 = P2 +

2

1rv22 + rgh2

Assume v2 ≈ 0 and P1 = P2 = P0 .

2

1rv12 + rgh1 = rgh2

v1 =√

2g(h2− h1) =√

(2)(9.81 m/s2)(66.0m) = 36.0 m/s

6. h2 − h1 = 3.00 × 102 m

g = 9.81 m/s2P1 +

2

1rv12 + rgh1 = P2 +

2

1rv22 + rgh2


2

1rv12 + rgh1 = rgh2

v1 =√

2g(h2− h1) =√

(2)(9.81 m/s2)(3.00× 102 m) = 76.7 m/s

7. h2 − h1 = 6.0 m

g = 9.81 m/s2P1 +

2

1rv12 + rgh1 = P2 +

2

1rv22 + rgh2


2

1rv12 + rgh1 = rgh2

v1 =√

2g(h2− h1) =√

(2)(9.81 m/s2)(6.0 m) = 11 m/s


II

1. V = 3.4 × 105 m3

T = 280 K

N = 1.4 × 1030 atoms

kB = 1.38 × 10−23 J/K

PV = NkBT

P = Nk

VBT =

P = 1.6 × 104 Pa

(1.4 × 1030 atoms)(1.38 × 10−23 J/K)(280 K)

3.4 × 105 m3

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Givens Solutions

2. V = 1.0 × 10−3 m3

N = 1.2 × 1013 molecules

T = 300.0 K

kB = 1.38 × 10−23 J/K

PV = NkBT

P = Nk

VBT =

P = 5.0 × 10−5 Pa

(1.2 × 1013 molecules)(1.38 × 10−23 J/K)(300.0 K)

1.0 × 10−3 m3

3. V = 3.3 × 106 m3


T = 360 K

kB = 1.38 × 10−23 J/K

P V = NkB T

P = Nk

VB T =

P = 2.3 × 105 Pa

(1.5 × 1032 molecules)(1.38 × 10−23 J/K)(360 K)

3.3 × 106 m3

4. N = 1.00 × 1027 molecules

T = 2.70 × 102 K

P = 36.2 Pa

kB = 1.38 × 10−23 J/K

P V = NkB T

V = Nk

PB T =

V = 1.03 × 105 m3

(1.00 × 1027 molecules)(1.38 × 10−23 J/K)(2.70 × 102 K)

36.2 Pa

5. V1 = 3.4 × 105 m2

T1 = 280 K

P1 = 1.6 × 104 Pa

T2 = 240 K

P2 = 1.7 × 104 Pa

P1

T

V

1

1 = P

T2 V

2

2

V2 = P1

P

V

2 T1

1

T2 =

V2 = 2.7 × 105 m3

(1.6 × 104 Pa)(3.4 × 105 m3)(240 K)

(1.7 × 104 Pa)(280 K)

6. A = 2.50 × 102 m2

T = 3.00 × 102 K

P = 101 kPa


kB = 1.38 × 10−23 J/K

P V = NkB T

V = Nk

PB T =

V =

V = lA

l = V

A =

1

2

.

.

7

5

8

0

××

1

1

0

0

6

2m

m

3

2 = 7.12 × 103 m

1.78 × 106 m3

(4.34 × 1031 molecules)(1.38 × 10−23 J/K)(3.00 × 102 K)

101 × 103 Pa

II

7. V = 7.36 × 104 m3

P = 1.00 × 105 Pa

N = 1.63 × 1030 particles

kB = 1.38 × 10−23 J/K

PV = NkBT

T = N

P

k

V

B =

T = 327 K

(1.00 × 105 Pa)(7.36 × 104 m3)(1.63 × 1030 particles)(1.38 × 10−23 J/K)

Givens Solutions

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serv

ed.


8. l = 3053 m

A = 0.040 m2


P = 105 kPa

kB = 1.38 × 10−23 J/K

P V = NkBT

T = P N

V

kB = P

N

A

kB

l

T = = 260 K(105 × 103 Pa)(0.040 m2)(3053 m)

(3.6 × 1027 molecules)(1.38 × 10−23 J/K)

9. P1 = 2.50 × 106 Pa

T1 = 495 K

V1 = 3.00 m3

V2 = 57.0 m3

P2 = 1.01 × 105 Pa

P1

T

V

1

1 = P

T2 V

2

2

T2 = P2

P

V

1 V2 T

1

1 =

T2 = 3.80 × 102 K

(1.01 × 105 Pa)(57.0 m3)(495 K)

(2.50 × 106 Pa)(3.00 m3)

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1. TC = 14°C T = TC + 273.15

T = (14 + 273.15) K

T =

TF = 95

TC + 32.0

TF = 95

(14) + 32.0°F = (25 + 32.0)°F

TF = 57°F

287 K


Givens Solutions

2. TF = (4.00 × 102)°F TC = 59

(TF − 32.0)

TC = 59

[(4.00 × 102) − 32.0]°C = 59

(368)°C

TC =

T = TC + 273.15

T = (204 + 273.15) K

T = 477 K

204°C

3. TC,1 = 117°C

TC,2 = −163°C

∆TC = TC,1 − TC,2 = 117°C − (−163°C)

∆TC = (2.80 × 102)°C

∆TF = 5

9∆TC

∆TF = 5

9(2.80 × 102)°F

∆TF = 504°F

10ChapterHeat

4. TF = 860.0°F TC = 59

(TF − 32.0)

TC = 9

5(860.0 − 32.0)°C = 9

5(828.0)°C

TC = 460.0°C

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5. ∆TF = 49.0°F

TC,2 = 7.00°C

TF = 95

TC + 32.0

TF,2 = 95

(7.00) + 32.0°F = (12.6 + 32.0)°F

TF,2 =

TF,1 = TF,2 − ∆TF = 44.6°F − 49.0°F

TF,1 =

TC,1 = 59

(TF,1 − 32.0)

TC,1 = 59

(−4.4 − 32.0)°C = 59

(−36.4)°C

TC,1 = −20.2°C

−4.4°F

44.6°F

Givens Solutions

6. ∆TC = 56°C

TC,2 = −49°C

TC,1 = TC,2 + ∆TC

TC,1 = −49°C + 56°C

TC,1 = 7°C

T1 = TC,1 + 273.15

T1 = (7 + 273.15) K

T1 =

T2 = TC,2 + 273.15

T2 = (−49 + 273.15) K

T2 = 2.24 × 102 K

2.80 × 102 K

1. mH = 3.05 × 105 kg

vi = 120.0 km/h

vf = 90.0 km/h

∆T = 10.0°C

k = m

∆

w

U

∆T =

(1.00 k

4

g

1

)

8

(

6

1.

J

00°C)

∆PE + ∆KE + ∆U = 0 ∆PE = 0

∆KE = 12

mHvf2 − 1

2mHvi

2 = m

2H(vf

2 − vi2)

∆U = −∆KE = m

2H(vi

2 − vf2)

mw = m∆w

U

∆T

∆∆

U

T =

1

k

∆∆

U

T =

2

m

k∆H

T(vi

2 − vf2)

mw = 120.

h

0 km

2

− 90.0

h

km

2

36

1

0

h

0 s

1

1

0

k

3

m

m

2

mw = (3.64 kg •s2/m2)(1110 m2/s2 − 625 m2/s2)

mw = (3.64 kg •s2/m2)(480 m2/s2)

mw = 1.7 × 103 kg

3.05 × 105 kg

(2

(1

)(

.0

4

0

18

k

6

g)

J

(

)

1

(1

.0

0

0

.0

°C

°C

)

)


7. TF = 116 °F T = TC + 273.15 = 59

(TF − 32.0) + 273.15

T = 59

(116 − 32.0) + 273.15 K = 59

(84) + 273.15 K = (47 + 273.15) K

T = 3.20 × 102 K


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2. h = 228 m

Ti = 0.0°C

g = 9.81 m/s2

fraction of MEf convertedto U = 0.500

k = (∆U/m) = energyneeded to melt ice =3.33 × 105 J/1.00 kg

∆PE = −mgh

When the ice lands, its kinetic energy is tranferred to the internal energy of theground and the ice. Therefore, ∆KE = 0 J.

∆U = (0.500)(MEf ) = −(0.500)(∆PE) = (0.500)(mgh)

∆m

U = (0.500)(gh)

f = ∆km

U =

(0.500

k

)(gh)

f =

f = 3.36 × 10−3

(0.500)(9.81 m/s2)(228 m)

(3.33 × 105 J/1.00 kg)

Givens Solutions

3. vi = 2.333 × 103 km/h

h = 4.000 × 103 m

g = 9.81 m/s2

fraction of ME convertedto U = 0.0100

k = m

∆∆U

T =

(1.00 k

(3

g

5

)(

5

1

J

.

)

00°C)

4. h = 8848 m

g = 9.81 m/s2


Ti = −18.0°C

∆U

∆T

/m =

4

1

4

.

8

00

J

°/k

C

g

∆PE = −mgh

MEf = −∆PE

When the hook lands, its kinetic energy is transferred to the internal energy of thehook and the ground. Therefore, ∆KE = 0 J.

∆U = (0.200)(MEf) = (−0.200)(∆PE) = (0.200)(mgh)

∆U/m = (0.200)(gh)

∆T = ∆U

∆T

/m

∆m

U = 414.800

J

°/k

C

g[(0.200)(gh)]

∆T = 414.800

J

°/k

C

g(0.200)(9.81 m/s2)(8848 m) = 38.7°C

Tf = Ti + ∆T = −18.0°C + 38.7°C

Tf = 20.7°C

(0.0100)(∆PE + ∆KE) + ∆U = 0

∆PE = PEf − PEi = 0 − mgh = −mgh

∆KE = KEf − KEi = 0 − 12

mvi2 = − 1

2mvi

2

∆U = −(0.0100)(∆PE + ∆KE) = −(0.0100)(m)−gh − 12

v2 = (0.0100)(m)gh + 12

v2

∆T = =

∆T =

∆T =

∆T = 7.02°C

(0.0100)[(3.92 × 104 m2/s2) + (2.100 × 105 m2/s2)]

355 m2/s2 •°C

(0.0100)(9.81 m/s2)(4.000 × 103 m) + 12

2.33

3

3

6

×00

10

s/

6

h

m/h

2

355 m2/s2 •°C

(0.0100)gh + 12

v2(1.00 k

3

g

5

)

5

(1

J

.00°C)

∆m

U

k


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7. ∆T = 0.230°C

g = 9.81 m/s2


∆

∆U

T

/m =

41

1

8

.0

6

0

J

°/

C

kg

∆m

U = ∆T ∆∆

U

T

/m =

(0.100

m

)(MEf ) =

(0.100

m

)(∆PE) = (0.100)(gh)

Because the kinetic energy at the bottom of the falls is converted to the internal en-ergy of the water and the ground, ∆KE = 0 J.

h =

h = 41

1

8

.0

6

0

J

°/

C

kg

h = 981 m

0.230°C(0.100)(9.81 m/s2)

∆T ∆

∆U

T

/m

(0.100)(g)

5. hi = 629 m

g = 9.81 m/s2

vf = 42 m/s


m = 3.00 g

Because hf = 0 m, PEf = 0 J. Further, vi = 0 m/s, so KEi = 0 J.

∆U = −(0.050)(∆ME) = −(0.050)(∆PE + ∆KE)

∆PE = PEf − PEi = 0 J − PEi = −mgh

∆KE = KEf − KEi = KEf − 0 J = 12

mvf2

∆U = (0.050)(−∆PE − ∆KE) = (0.050)(mgh − 12

mvf2) = (0.050)(m)(gh − 1

2vf

2)

∆U = (0.050)(3.00 × 10−3 kg)[(9.81 m/s2)(629 m) − (0.5)(42 m/s)2]

∆U = (0.050)(3.00 × 10−3 kg)(6170 m2/s2 − 880 m2/s2)

∆U = (0.050)(3.00 × 10−3 kg)(5290 m2/s2)

∆U = 0.79 J

Givens Solutions

6. h = 2.49 m

g = 9.81 m/s2

m = 312 kg

v = 0.50 m/s


(0.500)(∆PE + ∆KE) + ∆U = 0

∆PE = PEf − PEi = 0 − mgh = −mgh

∆KE = KEf − KEi = 0 − 2

1mv2 = −m

2

v2

∆U = −(0.500)(∆PE + ∆KE) = −(0.500)−mgh − m

2

v2

= (0.500)mgh + m

2

v2

∆U = (0.500)[(312 kg)(9.81 m/s2)(2.49 m) + (0.5)(312 kg)(0.50 m/s)2]

∆U = (0.500)[(7.62 × 103 J) + 39 J]

∆U = 3.83 × 103 J


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1. Q = (2.8 × 109 W)(1.2 s)

cp,c = 387 J/kg •°C

Tc = 26°C

Tf = 21°C

Q = mccp,c(Tc − Tf )

mc = cp,c(T

Q

c − Tf )

mc =

mc = = 1.7 × 106 kg(2.8 × 109 W)(1.2 s)(387 J/kg•°C)(5°C)

(2.8 × 109 W)(1.2 s)(387 J/kg•°C)(26°C − 21°C)

2. mw = 14.3 × 103 kg

Tw = 20.0°C

Tx = temperature ofburning wood = 280.0°C

Tf = 100.0°C

cp,x = specific heat capacity of wood =1.700 × 103 J/kg •°C

cp,w = 4186 J/kg •°C

cp,xmx (Tx − Tf ) = cp,wmw(Tf − Tw)

mx =

mx =

mx = 1.56 × 105 kg

k4g

18

•°6

C

J(143 × 103 kg)(100.0 − 20.0) °C

(1.700 × 103 J/kg •°C)(280.0 − 100.0) °C

cp,wmw(Tf − Tw)

cp,x(Tm − Tf )


Givens Solutions

3. ∆U = Q = (0.0100)(1.450 GW)(1.00 year)

cp,x = specific heat capacityof iron = 448 J/kg •°C

mx = mass of steel =25.1 × 109 kg

Q = cp,x mx ∆T

∆T = cp,x

Q

mx

∆T = 365

1

.2

y

5

ea

d

r

ays12d

4

a

h

y36

1

0

h

0 s

∆T = 40.7°C

(0.0100)(1.450 × 109 W)(1.00 year)

(448 J/kg •°C)(25.1 × 109 kg)

4. ml = 2.25 × 103 kg

Tl,i = 28.0°C

cp,l = cp,w = 4186 J/kg •°C

mi = 9.00 × 102 kg

Ti,i = −18.0°C

cp,i = 2090 J/kg •°C

Ti,f = 0.0°C

mlcp,l (Tl,i − Tl,f) = micp,i(Ti,f − Ti,i)

Tl,f = Tl,i − m

m

l

ic

c

p

p

,

,

l

i(Ti,f − Ti,i)

Tl,f = (28.0°C) − 29.

.

2

0

5

0

××

1

1

0

03

2

k

k

g

g [(0.0°C) − (−18.0°C)]

Tl,f = 28.0°C − 3.59°C = 24.4°C

2090 J/kg •°C4186 J/kg •°C

5. mw = 1.33 × 1019 kg

Tw = 4.000°C

cp,w = 4186 J/kg •°C

P = 1.33 × 1010 W

∆t = 1.000 × 103 years

P∆t = Q = mwcp,w(Tf − Tw)

Tf = mP

w

∆cp

t

,w + Tw

Tf = + 4.000°C

Tf = (7.54 × 10−3)°C + 4.000°C

Tf = 4.008°C

(1.33 × 1010 W)(1.000 × 103 years)36

1

5.

y

2

e

5

a

d

r

ay12d

4

a

h

y36

1

0

h

0 s

(1.33 × 1019 kg)(4186 J/kg •°C)


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7. Tm = −62.0°C

Tw = 38.0°C

mm = 180 g

mw = 0.500 kg

Tf = 36.9°C

cp,w = 4186 J/kg •°C

mmcp,m(Tf − Tm) = mwcp,w(Tw − Tf )

cp,m = m

m

m

w(cp,w)TT

f

w

−−T

T

m

fcp,m = 00.5.1080k

k

g

g(4186 J/kg •°C)

cp,m =

The metal could be gold (cp = 129 J/kg •°C) or lead (cp = 128 J/kg •°C).

1.3 × 102 J/kg •°C

(38.0°C) − (36.9°C)(36.9°C) − (−62.0°C)

Givens Solutions

6. Ti = 18.0°C

Tf = 32.0°C

Q = 20.8 kJ

mx = 0.355 kg

Q = mxcp,x∆T

cp,x = mx

Q

∆T =

mx(T

Q

f − Ti)

cp,x =

cp,x = 4190 J/kg •°C

20.8 × 103 J(0.355 kg)(32.0°C − 18.0°C)

1. mw,S = 1.20 × 1016 kg

mw,E = 4.8 × 1014 kg

TE = 0.0°C

TS = 100.0°C

cp,w = 4186 J/kg •°C

Lf of ice = 3.33 × 105 J/kg

QS = energy transferred by heat from Lake Superior = cp,wmw,S(TS − Tf)

QE = energy transferred by heat to Lake Erie = mw,ELf + mw,Ecp,w(Tf − TE)

From the conservation of energy,

QS = QE

cp,wmw,S(TS − Tf ) = mw,ELf + mw,Ecp,w(Tf − TE)

(mw,Ecp,w + cp,wmw,S)Tf = cp,wmw,STS + mw,Ecp,wTE − mw,ELf

Tf = , where TE = 0.0°C

Tf =

Tf =

Tf =

Tf = 92.9°C

(5.02 × 1021 J) − (1.6 × 1020 J )(4186 J/kg •°C)(1.25 × 1016 kg)

(4186 J/kg •°C)(1.20 × 1016 kg)(100.0°C) − (4.8 × 1014 kg)(3.33 × 105 J/kg)

(4186 J/kg•°C)[(4.8 × 1014 kg) + (1.20 × 1016 kg)]

cp,wmw,STS − mw,ELfcp,w (mw,E + mw,S)

cp,w(mw,STS + mw,ETE) − mw,ELfcp,w (mw,E + mw,S)


2. Tf = −235°C

Tfreezing = 0.0°

mw = 0.500 kg

cp,w = 4186 J/kg •°C

cp,ice = cp,i = 2090 J/kg •°C

Lf of ice = 3.33 × 105 J/kg

Qtot = 471 kJ

Qtot = cp,wmw(Ti − 0.0°) + Lf mw + mwcp,i(0.0°− Tf) = cp,wmwTi + Lf mw − cp,imwTf

Ti =

Ti =

Ti =

Ti = 28°C

(4.71 × 105 J) − (1.66 × 105 J) − (2.46 × 105 J)

2093 J/°C

(4.71 × 105 J) − (3.33 × 105 J/kg)(0.500 kg) + (−235°C)(2090 J/kg •°C)(0.500 kg)

(4186 J/kg •°C)(0.500 kg)

Qtot − Lf mw + cp,imwTf

cp,wmw


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3. Ti = 0.0°C

mi = 4.90 × 106 kg

Lf of ice = 3.33 × 105 J/kg

Ts = 100.0°C

Lv of steam = 2.26 × 106 J/kg

msLv = miLf

ms = m

L

i

v

Lf

ms =

ms = 7.22 × 105 kg

(4.90 × 106 kg)(3.33 × 105 J/kg)

2.26 × 106 J/kg

Givens Solutions

4. ms = 1.804 × 106 kg

Lf of silver = Lf,s =8.82 × 104 J/kg

Lf of ice = Lf,i =3.33 × 105 J/kg

msLf,s = miLf,i

mi = m

L

sL

f,i

f,s

mi =

mi = 4.78 × 105 kg

(1.804 × 106 kg)(8.82 × 104 J/kg)

3.33 × 105 J/kg

5. mg = 12.4414 kg

Tg,i = 5.0°C

Q = 2.50 MJ

cp,g = 129 J/kg •°C

Tg,f = 1063°C

Q = mgcp,g(Tg,f − Tg,i) + mgLf

Lf =

Lf =

Lf =

Lf = 6.4 × 104 J/kg

(2.50 × 106 J) − (1.70 × 106 J)

12.4414 kg

(2.50 × 106 J) − (12.4414 kg)(129 J/kg •°C)(1063°C − 5.0°C)

12.4414 kg

Q − mgcp,g(Tg, f − Tg,i)mg

6. Vp = 7.20 m3

Vc = (0.800)(Vp)

rc = 8.92 × 103 kg/m3

Lf = 1.34 × 105 J/kg

a. mc = rcVc

mc = (8.92 × 103 kg/m3)(0.800)(7.20 m3)

mc =

b. fraction of melted coins = m

Q

cLf =

1

1

0

5

0

Q = 1

1

0

5

0 mcLf =

Q = 1.03 × 109 J

15(5.14 × 104 kg)(1.34 × 105 J/kg)

