Regent Physics by Abhishek Jaguessar

8/4/2019 Regent Physics by Abhishek Jaguessar

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Regent Physics

ABHISHEK JAGUESSAR


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Physics Units

I. Physics Skills II. Mechanics

III. Energy

IV. Electricity and MagnetismV. Waves

VI. Modern Physics


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I. Physics Skills

A. Scientific Notation B. Graphing

C. Significant Figures

D. Units E. Prefixes

F. Estimation


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A. Scientific Notation

Use for very large or very smallnumbers

Write number with one digit tothe left of the decimal followedby an exponent (1.5 x 105)

Examples: 2.1 x 103 represents2100 and 3.6 x 10-4 represents

0.00036


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Scientific NotationProblems

1. Write 365,000,000 inscientific notation

2. Write 0.000087 in scientificnotation


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Answers

1.) 3.65 x 108

2.) 8.7 x 10-5


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B. Graphing

Use graphs to make a pictureof scientific data

independent variable, theone you change in your

experiment is graphed on thex axis and listed first in atable

dependent variable, the one

changed by your experiment isgraphed on the y axis andlisted second in a table


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Best fit line or curve isdrawn once points are plotted.Does not have to go throughall points. Just gives you the

trend of the points

The slope of the line is givenas the change in the y value

divided by the change in thex value


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Types of Graphs

1. Direct Relationship meansan increase/decrease in onevariable causes anincrease/decrease in the

other

Example below


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2. Inverse(indirect)relationship means that anincrease in one variablecauses a decrease in theother variable and vice versa

Examples


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3. Constant proportionmeans that a change in onevariable doesnt affect theother variable

Example;


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4. If either variable issquared(whether therelationship is direct orindirect), the graph will curve

more steeply.


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C. Significant figures

Uncertainty in measurements isexpressed by using significantfigures

The more accurate or precise a

measurement is, the moredigits will be significant


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Significant Figure Rules

1. Zeros that appear before anonzero digit are not significant(examples: 0.002 has 1significant figure and 0.13 has

2 significant figures) 2. Zeros that appear between

nonzero digits are significant

(examples: 1002 has 4significant figures and 0.405has 3 significant figures)


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Significant figuresrules(cont.)

3. zeros that appear after anonzero digit are significantonly if they are followed by adecimal point (20. has 2 sig

figs) or if they appear to theright of the decimal point (35.0has 3 sig figs)


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Sig Fig problems

1. How many significant figuresdoes 0.050900 contain?

2. How many significant figuresdoes 4800 contain?


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Answers

1. 5 sig figs

2. 2 sig figs


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D. Units

1. Fundamental units are unitsthat cant be broken down

2. Derived units are made upof other units and then

renamed

3. SI units are standardizedunits used by scientists

worldwide


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Fundamental Units

Meter (m) length, distance,displacement, height, radius,elongation or compression of aspring, amplitude, wavelength

Kilogram (kg) mass Second (s) time, period

Ampere (A) electric current

Degree (

o

) angle


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Derived Units

Meter per second (m/s)speed, velocity

Meter per second squared(m/s2) acceleration

Newton (N) force Kilogram times meter per

second (kg.m/s) momentum

Newton times second (N.s)--impulse


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Derived Units (cont.)

Joule (J) work, all types ofenergy

Watt (W) power

Coulomb (C) electric charge

Newton per Coulomb (N/C)electric field strength(intensity)

Volt (V)- potential difference(voltage)

Electronvolt (eV) energy(small amounts)


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Derived Units (cont.)

Ohm () resistance Ohm times meter (.m)

resistivity

Weber (Wb) number ofmagnetic field (flux) lines

Tesla (T) magnetic field (flux)density

Hertz (Hz)-- frequency


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E. Prefixes

Adding prefixes to base unitsmakes them smaller or largerby powers of ten

The prefixes used in Regents

Physics are tera, giga, mega,kilo, deci, centi, milli, micro,nano and pico


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Prefix Examples

A terameter is 10

12

meters,so 4 Tm would be 4 000 000000 000 meters

A gigagram is 109 grams, so

9 Gg would be 9 000 000 000grams

A megawatt is 106 watts, so100 MW would be 100 000 000

wattsA kilometer is 103 meters, so

45 km would be 45 000 meters


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Prefix examples (cont.)

A decigram is 10-1

gram, so15 dg would be 1.5 grams

A centiwatt is 10-2 watt, so 2dW would be 0.02 Watt

A millisecond is 10-3 second,so 42 ms would be 0.042second


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Prefix examples (cont.)

A microvolt is 10-6

volt, so 8V would be 0.000 008 volt

A nanojoule is 10-9joule, so530 nJ would be 0.000 000 530

joule

A picometer is 10-12 meter, so677 pm would be 0.000 000

000 677 meter


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Prefix Problems

1.) 16 terameters would be howmany meters?

2.) 2500 milligrams would be how

many grams?

3.) 1596 volts would be how manygigavolts?

4.) 687 amperes would be howmany nanoamperes?


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Answers

1.) 16 000 000 000 000 meters

2.) 2.500 grams

3.) 1596 000 000 000 gigavolts

4.) 0.000 000 687 amperes


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F. Estimation

You can estimate an answer toa problem by rounding theknown information

You also should have an ideaof how large common units are


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Estimation (cont.)