100

5.14 × 104 kg


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Givens Solutions

7. mw = 3.5 × 1019 kg

Ti = 10.0°C

Tf = 100.0°C

cp,w = 4186 J/kg•°C

Lv of water = 2.26 × 106 J/kg

P = 4.0 × 1026 J/s

a. Q = mwcp,w(Tf − Ti) + mwLv

Q = (3.5 × 1019 kg)(4186 J/kg •°C)(100.0°C − 10.0°C) + (3.5 × 1019 kg)(2.26 × 106 J/kg)

Q = 1.3 × 1025 + 7.9 × 1025 J =

b. Q = P∆t

∆t = Q

P =

4

9

.0

.2

××1

1

0

02

2

6

5

J/

J

s = 0.23 s

9.2 × 1025 J

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1. P = 5.1 kPa

W = 3.6 × 103 J

Vi = 0.0 m3

W = P∆V

∆V = W

P =

5

3

.1

.6

××1

1

0

03

3

P

J

a = 7.1 × 10−1 m3

V = Vi + ∆V = 0.0 m3 + 7.1 × 10−1 m3 = 7.1 × 10−1 m3


Givens Solutions

2. m = 207 kg

g = 9.81 m/s2

h = 3.65 m

P = 1.8 × 106 Pa

W = mgh = −P∆V

∆V = m

−g

P

h

∆V = = −4.1 × 10−3 m3(207 kg)(9.81 m/s2)(3.65 m)

−(1.8 × 106 Pa)

3. rf = 1.22 m

ri = 0.0 m

W = 642 kJ

W = P∆V

P = ∆W

V

∆V = 3

4p(rf3 − ri

3)

P = = = 8.44 × 104 Pa642 × 103 J

3

4p (1.22 m)3 − (0.0 m)3W

3

4p(rf3 − ri

3)

4. rf = 7.0 × 105 km

= 7.0 × 108 m

ri = 0.0 m

W = 3.6 × 1034 J

W = P∆V

V = 3

4pr3

∆V = Vf − Vi = 3

4p(rf3 − ri

3)

W = (P)( 3

4p)(rf3 − ri

3)

P = W

(3

4p)(rf3 − ri

3)

5. P = 87 kPa

∆V = −25.0 × 10−3 m3

W = P∆V

W = (87 × 103 Pa)(−25.0 × 10−3 m3) = −2.2 × 103 J

11ChapterThermodynamics

P = = 2.5 × 107 Pa(3.6 × 1034 J)

(

3

4p )[(7.0 × 108 m)3 − (0.0 m)3]

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6. rf = 29.2 cm

ri = 0.0 m

P = 25.0 kPa

m = 160.0 g

W = P∆V

∆V = 3

4p(rf3 − ri

3)

W = 12

mv2

12

mv2 = (P)43

p (rf3 − ri

3)

v = 8

3

pm

P (rf

3 − ri3)

v = (29.2 × 10−2m)3 − (0.0m)3v = 181 m/s

(8p)(25.0 × 103 Pa)(3)(160.0 × 10−3 kg)

Givens Solutions

1. m = 227 kg

h = 8.45 m

g = 9.81 m/s2

Ui = 42.0 kJ

Q = 4.00 kJ

∆U = Uf − Ui = Q − W

W = mgh

Uf = Ui + Q − W = Ui + Q − mgh

Uf = (42.0 × 103 J) + (4.00 × 103 J) − (227 kg)(9.81 m/s2)(8.45 m)

Uf = (42.0 × 103 J) + (4.00 × 103 J) − (18.8 × 103 J)

Uf = 27.2 × 103 J = 27.2 kJ


2. m = 4.80 × 102 kg

Q = 0 J

v = 2.00 × 102 m/s

Uf = 12.0 MJ

a. Assume that all work is transformed to the kinetic energy of the cannonball.

W = 12

mv2

W = 12

(4.80 × 102 kg)(2.00 × 102 m/s)2

W =

b. ∆U = Q − W = −W

Uf − Ui = −W

Ui = Uf + W

Ui = (12.0 × 106 J) + (9.60 × 106 J)

Ui = 21.6 × 106 J = 21.6 MJ

9.60 × 106 J = 9.60 MJ

3. m = 4.00 × 104 kg

cp = 4186 J/kg •°C

∆T = −20.0°C

W = 1.64 × 109 J

a. Q = mcp∆T

Q = (4.00 × 104 kg)(4186 J/kg •°C)(−20.0°C)

Q =

b. Q of gas = −Q of jelly = −(−3.35 × 109 J) = 3.35 × 109 J

∆U = Q − W

∆U = (3.35 × 109 J) − (1.64 × 109 J)

∆U = 1.71 × 109 J

−3.35 × 109 J


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6. P = ∆Q

t = 5.9 × 109 J/s

∆t = 1.0 s

∆U = 2.6 × 109 J

∆U = Q − W = P∆t − W

W = P∆t − ∆U

W = (5.9 × 109 J/s)(1.0 s) − (2.6 × 109 J)

W = 3.3 × 109 J

Givens Solutions

5. m = 5.00 × 103 kg

v = 40.0 km/h

Ui = 2.50 × 105 J

Uf = 2Ui

W = 12

mv2

∆U = Uf − Ui = Q − W

Uf = 2Ui

∆U = 2Ui − Ui = Ui

Q = ∆U + W = Ui + 12

mv2

Q = (2.50 × 105 J) + 12

(5.00 × 103 kg)40.0

h

km36

1

0

h

0 s 110

k

3

m

m

2

Q = (2.50 × 105 J) + (3.09 × 105 J)

Q = 5.59 × 105 J

4. m = 1.64 × 1015 kg

h = 75.0 m

g = 9.81 m/s2

Ti = 6.0°C

Tf = 100.0°C

cp,w = 4186 J/kg •°C

Lv of water = 2.26 × 106 J/kg

∆U = (−0.900)(Ui)

W = mgh = (1.64 × 1015 kg)(9.81 m/s2)(75.0 m)

W = 1.21 × 1018 J

Q = −mcp,w(Tf − Ti) − mLv = −m[cp,w(Tf − Ti) + Lv]

Tf − Ti = 100.0°C − 6.0°C = 94.0°C

Q = −(1.64 × 1015 kg)[(4186 J/kg •°C)(94.0°C) + (2.26 × 106 J/kg)]

Q = −4.35 × 1021 J

∆U = Uf − Ui = (−0.900)(Ui) = Q − W

Uf = (1 − 0.900)Ui = (0.100)(Ui)

−∆U = Ui − Uf = 0.

U

10

f

0 − Uf = (0.900)0.

U

10

f

0 = W − Q

Uf = 00.

.

1

9

0

0

0

0(W − Q)

Uf = 00.

.

1

9

0

0

0

0[(1.21 × 1018 J) − (−4.35 × 1021 J)]

Uf =

(Note: Nearly all of the energy is used to increase the temperature of the water and to vaporize the water.)

4.83 × 1020 J


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7. h = 1.00 × 102 m

v = 141 km/h

Ui = 40.0 ΜJ

m = 76.0 kg

a. ∆U = Q − W

All energy is transferred by heat, so W =

b. ∆U = Q = 12

mv2

∆U

U

i × 100 =

m

2U

v2

i (100)

∆U

U

i × 100 =

∆U

U

i × 100 = 0.146 percent

(76.0 kg)141

h

km

2

36

1

0

h

0 s

2

110k

3

m

m

2

(100)

(2)(40.0 × 106 J)

0 J

Givens Solutions

1. eff = 8 percent = 0.080

Qh = 2.50 kJeff =

Qh

Q

−

h

Qc = 1 − Q

Q

h

c

−Qc = Qh(eff − 1)

Qc = Qh(1 − eff )

Qc = (2.50 kJ)(1 − 0.08) = (2.50 kJ)(0.92)

Qc =

W = Qh − Qc

W = 2.50 kJ − 2.3 kJ

W = 0.2 kJ

2.3 kJ


2. Pnet = 1.5 MW

eff = 16 percent = 0.16

Qh = 2.0 × 109 J

eff = W

Qn

h

et = Pn

Qet

h

∆t

∆t = (eff

P

)

n

(

e

Q

t

h)

∆t =

∆t = 2.1 × 102 s

(0.16)(2.0 × 109 J)

1.5 × 106 W

3. Pnet = 19 kW

eff = 6.0 percent = 0.060

∆t = 1.00 h

Wnet = Pnet∆t

eff = W

Qn

h

et

Qh = W

efn

fet =

Pn

ee

ft

f

∆t

Qh =

Qh = 1.1 × 106 kJ = 1.1 × 109 J

(19 kW)(1.00 h)3.6 ×1 h

103 s

(0.060)


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4. Pnet = 370 W

∆t = 1.00 min = 60.0 s

eff = 0.19

eff = W

Qn

h

et = Pn

Qet

h

∆t

Qh = Pn

ee

ft

f

∆t

Qh =

Qh = 1.2 × 105 J = 120 kJ

(370 W)(60.0 s)

0.19

Givens Solutions

5. Wnet = 2.6 MJ

Qh/m = 32.6 M

kg

J

m = 0.80 kg

eff = W

Qn

h

et =

eff =

eff = 0.10 = 10 percent

2.6 MJ

32.6 M

kg

J(0.80 kg)

Wnet

Q

mh(m)

6. m = 3.00 × 104 kg

g = 9.81 m/s2

h = 1.60 × 102 m

Qc = 3.60 × 108 J

Wnet = mgh = Qh − Qc

Qh = Wnet + Qc

eff = W

Qn

h

et = Wn

W

et

n

+et

Qc =

mg

m

h

g

+h

Qc

eff =

eff = 0.12

(3.00 104 kg)(9.81 m/s2)(1.60 102 m)(3.00 104 kg)(9.81 m/s2)(1.60 102 m) + 3.60 108 J

Chapter 12Vibrations and Waves

II

1. m = 0.019 kg

g = 9.81 m/s2

k = 83 N/m

k = x

F =

m

x

g

x =

x = 2.25 × 10−3 m

(0.019 kg)(9.81 m/s2)

83 N/m


Givens Solutions

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2. m = 187 kg

k = 1.53 × 104 N/m

g = 9.81 m/s2

k = x

F =

m

x

g

x = (18

1

7

.5

k

3

g

×)(

1

9

0

.841

N

m

/m

/s2) = 0.120 m

3. m1 = 389 kg

x2 = 1.2 × 10−3 m

m2 = 1.5 kg

x

F

1

1 = F

x2

2

x1 = F

F1

2

x2 = m

m1g

2

x

g2

x1 = = 0.31 m(389 kg)(1.2 × 10−3 m)

(1.5 kg)

4. m = 18.6 kg

x = 3.7 m

g = 9.81 m/s2

k = x

F =

m

x

g =

k = 49 N/m

(18.6 kg)(9.81 m/s2)

(3.7 m)

5. h = 533 m

x1 = 13

h

m = 70.0 kg

x2 = 23

h

g = 9.81 m/s2

k = x

F =

(x2

m

−g

x1) =

3m

h

g

k = = 3.87 N/m3(70.0 kg)(9.81 m/s2)

(533 m)

6. k = 2.00 × 102 N/m

x = 0.158 m

g = 9.81 m/s2

a. F = kx = (2.00 × 102 N/m)(0.158 m) =

b. m = F

g =

k

g

x

m =

m = 3.22 kg

(2.00 × 102 N/m)(0.158 m)

(9.81 m/s2)

31.6 N



II

7. h = 1.02 × 104 m

L = 4.20 × 103 m

k = 3.20 × 10−2 N/m

F = kx = k(h − L)

F = (3.20 × 10−2 N/m)(6.0 × 103 m) = 190 N

Givens Solutions

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1. L = 6.7 m

g = 9.81 m/s2T = 2π

L

g = 2π

(9(.8

61

.7m

m/

)

s2) = 5.2 s


2. L = 0.150 m

g = 9.81 m/s2 T = 2pL

g = 2p = 0.777 s

(0.150 m)(9.81 m/s2)

3. x = 0.88 m

g = 9.81 m/s2 T = 2p4

g

x = 2p

T = 3.8 s

4(0.88 m)(9.81 m/s2)

4. f = 6.4 × 10−2 Hz

g = 9.81 m/s2 T = 1

f = 2p

L

g

L = 4p

g2f 2 =

L = 61 m

(9.81 m/s2)4π2(6.4 × 10−2 Hz)2

5. t = 3.6 × 103 s

N = 48 oscillations

g = 9.81 m/s2

T = 2pL

g =

N

t

L = =

L = 1.4 × 103 m

(3.6 × 103 s)2(9.81 m/s2)

4p2 (48)2

N

t

2

g

4p2

8. h = 348 m

L = 2.00 × 102 m

k = 25.0 N/m

g = 9.81 m/s2

F = kx = k(h − L) = mg

m = = 377 kg(25.0 N/m)(148 m)

(9.81 m/s2)

6. L = 1.00 m

T = 10.5 sg =

4pT

2

2

L = = 0.358 m/s24p2(1.00 m)

(10.5 s)2


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1. f1 = 90.0 Hz

k = 2.50 × 102 N/mT = 2p

m

k =

3.00 ×1

10−2f1

m = =

m = 0.869 kg

(2.50 × 102 N/m)4p 2(3.00 × 10−2)2(90.0 Hz)2

k4p 2(3.00 × 10−2)2f1

2


Givens Solutions

2. m1 = 3.5 × 106 kg

f = 0.71 Hz

k = 1.0 × 106 N/m

T = 1

f = 2p

m1k

+ m2m2 =

4pk2f 2 − m1

m2 = − 3.5 × 104 kg = 1.5 × 104 kg(1.0 × 106 N/m)4p2(0.71 Hz)2

3. m = 20.0 kg

f = 4

6

2

0

.7

s = 0.712 Hz

k = 4p

T

2

2m

= 4p2mf 2

k = 4p2 (20.0 kg)(0.712 Hz)2

k = 4.00 × 102 N/m

4. m = 2.00 × 105 kg

T = 1.6 sT = 2p

m

k

k = 4p

T

2

2

m = = 3.1 × 106 N/m

4p 2 (2.00 × 105 kg)

(1.6 s)2

5. m = 2662 kg

x = 0.200 m

g = 9.81 m/s2

T = 2pm

k = 2p

m

mx

g

T = 2p(

(

90

.8

.2

1

0 m

0 m

/s2

)) = 0.897 s

1. f = 2.50 × 102 Hz

v = 1530 m/sl =

v

f =

2.5

1

0

53

×0

1

m

02/s

Hz = 6.12 m


6. m = 10.2 kg

k = 2.60 × 102 N/mT = 2p

m

k = 2p = 1.24 s

(10.2 kg)(2.60 × 102 N/m)


II

2. f = 123 Hz

v = 334 m/sl =

v

f =

(

(

3

1

3

2

4

3

m

H

/

z

s

)

) = 2.72 m

Givens Solutions

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3. l = 2.0 × 10−2 m

v = 334 m/sf =

lv

= (2.

(

0

33

×4

1

m

0−/2

s)

m) = 1.7 × 104 Hz

4. l = 2.54 m

v = 334 m/sf =

lv

= (

(

3

2

3

.5

4

4

m

m

/s

)

) = 131 Hz

5. f = 73.4 Hz

l = 4.50 mv = fl = (73.4 Hz)(4.50 m) = 3.30 × 102 m/s

6. f = 2.80 × 105 Hz

l = 5.10 × 10−3 m

∆x = 3.00 × 103 m

v = fl = (2.80 × 105 Hz)(5.10 × 10−3 m)

v =

∆t = ∆v

x =

(

(

1

3

.4

.0

3

0

××1

1

0

03

3

m

m

/s

)

)

∆t = 2.10 s

1.43 × 103 m/s

Chapter 13Sound

II

1. Intensity = 3.0 × 10−3 W/m2

r = 4.0 mIntensity =

4pP

r 2

P = 4pr2(Intensity) = 4p (4.0 m)2(3.0 × 10−3 W/m2)

P = 0.60 W


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2. r = 8.0 × 103 m

Intensity = 1.0 × 10−12 W/m2Intensity =

4pP

r 2

P = 4pr2(Intensity)

P = 4p(8.0 × 103 m)2(1.0 × 10−12 W/m2) = 8.0 × 10−4 W

3. Intensity = 1.0 × 10−12 W/m2

P = 2.0 × 10−6 WIntensity =

4pP

r 2

r = 4p(In

Ptensity)

r = = 4.0 × 102 m2.0 × 10−6 W

4p (1.0 × 10−12 W/m2)

4. Intensity = 1.1 × 10−13 W/m2

P = 3.0 × 10−4 Wr 2 =

4p In

P

tensity

r = = = 1.5 × 104 m(3.0 × 10−4 W)

4p(1.1 × 10−13 W/m2)

P4p Intensity

5. P = 1.0 × 10−4 W

r = 2.5 m

Intensity = 4p

P

r2

Intensity = = 1.3 × 10−6 W/m2(1.0 × 10−4 W)

4p(2.5 m)2

6. Intensity = 2.5 × 10−6 W/m2

r = 2.5 m

P = 4pr2(Intensity)

P = 4p(2.5 m)2(2.5 × 10−6 W/m2)

P = 2.0 × 10−4 W


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1. f15 = 26.7 Hz

v = 334 m/s

n = 15

L = 4

15

f1

v

5 =

L = 46.9 m

15(334 m/s)4(26.7 Hz)


Givens Solutions

2. l = 3.47 m

vs = 5.00 × 102 m/s

n = 3

va = 334 m/s

L = l2

v

vs

a

n =

L = 7.79 m

(3.47 m)(5.00 × 102 m/s)(3)

2(334 m/s)

3. n = 19

L = 86 m

v = 334 m/s

f19 = 2

n

L

v =

19

2

(

(

3

8

3

6

4

m

m

)

/s)

f19 = 37 Hz

4. L = 3.50 × 102 m

f75 = 35.5 Hz

n = 75

f75 = 7

2

5

L

v

v = 2L

7

f

575 =

v = 331 m/s

2(3.50 × 102 m)(35.5 Hz)

75

5. L = 4.7 × 10−3 m

l = 3.76 × 10−3 m

ln = 4

n

L

n = l4L

n = = 5

4(4.7 × 10−3 m)(3.76 × 10−3 m)


Chapter 14Light and Reflection

II

1. f = 9.00 × 108 Hz

d = 60.0 m

c = 3.00 × 108 m/s

c = fl

ld

= = d

c

f

ld

=

ld

= 1.80 × 102 wavelengths

(60.0 m)(9.00 × 108 Hz)

(3.00 × 108 m/s)

d

f

c


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2. f = 5.20 × 1014 Hz

d = 2.00 × 10−4 m

c = 3.00 × 108 m/s

c = fl

ld

= = d

c

f

ld

=

ld

= 347 wavelengths

(2.00 × 10−4 m)(5.20 × 1014 Hz)

(3.00 × 108 m/s)

d

f

c

3. f = 2.40 × 1010 Hz

c = 3.00 × 108 m/s

c = fl

l = c

f

l =

l = 1.25 × 10−2 m = 1.25 cm

(3.00 × 108 m/s)(2.40 × 1010 Hz)

4. l = 1.2 × 10−6 m

c = 3.00 × 108 m/s

c = fl

f = lc

f =

f = 2.5 × 1014 Hz = 250 TH

(3.00 × 108 m/s)(1.2 × 10−6 m)



II

5. l1 = 2.0 × 10−3 m

l2 = 5.0 × 10−3 m

c = 3.00 × 108 m/s

c = fl

f1 = lc

1 =

f1 = 1.5 × 1011 Hz = 15 × 1010 Hz

f2 = lc

2 =

f2 = 6.0 × 1010 Hz

6.0 × 1010 Hz < f < 15 × 1010 Hz

(3.00 × 108 m/s)(5.0 × 10−3 m)

(3.00 × 108 m/s)(2.0 × 10−3 m)

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6. f = 10.0 Hz

c = 3.00 × 108 m/s

c = fl

l = c

f

l =

l = 3.00 × 107 m = 3.00 × 104 km

(3.00 × 108 m/s)

(10.0 Hz)

1. p = 3.70 × 105 m

f = 2.50 × 105 mq

1 =

1

f −

p

1

q

1 =

(2.50 ×1

105 m) −

(3.70 ×1

105 m) =

(4.00

1

×m

10−6) −

(2.70

1

×m

10−6)

q = 1.30

1

×m

10−6

−1

= 7.69 × 105 m =

M = − p

q =

−3

7

.7

.6

0

9

××

1

1

0

05

5

m

m

M = −2.08

769 km


2. h = 8.00 × 10−5 m

f = 2.50 × 10−2 m

q = −5.9 × 10−1 m

p

1 +

1

q =

1

f

p

1 =

1

f −

1

q

p

1 =

(2.50 ×1

10−2 m) −

(−5.9 ×1

10−1 m) =

4

1

0

m

.0 −

1

1

.6

m

9 =

4

1

1

m

.7

p =

M = h

h

= −

p

q = −

(

(

−2.