2 cans of Progresso soup arejust about the mass of1 kilogram

1 medium apple weighs1 newton

The length of an averagePhysics students leg is 1 meter


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Estimation Problems

1.) Which object weighsapproximately one newton?Dime, paper clip, student, golfball

2.) How high is an averagedoorknob from the floor?

101

m, 100

m, 101

m, 10-2

m


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Answers

1.) golf ball

2.) 100m


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II. Mechanics

A. Kinematics; vectors,velocity, acceleration

B. Kinematics; freefall

C. Statics

D. Dynamics E. 2-dimensional motion

F. Uniform Circular motion

G. Mass, Weight, Gravity H. Friction

I. Momentum and Impulse


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A. Kinematics; vectors,velocity, acceleration

In physics, quantities can bevector or scalar

VECTOR quantities have a

magnitude (a number), a unitand a direction

Example; 22m(south)


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SCALAR quantities only have amagnitude and a unit

Example; 22m


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VECTOR quantities;displacement, velocity,acceleration, force, weight,momentum, impulse, electric

field strength

SCALAR quantities; distance,

mass, time, speed,work(energy), power


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Distance vs.Displacement

Distance is the entire pathwayan object travels

Displacement is the shortestpathway from the beginning tothe end


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Distance/DisplacementProblems

1.) A student walks 12m duenorth and then 5m due east.What is the students resultantdisplacement? Distance?

2.) A student walks 50m duenorth and then walks 30m due

south. What is the studentsresultant displacement?Distance?


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Answers

1.) 13m (NE) for displacement

17 m for distance

2.) 20m (N) for displacement80 m for distance


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Speed vs. Velocity

Speed is the distance an objectmoves in a unit of time

Velocity is the displacement ofan object in a unit of time


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Average Speed/VelocityEquations

t

dv !

2

if vvv


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Symbols

timet

distanceorntdisplacemed

speedorvelocityinitialv

speedorvelocityfinalv

speedorvelocityaveragev

i

f

!

!

!

!

!


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Speed/VelocityProblems

1.) A boy is coasting down ahill on a skateboard. At 1.0s heis traveling at 4.0m/s and at4.0s he is traveling at 10.0m/s.

What distance did he travelduring that time period? (In allproblems given in RegentsPhysics, assume acceleration is

constant)


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Answers

1.) You must first find the boysaverage speed before you areable to find the distance


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Answers (cont.)

21mwasdtravellehedistanceeso......th3.0st

vuse

/0.7vso..../0.4

/0.10v

2use

f

!

!

!!

!

!

t

dthen

smsmv

sm

vvv

i

if


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Acceleration

The time rate change ofvelocity is acceleration (howmuch you speed up or slowdown in a unit of time)

We will only be dealing withconstant (uniform) acceleration


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Constant AccelerationEquations

adv

att

atvv

t

va

i

if

2v

2

1vd

22

f

2

i

!

!

!

(!


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Constant AccelerationProblems

1.) A car initially travels at20m/s on a straight, horizontalroad. The driver applies thebrakes, causing the car to slow

down at a constant rate of2m/s2 until it comes to a stop.What was the cars stoppingdistance? (Use two different

methods to solve the problem)


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Answers

First Method

vi=20m/s

vf=0m/s

a=2m/s2

Use vf2=vi

2+2ad to find d

d=100m


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Answer (cont.)

Second Method

100md

d""indto2

1vitdthen...use

t""indtoaUse

2

!

!

(!

at

t

v


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B. Kinematics: freefall

In a vacuum (empty space),objects fall freely at the samerate

The rate at which objects fall is

known as g, the accelerationdue to gravity

On earth, the g is 9.81m/s2


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Solving FreefallProblems

To solve freefall problems usethe constant accelerationequations

Assume a freely falling object

has a vi=0m/sAssume a freely falling object

has an a=9.81m/s2


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Freefall Problems

1.) How far will an object near

Earths surface fall in 5s?

2.) How long does it take for a

rock to fall 60m? How fast willit be going when it hits theground

3.) In a vacuum, which will hitthe ground first if droppedfrom 10m, a ball or a feather?


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Answers

1.)

mdso

st

sma

sm

attvi

6.122.....

5

/81.9

/0v

..ith....2

1

dUse

2

i

2

!

!

!

!

!


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Answers (cont.)

2.)

smvso

atv

stso

mdsma

sm

att

f

i

/3.34....

v

use......enth5.3....

60/81.9

/0v

.with....2

1vdse

f

2

i

2

i

!

!

!

!!

!

!


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Answers (cont.)

3.) Both hit at the same time

because g the accelerationdue to gravity is constant. Itdoesnt depend on mass ofobject because it is a ratio

m

F

gg

!


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Solving Anti??? FreefallProblems

If you toss an object straightup, that is the opposite offreefall.

So

vf is now 0m/s a is -9.8lm/s2

Because the object is slowing

down not speeding up


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Antifreefall Problems

1.) How fast do you have to

toss a ball straight up if youwant it reach a height of 20m?

2.) How long will the ball inproblem #1 take to reach the20m height?


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Answers

1.)

sm

md

sma

sm

adv

f

i

/8.19v

so........20

/81.9

/0v

...when....2vse

i

2

22f

!

!

!

!

!


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Answers (cont.)

2.)

stso

md

sma

smv

smv

attatv

f

i

i

0.2.......

20

/81.9

/0

/8.19

2

1vdorvUse

2

2

i

!