5

4

.

0

9

××

1

1

0

0−

−

2

1

m

m

)

) = 24.6

2.40 × 10−2 m

II

3. h = −28.0 m

h = 7.00 m

f = 30.0 m

Image is real, so q > 0 andh < 0.

M = − p

q =

h

h

q = − p

h

h

p

1 +

q

1 =

1

f

p

1 + =

1

f

p

11 −

h

h

=

1

f

p = f 1 − h

h

p = (30.0 m)1 + 7

2

.

8

0

.

0

0

m

m

p = 37.5 m

1

−p

h

h

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4. h = 67.4 m

h = 1.69 m

R = 12.0 m

(h > 0, q < 0)

M = − p

q =

h

h

q = − p

h

h

p

1 +

q

1 =

R

2

p

1 + =

R

2

p

11 −

h

h

=

R

2

p = R

21 −

h

h

p = (12.

2

0 m) 1 −

1

6

.

7

6

.

9

4

m

m = (6.00 m)(0.975)

p =

Image is virtual and therefore upright.

5.85 m

1

−p

h

h

II

5. h = 32 m

f = 120 m

p = 180 m

p

1 +

q

1 =

1

f

q

1 =

1

f −

p

1

q

1 =

(120

1

m) −

(180

1

m) =

0.

1

00

m

83 −

0.

1

00

m

56 =

0.

1

00

m

27

q =

M = h

h

= −

p

q

h= − q

p

h

h= −(37

(

0

18

m

0

)

m

(3

)

2 m)

h =

The image is inverted (h < 1)and real (q > 0)

−66 m

3.7 × 102 m

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6. h = 0.500 m

R = 0.500 m

p = 1.000 m

q

1 =

R

2 −

p

1

q

1 =

(0.50

2

0 m) −

(1.00

1

0 m) =

4

1

.0

m

0 −

1

1

.0

m

00 =

3

1

.0

m

0

q =

M = − p

q =

h

h

h = − q

p

h =

h =

The image is real (q > 0).

−0.166 m = −166 mm

−(0.333 m)(0.500 m)

(1.000 m)

0.333 m = 333 mm

II

7. p = 1.00 × 105 m

h = 1.00 m

h = −4.00 × 10−6 m

(h < 0 because image is inverted)

M = − p

q =

h

h

q = − p

h

h

p

1 +

q

1 =

R

2

R = =

R = = (

(

1

2.

+00

2.

×50

10

×

5

1

m

05)

)

R = 0.800 m = 80.0 cm

2(1.00 × 105 m)

1 + (4.0

(

0

1.

×00

10

m−6

)

m)

2p

1 − h

h

2

p

1 −

p

h

h

Givens Solutions

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8. h = 10.0 m

p = 18.0 m

h = −24.0 m

Image is real, so q > 0, andh must be negative.

M = − p

q =

h

h

q = − h

h

p

p

1 +

q

1 =

R

2

R = =

R = = (1

3

+6.

0

0

.4

m

17) =

(

(

3

1

6

.

.

4

0

1

m

7)

)

R = 25.4 m

2(18.0 m)

1 + 1

2

0

4

.

.

0

0

m

m

2p

1 − h

h

2

p

1 −

p

h

h


2. p = 553 m

R = −1.20 × 102 mp

1 +

q

1 =

R

2

q

1 =

R

2 −

p

1

q

1 =

(−1.20 ×2

102 m) −

(553

1

m) = −

0.

1

01

m

67 −

0.0

1

0

m

181 = −

0.

1

01

m

85

q = −54.1 m

M = − p

q =

−(

(

−5

5

5

4

3

.1

m

m

)

)

M = 9.78 × 10−2

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1. R = −6.40 × 106 m

p = 3.84 × 108 m

h = 3.475 × 106 m

p

1 +

q

1 =

R

2

q

1 =

R

2 −

p

1

q

1 =

(−6.40 ×2

106 m) −

(3.84 ×1

108 m) = −

(3.13

1

×m

10−7) −

(2.60

1

×m

10−9)

q = −3.16

1

×m

10−7

−1

= −3.16 × 106 m = −3.16 × 103 km

M = − p

q =

h

h

h = − q

p

h

h =

h = 2.86 × 104 m = 28.6 km

−(−3.16 × 106 m)(3.475 × 106 m)

(3.84 × 108 m)


Givens Solutions

3. R = −35.0 × 103 m

p = 1.00 × 105 m

p

1 +

q

1 =

R

2

q

1 =

R

2 −

p

1

q

1 =

(−35.0 ×2

103 m) −

(1.00 ×1

105 m) = −

(5.71

1

×m

10−5) −

(1.00

1

×m

10−5)

q = −6.71

1

×m

10−5

−1

= −1.49 × 104 m = −14.9 km


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4. h = 1.4 × 106 m

h = 11.0 m

R = −5.50 m

M = h

h

M = (1.4

11

×.0

10

m6 m)

M =

q = −pM

p

1 +

q

1 =

R

2

p

11 −

M

1 =

R

2

p = R

21 −

M

1

p = −5.5

2

0 m 7.9

1

×−

1

1

0−6p = 3.5 × 105 m = 3.5 × 102 km

Scale is 7.9 × 10−6:1

7.9 × 10−6

5. scale factor = 1:1400

f = −20.0 × 10−3 m

M = 14

1

00

M = − p

q

q = −Mp

p

1 +

q

1 =

1

f

p

11 −

M

1 =

1

f

p = f 1 − M

1

p = (−20.0 × 10−3 m)(1 − 1400)

p = 28 m


7. h = 4.78 × 10−3 m

h = 12.8 × 10−2 m

f = −64.0 × 10−2 m

M = − p

q =

h

h

p = − q

h

h

p

1 +

q

1 =

1

f

+ q

1 =

1

f

q

1

−h

h + 1 =

1

f

q = f 1 − h

h

q = (−64.0 × 10−2 m) 1 − 4

1

.

2

7

.

8

8

××

1

1

0

0

−

−

3

2

m

m = (−64.0 × 10−2 m)(0.963)

q = −61.6 × 10−2 m = −61.6 cm

1

−h

q

h

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6. h = 1.38 m

p = 6.00 m

h = 9.00 × 10−3 m

M = − p

q =

h

h

q = − p

h

h

p

1 +

q

1 =

R

2

p

1 −

p

h

h =

R

2

R =

R = = (1

12

−.0

15

m

3)

R = −7.89 × 10−2 m = −7.89 cm

2(6.00 m)

1 − 9.00

1.

×38

10

m−3 m

2p

1 − h

h

II

8. h = 0.280 m

h = 2.00 × 10−3 m

q = −50.0 × 10−2 m

M = h

h

= −

p

q

p = − q

h

h

p

1 +

q

1 =

1

f

+ q

1 =

1

f

q

1

−h

h + 1 =

1

f

f =

f = = (−50.

(

0

0.

×99

1

3

0

)

−2 m)

f = −50.4 × 10−2 m = −50.4 cm

(−50.0 × 10−2 m)

1 − (2.0

(0

0

.2

×8

1

0

0

m

−3

)

m)

q

1 − h

h

1

−h

q

h

Givens Solutions

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1. qi = 72°

θr = 34°

ni = 1.00

nr = ni (

(

s

s

i

i

n

n

θθ

r

i)

) = (1.00)

(

(

s

s

i

i

n

n

7

3

2

4

°°)

) = 1.7


Givens Solutions

2. θi = 47.9°

θr = 29.0°

ni = 1.00

nr = ni (

(

s

s

i

i

n

n

θθ

r

i)

) = (1.00)

(

(

s

s

i

i

n

n

4

2

7

9

.

.

9

0

°°)

) = 1.53

3. θr = 17°

ni = 1.5

nr = 1.33

θr = 15°

nr = 1.5

ni = 1.00

glass to water:

θi = sin−1n

nr

i(sin θr) = sin−1

1

1

.3

.5

3(sin 17°) =

air to glass:

θi = sin−1n

nr

i(sin θr) = sin−1

1

1

.

.

0

5

0(sin 15°) = 23°

15°

4. θi = 55.0°

θr = 53.8°

nr = 1.33

ni = nr (

(

s

s

i

i

n

n

θθ

r

i)

) = 1.33

(

(

s

s

i

i

n

n

5

5

3

5

.

.

8

0

°°)

) = 1.31

5. θi = 48°

ni = 1.00

nr = 1.5

θi = 3.0 × 101 °

ni = 1.5

nr = 1.6

θi = 28°

ni = 1.6

nr = 1.7

air to glass 1:

θr = sin−1n

n

r

i(sin θi) = sin−11

1

.0

.5

0(sin 48°) =

glass 1 to glass 2:

θr = sin−1n

n

r


1

.

.

5

6(sin 3.0°) =

glass 2 to glass 3:

θr = sin−1n

n

r


1

.

.

6

7(sin 28°) = 26°

28°

3.0 × 101 °

15ChapterRefraction

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1. f = 8.45 m

q = −25 m


Givens Solutions

2. h = 1.50 m

q = −6.00 m

f = −8.58 m

p

1 =

1

f −

1

q =

−8.5

1

8 m −

−6.0

1

0 m

p

1 =

−0

1

.1

m

17 +

0

1

.1

m

67 =

0

1

.0

m

50

p =

h = −h

q

′p = −

(1.5

(

0

−m

6.0

)(

0

2

m

0.0

)

m) = 5.00 m

20.0 m

3. h = 7.60 × 10−2 m

h = 4.00 × 10−2 m

f = −14.0 × 10−2 m

M = h

h

= −

p

q

q = − p

h

h

p

1 +

q

1 =

1

f

p

1 + =

−f

1

p

11 −

h

h

=

1

f

p = f 1 − h

h

p = (−14.0 × 10−2 m)1 − 7

4

.

.

6

0

0

0

××

1

1

0

0

−

−

2

2

m

m = (−14.0 × 10−2 m)(0.90)

p =

q = − p

h

h = −

q = −6.84 × 10−2 m = −6.84 cm

(1.3 × 10−1 m)(4.00 × 10−2 m)

(7.60 × 10−2 m)

1.30 × 10−1 m = 13.0 cm

1

−p

h

h′

p

1 +

q

1 =

1

f

p

1 =

1

f −

q

1

p

1 =

(8.4

1

5 m) −

(−1

25) =

0

1

.1

m

18 +

0

1

.0

m

40 =

0

1

.1

m

58

p = 6.3 m

M = − p

q =

−(

6

−.

2

3

5

m

m)

M = 4.0


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4. h = 28.0 m

h = 3.50 m

f = −10.0 m

M = h

h

= −

p

q

q = − h

h

p

p

1 +

q

1 =

1

f

p

1 + =

1

f

p

11 −

h

h

=

1

f

p = f 1 − h

h

p = (−10.0 m)1 − 2

3

8

.5

.0

0

m

m

p = 70.0 m

1

−h

h

p

Givens Solutions

5. h = 1.40 cm

q = −19.0 cm

f = 20.0 cm

p

1 =

1

f −

q

1 =

−20.

1

0 cm −

−19.

10 cm

p

1 =

−0

1

.0

cm

500 +

0

1

.0

c

5

m

26 =

2.6

1

0

c

×m

103

p = 385 cm =

h = − p

q

h = −

(385

(−cm

19

)

.

(

0

1

c

.4

m

0

)

cm) = 28.4 cm

3.85 m

6. h = 1.3 × 10−3 m

h = 5.2 × 10−3 m

f = 6.0 × 10−2 m

M = h

h

= −

p

q

q = − p

h

h

p

1 +

q

1 =

1

f

p

1 + =

1

f

p

11 −

h

h

=

1

f

p = f 1 − h

h

p = (6.0 × 10−2 m)1 − (

(

1

5

.

.

3

2

××

1

1

0

0

−

−

3

3

m

m

)

)

p =

q = − p

h

h =

q = −0.18 m = −18 cm

−(4.5 × 10−2 m)(5.2 × 10−3 m)

(1.3 × 10−3 m)

4.5 × 10−2 m = 4.5 cm

1

− p

h

h


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7. f = 26.7 × 10−2 m

p = 3.00 m

Image is real, so h< 0.

p

1 +

q

1 =

1

f

q

1 =

1

f −

p

1

q

1 =

(26.7 ×1

10−2 m) −

(3.0

1

0 m) =

3

1

.7

m

5 −

0

1

3

m

33 =

3

1

.4

m

2

q =

M = − p

q

M = − (

(

0

3

.2

.0

9

0

2

m

m

)

)

M = −9.73 × 10−2

0.292 m = 29.2 cm

Givens Solutions

8. h = 2.25 m

p = 12.0 m

f = −5.68 m

9. h = 0.108 m

p = 4h = 0.432 m

f = −0.216 m

p

1 +

q

1 =

1

f

1

q =

1

f −

p

1

q

1 =

(−0.2

1

16 m) −

(0.43

1

2 m) = −

4

1

.6

m

3 −

2

1

.3

m

1 = −

6

1

.9

m

4

q =

h

h

= −

p

q

h = − q

p

h

h =

h = 0.0360 m = 36.0 mm

− (−0.144 m)(0.108 m)

(0.432 m)

− 0.144 m = −144 mm

1

q =

1

f −

p

1 =

−5.6

1

8 m −

12.

1

0 m

1

q =

−0

1

.1

m

76 −

0

1

.0

m

83 =

−0

1

.2

m

59

q =

h = − h

q

′p = −

(2.25

−m

3.8

)(

6

1

m

2.0 m) = 6.99 m

−3.86 m


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10. p = 117 × 10−3 m

M = 2.4

M = − p

q

q = −pM = −(117 × 10−3 m)(2.4)

q = − 0.28 m

p

1 +

q

1 =

1

f

1

f =

(117 ×1

10−3 m) −

(0.2

1

8 m) =

8

1

.5

m

5 −

1

3.

m

6 =

1

5.

m

0

f = 0.20 m = 2.0 × 102 mm

Givens Solutions

11. Image is real, and thereforeinverted.

h

h

= M = −64

q = 12 m

p = − M

q

p

1 +

q

1 =

1

f

− M

q +

q

1 =

1

f

f = (1 −

q

M)

f = [1

(

−12

(−m

6

)

4)]

f = Image is inverted0.18 m = 18 cm

12. h = −0.55 m

h = 2.72 m

p = 5.0 m

− p

q =

h

h

q = − p

h

h

p

1 +

q

1 =

1

f

1

f =

p

1 −

p

h

h

1

f =

p

11 −

h

h

f =

f = = 5.

5

0

.9

m

f = 0.85 m

5.0 m

1 − (

(

−2

0

.7

.5

2

5

m

m

)

)

p

1 − h

h


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Givens Solutions

13. p = 12.0 × 10−2 m

M = 3.0

M = − p

q

q = − Mp

p

1 +

q

1 =

1

f

p

1 −

M

1

p =

1

f

p

1 1 −

M

1 =

1

f

f =

f =

f = 1.8 × 10−1 m = 18 cm

(12.0 × 10−2 m)

1 − 3

1

.0

p

1 − M

1

14. h = 7.60 × 10−2 m

p = 16.0 × 10−2 m

f = −12.0 × 10−2 m

p

1 +

q

1 =

1

f

M = − p

q =

h

h

q = − p

h

h

p

1 + =

1

f

p

11 −

h

h

=

1

f

1 − h

h

=

p

f

h

h

= 1 −

p

f

h =

h = = (7.60

(

×2.3

1

3

0

)

−2 m)

h = 3.26 × 10−2 m = 3.26 cm

(7.60 × 10−2 m)

1 − (

(

−1

1

6

2

.0

.0

××

1

1

0

0

−

−

2

2

m

m

)

)

h

1 − p

f

1

−p

h

h


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15. h = 48 m

f = 1.1 × 10−1 m

p = 120 m

p

1 +

q

1 =

1

f

q

1 =

1

f −

p

1

q

1 =

(1.1 × 1

1

0−1 m) −

(120

1

m) =

1

9.

m

1 −

(8.3

1

×m

10−3) =

1

9.

m

1

q = 1.1 × 10−1m

M = h

h

= −

p

q

h = − q

p

h

h =

h = 4.4 × 10−2 m = − 4.4 cm

−(1.1 × 10−1m)(48 m)

(120 m)

Givens Solutions

16. f = −0.80 m

h = 0.50 × 10−3 m

h = 0.280 m

M = h

h

= −

p

q

q = − p

h

h

p

1 +

q

1 =

1

f

p

1 + =

1

f

p

11 −

h

h

=

1

f

p = f 1 − h

h

p = (− 0.80 m)1 − (0.5

(0

0

.2

×8

1

0

0

m−3

)

m)

p =

q = − p

h

h =

q = −0.80 m

−(4.5 × 102 m)(0.50 × 10−3 m)

(0.280 m)

4.5 × 102 m

1

−p

h

h

1. θc = 46°

ni = 1.5nr = nisinθc = (1.5)(sin 46°) = 1.1



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2. ni = 1.00

qi = 75.0°

qr = 23.3°

ni = 2.44

nr = 1.00

nr = ni (

(

s

s

i

i

n

n

θθ

r

i)

) = (1.00)

(

(

s

s

i

i

n

n

7

2

5

3

.

.

0

3

°°)

) = 2.44

θc = sin = −1n

nr

i

θc = sin−11

2

.

.

0

4

0

4 = 24.2°

Givens Solutions

3. θc = 42.1°

nr = 1.00ni =

sin

nr

θc =

sin

1.

4

0

2

0

.1° = 1.49

4. ni = 1.56

nr = 1.333sin qc =

n

n

i

r

θc = sin−1n

nr

i = sin−111.3.5363 = 58.7°

5. ni = 1.52

h = 0.025 mm

nr = 1.00

θc = sin−1n

nr

i = sin−1

1

1

.

.

0

5

0

2 =

∆x = h(tan qc) where tan qc = n

nr

i

∆x = hn

nr

i = (0.025 mm)

1

1

.

.

0

5

0

2 = 0.0160 mm

d = 2∆x = 2(0.0160 mm) = 0.0320 mm

41.1°


Chapter 16Interference and Diffraction

II

HR

W m

ater

ial c

opyr

ight

ed u

nder

not

ice

appe

arin

g ea

rlier

in th

is b

ook.


Givens Solutions

1. d = 1.20 × 10−6 m

l = 156.1 × 10−9 m

m = 5; constructive interference

For constructive interference,

d(sin q) = ml

sin q = m

d

l

q = sin−1m

d

l

q = sin−1(5

(

)

1

(1

.2

5

0

6.

×1

1

×0

1−06

−

m

9

)

m)

q = 40.6°

2. d = 6.00 × 10−6 m

l = 6.33 × 10−7 m

m = 0; destructive interference

For destructive interference,

d(sin q) = m + 12

l

sin q =

q = sin−1 q = sin−1 q = 3.02°

0 + 12

(6.33 × 10−7 m)

(6.00 × 10−6 m

m + 12

l

d

m + 12

l

d

3. d = 0.80 × 10−3 m


q = 1.6°


d(sin q) = m + 12

l

l =

l =

l = 6.4 × 10−6 m = 6.4 mm

(0.80 × 10−3 m)[sin (1.6°)]

3 + 12

d(sin q)m + 1

2


4. d = 15.0 × 10−6 m


q = 19.5°


d(sin q) = ml

l = d(s

m

in q)

l =

l = 2.50 × 10−6 m = 2.50 mm

(15.0 × 10−6 m)[sin(19.5°)]

2

6. f = 60.0 × 103 Hz

c = 3.00 × 108 m/s


q = 52.0°


d(sin q) = ml = m

f

c

d = f (s

m

in

c

q)

d =

d = 2.54 × 104 m = 25.4 km

(4)(3.00 × 108 m/s)(60.0 × 103 Hz)[sin (52.0°)]

7. f = 137 × 106 Hz

c = 3.00 × 108 m/s


q = 60.0°


d(sin q) = ml = m

f

c

d = f (s

m

in

c

q)

d =

d = 5.06 m

(2)(3.00 × 108 m/s)(137 × 106 Hz)[sin(60.0°)]

mmax = d[sin(

l90.0°)] =

ld

= d

c

f

mmax = = 2.31

The second-order maximum (m = 2) is the highest observable with this apparatus.