!

!

!

!

!!


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C. Statics

The study of the effect of

forces on objects at rest

Force is a push or pull

The unit of force is thenewton(N) (a derived vector

quantity)


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Adding forces

When adding concurrent

(acting on the same object atthe same time) forces followthree rules to find the resultant(the combined effect of the

forces)

1.) forces at 00, add them

2.) forces at 1800

, subtract them3.) forces at 900, use

Pythagorean Theorem


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Force Diagrams

Forces at 00

Forces at 1800

Forces at 900

Composition of Forces


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Composition of ForcesProblems

1.) Find the resultant of two5.0N forces at 00? 1800? And

900?


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Answers

1.) 00

5N 5N = 10N


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Answers (cont.)

1.) 1800

5N 5N = 0N


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Answers (cont.)

1.) 900

5N =

5N 7.1N


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Resolution of forces

The opposite of adding

concurrent forces.

Breaking a resultant force into

its component forces

Only need to know

components(2 forces) at a 900

angle to each other

Resolving forces using


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Resolving forces usingGraphical Method

To find the component forces

of the resultant force 1.) Draw x and y axes at the tail

of the resultant force

2.) Draw lines from the head ofthe force to each of the axes

3.) From the tail of the resultantforce to where the lines intersect

the axes, are the lengths of thecomponent forces


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Resolution Diagram

Black arrow=resultant force

Orange lines=reference linesGreen arrows=component forces

y

x

Resolving Forces Using


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Resolving Forces UsingAlgebraic Method

AandAbet eenangle

orceresultanttheA

componentA

componentA

componentverticalindtosinA

componenthorizontalindtocosA

use...lly,algebraicacomponentsindTo

x

y

x

y

x

!

!

!

!

!

!

U

U

U

vertical

horizontal

A

A


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Equilibrium

Equilibrium occurs when the

net force acting on an object iszero

Zero net force means thatwhen you take into account allthe forces acting on an object,they cancel each other out


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Equilibrium (cont.)

An object in equilibrium can

either be at rest or can bemoving with constant(unchanging) velocity

An equilibrant is a force equaland opposite to the resultantforce that keeps an object inequilibrium


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Equilibrium Diagram

Black arrows=components

Blue arrow=resultant

Red arrow=equilibrant

=

+


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Problems

1.) 10N, 8N and 6N forces act

concurrently on an object thatis in equilibrium. What is theequilibrant of the 10N and 6Nforces? Explain.

2.) A person pushes alawnmower with a force of300N at an angle of 600 to theground. What are the vertical

and horizontal components ofthe 300N force?


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Answers

1.) The 8N force is the

equilibrant (which is also equalto and opposite the resultant)The 3 forces keep the object in

equilibrium, so the third forceis always the equilibrant of theother two forces.


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Answers (cont.)

2.)

NNAand

NNAso

A

A

y

x

y

26060sin300....

15060cos300......

60

300NA

componentverticalindtosinAand

componenthorizontalindtocosAUse

0

0

0

x

!!

!!

!

!

!

!

U

U

U


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C. Dynamics

The study of how forces affect

the motion of an object

Use Newtons Three Laws ofMotion to describe Dynamics

Newtons 1stLaw of


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Motion

Also called the law of inertia

Inertia is the property of anobject to resist change. Inertiais directly proportional to the

objects massAn object will remain in

equilibrium (at rest or movingwith constant speed) unless

acted upon by an unbalancedforce

Newtons Second Law of


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Motion

When an unbalanced (net) force

acts on an object, that objectaccelerates in the direction of theforce

How much an object acceleratesdepends on the force exerted on itand the objects mass (Seeequation)

m

Fneta !


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Symbols

Fnet=the net force exerted on

an object (the resultant of allforces on an object) in newtons(N)

m=mass in kilograms (kg)

a=acceleration in m/s2

Newtons Third Law of


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Motion

Also called law of action-reaction

When an object exerts a force onanother object, the second objectexerts a force equaland opposite

to the first force

Masses of each of the objects dontaffect the size of the forces (willaffect the results of the forces)


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Free-Body Diagrams

A drawing (can be to scale)

that shows all concurrentforces acting on an object

Typical forces are the force of

gravity, the normal force, theforce of friction, the force ofacceleration, the force oftension, etc.


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Free-body Forces

Fg is the force of gravity or

weight of an object (alwaysstraight down)

FN is the normal force (the

force of a surface pushing upagainst an object)

Ff is the force of friction whichis always opposite the motion

Fa is the force of accelerationcaused by a push or pull


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Free-Body Diagrams

If object is moving with

constant speed to the right. Black arrow=Ff Green arrow=FgYellow arrow=FN Blue arrow=Fa

Free-Body Diagrams on

Sl


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a Slope

When an object is at rest or

moving with constant speed ona slope, some things about theforces change and some dont

1.) Fg is still straight down 2.) Ff is still opposite motion

3.) FN is no longer equal andopposite to F

g 4.) Ff is still opposite motion

More

Free-Body Diagrams on

Sl


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a Slope

Fa=Ff=Ax=Acos

FN=Ay=AsinFg=mg (still straight down)

On a horizontal surface, force ofgravity and normal force areequal and opposite

On a slope, the normal force is

equal and opposite to the ycomponent of the force ofgravity

Free-Body Diagram on a

Sl


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Slope

Green arrow=FN

Red arrow=FgBlack arrows=Fa and FfOrange dashes=AyPurple dashes=A

x

D i P bl


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Dynamics Problems

1.) Which has more inertia a

0.75kg pile of feathers or a0.50kg pile of lead marbles?

2.) An unbalanced force of10.0N acts on a 20.0kg massfor 5.0s. What is theacceleration of the mass?

A


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Answers

1.) The 0.75kg pile of feathers

has more inertia because it hasmore mass. Inertia isdependent on the mass ofthe object

Answers (cont )


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Answers (cont.)

2.)

2

net

net

0.50m/sso......a

this)needt(don'0.5

0.20

0.10F

h.......witm

Fase

!

!

!

!

!

st

kgm

N

More Problems


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More Problems

3.) A 10N book rests on a

horizontal tabletop. What is theforce of the tabletop on thebook?

4.) How much force would it taketo accelerate a 2.0kg object5m/s2? How much would that

same force accelerate a 1.0kgobject?

Answers


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Answers

1.) The force of the tabletop onthe book is also 10N

(action/reaction)

Answers (cont )


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Answers (cont.)

2.)

2

net

net

2

net

10m/sso.....a

part)firstinas(sameN10

1.0kgm

:partSecond

10so....2.0kgm5m/sa

:partirst

questiontheofpartsbothforaUse

!

!

!

!!!

!

N

m

E 2 Dimensional Motion


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E. 2-Dimensional Motion

To describe an object moving

2-dimensionally, the motionmust be separated into a

horizontal component and avertical component (neitherhas an effect on the other)

Assume the motion occurs in aperfect physics world; a

vacuum with no friction

Types of 2 D Motion


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Types of 2-D Motion

1.) Projectiles fired horizontally

an example would be abaseball tossed straighthorizontally away from you

Projectile Fired

Horizontally


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Horizontally

Use the table below to

solve these type of 2-Dproblems

Quantities Horizontal Vertical

vi Same as vf 0m/s

vf Same as vi

a 0m/s2 9.81m/s2

d

t Same asvertical time

Same ashorizontal

time

Types of 2-D Motion


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Types of 2-D Motion

2.) Projectiles fired at an angle

an example would be asoccer ball lofted into the airand then falling back onto theground

Projectile Fired at an

Angle


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Angle

Use the table below to solve thesetype of 2-D problems

Ax=Acos and Ay=Asin


vi Ax Ay

vf Ax 0m/s

a 0m/s2 - 9.81m/s2

d

t twicevertical time

2-D Motion Problems


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2 D Motion Problems

1.) A girl tossed a ball

horizontally with a speed of10.0m/s from a bridge 7.0mabove a river. How long did theball take to hit the river? How

far from the bottom of thebridge did the ball hit the river?

Answers


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Answers

1.) In this problem you areasked to find time and

horizontal distance (see tableon the next page)

Answers (cont )


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Answers (cont.)


vi 10.0m/s 0.0m/s

vf 10.0m/s Dont need

a 0.0m/s 9.81m/s2

d ? 7.0m

t ? ?

Answers (cont.)


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Answers (cont.)

Use

14.3mhorizontalso......d

vertical)assame(43.1

/0.10v

d""horizontalfindn toinformatiohorizontalithvUse

43.1......

/81.9

/0.0v

7.0md

t""findn toinformatioticalith ver2

1vdUse

2

i

2

i

!

!

!

!

!

!

!

!

!

st

smt

d

stso

sma

sm

att

More 2-D Motion

Problems


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Problems

2.) A soccer ball is kicked at an

angle of 600 from the groundwith an angular velocity of10.0m/s. How high does thesoccer ball go? How far away

from where it was kicked doesit land? How long does its flighttake?

Answers


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Answers

2.) In this problem you are

asked to find vertical distance,horizontal distance andhorizontal time. Finding verticaltime is usually the best way to

start. (See table on next page)

Answers (cont.)


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s s ( )


vi Ax=5.0m/s Ay=8.7m/s

vf Ax=5.0m/s 0.0m/s

a 0.0m/s - 9.81m/s2

d ? ?

t ? Need to find

Answers (cont.)


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( )

Find vertical t first using

vf=vi+at with.

vf=0.0m/s

vi=8.7m/sa=-9.81m/s2

Sovertical t=0.89s and

horizontal t is twice thatand equals 1.77s

Answers (cont.)


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( )

Find horizontal

d using

8.87mhorizontaldso........77.1

/0.5v

...hen.v

!!

!

!

st

sm

t

d

Answers (cont.)


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( )

Find vertical d

by using

mdso

sma

st

smv

att

i

85.3.......

/81.9

89.0

/7.8

...hen...

2

1vd

2

2

i

!

!

!

!

!

F. Uniform Circular

Motion


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When an object moves with

constant speed in a circularpath

The force (centripetal) will beconstant towards the center

Acceleration (centripetal) willonly be a direction changetowards the center

Velocity will be tangent to the

circle in the direction ofmovement

Uniform Circular Motion

Symbols


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y

Fc=centripetal force, (N)

v=constant velocity (m/s)

ac=centripetal acceleration (m/s2)

r=radius of the circular pathway (m)

m=mass of the object in motion (kg)


Diagram


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Equations


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r

mv

r

v

2

c

2

ca

!

!