(5.06 m)(137 × 106 Hz)

(3.00 × 108 m/s)

Givens Solutions

5. l = 443 × 10−9 m


q = 2.27°


d(sin q) = m + 12

l

d =

d =

d = 5.03 × 10−5 m

4 + 12

(443 × 10−9 m)

[sin (2.27°)]

m + 12

l(sin q)

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4. l = 40.0 × 10−9 m

d = 150.0 × 10−9 m

m = 2

d(sin q) = ml

q = sin−1m

d

l

q = sin−1 q = 32.2°

2(40.0 × 10−9 m)(150.0 × 10−9 m)


Givens Solutions

1. d = 1.00 × 10

12 lines/m

m = 1

q = 30.0°

c = 3.00 × 108 m/s

d(sin q) = ml

l = d(si

m

n q)

l =

l =

f = lc

= (

(

3

5

.

.

0

0

0

0

××

1

1

0

0

8

−3

m

m

/s

)

)

f = 6.00 × 1010 Hz = 60.0 Ghz

5.00 × 10−3 m = 5.00 mm

[sin(30.0°)](1.00 × 102 lines/m)(1)

3. l = 714 × 10−9 m

m = 3

q = 12.0°

d(sin q) = ml

d = (s

m

in

lq)

d = (3)

[

(

s

7

i

1

n

4

(

×12

1

.0

0

°

−

)

9

]

m)

d =

or

9.71 × 104 lines/m

1.03 × 10−5 m between lines

2. d = 2.0 × 10−8 m

m = 3

q = 12°

d(sin q) = ml

l = d(si

m

n q)

l =

l = 1.4 × 10−9 m = 1.4 nm

(2.0 × 10−8 m)[sin(12°)]

3


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6. l = 2.2 × 10−6 m

d = 6.4 × 10

14 lines/m

q = 34.0°

d(sin q) = ml

m = d(si

ln q)

m = = 4.0

m = 4.0

[sin (34.0°)](6.4 × 104 lines/m)(2.2 × 10−6 m)

5. f = 1.612 × 109 Hz

c = 3.00 × 108 m/s

d = 45.0 × 10−2 m

m = 1

d(sin q) = ml = m

f

c

q = sin−1m

df

c

q = sin−1 q = 24.4°

(1)(3.00 × 108 m/s)(45.0 × 10−2 m)(1.612 × 109 Hz)

Givens Solutions

7. d = 25 × 104

1

lines/m

l = 7.5 × 10−7 m

q = 48.6°

d(sin q) = ml

m = d(si

ln q)

m = = 4.0

m = 4.0

[sin(48.6°)](25 × 104 lines/m)(7.5 × 10−7 m)

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1. q1 = 0.085 C

r = 2.00 × 103 m

Felectric = 8.64 × 10−8 N

kC = 8.99 × 109 N •m2/C2

Felectric = kC q

r1q

22

q2 = Fel

ke

C

ctr

qic

1

r2

q2 = = 4.5 × 10−10 C(8.64 × 10−8 N)(2.00 × 103 m)2

(8.99 × 109 N •m2/C2)(0.085 C)


Givens Solutions

2. q1 = q

q2 = 3q

Felectric = 2.4 × 10−6 N

r = 3.39 m

kC = 8.99 × 109 N •m2/C2

F = kC q

r1q

22 = kC

3

r

q2

2

q = 3

F

kr

C

2

= q = 3.2 × 10−8 C

(2.4 × 10−6 N)(3.39 m)2

(3)(8.99 × 109 N •m2/C2)

3. Felectric = 1.0 N

r = 2.4 × 1022 m

kC = 8.99 × 109 N •m2/C2

F = kC N2

r

(q2

e)2

qe = F

kr

C

2

= rk

F

C

qe = (2.4 × 1022 m)qe = 2.5 × 1017 C

1.0 N8.99 × 109 N •m2/C2

4. r = 1034 m

q1 = 2.0 × 10−9 C

q2 = −2.8 × 10−9 C

kC = 8.99 × 109 N •m2/C2

r2 = 2r

Felectric = kC q

r1q

22 =

Felectric =

r2 = 2r = (2)(1034 m) = 2068 m

q = Felec

kt

C

ricr22

= q = 4.7 × 10−9 C

(4.7 × 10−14 N)(2068 m)2

8.99 × 109 N •m2/C2

4.7 × 10−14 N

(8.99 × 109 N •m2/C2)(2.0 × 10−9 C)(2.8 × 10−9 C)

(1034 m)2

5. q1 = 1.0 × 105 C

q2 = −1.0 × 105 C

r = 7.0 × 1011 m

kC = 8.99 × 109 N •m2/C2

F = kC q

r1q

22

F = (8.99 × 109 N •m2/C2)((71.0.0××

1

1

0

01

5

1

C

m

)

)

2

2F = 1.8 × 10−4 N

17ChapterElectric Forces and Fields

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6. N = 2 000 744

qp = 1.60 × 10−19 C

r = 1.00 × 103 m

kC = 8.99 × 109 N •m2/C2

q = N

2

qp = =

Felectric = kC q

r2

2

Felectric = (8.99 × 109 N •m2/C2)((11.6.000××1

1

0

0

−

3

13

m

C

)

)2

2

Felectric = 2.30 × 10−22 N

1.60 × 10−13 C(2 000 744)(1.60 × 10−19 C)

2

Givens Solutions

7. N1 = 4.00 × 103

N2 = 3.20 × 105

q = 1.60 × 10−19 C

r = 1.00 × 103 m

kC = 8.99 × 109 N •m2/C2

8. Felectric = 2.0 × 10−28 N

N = 111

qp = 1.60 × 10−19 C

kC = 8.99 × 109 N •m2/C2

9. q = 1.00 C

Felectric = 4.48 m × 104 N

kC = 8.99 × 109 N •m2/C2

r = kCFel

qec

2

tr

ic = = 448 m

(8.99 × 109 N •m2/C2)(1.0 C)2

4.48 × 104 N

10. Felectric = 1.18 × 10−11 N

q1 = 5.00 × 10−9 C

q2 = −2.50 × 10−9 C

kC = 8.99 × 109 N •m2/C2

Felectric = kC q

r1q

22

r = F

kel

Cec

qt

2

r

ic =

r =

L = r cos q = (97.6 m)cos 45° = 69.0 m

97.6 m

(8.99 × 109 N •m2/C2)(5.00 × 10−9 C)(2.50 × 10−9 C)

1.18 × 10−11 N

Felectric = kC

r

q21q2 =

kCN

r1N

22q2

Felectric =

Felectric =

Felectric = kC N2

r

2

2

q2

=

Felectric = 2.36 × 10−23 N

(8.99 × 109 N •m2/C2)(3.20 × 105)2(1.60 × 10−19 C)2

(1.00 × 103 m)2

2.95 × 10−25 N

(8.99 × 109 N •m2/C2)(4.00 × 103)(3.20 × 105)(1.60 × 10−19 C)2

(1.00 × 103 m)2

Felectric = kC q

r2

2

= kC N

r

2q2

p2

r = k

F

Ce

N

le

c

2

t

qri

pc

2

= r = 1.2 × 102 m

(8.99 × 109 N •m2/C2)(111)2(1.60 × 10−19 C)2

2.0 × 10−28 N


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2. q1 = 2.0 × 10−9 C

q2 = 3.0 × 10−9 C

q3 = 4.0 × 10−9 C

q4 = 5.5 × 10−9 C

r1,2 = 5.00 × 102 m

r1,3 = 1.00 × 103 m

r1,4 = 1.747 × 103 m

kC = 8.99 × 109 N •m2/C2

3. w = 7.00 × 10−2 m

L = 2.48 × 10−1 m

q = 1.0 × 10−9 C

kC = 8.99 × 109 N •m2/C2

F = kC q

r1q

22

Fx = F1 + F2(cos q) = F1 + F2 Fx = kCq2

L

12 +

(w 2 +L

L2)3/2Fx = kCq2(2.48 × 1

1

0−1 m)2 + Fx = (8.99 × 109 N •m2/C2)(1.0 × 10−9 C)2(30.8/m2) = 2.8 × 10−7 N

Fy = F3 + F2(sin q) = F3 + F2 Fy = kCq2

w

12 +

(w2 +w

L2)3/2Fy = kCq2(7.00 × 1

1

0−2 m)2 + Fy = (8.99 × 109 N •m2/C2)(1.0 × 10−9 C)2(2.00 × 102/m2) = 1.8 × 10−6 N

Fnet =√

Fx2 + Fy

2 =√

(2.8 × 10−7N)2 + (1.8× 10−6N)2Fnet = 1.8 × 10−6 N

q = tan−1F

F

x

y = tan−112.

.

8

8

××

1

1

0

0

−

−

6

7

N

N = 81°

Fnet = 1.8 × 10−6 N, 81° above the positive x-axis

7.00 × 10−2 m[(7.00 × 10−2 m)2 + (2.48 × 10−1 m)2]3/2

w√

w2+ L2

2.48 × 10−1 m[(7.00 × 10−2 m)2 + (2.48 × 10−1 m)2]3/2

L√

w2+ L2

1. q1 = 2.80 × 10−3 C

q2 = −6.40 × 10−3 C

q3 = 4.80 × 10−2 C

r1,3 = 9740 m

r1,2 = 892 m

kC = 8.99 × 109 N •m2/C2

F = kC q

r1q

22

F1,2 = = 2.02 × 10−1 N

F1,3 = = 1.27 × 10−2 N

F1,tot = F1,2 + F1,3 = −(2.02 × 10−1 N) + (1.27 × 10−2 N) = −0.189 N

F1,tot = 0.189 N downward

(8.99 × 109 N •m2/C2)(2.80 × 10−3 C)(4.80 × 10−2 C)

(9740 m)2

(8.99 × 109 N •m2/C2)(2.80 × 10−3 C)(6.40 × 10−3 C)

(892 m)2


Givens Solutions

F = kC q

r1q

22

F1,2 = = 2.2 × 10−13 N

F1,3 = = 7.2 × 10−14 N

F1,4 = = 3.2 × 10−14 N

F1,tot = F1,2 + F1,3 + F1,4 = (2.2 × 10−13 N) + (7.2 × 10−14 N) + (3.2 × 10−14 N)

F1,tot = 3.2 × 10−13 N down the rope

(8.99 × 109 N •m2/C2)(2.0 × 10−9 C)(5.5 × 10−9 C)

(1.747 × 103 m)2

(8.99 × 109 N •m2/C2)(2.0 × 10−9 C)(4.0 × 10−9 C)

(1.00 × 103 m)2

(8.99 × 109 N •m2/C2)(2.0 × 10−9 C)(3.0 × 10−9 C)

(5.00 × 102 m)2


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4. L = 10.7 m

w = 8.7 m

q1 = −1.2 × 10−8 C

q2 = 5.6 × 10−9 C

q3 = 2.8 × 10−9 C

q4 = 8.4 × 10−9 C

kC = 8.99 × 109 N •m2/C2

5. d = 1.2 × 103 m

q1 = 1.6 × 10−2 C

q2 = 2.4 × 10−3 C

q3 = −3.2 × 10−3 C

q4 = −4.0 × 10−3 C

kC = 8.99 × 109 N •m2/C2

∆x = ∆y = = = 8.5 × 102 m

F = kC

r

q21q2

Fx = −F2 + F3(cos 45°) = kCq1− ∆q

x22 +

q3(co

d

s2

45°)

Fx = (8.99 × 109 N•m2/C2)(1.6 × 10−2 C)− (8

2

.

.

5

4

××

1

1

0

02

−3

m

C

)2 + Fx = −0.24 N

Fy = −F4 − F3(sin 45°) = kCq1∆q

y42 +

q3(si

d

n2

45°)

Fy = −(8.99 × 109 N•m2/C2)(1.60 × 10−2 C)(8

4

.

.

5

0

××

1

1

0

02

−3

m

C

)2 + Fy = −1.0 N

Fnet =√

Fx2 + Fy

2 =√

(0.24N)2 + (1.0N)2 = 1.0 N

q = tan−1F

F

x

y = tan−1((01..204N

N

)

) = 77°

Fnet = 1.0 N, 77° below the negative x-axis

(3.2 × 10−3 C)(sin 45°)

(1.2 × 103 m)2

(3.2 × 10−3 C)(cos 45°)

(1.2 × 103 m)2

1.2 × 103 m√

2

d√

2

Givens Solutions

F = kC q

r1q

22

Fx = F4 + F3(cos q)

Fy = F2 + F3(sin q)

Fx = kCq1L

q42 +

(L2 +q3

w

L2)3/2

Fx = (8.99 × 109 N•m2/C2)(1.2 × 10−8 C)(8.

(

4

10

×.7

10

m

−9

)2

C) +

Fx = 9.1 × 10−9 N

Fy = kCq1w

q22 +

(L2 +q3

w

w2)3/2

Fy = (8.99 × 109 N •m2/C2)(1.2 × 10−8 C)(5.6

(8

×.7

1

m

0−

)

9

2

C) +

Fy = 9.0 × 10−9 N

Fnet =√

Fx2 + Fy

2 =√

(9.1 × 10−9N)2 + (9.0× 10−9N)2 = 1.28 × 10−8 N

q = tan−1F

F

x

y = tan−199..10××

1

1

0

0−

−

9

9

N

N

) = 45°

Fnet = 1.28 × 10−8 N, 45° above the positive x-axis

(2.8 × 10−9 C)(8.7 m)[(10.7 m)2 + (8.7 m)2]3/2

(2.8 × 10−9 C)(10.7 m)[(10.7 m)2 + (8.7 m)2]3/2


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Givens Solutions

1. q1 = 2.5 × 10−9 C

q3 = 1.0 × 10−9 C

r2,1 = 5.33 m

r3,1 = 1.90 m

r3,2 = r2,1 − r3,1 = 5.33 m − 1.90 m = 3.43 m

F3,1 = F3,2 = kC(q

r3

3

,

q

1)12 = kC(q

r3

3

,

q

2)22

q2 = q1r

r

3

3

,

,

1

22

q2 = (2.50 × 10−9 C) 31.

.

4

9

3

0

m

m

2= 8.15 × 10−9 C

6. d = 228.930

3

× 103 m =

7.631 × 104 m

q1 = 8.8 × 10−9 C

q2 = −2.4 × 10−9 C

q3 = 4.0 × 10−9 C

kC = 8.99 × 109 N •m2/C2

q = 60.0°

F = kC

r

q21q2

Fx = F2 − F3(cos 60.0°)

Fy = F3(sin 60.0°)

Fx = kCq1d

q22 −

q3(co

d

s2

60.0°)

Fx = (8.99 × 109 N•m2/C2)(8.8 × 10−9 C)(7.

2

6

.

3

4

1

××1

1

0

0

−

4

9

m

C

)2 − Fx = 5.5 × 10−18 N

Fy = − kCq1q3(

r

si2

n 60.0°)

Fy = −

Fy = −4.7 × 10−17 N

Fnet =√

Fx2 + Fy

2 =√

(5.5 × 10−18 N)2 + (4.7× 10−17 N)2 = 4.7 × 10−17 N

q = tan−1F

F

x

y = tan−145.

.

7

5

××

1

1

0

0

−

−

1

1

7

8

N

N = 83°

Fnet = 4.7 × 10−18 N, 83° below the positive x-axis

(8.99 × 109 N •m2/C2)(8.8 × 10−9 C)(4.0 × 10−9 C)(sin 60.0°)

(7.631 × 104 m)2

(4.0 × 10−9 C)(cos 60.0°)

(7.631 × 104 m)2


2. q1 = 7.5 × 10−2 C

q3 = 1.0 × 10−4 C

r2,1 = 6.00 × 102 km

r3,1 = 24 km

r3,2 = r2,1 − r3,1 = 6.00 × 102 km − 24 km = 576 km

F3,1 = F3,2 = kC(q

r3

3

,

q

1)12 = kC(q

r3

3

,

q

2)22

q2 = q1 = r

r

3

3

,

,

1

22

q2 = (7.5 × 10−2 C) 52746k

k

m

m

2= 43 C


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3. mE = 6.0 × 1024 kg

mm = 7.3 × 1022 kg

G = 6.673 × 10−11 N•m2/kg2

kC = 8.99 × 109 N •m2/C2

Fg = Felectric

Gm

rE2

mm = kC

r2

q2

q = Gm

kE

C

mm = q = 5.7 × 1013 C

(6.673 × 10−11 N •m2/kg2)(6.0 × 1024 kg)(7.3 × 1022 kg)

8.99 × 109 N •m2/C2

4. m = 17.23 kg

r = 0.800 m

Fnet = 167.6 N

g = 9.81 m/s2

kC = 8.99 × 109 N •m2/C2

Fnet = Fg − Felectric

Fnet = mg − kC

r2

q2

q = q = q = 1.0 × 10−5 C

(0.800 m)2[(17.23 kg)(9.81 m/s2) − (167.6 N)]

8.99 × 109 N •m2/C2

r2(mg − Fnet)kC

5. m1 = 9.00 kg

m2 = 8.00 kg

r = 1.00 m

kC = 8.99 × 109 N •m2/C2

g = 9.81 m/s2

Fg,1 = Fg,2 + Felectric

g(m1 − m2) = kC

r2

q2

q = gr

2(mk1

C − m2) =

q = 3.30 × 10−5 C

(9.81 m/s2)(1.00 m)2(9.00 kg − 8.00 kg)

8.99 × 109 N •m2/C2

6. m = 9.2 × 104 kg

l1 = 1.00 m

g = 9.81 m/s2

l2 = 8.00 m

r = 2.5 m

kC = 8.99 × 109 N •m2/C2

t1 = t2

mgl1 = kC

r

q2

2l2

q = r 2k

m

Cl

g

2l1

q = = 8.9 × 10−3 C(2.5 m)2(9.2 × 104 kg)(9.81 m/s2)(1.00 m)

(8.99 × 109 N •m2/C2)(8.00 m)

Givens Solutions

7. q1 = 2.0 C

q2 = 6.0 C

q3 = 4.0 C

L = 2.5 × 109 m

Fnet = 0 = F1 + F2 kC q

x1q

23 =

(

k

LC

−q2

x

q

)32

x

q21 =

(L

q

−2

x)2 (L − x)√

q1 = x√

q2

L

x −

x

x =

q

q2

1

L

x =

q

q2

1 + 1

x = = = 9.3 × 108 m2.5 × 109 m

6

2

.