Problems


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1.) A 5kg cart travels in a circle

of radius 2m with a constantvelocity of 4m/s. What is thecentripetal force exerted on thecart that keeps it on its circular

pathway?

Answers


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1.)

N40so.....

forcelcentripetafindtouseThen

8m/sso.....a

onacceleratilcentripetafindtoauseirst

c

c

2

c

2

c

!

!

!

!

cma

r

v

G. Mass, Weight and

Gravity


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Mass is the amount of matter

in an object

Weight is the force of gravity

pulling down on an object

Gravity is a force of attractionbetween objects

Mass


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Mass is measured in kilograms

(kg)

Mass doesnt change with

location (for example, if youtravel to the moon your massdoesnt change)

Weight


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Weight is measured in newtons

(N)

Weight does change with

location because it isdependent on the pull ofgravity

Weight is equal to mass times

the acceleration due to gravity

Weight/Force of Gravity

Equations


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earth)onm/s(9.81gravitytodueonacceleratithetimesmassitsto

equalisobjectanof)(or eightgravityofforcethemeanshich

mg

objects2bet eengravityof

forcethetorelated(squared)indirectlyisanceand...dist

objects2bet eengravityofforce

thetorelateddirectlyismassthatmeanshich

2

g

221g

!

!r

mmG

Gravitational Field

Strength


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g=acceleration due to gravity butit is also equal to gravitational

field strength The units of acceleration due to

gravity are m/s2

The units of gravitational field

strength are N/kg Both quantities are found from

the equation:

m

Fg

g!

Mass, Weight, Gravity

Problems


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1.) If the distance between two

masses is doubled, whathappens to the gravitationalforce between them?

2.) If the distance between twoobjects is halved and the massof one of the objects isdoubled, what happens to the

gravitational force betweenthem?

Answersmm


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1.) Distance has an inverse squaredrelationship with the force ofgravity.

Sosince r is multiplied by 2 in theproblem, square 2so.22=4,then take the inverse of thatsquare which equals .so.the

answer is the original Fg

2

21

r

mmGFg !

Answers (cont.)


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2.) Mass has a direct

relationship with Fg anddistance has an inversesquared relationship with Fg.

Firstsince m is doubled so is

Fg and since r is halved, square , which is and then takethe inverse which is 4.

Thencombine 2x4=8 Soanswer is 8 times Fg

More Problems


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3.) Determine the force ofgravity between a 2kg and a3kg object that are 5m apart.

4.) An object with a mass of

10kg has a weight of 4N onPlanet X. What is theacceleration due to gravity onPlanet X? What is the

gravitational field strength onPlanet X?

Answers


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3.)

4.)

strengthfieldnalgravitatiofor0.4N/kgg

gravitytodueonacceleratifor0.4m/sg

m

gUsing

106.1

/6.67x10G

.ith...Using

2

g

11

2211-

2

1

g

2

!

!

!

!

y!

!

NxF

kgmNr

mmG

g

H. Friction


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The force that opposes motion

measured in newtons (N)Always opposite direction of

motion

Static Friction is the forcethat opposes the start ofmotion

Kinetic Friction is the force of

friction between objects incontact that are in motion

Coefficient of Friction


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The ratio of the force of friction

to the normal force (no unit,since newtons cancel out)

Equation

Ff

= FN

=coefficient of friction

Ff

=force of friction

FN=normal force



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The smaller the coefficient, the

easier the surfaces slide overone another

The larger the coefficient, theharder it is to slide the surfacesover one another

Use the table in the reference

tables


Problems


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1.) A horizontal force is used to

pull a 2.0kg cart at constantspeed of 5.0m/s across atabletop. The force of frictionbetween the cart and the

tabletop is 10N. What is thecoefficient of friction betweenthe cart and the tabletop? Isthe friction force kinetic orstatic? Why?

Answers


131/343

1.)

The friction force is kineticbecause the cart is moving

over the tabletop

51.0

FFsingso.......u10

6.19/81.90.2

f

2

!

!!

!y!!!

Q

QNF

NsmkgmgFF

f

g

I. Momentum and

Impulse

M i i


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Momentum is a vector quantity

that is the product of mass andvelocity (unit is kg.m/s)

Impulse is the product of theforce applied to an object andtime (unit is N.s)

Momentum and Impulse

Symbols

t


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p=momentum

p=change in momentum=(usually) m(vf-vi)

J=impulse


Equations


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p=mv

J=Ft

J=p

pbefore=pafter


Problems

1 ) A 5 0kg cart at rest experiences


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1.) A 5.0kg cart at rest experiencesa 10N.s (E) impulse. What is thecarts velocity after the impulse?

2.) A 1.0kg cart at rest is hit by a

0.2kg cart moving to the right at10.0m/s. The collision causes the1.0kg cart to move to the right at3.0m/s. What is the velocity of the0.2kg cart after the collision?

Answer

1 ) Use J=p so


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1.) Use J=p so

J=10N.s(E)=p=10kg.m/s(E)and since original p was 0kg.m/s

and p=10kg.m/s(E),

new p=10kg.m/s(E)then use.. p=mv so..

10kg.m/s(E)=5.0kg x v so.

v=2m/s(E)

Answers (cont.)

2 ) Use p =p


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2.) Use pbefore=pafter

Pbefore=0kg.m/s + 2kg.m/s(right)=2kg.m/s(right)

Pafter=2kg.m/s(right)=3kg.m/s +P(0.2kg cart) so.p of 0.2kg cart

must be -1kg.m/s or1kg.m/s(left)

more..

Answers (cont.)

So if p after collision of 0 2kg


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So if p after collision of 0.2kg

cart is 1kg.m/s(left) andp=mv

1kg.m/s(left)=0.2kg x v

And v=5m/s(left)

III. Energy

A Work and Power


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A. Work and Power

B. Potential and Kinetic Energy C. Conservation of Energy

D. Energy of a Spring

A. Work and Power

Work is using energy to move


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Work is using energy to move

an object a distance Work is scalar

The unit of work is the Joule(J)

Work and energy aremanifestations of the samething, that is why they have

the same unit of Joules

Work and Power (cont.)

Power is the rate at which work


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Power is the rate at which work

is done so there is a timefactor in power but not in work

Power and time are inverselyproportional; the less time it

takes to do work the morepower is developed

The unit of power is the watt

(W) Power is scalar

Work and Power

Symbols W=work in Joules (J)


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F=force in newtons

(N) d=distance in meters

(m)

ET=change in total

energy in Joules (J) P=power in watts (W)

t=time in seconds (s)

(m/s)inspeedaverage!v

Work and Power

EquationsW=Fd=ET


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***When work is done vertically,F can be the weight of theobject Fg=mg

vFt

Fd

t

WP !!!

Work and Power

Problems

1.) A 2.5kg object is moved 2.0m in


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g j2.0s after receiving a horizontal

push of 10.0N. How much work isdone on the object? How muchpower is developed? How muchwould the objects total energy

change? 2.) A horizontal 40.0N force causes

a box to move at a constant rate of5.0m/s. How much power is

developed? How much work is doneagainst friction in 4.0s?

Answers

1.) to find work use W=Fd


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)

SoW=10.0N x 2.0m=20.0J

To find power use P=W/t

SoP=20.0J/2.0s=10.0W

To find total energy change itsthe same as work done so

ET=W=20.0J

Answers (cont.)

2.) To find power use vF!


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So P=40.0N x 5.0m/s=200W

To find work use P=W/t

so200W=W/4.0s So.W=800J

More problems

3.) A 2.0kg object is raised


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vertically 0.25m. What is thework done raising it?

4.) A lift hoists a 5000N object

vertically, 5.0 meters in the air.How much work was donelifting it?

Answers

3.) to find work use W=Fd with


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F equal to the weight of theobject

So..W=mg x d

So...W=2.0kgx9.81m/s2x0.25m

SoW=4.9J

Answers (cont.)

4.) to find work use W=Fd


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Even though it is verticalmotion, you dont have tomultiply by g because weightis already given in newtons

SoW=Fd=5000N x 5.0m

And W=25000J

B. Potential and Kinetic

Energy

Gravitational Potential Energyis energy of position above the


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is energy of position above the

earth Elastic Potential Energy is

energy due to compression orelongation of a spring

Kinetic Energy is energy due tomotion

The unit for all types of energy

is the same as for work theJoule (J). All energy is scalar

Gravitational PotentialEnergy Symbols and

Equation

PE=change in gravitationalpotential energy in Joules (J)


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potential energy in Joules (J)

m=mass in kilograms (kg) g=acceleration due to gravity in

(m/s2)

h=change in height in meters (m)

Equation PE=mgh

***Gravitational PE only

changes if there is a change invertical position

Gravitational PE

Problems

1.) How much potential energyi i d b 5 2k bj t


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is gained by a 5.2kg objectlifted from the floor to the topof a 0.80m high table?

2.) How much work is done inthe example above?

Answers

1.) Use PE=mgh to find

t ti l i d


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potential energy gained soPE=5.2kgx9.81m/s2x0.80m

SoPE=40.81J

2.) W=ET so..W is also 40.81J

Kinetic Energy Symbols

and Equation KE=kinetic energy in Joules (J)


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m=mass in kilograms (kg)

v=velocity or speed in (m/s)

2

21KE mv!

Kinetic Energy Problems

1.) If the speed of a car isdoubled what happens to its


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doubled, what happens to itskinetic energy?

2.) A 6.0kg cart possesses 75J

of kinetic energy. How fast is itgoing?

Answers

1.) Using KE=1/2mv2 if v isdoubled because v if squared


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doubled, because v if squaredKE will be quadrupled.

2.) Use KE=1/2mv2 so..

75J=1/2 x 6.0kg x v

And..v=5m/s

C. Conservation of

Energy

In a closed system the totalamount of energy is conserved


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amount of energy is conserved

Total energy includes potentialenergy, kinetic energy andinternal energy

Energy within a system can betransferred among differenttypes of energy but it cant be

destroyed

Conservation of Energy in

a Perfect Physics World

In a perfect physics world sincethere is no friction there will be


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there is no friction there will beno change in internal energy soyou dont have to take that intoaccount

In a perfect physics worldenergy will transfer betweenPE and KE

In the Real World

In the real world there isfriction so the internal energy


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friction so the internal energyof an object will be affected bythe friction (such as airresistance)

Conservation of Energy

Symbols

ET=total energy of a system


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PE=potential energy KE=kinetic energy

Q=internal energy

***all units are Joules (J)


Equations


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In a real world situation,ET=KE+PE+Q because frictionexists and may cause anincrease in the internal energy

of an object In a perfect physics world

ET=KE+PE with KE+PE equal

to the total mechanical energyof the system object

Conservation of EnergyExamples (perfect physics

world)

position #1


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position #1

position #2

position #3 more..