.0

0C

C + 1

L

q

q2

1 + 1


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Givens Solutions

8. q1 = 55 × 10−6 C

q2 = 137 × 10−6 C

q3 = 14 × 10−6 C

L = 87 m

Fnet = 0 = F1 + F2

kC q

x1q

23 =

(

k

LC

−q2

x

q

)32

x

q21 =

(L

q

−2

x)2

(L − x) √

q1 = x√

q2

L

x −

x

x =

q

q2

1

L

x =

q

q2

1 + 1

x = = = 34 m

87 m

1

5

35

7×× 1

10

0−−

6

6

C

C + 1

L

q

q2

1 + 1

9. F = 1.00 × 108 N

q1 = 1.80 × 104 C

q2 = 6.25 × 104 C

kC = 8.99 × 109 N •m2/C2

F = kC

r

q21q2

r = kCq

F1q2

r = r = 3.18 × 105 m

(8.99 × 109 N •m2/C2)(1.80 × 104 C)(6.25 × 104 C)

1.00 × 108 N

10. m = 5.00 kg

q = 4.00 × 10−2 C

kC = 8.99 × 109 N •m2/C2

g = 9.81 m/s2

Fg = Felectric

mg = kC

h

q2

2

h = k

mCq

g

2

h = = 542 m(8.99 × 109 N •m2/C2)(4.00 × 10−2 C)2

(5.00 kg)(9.81 m/s2)

11. m = 1.0 × 10−19 kg

r = 1.0 m

q = 1.60 × 10−19 C

kC = 8.99 × 109 N •m2/C2

Fres = Felectric = kC

r

q2

2

Fres =

Fres = 2.3 × 10−28 N

(8.99 × 109 N •m2/C2)(1.60 × 10−19 C)2

(1.0 m)2

12. m = 5.0 × 10−6 kg

q = 2.0 × 10−15 C

r = 1.00 m

kC = 8.99 × 109 N •m2/C2

G = 6.673 × 10−11 N •m2/kg2

Fnet = Felectric + Fg

Felectric = kC

r

q2

2

=

Fg = G

r

m2

2

=

Fnet = 3.6 × 10−20 N + 1.7 × 10−21 N = 3.8 × 10−20 N

(6.673 × 10−11 N •m2/kg2)(5.0 × 10−6 kg)2

(1.00 m)2

(8.99 × 109 N •m2/C2)(2.0 × 10−15 C)2

(1.00 m)2


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Givens Solutions

13. m = 2.00 × 10−2 kg

q1 = 2.0 × 10−6 C

q2 = −8.0 × 10−6 C

r = 1.7 m

kC = 8.99 × 109 N •m2/C2

g = 9.81 m/s2

Felectric = Ffriction

kC

r

q21q2 = mkmg

mk = k

mCq

g1

r

q2

2

mk = = 0.25(8.99 × 109 N •m2/C2)(2.0 × 10−6 C)(8.0 × 10−6 C)

(2.00 × 10−2 kg)(9.81 m/s2)(1.7 m)2

1. r = 3.72 m

E = 0.145 N/C

kC = 8.99 × 109 N •m2/C2

q = 60.0°


2. ∆y = 190 m

q1 = 1.2 × 10−8 C

∆x = 120 m

Ex = 1.60 × 10−2 N/C

kC = 8.99 × 109 N •m2/C2

E = k

rC2q

Ex = E1 + E2(cos q) = k

∆C

x

q21 +

q2 = Ex − k

∆C

x

q21(∆x2

k

+

C∆∆

x

y2)3/2

Ex −

k

∆C

x

q21 = 1.60 × 10−2 N/C −

= 8.5 × 10−3 N/C

(∆x2

k

+

C∆∆

x

y2)3/2

= = 1.0 × 10−5 C2/N

q2 = (8.5 × 10−3 N/C)(1.0 × 10−5 C2/N) = 8.5 × 10−8 C

[(120 m)2 + (190 m)2]3/2

(8.99 × 109 N •m2/C2)(120 m)

(8.99 × 109 N •m2/C2)(1.2 × 10−8 C)

(120 m)2

kCq2(∆x)(∆x2 + ∆y2)

√∆x2+ ∆y2

E = k

rC2

q

Ex = k

rC2

q(cos 60.0°) −

k

rC2

q(cos 60.0°) = 0 N/C

Because Ex = 0 N/C, the electric field points directly upward.

Ey = 2kCq(s

r

in2

60.0°)

q = 2kC(s

E

in

yr

6

2

0.0°) = = 1.29 × 10−10 C

(0.145 N/C)(3.72 m)2

(2)(8.99 × 109 N •m2/C2)(sin 60.0°)

3. q1 = 1.80 × 10−5 C

q2 = −1.20 × 10−5 C

Enet = 22.3 N/C toward q2

kC = 8.99 × 109 N •m2/C2

Enet = k

rC2(q1 + q2) r2 =

E

k

n

C

et(q1 + q2)

r = kC(q

E1

n

e

+

t q2)

r = r = 1.10 × 102 m

(8.99 × 109 N •m2/C2)[(1.80 × 10−5 C) + (1.20 × 10−5 C)]

22.3 N/C toward q2


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Givens Solutions

4. d = 86.5 m

q1 = 4.8 × 10−9 C

q2 = 1.6 × 10−8 C

5. q = 3.6 × 10−6 C

L = 960 m

w = 750 m

kC = 8.99 × 109 N •m2/C2

E = k

rC2

q

Ey = E1 + E2(sin q) = k

wC

2

q +

Ey = kCqw

12 +

(w 2 +w

L2)3/2Ey = (8.99 × 109 N •m2/C2)(3.6 × 10−6 C)(750

1

m)2 + Ey = 7.1 × 10−2 N/C

Ex = E3 + E2(cos q) = k

LC

2

q +

Ex = kCqL

12 +

(w 2 +L

L2)3/2Ex = (8.99 × 109 N •m2/C2)(3.6 × 10−6 C)(960

1

m)2 + Ex = 5.2 × 10−2 N/C

Enet =√

Ey2 + Ex

2 =√

(7.1 × 10−2N/C)2 + (5.2× 10−2N/C)2 = 8.8 × 10−2 N/C

q = tan−1E

E

y

x = tan−175.

.

1

2

××

1

1

0

0

−

−

2

2N

N

/

/

C

C = 54°

Enet = 8.8 × 10−2 N/C, 54° above the horizontal

960 m[(750 m)2 + (960 m)2]3/2

kCqL√

w2+ L2(w2 + L2)

750 m[(750 m)2 + (960 m)2]3/2

kCqw√

w2+ L2(w2 + L2)

Enet = E1 + E2 = 0

E1 = E2

x

q12 =

(d −q2

x)2

(d − x)√

q1 = x√

q2

x √

q1 +√

q2 = d√

q1

x = =

x = 3.0 × 101 m

(86.5 m)√

4.8× 10−9C√

(4.8 × 10−9C) +√

(1.6 × 10−8C)d√

q1√

q1 +√

q2


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6. w = 218 m

h = 50.0 m

q = 6.4 × 10−9 C

kC = 8.99 × 109 N •m2/C2

q1 = q2 = q

q3 = 3q

q4 = 2q

r = = = 112 m

q = tan−1w

h = tan−1520

1

.

8

0

m

m = 12.9°

The electric fields of charges on opposite corners of the rectangle cancel to give 2qon the lower left corner and q on the lower right corner.

E = k

rC2

q

Ex = kC

r2

2q −

k

rC2

q(cos q) =

kCq(

r

c2

os q)

Ex = = 4.5 × 10−3 N/C

Ey = kC

r2

2q +

k

rC2

q(sin q) =

3kCq

r

(2

sin q)

Ey = = 3.1 × 10−3 N/C

Enet =√

Ex2 + Ey

2 =√

(4.5 × 10−3N/C)2 + (3.1× 10−3N/C)2

Enet = 5.5 × 10−3 N/C

q = tan−1E

E

x

y = tan−134.

.

1

5

××

1

1

0

0

−

−

3

3N

N

/

/

C

C = 35°

Enet = 5.5 × 10−3 N/C, 35° above the positive x-axis

(3)(8.99 × 109 N •m2/C2)(6.4 × 10−9 C)(sin 12.9°)

(112 m)2

(8.99 × 109 N •m2/C2)(6.4 × 10−9 C)(cos 12.9°)

(112 m)2

√(50.0m)2 + (218m)2

2

√h2+ w2

2

Givens Solutions


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1. PEelectric = 0.868 J

r = 15.4 × 103 m

kC = 8.99 × 109 N •m2

C2

PEelectric = kC q1

r

q2

q1q2 = q2 = (PEele

kc

C

tric)(r)

q = (PEele

kcC

tric)(r)

q = q = q1 = q2 = 1.22 × 10−3 C

(0.868 J)(15.4 × 103 m)(8.99 × 109 N •m2/C2)


Givens Solutions

2. r = 281 m

q1 = 2.40 × 10−7 C

PEelectric = −2.0 × 10−5 J

kC = 8.99 × 109 N •m2

C2

PEelectric = kCq

r1q2

q2 = (P

(

E

ke

C

le

)ct

(r

qic

1

)

)

(r)

q2 =

q2 = −2.6 × 10−6 C

(−2.0 × 10−5 J)(281 m)(8.99 × 109

N

C

•m2

2

)(2.40 × 10−7 C)

3. PEelectric = 4.80 × 10−4 J

d = 2365 m

E = −1.50 × 102 N/C

PEelectric = −qEd

q = −PE

Ee

dlectric

q =

q = 1.35 × 10−9 C

−(4.80 × 10−4 J)(−1.50 × 102 N/C)(2365 m)

4. q1 = 44 × 10−6 C

q2 = 44 × 10−6 C

PEelectric = 1.083 × 10−2 J

kC = 8.99 × 109 N

C

•m2

2

PEelectric = kC q1

r

q2

r = P

k

EC

e

q

le

1

c

q

tr

2

ic

r = 8.99 × 109 N

C

•m2

2

((14.048

×3

1

×0

1

−6

0−C2

)

J

2

)

r = 1.6 × 103 m = 1.6 km

18ChapterElectrical Energy and Capacitance


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5. q1 = −16.0 × 10−3 C

q2 = 24.0 × 10−3 C

W = 2.8 × 10−4 J

rf = ∞

kC = 8.99 × 109 N •m2

C2

W = ∆PEelectric = PEelectric,f − PEelectric,i

W = kCq1q2r

1

f −

r

1

i =

−kC

r

q

i

1q2

ri = =

ri = 122 m

−(8.99 × 109 N •m2/C2)(−16.0 × 10−3 C)(24.0 × 10−3 C)

2.8 × 10−4 J

−kCq1q2W

Givens Solutions

6. d = 1410 m

E = 380 N/C

q = −1.60 × 10−19 C

∆PEelectric = −qEd

∆PEelectric = −(−1.60 × 10−19 C)(380 N/C)(1410 m)

∆PEelectric = 8.6 × 10−14 J

7. d = 275 m

q = 12.5 × 10−9 C

E = 1.50 × 102 N/C

PEelectric = −qEd

PEelectric = −(12.5 × 10−9 C)(1.50 × 102 N/C)(275 m)

PEelectric = −5.16 × 10−4 J

8. P = 1.5 × 105 W

vf = 2.50 × 102 km/h

vi = 0 km/h

mc = 5.00 × 102 kg

mp = 2.000 × 103 kg

kC = 8.99 × 109 N

C

•m2

2

a. W = ∆KE = KEf − KEi = 12

mvf2 − 1

2mvi

2 = 12

mvf2

W = 12

(mc + mp)vf2

W = 12

[(5.00 × 102 kg) + (2.000 × 103 kg)]2.50 × 102 k

h

m36

1

0

h

0 s110k

3

m

m

2

W =

b. W = P∆t

∆x = (vf

2

+ vi) ∆t

∆x = (vf

2

+ vi) W

P =

v

2

f W

P

∆x =

∆x =

c. PEelectric = kC q1

r

q2

PEelectric = W

r = ∆x

q1 = q2 = q = (PEele

kcC

tric)(r) =

(Wk

)(

C∆x)

q = q = 0.97 C

(6.03 × 106 J)(1.4 × 103 m)

8.99 × 109 N

C

•m2

2

1.4 × 103 m = 1.4 km

(2.50 × 102 km/h)36

1

0

h

0 s110k

3

m

m(6.03 × 106 J)

2(1.5 × 105 W)

6.03 × 106 J

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Givens Solutions

1. q1 = −12.0 × 10−9 C

q2 = −68.0 × 10−9 C

V = −25.3 V

r1 = 16.0 m

r2 = d − r1

kC = 8.99 × 109 N

C

•m2

2

V = kCq

r1

1 + q

r2

2 = kCq

r1

1 + (d

q

−2

r1)

k

V

C −

q

r1

1 + (d

q

−2

r1)

d = + r1

d = +16.0 m

d = 33.0 m + 16.0 m = 49.0 m

−68.0 × 10−9 C

8.99 ×−12

059.3

N

V

•m2/C2 − (−12.

1

0

6

×.0

1m

0−9 C)

q2

k

V

C −

q

r1

1


2. q1 = 18.0 × 10−9 C

q2 = 92.0 × 10−9 C

V = 53.3 V

r1 = d − r2

d = 97.5 m

kC = 8.99 × 109 N

C

•m2

2

V = kCq

r = kC

q

r1

1 + q

r2

2V = kCd −

q1

r2 +

q

r2

2k

V

C =

(q1r

(2

d

+−

q

r2

2

d

)(

−r2

q

)2r2)

−k

V

Cr2

2 + k

V

Cdr2 = (q1 − q2)r2 + q2d

k

V

Cr2

2 + q1 − q2 − V

kC

d r2 + q2d = 0

Solve using the quadratic formula:

r2 =

q1 − q2 − V

kC

d = 18.0 × 10−9 C − 92.0 × 10−9 C −

q1 − q2 − V

kC

d = −652 × 10−9 C

4V

k

q

C

2d = = 2.13 × 10−13 C2

2

k

V

C = = 11.9 × 10−9

m

C

r2 =

r2 = 652

11

±.9

460 m

−(−652 × 10−9 C) ±√

(−652× 10−9C)2 − (2.13 × 10−13 C2)

(11.9 × 10−9 C/m)

2(53.3 V )(8.99 × 109 N •m2/C2)

4(53.3 V)(92.0 × 10−9 C)(97.5 m)

(8.99 × 109 N •m2/C2)

(53.3 V)(97.5 m)

8.99 × 109 N

C

•m2

2

−q1 − q2 − V

kC

d ± q1− q2−

V

kC

d

2 − 4Vk

q

C2d

2

k

V

C


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3. V = 1.0 × 106 V

r = 12 × 10−2 m

kC = 8.99 × 109 N

C

•m2

2

V = kC q

r

q = V

kC

r

q =

q = 1.3 × 10−5 C

(1.0 × 106 V)(12 × 10−2 m)

8.99 × 109 N

C

•m2

2

4. ME = 5.98 × 1024 kg

G = 6.673 × 10−11 N

k

•

g

m2

2

kC = 8.99 × 109 N

C

•m2

2

m = 1.0 kg

q = 1.0 C

mVgravity = qVelectric

mM

rEG =

qQ

rEkC

QE = m

q

M

kC

EG

QE =

QE = 4.44 × 104 C

(1.0 kg)(5.98 × 1024 kg)6.673 × 10−11 N

k

•

g

m2

2

(1.0 C)8.99 × 109 N

C

•m2

2

Givens SolutionsOf the two roots, the one that yields the correct answer is

r2 = (652

11

−.9

460) m

r2 = 16.1 m

5. msun = 1.97 × 1030 kg

mH = mass of hydrogenatom = 1.67 × 10−27 kg

q1 = charge of proton = +1.60 × 10−19 C

q2 = charge of electron = −1.60 × 10−19 C

r1 = 1.1 × 1011 m

r2 = 1.5 × 1011 m − 1.1 ×1011 m = 4.0 × 1010 m

kC = 8.99 × 109 N

C

•m2

2

a. Q+ = charge of proton cloud = (number of protons)q1 = m

msu

H

nq1

Q+ =

Q+ =

Q− = charge of electron cloud = m

msun

H

q2

Q− =

b. V = kCq

r = kC

Q

r1

+ + Q

r2

−V = 8.99 × 109

N

C

•m2

2

11..819××1

1

0

01

3

1

8

m

C −

1

4

.

.

8

0

9

××1

1

0

01

3

0

8

m

C

V = −2.7 × 1037 V

−1.89 × 1038 C

1.89 × 1038 C

(1.97 × 1030 kg)(1.60 × 10−19 C)

(1.67 × 10−27 kg)


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Givens SolutionsGivens Solutions

6. r = r1 = r2 = r3 = r4 =

2

x2+

2

y2

x = 292 m

y = 276 m

q = 64 × 10−9 C

q1 = 1.0q

q2 = −3.0q

q3 = 2.5q

q4 = 4.0q

kC = 8.99 × 109 N

C

•m2

2

V = kCq

r = kC

q

r1

1 + q

r2

2 + q

r3

3 + q

r4

4V =

V =

V = 13 V

8.99 × 109 N

C

•m2

2

(64 × 10−9 C)(4.5)

2922m

2 + 276

2m

2

kC q(1.0 − 3.0 + 2.5 + 4.0)

2

x2+

2

y2

7. q1 = q2 = q3 = q= 7.2 × 10−2 C

l = 1.6 × 107 m

r1 = r2 = 2

l

r3 = l 2 − 2

l2

kC = 8.99 × 109 N

C

•m2

2

V = kCq

r = kC

q

r1

1 + q

r2

2 + q

r3

3V = kC + + = 2 + 2 +

V = 4 + V = 2.1 × 102 V

1

34

8.99 × 109

N

C

•m2

2

(0.072 C)

(1.6 × 107 m)

1

1− 122

kCq

l

q

l 2 − 2

l2

q

2

lq

2

l

8. q1 = q2 = q3 = q= 25.0 × 10−9 C

r1 = r2 = l

r3 =√l 2 + l 2

l = 184 m

kC = 8.99 × 109 N

C

•m2

2

V = kCq

r = kC

q

r1

1 + q

r2

2 + q

r3

3V = kCq + + = 1 + 1 +

V = (2.707)

V = 3.31 V

8.99 × 109 N

C

•m2

2

(25.0 × 10−9 C)

(184 m)

1√

2kCq

l1√

l 2 + l 2

1l

1l

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Givens Solutions

1. ∆V = 3.00 × 102 V

PEelectric = 17.1 kJ

PEelectric = 12

C(∆V )2

C = 2P

(∆E

Vele

)ct2ric

C = (

2

3

(

.

1

0

7

0

.1

××10

12

03

V

J

)

)2

C = 3.80 × 10−1 F

2. PEelectric = 1450 J

∆V = 1.0 × 104 V

PEelectric = 12

C(∆V )2

C = 2P

(∆E

Vele

)ct2ric

C = (1.

2

0

(

×14

1

5

0

04

J

V

)

)2

C = 2.9 × 10−5 F

3. Emax = 3.0 × 106 V/m

d = 0.2 × 10−3 m

A = 6.7 × 103 m2

e0 = 8.85 × 10−12 C2/N•m2

∆Vmax = Emaxd

∆Vmax = Q

Cmax =

Emaxd =

Qmax = Emaxe0A

Qmax = (3.0 × 106 V/m)(8.85 × 10−12 C2/N •m2)(6.7 × 103 m2)

Qmax = 0.18 C

Qmax

e0

d

A

Qmax

e0

d

A

4. r = 3.1 m

d = 1.0 × 10−3 m

Emax = 3.0 × 106 V/m

e0 = 8.85 × 10−12 C2/N •m2

Qmax = C∆Vmax = CEmaxd

C = e0

d

A

Qmax = e0AEmax = e0pr2Emax

Qmax = (8.85 × 10−12 C2/N •m2)(p)(3.1 m)2(3.0 × 106 V/m)

Qmax = 8.0 × 10−4 C = 0.80 mC


5. P = 5.0 × 1015 W

∆t = 1.0 × 10−12 s

C = 0.22 F

PEelectric = 12

C(∆V)2

PEelectric = P∆t

P∆t = 12

C(∆V )2

∆V = 2PC

∆t

∆V = ∆V = 210 V

2(5.0 × 1015 W)(1.0 × 10−12 s)

(0.22 F)



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6. A = 2.32 × 105 m2

d = 1.5 × 10−2 m

Q = 0.64 × 10−3 C

e0 = 8.85 × 10−12 C2/N •m2

PEelectric = 12

Q

C

2

C = e0

d

A

PEelectric = 12

Q

e0

2

A

d

PEelectric =


(0.64 × 10−3 C)2(1.5 × 10−2 m)(8.85 × 10−12 C2/N •m2)(2.32 × 105 m2)

12

Givens Solutions

7. r = 18.0 m

∆V = 575 V

PEelectric = 3.31 J

PEelectric = 12

C(∆V)2

C = 2 P

(∆E

Vele

)c2tric =

(

2

5

(

7

3

5

.3

V

1

)

J2)

C =

d = e0

C

A =

e0

C

πr2

d =

d = 4.5 × 10−4 m = 0.45 mm

(8.85 × 10−12 C2/N•m2)(π)(18.0 m)2

(2.00 × 10−5 F)

2.00 × 10−5 F

8. di = 5.00 × 10−3 m

df = 0.30 × 10−3 m

e0 = 8.85 × 10−12 C2/N •m2

A = 1.20 × 10−4 m2

∆C = Cf − Ci = ed0A

f −

ed0A

i

∆C = e0Ad

1

f −

d

1

i

∆C = 8.85 × 10−12 N

C

•m

2

2(1.20 × 10−4 m2)0.30 ×1

10−3 m −

5.00 ×1

10−3 m

∆C = 3.3 × 10−12 F = −3.3 pF

9. A = 98 × 106 m2

C = 0.20 F

e0 = 8.85 × 10−12 C2/N •m2

C = e0

d

A

d = e0

C

A

d =

d = 4.3 × 10−3 m = 4.3 mm

(8.85 × 10−12C2/N •m2)(98 × 106 m2)

(0.20 F)


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Givens Solutions

11. A = 44 m2

e0 = 8.85 × 10−12 C2/N •m2

Q = 2.5 × 10− 6 C

∆V = 30.0 V

a. C = ∆Q

V

C = (2.5

(3

×0.

1

0

0

V

−6

)

C)

C =

b. C = e0

d

A

d = e0

C

A

d =

d =

c. PEelectric = 12

Q∆V

PEelectric = 12

(2.5 × 10−6 C)(30.0 V)


4.7 × 10−3 m

(8.85 × 10−12 C2/N •m2)(44 m2)

(8.3 × 10−8 F)

8.3 × 10−8 F = 83 nF

10. A = 7.0 m × 12.0 m

d = 1.0 × 10−3 m

e0 = 8.85 × 10−12 C2/N •m2

PEelectric = 1.0 J

a. C = e0

d

A

C =

C =

b. PEelectric = 12

C(∆V )2

∆V = 2PE

Celectric∆V =

(7.4

2 (

×1.

10

0J−)7 F)

∆V = 1.6 × 103 V = 1.6 kV

7.4 × 10−7 F = 0.74 mF

(8.85 × 10−12 C2/N •m2)(7.0 m)(12.0 m)

(1.0 × 10−3 m)


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1. I = 3.00 × 102 A

∆t = 2.4 min

∆Q = I∆t

∆Q = (3.00 × 102 A)(2.4 min)16

m

0

i

s

n

∆Q = 4.3 × 104 C


Givens Solutions

2. ∆t = 7 min, 29 s

I = 0.22 A

∆Q = I∆t

∆Q = (0.22 A)(7 min)16

m

0

i

s

n + 29 s = (0.22 A)(449 s)

∆Q = 99 C

3. ∆t = 3.3 × 10−6 s

I = 0.88 A

q = e = 1.60 × 10−19 elec

C

tron

∆Q = I∆t = nq

n = I∆q

t

n =

n = 1.8 × 1013 electrons

(0.88 A)(3.3 × 10−6 s)(1.60 × 10−19 C/electron)

4. ∆t = 3.00 h

∆Q = 1.51 × 104 CI =

∆∆Q

t

I =

I = 1.40 A

(1.51 × 104 C)

(3.00 h)3.60

1

×h

103 s

5. ∆Q = 1.8 × 105 C

∆t = 6.0 minI =

∆∆Q

t

I =

I = 5.0 × 102 A

(1.8 × 105 C)

(6.0 min)16

m

0

i

s

n

19ChapterCurrent and Resistance


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6. I = 13.6 A

Q = 4.40 × 105 C∆t =

∆I

Q

∆t = (4.4

(1

0

3

×.6

1

A

05

)

C)

∆t = 3.24 × 104 s = 9.00 h

Givens Solutions

1. ∆V = 440 V

I = 0.80 AR =

∆I

V

R = (

(

0

4

.

4

8

0

0

V

A

)

)

R = 5.5 × 102 Ω


2. ∆V = 9.60 V

I = 1.50 AR =

∆I

V

R = (

(

9

1

.

.

6

5

0

0

V

A)

)

R = 6.40 Ω

3. ∆V = 312 V

∆Q = 2.8 × 105 C

∆t = 1.00 h

I = ∆∆Q

t

R = ∆

I

V = =

∆∆V

Q

∆t

R =

R = 4.0 Ω

(312 V)(1.00 h)3.60

1

×h

103 s

(2.8 × 105 C)

∆V

∆∆Q

t

4. I = 3.8 A

R = 0.64 Ω

∆V = IR

∆V = (3.8 A)(0.64 Ω)

∆V = 2.4 V

5. R = 0.30 Ω

I = 2.4 × 103 A∆V = IR = (2.4 × 103 A)(0.30 Ω) = 7.2 × 102 V


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6. ∆V = 3.0 V

R = 16 ΩI =

∆R

V

I = (

(

3

1

.

6

0

ΩV

)

)

I = 0.19 A

Givens Solutions

7. ∆V = 6.00 × 102 V

R = 4.4 ΩI =

∆R

V = = 1.4 × 102 A

(6.00 × 102 V)

(4.4 Ω)

1. P = 12 × 103 W

R = 2.5 × 102 Ω

P = I 2R

I = R

P

I = (

(

2

1.

2

5×× 1

10

0

3

2

WΩ)

)

I = 6.9 A


2. P = 33.6 × 103 W

∆V = 4.40 × 102 V

P = I∆V

I = ∆P

V

I = (

(

3

4

3

.4

.6

0

××

1

1

0

0

3

2

W

V)

)

I = 76.4 A

3. P = 850 W

V = 12.0 V

P = I∆V

I = ∆P

V

I = 8

1

5

2

0

.0

W

V

I = 70.8 A

4. P = 41.

.

2

1

××

1

1

0

0

1

3

0

h

J

R = 40.0 Ω

P = (∆V

R

)2

∆V =√

PR

∆V = 41.

.

2

1 ×× 1

10

0

1

3

0

h

J

361

0h

0 s(40.0 Ω)

∆V = 6.5 × 102 V


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Givens Solutions

5. P = 6.0 × 1013 W

∆V = 8.0 × 106 VP =

(∆R

V )

2

R = (∆

P

V )

2

R = (

(

6

8

.

.

0

0

××

1

1

0

01

6

3

V

W

)2

)

R = 1.1 Ω

6. I = 6.40 × 103 A

∆V = 4.70 × 103 V

P = I∆V

P = (6.40 × 103 A)(4.70 × 103 V)

P = 3.01 × 107 W = 30.1 MW

1. P = 8.8 × 106 kW

total cost = $1.0 × 106

cost of energy =$0.081/kW•h

total cost of electricity = P∆t (cost of energy)

∆t =

∆t =

∆t = 1.4 h

$1.0 × 106

(8.8 × 106 kW)($0.081/kW•h)

total cost of electricity

P(cost of energy)


2. P = 104 kW


purchase power = $18 000

energy that can be purchased = p

c

u

o

r

s

c

t

h

o

a

f

se

en

p

e

o

r

w

gy

er = P∆t

∆t = (c

p

o

u

st

rc

o

h

f

a

e

s

n

e

e

p

rg

o

y

w

)

e

(

r

P)

∆t =

∆t = 1.4 × 103 h = 6.0 × 101 days

$18 000($0.120/kW •h)(104 kW)

3. ∆t = 1.0 × 104 h


total cost = $23

total cost of electricity = P∆t (cost of energy)

P =

P =

P = 2.7 × 10−2 kW

$23(1.0 × 104 h)($0.086 kW•h)

total cost of electricity

∆t(cost of energy)


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4. ∆V = 110 V

R = 80.0 Ω (for maximumpower)

∆t = 24 h

cost of energy =$0.086/kW •h

P = (∆

R

V)2

total cost of electricity = P∆t(cost of energy)

total cost = (∆V

R

)2(∆t) (cost of energy)

total cost = (11

(

0

8

V

0.

)

0

2(

Ω2

)

4 h) 1$0

k

.

W

08

•

6

h11

00

k

0

W

W

total cost = $0.31

Givens Solutions

5. 15.5 percent of solar energyconverted to electricity


purchase power = $1000.00

(0.155)Esolar = p

c

u

o

r

s

c

t

h

o

a

f

se

en

p

e

o

r

w

gy

er

Esolar =

Esolar = 8.1 × 104 kW •h = 2.9 × 1011 J

($1000.00)(0.155)($0.080/kW•h)


Chapter 20Circuits and Circuit Elements

II

1. R = 160 kΩ

R1 = 2.0R

R2 = 3.0R

R3 = 7.5R

Req = R1 + R2 + R3 = 2.0R + 3.0R + 7.5R = 12.5R

Req = (12.5)(160 kΩ) = 2.0 × 103 kΩ


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2. R = 5.0 × 108 Ω

R1 = 13

R

R2 = 27

R

R3 = 15

R

Req = R1 + R2 + R3 = 13

R + 27

R + 15

R

Req = 35 +

1

3

0

0

5

+ 21 R =

1

8

0

6

5R =

1

8

0

6

5 (5.0 × 108 Ω) = 4.1 × 108 Ω

3. R1 = 16 kΩ

R2 = 22 kΩ

R3 = 32 kΩ

Req = 82 kΩ

R4 = Req − R1 − R2 − R3 = 82 kΩ − 16 kΩ − 22 kΩ − 32 kΩ = 12 kΩ

4. R1 = 3.0 kΩ

R2 = 4.0 kΩ

R3 = 5.0 kΩ

P = (0.0100)(3.2 MW) =0.032 MW

Req = R1 + R2 + R3 = 3.0 kΩ + 4.0 kΩ + 5.0 kΩ = 12.0 kΩ

P = (∆V

R

)2

∆V =√

PReq =√

(3.2 × 104 W)(1.20 × 104 Ω) = 2.0 × 104 V

5. R1 = 4.5 Ω

R2 = 4.0 Ω

R3 = 16.0 Ω

R12 = R1 + R2 = 4.5 Ω + 4.0 Ω =

R13 = R1 + R3 = 4.5 Ω + 16.0 Ω =

R23 = R2 + R3 = 4.0 Ω + 16.0 Ω = 20.0 Ω

20.5 Ω

8.5 Ω

6. R1 = 2.20 × 102 Ω

∆Vi = 1.20 × 102 V

∆Vf = 138 V

Because the current is unchanged, the following relationship can be written.

R

V

1

i = R1

V

+f

R2

R2 = Vf R1

V

−

i

Vi R1 =

R2 = = 40

1

0

2

0

0

V

V

•Ω = 33 Ω

30 400 V •Ω − 26 400 V •Ω

120 V

(138 V)(220 Ω) − (120 V)(220 Ω)

120 V

Holt Physics Solution Manual

II

7. R1 = 3.6 × 10−5 Ω

R2 = 8.4 × 10− 6 Ω

I = 280 A

Req = R1 + R2 = 3.6 × 10−5 Ω + 8.4 × 10−6 Ω = 4.4 × 10−5 Ω

P = I 2Req = (280 A)2(4.4 × 10−5 Ω) = 3.4 W

Givens Solutions

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II Ch. 20–2

1. R1 = 1.8 Ω

R2 = 5.0 Ω

R3 = 32 Ω

Req = R

1

1 +

R

1

2 +

R

1

3

−1= 1.8

1

Ω +

5.0

1

Ω +

32

1

Ω

−1

Req = 0.55 Ω1

+ 0.20 Ω1

+ 0.031 Ω1

−1

= 0.78 Ω1

−1

= 1.3 Ω


2. R = 450 Ω

R1 = R

R2 = 2.0R

R3 = 0.50R

Req = R

1

1 +

R

1

2 +

R

1

3

−1= 450

1

Ω +

900

1

Ω +

220

1

Ω

−1

Req = 0.0022 Ω1

+ 0.0011 Ω1

+ 0.0045 Ω1

−1

= 0.0078 Ω1

−1

= 1.3 × 102 Ω

3. R1 = 2.48 × 10−2 Ω

Req = 6.00 × 10−3 ΩR2 =

R

1

eq −

R

2

1

−1= 6.00 ×

1

10−3 Ω −

2.48 ×2

10−2 Ω

−1

R2 = 167 Ω1

− 80.6 Ω1

−1

= 86 Ω1

−1

= 0.012 Ω

4. R1 = R

R2 = 3R

R3 = 7R

R4 = 11R

Req = 6.38 × 10−2 Ω

Req = R

1

1 +

R

1

2 +

R

1

3 +

R

1

4

−1=

R

1 +

3

1

R +

7

1

R +

11

1

R

−1

Req = 231 + 7

2

7

31

+R

33 + 21

−1= 233612R

−1=

1.

R

57

−1

R = 1.57Req = 1.57(6.38 × 10−2 Ω) = 0.100 Ω

5. ratio = 1.22 × 10−2 Ω/ml = 1813 km

R1 = 12

R

R2 = 14

R

R3 = 15

R

R4 = 2

1

0 R

a. R = (ratio)(l ) = (1.22 × 10−2 Ω/m)(1.813 × 106 m) =

b. Req = R

1

1 +

R

1

2 +

R

1

3 +

R

1

4

−1=

R

2 +

R

4 +

R

5 +

2

R

0

−1

Req = 3

R

1

−1= 1.00 ×

3

1

1

010 Ω

−1= 3.23 × 108 Ω

2.21 × 104 Ω

6. ∆V = 14.4 V

P = 225 WP =

(∆R

V)2

R = (∆

P

V)2

= (1

2

4

2

.

5

4

W

V)2

=

Req = R

4

−1=

R

4 =

0.92

4

2 Ω = 0.230 Ω

I = R

∆

e

V

q =

0

1

.2

4

3

.4

0

V

Ω = 62.6 A

0.922 Ω


II

7. L = 3.22 × 105 km

l = 1.00 × 103 km

ratio = 1.0 × 10−2 Ω/m

∆V = 1.50 V

Req = NR

1 where N = and R = (ratio)l

Req = −1

= −1

= 31 Ω

I = R

∆

e

V

q =

1

3

.5

1

0

ΩV

= 0.048 A

3.22 × 108 m(1.0 × 10−2 Ω/m)(1.00 × 106 m)2

L(ratio)l 2

Ll

Givens Solutions

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1. R1 = 6.60 × 102 Ω

R2 = 2.40 × 102 Ω

R3 = 2.00 × 102 Ω

R4 = 2.00 × 102 Ω

R12 = R1 + R2 = 660 Ω + 240 Ω = 900 Ω

R123 = R

1

12 +

R

1

3

−1= 900

1

Ω +

200

1

Ω

−1

R123 = 0.00111 Ω1

+ 0.00500 Ω1

−1

= 0.00611 Ω1

−1

= 164 Ω

Req = R123 + R4 = 164 Ω + 200 Ω = 364 Ω


3. R1 = 2.5 Ω

R2 = 3.5 Ω

R3 = 3.0 Ω

R4 = 4.0 Ω

R5 = 1.0 Ω

∆V = 12 V

R12 = R1 + R2 = 2.5 Ω + 3.5 Ω = 6.0 Ω

R123 = R

1

12 +

R

1

3

−1= 6.0

1

Ω +

3.0

1

Ω

−1

R123 = 0.17 Ω1

+ 0.33 Ω1

−1

= 0.50 Ω1

−1

= 2.0 Ω

R45 = R

1

4 +

R

1

5

−1= 4.0

1

Ω +

1.0

1

Ω

−1

R45 = 0.25 Ω1

+ 1.0 Ω1

−1

= 1.2 Ω1

−1

= 0.83 Ω

Req = R123 + R45 = 2.0 Ω + 0.83 Ω =

I = ∆R

V =

2

1

.

2

8

V

Ω = 4.3 A

2.8 Ω

2. ∆V = 24 V

R1 = 2.0 Ω

R2 = 4.0 Ω

R3 = 6.0 Ω

R4 = 3.0 Ω

R12 = R1 + R2 = 2.0 Ω + 4.0 Ω = 6.0 Ω

R34 = R

1

3 +

R

1

4

−1= 6.0

1

Ω +

3.0

1

Ω

−1

R34 = 0.17 Ω1

+ 0.33 Ω1

−1

= 0.50 Ω1

−1

= 2.0 Ω

Req = R

1

12 +

R

1

34

−1= 6.0

1

Ω +

2.0

1

Ω

−1

Req = 0.17 Ω1

+ 0.50 Ω1

−1

= 0.67 Ω1

−1

= 1.5 Ω

I = R

∆

e

V

q =

1

2

.

4

5

V

V = 16 A

Holt Physics Solutions ManualII Ch. 20–4

II

4. ∆V = 1.00 × 103 V

R1 = 1.5 Ω

R2 = 3.0 Ω

R3 = 1.0 Ω

R12 = R

1

1 +

R

1

2

−1= 1.5

1

Ω +

3.0

1

Ω

−1

R12 = 0.67 Ω1

+ 0.33 Ω1

−1

= 1.00 Ω1

−1

= 1.00 Ω

Req = R12 + R3 = 1.00 Ω + 1.0 Ω =

P = (∆

R

V

eq

)2

= (1.00

2

×.0

1

Ω03 V)2

= 5.0 × 105 W

2.0 Ω

Givens Solutions

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5. ∆V = 2.00 × 103 V

I = 1.0 × 10−8 A

R1 = r

R2 = 3r

R3 = 2r

R4 = 4r

Req = ∆I

V =

2

1

.

.

0

0

0

××1

1

0

0−

3

8 A

V =

R12 = R1 + R2 = r + 3r = 4r

R34 = R3 + R4 = 2r + 4r = 6r

Req = R

1

12 +

R

1

34

−1=

4

1

r +

6

1

r

−1

Req = 31+2r

2

−1=

1

5

2r

−1=

1

5

2 r

r = 1

5

2Req =

1

5

2 (2.0 × 1011 Ω) = 8.3 × 1010 Ω

2.0 × 1011 Ω

1. R = 8.1 × 10−2 Ω

Req = 0.123 Ω

∆V = 220 V

R12 = R45 = 0.16 Ω

R12345 = 0.042 Ω

a. I = R

∆

e

V

q =

0

2

.1

2

2

0

3

V

Ω = 1800 A

∆V12345 = IR12345 = (1800 A)(0.042 Ω) = 76 V

∆V3 = ∆V12345 =

I3 = ∆R

V

3

3 = 8.1 ×

76

10

V−2 Ω

= 9.4 × 102 A

76 V


6. P = 6.0 × 105 W

∆V = 220 V

R = (∆

P

V)2

= 6.

(

0

22

×0

1

V

05

)2

W =

R12 = R45 = 2R = 2(0.081 Ω) = 0.16 Ω

R12345 = R

1

12 +

R

1

3 +

R

1

45

−1= 0.1

1

6 Ω +

0.08

1

1 Ω +

0.1

1

6 Ω

−1

R12345 = 6.2 Ω1

+ 12 Ω1

+ 6.2 Ω1

−1

= 24 Ω1

−1

= 0.042 Ω

Req = R12345 + R6 = 0.042 Ω + 0.081 Ω = 0.123 Ω

P = (∆

R

V

eq

)2

= (

0

2

.

2

1

0

23

V

Ω)2

= 3.9 × 105 W

8.1 × 10−2 Ω


II

b. ∆V12 = ∆V12345 = 76 V

I12 = ∆R

V

12

12 = 0

7

.1

6

6

V

Ω = 4.8 × 102 A

I2 = I12 =

∆V2 = I2R2 = (4.8 × 102 A)(8.1 × 10−2 Ω) =

c. Same as part b:

I4 =

∆V4 = 39 V

4.8 × 102 A

39 V

4.8 × 102 A

Givens Solutions

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2. ∆V = 12 V

R1 = 2.5 Ω

R3 = 3.0 Ω

R4 = 4.0 Ω

R5 = 1.0 Ω

R12 = 6.0 Ω

R123 = 2.0 Ω

R45 = 0.83 Ω

Req = 2.8 Ω

I = 4.3 A

a. ∆V45 = IR45 = (4.3 A)(0.83 Ω) = 3.6 V

∆V5 = ∆V45 =

I5 = ∆R

V

5

5 = 1

3

.

.

0

6

ΩV

=

b. ∆V123 = IR123 = (4.3 A)(2.0 Ω) = 8.6 V

∆V12 = ∆V123 = 8.6 V

I1 = I12 = ∆R

V

1

1

2

2 = 6

8

.

.

0

6

ΩV

=

∆V1 = I1R1 = (1.4 A)(2.5 Ω) =

c. I45 = I = 4.3 A

∆V45 = I45R45 = (4.3 A)(0.83 Ω) = 3.6 V

V4 = ∆V45 =

I4 = ∆R

V

4

4 = 4

3

.

.

0

6

ΩV

=

d. ∆V3 = ∆V123 =

I3 = ∆R

V

3

3 = 3

8

.

.

0

6

ΩV

= 2.9 A

8.6 V

0.90 V

3.6 V

3.5 V

1.4 A

3.1 A

3.6 V

II

3. R1 = 15 Ω

R2 = 3.0 Ω

R3 = 2.0 Ω

R4 = 5.0 Ω

R5 = 7.0 Ω

R6 = 3.0 Ω

R7 = 3.0 × 101 Ω

∆V = 2.00 × 103 V

R23 = R2 + R3 = 3.0 Ω + 2.0 Ω = 5.0 Ω

R234 = R

1

23 +

R

1

4

−1= 5.0

1

Ω +

5.0

1

Ω

−1

R234 = 0.40 Ω1

−1

= 2.5 Ω

R56 = R5 + R6 = 7.0 Ω + 3.0 Ω = 10.0 Ω

R567 = R

1

56 +

R

1

7

−1= 10.