Examples (cont.)


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Examples (cont.)

Position #1


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Position #2

Position #3


(perfect physics world) Position #1 the ball/bob has

not starting falling yet so the


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total energy is all ingravitational potential energy

Position #2 the ball/bob ishalfway down, so total energyis split evenly between PE andKE

Position #3 the ball/bob is atthe end of its fall so totalenergy is all in KE


Problems 1.) A 2.0kg block starts at rest and

then slides along a frictionlesstrack. What is the speed of the


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pblock at point B?

A

7.0m

B

Answer Since there is no friction, Q

does not need to be included


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Souse ET=PE+KE

At position B, the total energyis entirely KE

Since you cannot find KEdirectly, instead find PE at thebeginning of the slide and that

will be equal to KE at the endof the slide more..

Answer (cont.) PE (at position A)

=mgh=2.0kgx9.81m/s2x7.0m


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=137.3J

KE (at position A) =0J becausethere is no speed

So ET (at position A)=137.3JAt position B there is no height

so the PE is 0J

More.

Answer (cont.)At position B the total energy

still has to be 137.3J because


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energy is conserved andbecause there is no friction noenergy was lost along the

slide So.ET(position B)=137.3J=0J+KE

SoKE also equals 137.3J atposition B

More

Answer (cont.) Use KE=1/2mv2

SoKE=137.3J=1/2x2.0kgxv2


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So v (at position B)=11.7m/s

More Conservation of

Energy Problems

position #1

1.) From whatheight must you

d h 0 5k b ll


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position #2

position #3

drop the 0.5kg ballso that the it will betraveling at 25m/s atposition #3(bottomof the fall)?

2.)How fast will itbe traveling atposition #2 (halfwaydown)?

*Assume no friction

Answers 1.) At position #3, total energy

will be all in KE because there

i h i ht d f i ti


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is no height and no friction

Souse ET=KE=1/2mv2

KE=1/2 x 0.5kg x (25m/s)2

SoKE=156.25J=PE (atposition#1)

SoPE=156.25J=mgh

And h=31.86m

Answers (cont.) 2.) Since position #2 is half

way down total energy will be

h lf i PE d h lf i KE


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half in PE and half in KE

SoKE at position #2 will behalf that at position #3

SoKE at position #2 is78.125J

Then use KE (at #2)=78.125J

=1/2 x 0.5kg x v2

v=17.68m/s at position #2

D. Energy of a Spring Energy stored in a spring is

called elastic potential energy

E i d i i


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Energy is stored in a springwhen the spring is stretched orcompressed

The work done to compress orstretch a spring becomes itselastic potential energy

Spring Symbols Fs=force applied to stretch or

compress the spring in

newtons (N)


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newtons (N)

k=spring constant in (N/m)***specific for each type of

spring x=the change in length in the

spring from the equilibriumposition in meters (m)

Spring Equations Fs=kx


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PEs=1/2kx2

Spring Diagrams


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Spring Problems 1.) What is the potential

energy stored in a spring that

stretches 0 25m from


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stretches 0.25m fromequilibrium when a 2kg mass ishung from it?

2.) 100J of energy are storedwhen a spring is compressed0.1m from equilibrium. Whatforce was needed to compress

the spring?

Answers 1.) Using PEs=1/2kx

2 youknow x but not k

You can find k using F =kx


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You can find k using Fs=kx

With Fs equal to the weight ofthe hanging mass

So Fs=Fg=mg=2kgx9.81m/s2

Fs=19.62N=kx=k x 0.25m

k=78.48N/m

More

Answers (cont.) Now use PEs=1/2kx

2

PEs

=1/2 x 78.48N/m x (0.25m)2

S PE 2 45J


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sSo PEs=2.45J

Answers (cont.) 2.) To find the force will use

Fs=kx, but since you only know

x you must find k also


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x you must find k also

Use PEs=1/2kx2 to find k

PEs=100J=1/2k(0.1m)2

k=20 000N/m

use Fs=kx=20 000N/m x 0.1m

Fs=2000N

Examples of Forms of

Energy 1.) Thermal Energy is heat energy

which is the KE possessed by the

particles making up an object


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particles making up an object 2.) Internal Energy is the total PE

and KE of the particles making upan object

3.) Nuclear Energy is the energyreleased by nuclear fission or fusion

4.) Electromagnetic Energy is theenergy associated with electric and

magnetic fields

IV. Electricity and

MagnetismA. Electrostatics/Electric Fields

B. Current Electricity

C Series Circuits


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C. Series Circuits

D. Parallel Circuits

E. Electric Power and Energy

F. Magnetism andElectromagnetism

A. ElectrostaticsAtomic Structurethe atom

consists of proton(s) and

neutron(s) in the nucleus and


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neutron(s) in the nucleus andelectrons outside the nucleus.