1

0 Ω +

30

1

Ω

−1

R567 = 0.100 Ω1

+ 0.033 Ω1

−1

= 0.133 Ω1

−1

= 7.52 Ω

Req = R1 + R234 + R567 = 15 Ω + 2.5 Ω + 7.52 Ω = 25 Ω

a. I = R

∆

e

V

q =

2.00

2

×5 Ω

103 V = 80 A

∆V234 = IR234 = (80 A)(2.5 Ω) = 2.0 × 102 V

∆V4 = ∆V234 =

I4 = ∆R

V

4

4 = 2

5

0

.0

0

ΩV

=

b. ∆V23 = ∆V234 = 200 V

I23 = ∆R

V

23

23 = 2

5

0

.0

0

ΩV

= 40 A

I3 = I23 =

∆V3 = I3R3 = (40 A)(2.0 Ω) =

c. I567 = I = 80 A

V567 = I567R567 = (80 A)(7.52 Ω) = 600 V

∆V56 = ∆V567 = 600 V

I56 = ∆R

V

56

56 = 1

6

0

0

.

0

0

V

Ω = 60 A

I5 = I56 =

∆V5 = I5R5 = (60 A)(7.0 Ω) =

d. ∆V7 = ∆V567 =

I7 = ∆R

V

7

7 = 6

3

0

0

0

ΩV

= 2.0 × 101 A

6.0 × 102 V

4.2 × 102 V

6.0 × 101 A

8.0 × 101 V

4.0 × 101 A

4.0 × 101 A

2.0 × 102 V

Givens Solutions

Cop

yrig

ht ©

by H

olt,

Rin

ehar

t and

Win

ston

.All

right

s re

serv

ed.

Holt Physics Solutions ManualII Ch. 20–6

Chapter 21Magnetism

II

1. B = 45 T

v = 7.5 × 106 m/s

q = e = 1.60 × 10−19 C

me = 9.109 × 10−31 kg

Fmagnetic = qvB

Fmagnetic = (1.60 × 10−19 C)(7.5 × 106 m/s)(45 T)



Givens Solutions

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rved

.

2. q = 12 × 10−9 C

v = 450 km/h

B = 2.4 T

Fmagnetic = qvB

Fmagnetic = (12 × 10−9 C)(450 km/h) 36

1

0

h

0 s 110k

3

m

m (2.4 T)


3. v = 350 km/h

q = 3.6 × 10−8 C

B = 7.0 × 10−5 T

q = 30.0°

Fmagnetic = qvB = q[v (sin q)]B

Fmagnetic = (3.6 × 10−8 C)(350 km/h)36

1

0

h

0 s 110k

3

m

m (sin 30.0°)(7.0 × 10−5 T)


4. v = 2.60 × 102 km/h


q = 1.60 × 10−19 C

Fmagnetic = qvB

B = Fma

q

g

v

netic

B =

B = 2.6 T

(3.0 × 10−17 N)

(1.60 × 10−19 C)(2.60 × 102 km/h)36

1

0

h

0 s 110k

3

m

m


II

5. q = 1.60 × 10−19 C

v = 60.0 km/h


Fmagnetic = qvB

B = Fma

q

g

v

netic

B =

B = 7.5 × 10−5 T

(2.0 × 10−22 N)

(1.60 × 10−19 C)(60.0 km/h) 36

1

0

h

0 s 110k

3

m

m

Givens Solutions

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.


6. q = 88 × 10−9 C

B = 0.32 T


Fmagnetic = qvB

v = Fm

q

ag

B

netic

v =

v = 44 m/s = 160 km/h

(1.25 × 10−6 N)(88 × 10−9 C)(0.32 T)

7. q = 1.60 × 10−19 C

B = 6.4 T

Fmagnetic = 2.76 × 10−16 N

∆x = 4.0 × 103 m

vi = 0 m/s

vf = 270 m/s

a. Fmagnetic = qvB

v = Fm

q

ag

B

netic

v =

v =

b. ∆x = (vf

2

+ vi) ∆t

∆t = (vf

2∆+

x

vi)

∆t = (2

2

7

(

0

4.

m

0

/

×s

1

+0

0

3

m

m

/

)

s)

∆t = 3.0 × 101 s

2.7 × 102 m/s = 9.7 × 102 km/h

(2.76 × 10−16 N)(1.60 × 10−19 C)(6.4 T)

8. B = 0.600 T

q = 1.60 × 10−19 C

v = 2.00 × 105 m/s

m1 = 9.98 × 10−27 kg

m2 = 11.6 × 10−27 kg

a. Fmagnetic = qvB

Fmagnetic = (1.60 × 10−19 C)(2.00 × 105 m/s)(0.600 T)

Fmagnetic =

b. Fc,1 = m

r1

1

v2

= Fmagnetic

Fc,2 = m

r2

2

v2

= Fmagnetic

r1 = Fm

m

a

1

g

v

ne

2

tic

r2 = Fm

m

a

2

g

v

ne

2

tic

1.92 × 10−14 N


II

r1 =

r1 = 2.08 × 10−2 m

r2 =

r2 = 2.42 × 10−2 m

r2 − r1 = 3.40 × 10−3 m = 3.4 mm

(11.6 × 10−27 kg)(2.00 × 105 m/s)2

(1.92 × 10−14 N)

(9.98 × 10−27 kg)(2.00 × 105 m/s)2

(1.92 × 10−14 N)

Givens Solutions

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.

1. B = 22.5 T

l = 12 × 10−2 m

I = 8.4 × 10−2 A

Fmagnetic = BIl

Fmagnetic = (22.5 T)(8.4 × 10−2 A)(12 × 10−2 m)

Fmagnetic = 0.23 N


2. l = 1066 m


I = 0.80 A

Fmagnetic = BIl

B =

B = (0

(

.

6

8

.

0

3

A

×)

1

(1

0

0

−

6

2

6

N

m

)

)

B = 7.4 × 10−5 T

Fmagnetic

I l

3. l = 5376 m

Fmagnetic = 3.1 N

I = 12 A

q = 38°

Fmagnetic = BI l = [B(sin q)]Il

B =

B =

B = 7.8 × 10−5 T

(3.1 N)(12 A)(5376 m)(sin 38.0°)

FmagneticI l (sin q)

4. l = 21.0 × 103 m

B = 6.40 × 10−7 T


Fmagnetic = BIl

I =

I =

I = 1.34 A

(1.80 × 10−2 N)(6.40 × 10−7 T)(21.0 × 103 m)

Fmagnetic

Bl


II

5. B = 2.5 × 10−4 T

l = 4.5 × 10−2 m


Fmagnetic = BIl

I =

I =

I = 3.2 10–2 A

(3.6 × 10−7 N)(2.5 × 10−4 T)(4.5 × 10−2 m)

Fmagnetic

Bl

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.

6. Fmagnetic = 5.0 × 105 N

B = 3.8 T

I = 2.00 × 102 A

Fmagnetic = BIl

l = Fma

B

g

I

netic

l =

l = 6.6 × 102 m

(5.0 × 105 N)(3.8 T)(2.00 × 102 A)

7. Fmagnetic = 16.1 N

B = 6.4 × 10−5 T

I = 2.8 A

Fmagnetic = BIl

l = Fma

B

g

I

netic

l =

l = 9.0 × 104 m

(16.1 N)(6.4 × 10−5 T)(2.8 A)

8. B = 0.040 T

I = 0.10 A

q = 45°

l = 55 cm 0.55 m

Fmagnetic = BIl = [B(sin q)]Il

Fmagnetic = (0.040 T)(sin 45°)(0.10 A)(0.55 m)

Fmagnetic = 1.6 × 10−3 Ν

9. B = 38 T

l = 2.0 m

m = 75 kg

g = 9.81 m/s2

Fmagnetic = BIl

Fg = mg

Fmagnetic = Fg

BIl = mg

I =

I = (75

(3

k

8

g

T

)(

)

9

(

.

2

8

.

1

0

m

m

/

)

s2)

I = 9.7 A

mgBl

II

10. l = 478 × 103 m

Fmagnetic = 0.40 N

B = 7.50 × 10−5 T

Fmagnetic = BIl

I =

I =

I = 1.1 × 10−2 A

(0.40 N)(7.50 × 10−5 T)(478 × 103 m)

Fmagnetic

Bl

Givens Solutions

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art a

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inst

on.A

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rese

rved

.


II

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rved

.


1. Ai = 6.04 × 105 m2

Af = 12

(6.04 × 105 m2)

B = 6.0 × 10−5 T

emf = 0.80 V

N = 1 turn

q = 0.0°

emf = −N∆[AB

∆(

t

cos q)]

∆t = −NB

e

(

m

co

f

s q) ∆A

∆t = −NB

e

(

m

co

f

s q) (Af − Ai)

∆t = (6.04 × 105 m2)12

− 1

∆t = 23 s

−(1)(6.0 × 10−5 T)(cos 0.0°)

(0.80 V)


Givens Solutions

4. Af = 3.2 × 104 m2

Ai = 0.0 m2

∆t = 20.0 min

B = 4.0 × 10−2 T

N = 300 turns

q = 0.0°

emf = −N∆[AB

∆(

t

cos q)] =

−NB(

∆c

t

os q) (Af − Ai)

emf = [(3.2 × 104 m2) − (0.0 m2)]

emf = −3.2 × 102 V

−(300)(4.0 × 10−2 T)(cos 0.0°)

(20.0 min)16

m

0

i

s

n

2. r = 100

2

.0 m = 50.0 m

Bi = 0.800 T

Bf = 0.000 T

q = 0.00°

emf = 46.7 V

N = 1 turn

emf = −N∆[AB

∆(

t

cos q)]

∆t =

∆t =

∆t = 135 s

−(1)(p)(50.0 m)2(cos 0.0°)(0.000 T − 0.800 T)

(46.7 V)

−N(pr2)(cos q)(Bf − Bi)emf

22ChapterInduction and Alternating Current

3. emf = 32.0 × 106 V

Bi = 1.00 × 103 T

Bf = 0.00 T

A = 4.00 × 10−2 m2

N = 50 turns

q = 0.00°

emf = −N∆[AB

∆(

t

cos q)]

∆t =

∆t =

∆t = 6.3 × 10−5 s

−(50)(4.00 × 10−2 m2)(cos 0.00°)[(0.00 T) − (1.00 × 103 T)]

(32.0 × 106 V)

−NA(cos q)(Bf − Bi)emf

II

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.


5. Bi = 8.0 × 10−15 T

Bf = 10 Bi = 8.0 × 10−14 T

∆t = 3.0 × 10−2 s

A = 1.00 m2

emf = −1.92 × 10−11 V

q = 0.0°

emf = −N∆[AB

∆(

t

cos q)]

N = A(B

−

f

(

−em

B

f

i

)

)

(

(

∆co

t)

s q)

N =

N = 8 turns

−(−1.92 × 10−11 V)(3.0 × 10−2 s)(1.00 m2)[(8.0 × 10−14 T) − (8.0 × 10−15 T)](cos 0.0°)

6. Bi = 0.50 T

Bf = 0.00 T

N = 880 turns

∆t = 12 s

emf = 147 V

q = 0.0°

emf = −N∆[AB

∆(

t

cos q)] = −NA(cos q)

∆∆

B

t

A = N(c

−o

(

s

e

qm

)

f

(

)

B

(∆

f

t

−)

Bi)

A =

A = 4.0 m2

−(147 V)(12 s)(880)(cos 0.0°)(0.00 T − 0.50 T)

Givens Solutions

1. f = 833 Hz

D = 5.0 cm 0.050 m

B = 8.0 × 10−2 T

maximum emf 330 V

maximum emf = NABw = NAB(2p f )

A pr2 p D

2

2

p 0.05

2

m

2

2.0 103 m2

N ma

A

xi

B

m

(

u

2pm

f)

emf

N = 4.0 × 102 turns

330 V (2.0 10–3 m2)(2p)(833 Hz)(8.0 10–2 T)


3. r = 19.

2

3 m = 9.65 m

w = 0.52 rad/s

maximum emf = 2.5 V

N = 40 turns

maximum emf = NABw

B = ma

N

xi

(

m

pu

r2

m

)wemf

B =

B = 4.1 × 10−4 T

2.5 V(40)(p)(9.65 m)2(0.52 rad/s)

2. w = 335 rad/s

maximum emf = 214 V

B = 8.00 × 10−2 T

A = 0.400 m2

maximum emf = NABw

N = maxim

AB

u

wm emf

N =

N = 20.0 turns

214 V(0.400 m2)(0.0800 T)(335 rad/s)


II

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rved

.4. maximum emf =

8.00 × 103 V

N = 236

A = (6.90 m)2

w = 57.1 rad/s

maximum emf = NABw

B = maxi

N

m

A

u

wm emf

B =

B = 1.25 × 10−2 T

8.00 × 103 V(236)(6.90 m)2(57.1 rad/s)

Givens Solutions

5. N = 1000 turns

A = 8.0 × 10−4 m2

B = 2.4 × 10−3 T

maximum emf = 3.0 V

maximum emf = NABw

w = maxim

NA

um

B

emf

w =

w = 1.6 × 103 rad/s

3.0 V(1000)(8.0 × 10−4 m2)(2.4 × 10−3 T)

6. N = 640 turns

A = 0.127 m2

maximum emf =24.6 × 103 V

B = 8.00 × 10−2 T

maximum emf = NABw

w = maxim

NA

um

B

emf

w =

w = 3.78 × 103 rad/s

24.6 × 103 V(640)(0.127 m2)(8.00 × 10−2 T)


7. f = 1.0 × 103 Hz

B = 0.22 T

N = 250 turns

r = 12 × 10−2 m

maximum emf = NABw = NAB(2p f ) = N(pr3)Bw = N(pr2)B(2p f )

maximum emf = (250)(p)(12 × 10−2 m)2(0.22 T)(2p)(1.0 × 103 Hz)

maximum emf = 1.6 × 104 V = 16 kV

1. ∆Vrms = 120 V

R = 6.0 × 10−2 Ω

= 0.7071

√2

a. Irms = ∆V

Rrms

Irms = (6.0

(1

×2

1

0

0

V−2

)

Ω)

Irms =

b. Imax = (Irms) √

2

Imax = (2.0

(0

×.7

1

0

0

7

3

)

A)

Imax =

c. P = (Irms)(∆Vrms)

P = (2.0 × 103 A)(120 V)

P = 2.4 × 105 W

2.8 × 103 A

2.0 × 103 A


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.

2. P = 10.0(Acoustic power)

Acoustic power =30.8 × 103 W

∆Vrms = 120.0 V

= 0.7071

√2

P = ∆Vrms Irms

Irms = ∆V

P

rms

Irms = I√ma

2x

I√ma

2x =

∆V

P

rms

Imax = ∆P

V

√

rm

2

s

Imax =

Imax = 3.63 × 103 A

(10.0)(30.8 × 103 W)

(120.0 V)(0.707)

3. P = 1.325 × 108 W

∆Vrms = 5.4 × 104 V

= 0.7071

√2

P = ∆VrmsI rms = (Irms)2R = (∆V

Rrms)

2

Irms = I√ma

2x

Imax =√

2 Irms = ∆

√

V

2

rm

P

s

Imax =

Imax =

R = (∆V

Prms)

2

R = (1

(5

.3

.4

25

××10

1

4

08

V

W

)2

)

R = 22 Ω

3.5 × 103 A

1.325 × 108 W(5.4 × 104 V)(0.707)

Givens Solutions

4. ∆Vrms = 1.024 × 106 V

Irms = 2.9 × 10−2 A

= 0.7071

√2

∆Vmax = ∆Vrms

√2

∆Vmax = (1.02

(0

4

.7

×0

1

7

0

)

6 V)

∆Vmax =

Imax = Irms

√2

Imax = (2.9

(0

×.7

1

0

0

7

−

)

2 A)

Imax = 4.1 × 10−2 A

1.45 × 106 V = 1.45 MV


II

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.

R = ∆I

V

m

m

ax

ax = ∆I

V

rm

rm

s

s

R = (

(

0

3

.

2

8

0

0

V

A

)

) =

(

(

0

2

.

3

5

0

7

V

A

)

)

R = 4.0 × 102 Ω

Givens Solutions

6. Imax = 75 A

R = 480 Ω

= 0.7071

√2

∆Vrms = ∆V√m

2ax

∆Vmax = (Imax)(R)

∆Vrms = Im√a

2xR

∆Vrms = (75 A)(480 Ω)(0.707)

∆Vrms = 2.5 × 104 V = 25 kV

7. Ptot = 6.2 × 107 W

Ptot = 24 P

R = 1.2 × 105 Ω

= 0.7071

√2

P = (Irms)2R = P

2t

4ot

P = 6.2 ×

2

1

4

07 W

P =

Irms =R

P

Irms = (

(

2

1.

.

6

2 ×× 1

10

0

6

5W

Ω)

)

Irms =

Imax =√

2 Irms

Imax = 0

4

.

.

7

7

0

A

7

Imax = 6.6 A

4.7 A

2.6 × 106 W = 2.6 MW

5. ∆Vmax = 320 V

Imax = 0.80 A

= 0.7071

√2

∆Vrms = ∆V√m

2ax

∆Vrms = (320 V)(0.707)

∆Vrms =

Irms = I√ma

2x

Irms = (0.80 A)(0.707)

Irms = 0.57 A

2.3 × 102 V


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rved

.

1. N1 = 5600 turns

N2 = 240 turns

∆V2 = 4.1 × 102 V

∆V1 = ∆V2 N

N1

2

∆V1 = (4.1 × 103 V) 5264000∆V1 = 9.6 × 104 V = 96 kV


Givens Solutions

2. N1 = 74 turns

N2 = 403 turns

∆V2 = 650 V

∆V1 = ∆V2 N

N1

2

∆V1 = (650 V)4

7

0

4

3

∆V1 = 120 V

4. ∆V1 = 765 × 103 V

∆V2 = 540 × 103 V

N1 = 2.8 × 103 turns

∆∆

V

V

1

2 = N

N2

1

N2 = ∆∆

V

V

1

2 N1

N2 = 574

6

0

5

××

1

1

0

0

3

3

V

V(2.8 × 103)

N2 = 2.0 × 103 turns

3. ∆V1 = 2.0 × 10−2 V

N1 = 400 turns

N2 = 3600 turns

∆V2 = 2.0 × 10−2 V

∆V2 = ∆V1 N

N2

1

∆V2 = (2.0 × 10−2 V) 3460000∆V2 =

∆V1 = ∆V2 N

N1

2

∆V1 = (2.0 × 10−2 V) 3460000∆V1 = 2.2 × 10−3 V

0.18 V


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hts

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.

N

N

1

2 = ∆∆

V

V2

1

N1 = N2∆∆

V

V1

2

N1 = (660)122

2

0

0

V

V

N1 = 360 turns

Givens Solutions

6. P = 20.0 W

∆V1 = 120 V

N

N1

2 = 0.36

a. P = (∆V1)(I1)

I1 = ∆

P

V1 =

(

(

2

1

0

2

.0

0

W

V)

)

I1 =

b. ∆∆

V

V

1

2 = N

N2

1

∆V2 = N

N2

1∆V1

∆V2 = 0.

1

36(120 V)

∆V2 = 3.3 × 102 V

0.17 A

5. ∆V1 = 230 × 103 V

∆V2 = 345 × 103 V

N1 = 1.2 × 104 turns

N

N

1

2 = ∆∆

V

V2

1

N2 = N1 ∆∆

V

V2

1

N2 = (1.2 × 104)324

3

5

0

××

1

1

0

0

3

3

V

V

N2 = 1.8 × 104 turns

7. ∆V1 = 120 V

∆V2 = 220 V

I2 = 30.0 A

N2 = 660 turns

P1 = P2

∆V1I1 = ∆V2I2

I1 = ∆∆

V

V2

1 I2

I1 = 212

2

0

0

V

V (30.0 A)

I1 = 55 A

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1. E = 1.29 × 10−15 J

C = 3.00 × 108 m/s

h = 6.63 × 10−34 J•s

l = h

E

c =

l = 1.54 × 10−10 m = 0.154 nm

(6.63 × 10−34 J•s)(3.00 × 108 m/s)

1.29 × 10−15 J


Givens Solutions

2. E = 6.6 × 10−19 J

C = 3.00 × 108 m/s

h = 6.63 × 10−34 J•s

l = h

E

c =

l = 3.0 × 10−7 m

(6.63 × 10−34 J•s)(3.00 × 108 m/s)

6.6 × 10−19 J

3. E = 5.92 × 10−6 eV

C = 3.00 × 108 m/s

h = 6.63 × 10−34 J•s

l = h

E

c =

l = 0.210 m

(6.63 × 10−34 J•s)(3.00 × 108 m/s)(5.92 × 10−6 eV)(1.60 × 10−19 J/eV)

23Chapter

4. E = 2.18 × 10−23 J

h = 6.63 × 10−34 J •s

E = hf

f = E

h

f = 6

2

.6

.1

3

8

××1

1

0

0−

−

34

23

J

J

•s

f = 3.29 × 1010 Hz

Atomic Physics

5. E = 1.85 × 10−23 J

h = 6.63 × 10−34 J •s

f = E

h =

6

1

.6

.8

3

5

××1

1

0

0−

−

3

2

4

3

J/

J

s = 2.79 × 1010 Hz

6. f = 9 192 631 770 s−1

h = 6.626 0755 × 10−34 J •s

1 eV = 1.602 117 33 × 10−19 J

E = hf

E =

E = 3.801 9108 × 10−5 eV

(6.626 0755 × 10−34 J •s)(9 192 631 770 s−1)

1.602 117 33 × 10−19 J/eV

7. l = 92 cm = 92 × 10−2 m

c = 3.00 × 108 m/s

h = 6.63 × 10−34 J •s

h = 4.14 × 10−15 eV •s

f = lc

f = 3.

9

0

2

0

××

1

1

0

0−

8

2

m

m

/s

f =

E = hf

3.3 × 108 Hz = 330 MHz

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Givens Solutions

8. v = 1.80 × 10−17 m/s

∆t = 1.00 year

l = ∆x

c = 3.00 × 108 m/s

h = 6.63 × 10−34 J •s

∆x = v∆t

∆x = (1.80 × 10−17 m/s)(1.00 year)365

1

.2

y

5

ea

d

r

ays124

da

h

y36

1

0

h

0 s

∆x =

E = hf = h

lc =

∆h

x

c

E =

E = 3.50 × 10−16 J

(6.63 × 10−34 J •s)(3.00 × 108 m/s)

5.68 × 10−10 m

5.68 × 10−10 m

E = (6.63 × 10−34 J •s)(3.3 × 108 Hz)

E =

E = (4.14 × 10−15 eV •s)(3.3 × 108 Hz)

E = 1.4 × 10−6 eV

2.2 × 10−25 J


1. hft = 4.5 eV

KEmax = 3.8 eV

h = 4.14 × 10−15 eV•s

f = [KEma

hx + hft] =

4

[3

.1

.8

4

e

×V

10

+−41.55

e

e

V

V

•

]

s = 2.0 × 1015 Hz

2. hft = 4.3 eV

KEmax = 3.2 eV

h = 4.14 × 10−15 eV •s

KEmax = hf − hft

f = KEma

hx + hft

f = 4.

3

1

.

4

2

×eV

10

+−415.3

e

e

V

V

•s

f = 1.8 × 1015 Hz

3. hft ,Cs = 2.14 eV

hft,Se = 5.9 eV

h = 4.14 × 10−15 eV •s

c = 3.00 × 108 m/s

KEmax = 0.0 eV for bothcases

a. KEmax = hf − hft = 0.0 eV = h

lc − hft

l = h

h

f

c

t

lCs = hf

h

t,

c

Cs =

lCs =

b. lSe = hf

h

t,

c

Se =

lSe = 2.1 × 10−7 m = 2.1 × 102 nm

(4.14 × 10−15 eV •s)(3.00 × 108 m/s)

5.9 eV

5.80 × 10−7 m = 5.80 × 102 nm

(4.14 × 10−15 eV •s)(3.00 × 108 m/s)

2.14 eV


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.4. l = 2.00 × 102 nm =

2.00 × 10−7 m

v = 6.50 × 105 m/s

me = 9.109 × 10−31 kg

c = 3.00 × 108 m/s

h = 4.14 × 10−15 eV •s

KEmax = 12

mev2 = hf − hft

12

mev2 = h

lc − hft

hft = h

lc − 1

2mev2

hft = −

hft = 6.21 eV − 1.20 eV

hft =

ft = 4.14 ×

5.0

1

1

0−e1V5 eV•s

= 1.21 × 1015 Hz

5.01 eV

(0.5)(9.109 × 10−31 kg)(6.50 × 105 m/s)2

1.60 × 10−19 J/eV

(4.14 × 10−15 eV •s)(3.00 × 108 m/s)

2.00 × 10−7 m

6. l = 2.00 × 102 nm =2.00 × 10−7 m

KEmax = 0.46 eV

h = 4.14 × 10−15 eV •s

c = 3.00 × 108 m/s

KEmax = hf − hft

hft = hf − KEmax = h

lc − KEmax

hft = − 0.46 eV

hft = 6.21 eV − 0.46 eV

hft =

ft = 4.14 ×

5.

1

8

0

e−V15 eV•s = 1.4 × 1015 Hz

5.8 eV

(4.14 × 10−15 eV •s)(3.00 × 108 m/s)

2.00 × 10−7 m

7. l = 589 nm = 589 × 10−9 m

hft = 2.3 eV

c = 3.00 × 108 m/s

h = 4.14 × 10−15 eV •s

KEmax = hf − hft = h

lc − hft

KEmax = − 2.3 eV

KEmax = 2.11 eV − 2.3 eV

KEmax =

No. The photons in the light produced by sodium vapor need 0.2 eV more energyto liberate photoelectrons from the solid sodium.

−0.2 eV

(4.14 × 10−15 eV •s)(3.00 × 108 m/s)

589 × 10−9 m

5. f = 2.2 × 1015 Hz

KEmax = 4.4 eV

h = 4.14 × 10−15 eV •s

KEmax = hf − hft

hft = hf − KEmax

hft = (4.14 × 10−15 eV •s)(2.2 × 1015 Hz) − 4.4 eV

hft = 9.1 eV − 4.4 eV =

ft = 4.14 ×

4

1

.7

0−e

1V5 eV •s

= 1.1 × 1015 Hz

4.7 eV

Givens Solutions


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8. hft = 2.3 eV

l = 410 nm = 4.1 × 10−7 m

h = 4.14 × 10−15 eV •s

c = 3.00 × 108 m/s

Givens Solutions

9. hft ,Zn = 4.3 eV

hft ,Pb = 4.1 eV

KEmax ,Zn = 0.0 eV

me = 9.109 × 10−31 kg

KEmax = hf − hft

KEmax ,Pb = hf − hft,Pb = (KEmax,Zn + hft,Zn) − hft,Pb

KEmax ,Pb = 12

mev2

12

mev2 = (KEmax,Zn + hft,Zn) − hft,Pb

v = v = v = 9.1(

0

29

)(

×0 .

1

20

e−V

31)

kg1.601

×e

1

V0−

19

J = 3 × 105 m/s

1.60 × 10−19 J

1 eV

(2)(0.0 eV + 4.3 eV − 4.1 eV)

9.109 × 10−31 kg

2(KEmax ,Zn + hft,Zn − hft,Pb)

me

1. v = 3.2 m/s

l = 3.0 × 10−32 m

h = 6.63 × 10−34 J•s

m = lh

v = =

2. l = 6.4 × 10−11 m

6.9 × 10−3 kg6.63 × 10−34 J•s

(3.0 × 10−32 m)(3.2 m/s)


v = 64 m/s

h = 6.63 × 10−34 J •smv =

lh

m = lh

v

m =

m = 1.6 × 10−25 kg

6.63 × 10−34 J •s(6.4 × 10−11 m)(64 m/s)

KEmax = h

lc − hft

KE = − 2.3 eV

KE = 3.03 eV − 2.3 eV = 0.7 eV

(4.14 × 10−15 eV •s)(3.00 × 108 m/s)

4.1 × 10−7 m

3. q = (2)(1.60 × 10−19 C) =3.20 × 10−19 C

∆V = 240 V

h = 6.63 × 10−34 J •s

l = 4.4 × 10−13 m

KE = q∆V = 12

mv2

m = 2q

V

∆2V

v = l

h

m = = 1.0 × 105 m/s

m =

m = 1.5 × 10−26 kg

2(3.20 × 10−19 C)(240 V)

(1.0 × 105 m/s)2

6.63 × 10−34 J •s(4.4 × 10−13 m)(1.5 × 10−26 kg)


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.4. l = 2.5 nm = 2.5 × 10−9 m

mn = 1.675 × 10−27 kg

h = 6.63 × 10−34 J •s

mv = lh

v = lm

h

n

v =

v = 1.6 × 102 m/s

6.63 × 10−34 J •s(2.5 × 10−9 m)(1.675 × 10−27 kg)

Givens Solutions

5. m = 7.65 × 10−70 kg

l = 5.0 × 1032 m

h = 6.63 × 10−34 J •s

mv = lh

v = lh

m

v =

v = 1.7 × 103 m/s

6.63 × 10−34 J •s(5.0 × 1032 m)(7.65 × 10−70 kg)

6. m = 1.6 g = 1.6 × 10−3 kg

v = 3.8 m/s

h = 6.63 × 10−34 J •s

mv = lh

l = m

h

v

l =

l = 1.1 × 10−31 m

6.63 × 10−34 J •s(1.6 × 10−3 kg)(3.8 m/s)

7. ∆x = 42 195 m

∆t = 3 h 47 min = 227 min

m = 0.080 kg

h = 6.63 × 10−34 J •s

v = ∆∆

x

t = =

lh

= mv

l = m

h

v

l =

l = 2.7 × 10−33 m

6.63 × 10−34 J •s(0.080 kg)(3.10 m/s)

3.10 m/s42 195 m

(227 min)16

m

0

i

s

n

Chapter 25Subatomic Physics

II

1. E = 610 TW•h

atomic mass of 12H =

2.014 102 u

atomic mass of 2656Fe =

55.934 940 u

atomic mass of 22688Ra =

226.025 402 u

a. ∆m = c

E2 =

∆m =

b. N = atomic

∆m

m

ass of 12H

=

N =

c. N = =

N =

d. N = =

N = 6.4 × 1025 22688Ra nuclei

24 kg(1.66 × 10−27 kg/u)(226.025 402 u)

∆matomic mass of 226

88Ra

2.6 × 10262656Fe nuclei

24 kg(1.66 × 10−27 kg/u)(55.934 940 u)

∆matomic mass of 26

56Fe

7.2 × 102712H nuclei

24 kg(1.66 × 10−27 kg/u)(2.014 102 u)

24 kg

(610 × 109 kW •h)(3.6 × 106 J/kW •h)

(3.00 × 108 m/s)2

Additional Practice 25A, p. 173

Givens Solutions

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ook.


2. m = 4.1 × 107 kg

h = 10.0 cm

Z = 26

N = 56 − 26 = 30

mH = 1.007 825 u

mn = 1.008 665 u

atomic mass of 2656Fe =

55.934 940 u

a. E = mgh = (4.1 × 107 kg)(9.81 m/s2)(0.100 m)

E =

b. Etot = = 2.5 × 1020 MeV

∆mtot = = 2.7 × 1017 u

∆m = Z(atomic mass of H) + Nmn − atomic mass of 2656Fe

∆m = 26(1.007 825 u) + 30(1.008 665 u) − 55.934 940 u

∆m = 0.528 46 u

Ebind = (0.528 46 u)931.50 M

u

eV

Ebind = 492.26 MeV

N = E

E

b

t

i

o

n

t

d = = 5.1 × 1017 reactions

mtot = (5.1 × 1017)(55.934 940 u)1.66 × 10−27 k

u

g = 4.7 × 10−8 kg

2.5 × 1020 MeV

492.26 MeV

2.5 × 1020 MeV931.50 MeV/u

(4.0 × 107 J)(1 × 10−6 MeV/eV)

(1.60 × 10−19 J/eV)

4.0 × 107 J


6. P = 42 MW = 42 × 106 W

atomic mass of 147N =

14.003 074 u

atomic mass of H =1.007 825 u

mn = 1.008 665 u

Z = 7

N = 14 − 7 = 7

∆t = 24 h

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3. E = 2.0 × 103 TW •h =2.0 × 1015 W •h

atomic mass of 23592U =

235.043 924 u


mn = 1.008 665 u

Z = 92

N = 235 − 92 = 143

∆m = Z(atomic mass of H) + Nmn − atomic mass of 23592U

∆m = 92(1.007 825 u) + 143(1.008 665 u) − 235.043 924 u = 1.915 071 u

Ebind = (1.915 071 u)931.50 M

u

eV = 1.7839 × 103 MeV = 1.7839 × 109 eV

Ebind = (1.7839 × 109 eV)1.60 × 10−19 e

J

V = 2.85 × 10−10 J

E = (2.0 × 1015 W •h)3.60 ×h

103 s = 7.2 × 1018 J

N = Eb

E

ind =

2

7

.8

.2

5

××

1

1

0

0

1

−

8

10

J

J = 2.5 × 1028 reactions

mtot = (2.5 × 1028)(235.043 924 u)1.66 × 10−27 k

u

g = 9.8 × 103 kg

4. E = 2.1 × 1019 J

atomic mass of 126C =

12.000 000 u


mn = 1.008 665 u

Z = 6

N = 12 − 6 = 6

∆m = Z(atomic mass of H) + Nmn − atomic mass of 126C

∆m = 6(1.007 825 u) + 6(1.008 665 u) − 12.000 000 u

∆m = 9.894 × 10−2 u

Ebind = (9.894 × 10−2 u)931.50 M

u

eV = 92.16 MeV

Ebind = (92.16 × 106 eV)1.60 × 10−19 e

J

V = 1.47 × 10−11 J

N = Eb

E

ind =

1

2

.4

.1

7

××

1

1

0

0

1

−

9

11

J

J = 1.4 × 1030 reactions

mtot = (1.4 × 1030)(12.000 000 u)1.66 × 10−27 k

u

g = 2.8 × 104 kg

∆m = Z(atomic mass of H) + Nmn − atomic mass of 147N

∆m = 7(1.007 825 u) + 7(1.008 665 u) − 14.003 074 u

∆m = 0.112 356 u

Ebind = (0.112 356 u)931.50 M

u

eV = 104.66 MeV

Ebind = (104.66 × 106 eV)1.60 × 10−19 e

J

V = 1.67 × 10−11 J

N = E

P

b

∆

in

t

d = = 2.2 × 1023 reactions

mtot = (2.2 × 1023)(14.003 074 u)1.66 × 10−27 k

u

g = 5.1 × 10−3 kg = 5.1 g

(42 × 106 W)(24 h)(3600 s/h)

1.67 × 10−11 J

5. Ptot = 3.9 × 1026 J/s

Z = 2

N = 4 − 2 = 2

atomic mass of 42He =

4.002 602 u


mn =1.008 665 u

∆m = Z(atomic mass of H) + Nmn − atomic mass of 42He

∆m = (2)(1.007 825 u) + (2)(1.008 665 u) − 4.002 602 u

∆m = 0.030 378 u

E = (0.030 378 u)(931.50 MeV/u)

E = 28.297 MeV

∆N

t =

P

Etot =

∆N

t = 8.6 × 1037 reactions/s

(3.9 × 1026 J/s)(1 × 10−6 MeV/eV)(1.60 × 10−19 J/eV)(28.297 MeV)


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7. P = 3.84 × 107 W

atomic mass of 126C =

12.000 000 u


mn = 1.008 665 u

Z = 6

N = 12 − 6 = 6

∆m = Z(atomic mass of H) + Nmn − atomic mass of 126C

∆m = 6(1.007 825 u) + 6(1.008 665 u) − 12.000 000 u

∆m = 9.894 × 10−2 u

Ebind = (9.894 × 10−2 u)931.50 × 106 e

u

V1.60 × 10−19

e

J

V

Ebind = 1.47 × 10−11 J

∆N

t =

Eb

P

ind =

1

3

.

.

4

8

7

4

××

1

1

0

0−

7

1

W1 J

= 2.61 × 1018 reactions/s

m

∆t

tot = (2.61 × 1018 s−1)(12.000 000 u)1.66 × 10−27

k

u

g = 5.20 × 10−8 kg/s

1. 23892U + 0

1n → X

X → 93993Np + −1

0e + v

23993Np → 239

94Pu + −10e + v

mass number of X = 238 + 1 = 239

atomic number of X = 92 + 0 = 92 (uranium)

X = 23992U

The equations are as follows:

23892U + 0

1n → 23992U

23992U → 939

93Np + −10e + v

23993Np → 239

94Pu + −10e + v

Additional Practice 25B, pp. 174–175

3. X → 13556Ba + −1

0e + v mass number of X = 135 + 0 = 135

atomic number of X = 56 + (−1) = 55 (cesium)

X = 13555 Cs

13555 Cs → 135

56Ba + −10e + v

2. X → Y + 42He

Y → Z + 42He

Z → 21283Bi + −1

0e + v

mass number of Z = 212 + 0 = 212

atomic number of Z = 83 − 1 = 82 (lead)

Z = 21282Pb

mass number of Y = 212 + 4 = 216

atomic number of Y = 82 + 2 = 84 (polonium)

Y = 21684Po

mass number of X = 216 + 4 = 220

atomic number of X = 84 + 2 = 86 (radon)

X = 22086Rn

The equations are as follows:

22086Rn → 216

84Po + 42He

21684Po → 212

82Pb + 42He

21282Pb → 212

83Bi + −10e + v


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Givens Solutions

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5. 22890Th → X + 4

2He + g mass number of X = 228 − 4 = 224

atomic number X = 90 − 2 = 88 (radium)

X = 22488Ra

22890Th → 224

88Ra + 42He + g

6. 11p + 7

3Li → X + 42He mass number of X = 1 + 7 − 4 = 4

atomic number of X = 1 + 3 − 2 = 2 (helium)

X = 42He

11p + 7

3Li → 42He + 4

2He

7. 21785At → X + 4

2He mass number of X = 217 − 4 = 213

atomic number of X = 85 − 2 = 83 (bismuth)

X = 21383Bi

21785At → 213

83Bi + 42He

1. T1/2 = 26 min, 43.53 sl =

0

T

.6

1

9

/2

3 =

l =

5 times the run time = 5 half-lives

percent of sample remaining = (0.5)5(100)

percent decayed = 100 − percent remaining = 100 − (0.5)5(100) = 96.875 percent

4.32 × 10−4 s−1

0.693(26 min)(60 s/min) + 43.53 s

Additional Practice 25C, pp. 176–177

4. 23592U + 1

0n → 14456 Ba +

8936Kr + X

mass number of X = 235 + 1 − 144 − 89 = 3

atomic number of X = 92 + 0 − 56 − 36 = 0 (neutron)

X = 3 10n

23592U + 1

0n → 14456Ba + 89

36Kr + 310n

23592U + 1

0n → 14054Xe + Y +

2 10n

mass number of Y = 235 + 1 − 140 − 2 = 94

atomic number of Y = 92 + 0 − 54 − 0 = 38 (strontium)

Y = 9438Sr

23592U + 1

0n → 14054Xe + 94

38Sr + 2 10n


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2. T1/2 = 1.91 years

decrease = 93.75 percent =0.9375

If 0.9375 of the sample has decayed, 1.0000 − 0.9375 = 0.0625 of the sample remains.

0.0625 = (0.5)4, so 4 half-lives have passed.

∆t = 4T1/2 = (4)(1.91 years) = 7.64 years

3. T1/2 = 11.9 s

Ni = 1.00 × 1013 atoms

Nf = 1.25 × 1012 atoms

∆N

N

i = = 0.875

If 0.875 of the sample has decayed, 1.000 − 0.875 = 0.125 of the sample remains.


∆t = 3T1/2 = (3)(11.9 s) = 35.7 s

1.00 × 1013 − 1.25 × 1012

1.00 × 1013

4. ∆t = 4800 years

T1/2 = 1600 yearsT

∆

1

t

/2 =

4

1

8

6

0

0

0

0

y

y

e

e

a

a

r

r

s

s = 3 half-lives

amount remaining after 3T1/2 = (0.5)3 = 0.125 = 12.5 percent

7. T = 4.4 × 10−22 sT

1 = l =

0

T

.6

1

9

/2

3

T = 0

T

.61

9/2

3

T1/2 = (0.693)(T ) = (0.693)(4.4 × 10−22 s) = 3.0 × 10−22 s

5. ∆t = 88 years

amount of sample

remaining = 1

1

6 = 0.0625


T1/2 = 14

∆t = 88 y

4

ears = 22 years

l = 0

T

.6

1

9

/2

3 =

22

0.

y

6

e

9

a

3

rs

l = 3.2 × 10−2 years−1

6. ∆t = 34 days, 6 h, 26 min

amount of sample

remaining = 5

1

12 = 1.95 × 10−3

∆t = (34 days)2

d

4

ay

h60

h

min + (6 h)60

h

min + 26 min

∆t = 4.9346 × 104 min

1.95 × 10−3 = (0.5)9, so 9 half-lives have passed.

T1/2 = 19

∆t = 4.9346 ×

9

104 min = 5482.9 min

l = 0

T

.6

1

9

/2

3 =

548

0

2

.6

.9

93

min

l = 1.26 × 10−4 min−1 = 0.182 days−1

PROBLEM WORKBOOK - Langlo...

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