The proton and neutron have

similar mass (listed inreference table)

The electron has very littlemass (also in reference table)

A proton has a positive chargethat is equal in magnitude but

opposite in sign to the


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opposite in sign to theelectrons negative charge

A neutron has no charge so it

is neutral

The unit of charge is thecoulomb (C)

Each proton or each electron


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phas an elementary charge (e)of 1.60x10-19C

The magnitude of the chargeon both an electron and aproton are the same, only thesigns are different

An object has a neutral chargeor no net charge if there are

equal numbers of protons and


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q pneutrons

An object will have a net

negative charge if there aremore electrons than protons

An object will have a netpositive charge if there are

less electrons than protons

Transfer of charge occurs onlythrough movement of electrons

If an object loses electrons, it


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jwill have a net positive charge

If an object gains electrons, it

will have a net negative charge

When the 2 spheres touch


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Next.

And then are pulled apart. Thisis what happens


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On the previous page, onlycharge is transferred

Total charge stays the same


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So charge is always conserved

There is CONSERVATION OF

CHARGE just like with energyand momentum

Transfer of Charge

Problem


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1.) Sphere #1 touches sphere #2and then is separated. Thensphere #2 touches #3 and isseparated. What are the finalcharges on each sphere?

Answer


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1.) #1 has a charge of +2e

#2 has a charge of +3e#3 has a charge of +3e

Electrostatic Symbols

and Constants Fe=electrostatic force in

newtons (N) can be attractive

or repulsivek l t t ti t t


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k=electrostatic constant8.99x109N.m2/C2

q=charge in coulombs (C) r=distance of separation in

meters (m)

E=electric field strength in(N/C)

Electrostatic Equations

2

21

r

qkq

Fe !


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r

q

FE e!

Electric FieldsAn electric field is an area

around a charged particle in

which electric force can bedetected


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Electric fields are detected andmapped using positive test

charges Field lines are the imaginary

lines along which a positivetest charge would move

(arrows show the direction)

Electric Field Line

DiagramsNegative charge Positive charge


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Parallel Plates

+

-

Electric Field Line

DiagramsNegative and positive charges


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Field lines

Electric Field Line

DiagramsTwo positive charges


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Field lines

Electric Field LineDiagrams

Two negative charges


200/343

Field lines

Electrostatics Problems

1.) What is the magnitude ofthe electric field strength when

an electron experiences a 5.0Nforce?


201/343

2.) Is a charge of 4.8x10-19C?possible? A charge of

5.0x10-19C? 3.) What is the electrostatic

force between two 5.0Ccharges that are 1.0x10-4mapart?

Answers

1.) Use

N

e

0.5F

withq

FE

!

!


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CxC

Cxq

N

19

19-

19

e

101.31.60x10

5.0so...E

electrononeoncharge

elementaryisthatbecause1060.1

0.5F

!!

!

Answers (cont.)

2.) Only whole number multiples ofthe elementary charge are possible.

To find out if charge is possible,divide by 1.60x10-19C.

4 8 10 19C i ibl b h


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4.8x10-19C is possible because whenit is divided you get 3, which is a

whole number. 5.0x10-19C is not possible because

when it is divided you get 3.125which is not a whole number.


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B. Current Electricity

Current is the rate at whichcharge passes through a closed

pathway (a circuit) The unit of current is ampere


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p(A)

Conditions needed for aCircuit

Must have a potentialdifference supplied by a

battery or power source Must have a pathway


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p ysupplied by wires

Can have a resistor(s) Can have meters (ammeters,

voltmeters, ohmmeters)

Current ElectricityQuantities

Current is the flow of chargepast a point in a circuit in a

unit of time. Measured inamperes (A)

Potential Difference is the work


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Potential Difference is the workdone to move a charge

between two points. Measuredin volts (V)

Resistance is the opposition tothe flow of current. Measured

in ohms ()

Current ElectricitySymbols

I=current in amperes (A)

q=charge in coulombs (C)

t=time in seconds (s) V=potential difference in volts (V)


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W=work in Joules (J)

R=resistance in ohms ()

=resistivity in (.m)

L=length of wire in meters (m)

A=cross-sectional area of wire in

square meters (m2)

Current ElectricityEquations

q

WV

t

qI !

(!


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A

LR

I

VR

V!!


210/343

Resistivity

Resistivity is an propertyinherent to a material (and its

temperature) that is directlyproportional to the resistanceof the material


211/343

Use the short, thick, cold rule

to remember what affectsresistance

Short, thick, cold wires havethe lowest resistance (best

conduction)

Current ElectricityProblems

1.) How many electrons pass apoint in a wire in 2.0s if the

wire carries a current of 2.5A? 2.) A 10 ohm resistor has 20C

f h i h h i i


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of charge passing through it in5s. What it the potentialdifference across the resistor?

Answers

1.)

1.60x101eusetablesreferencethefrom

n......the5.0Cqso...0.2

2.5Aith

t

qIUse

19-!

!(

(!

(!

C

s

q


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electrons3.13x10electronsofnumber.... 19!so

Answers (cont.)

2.)

/

VRgetto

together

t

qIand

VRUse

(!

(!!

tq

I


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V40....

20C/5s

V

10....

!

!;

Vso

so

Resistivity Problem

3.) What is the resistance of a5.0m long aluminum wire with

a cross-sectional area of2.0x10-6m2 at 200C?


215/343

Answer

3.)

y;

!

)0.5)((2.82x10R

tablesreferencefrom""hwitRse

8- mm

VV


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;!

!

07.0...

100.2

))((so....R

26

Rso

mx

C. Series Circuits

Series circuit is a circuit with asingle pathway for the current

Current stays the samethroughout the circuit

Resistors share the potential


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Resistors share the potentialdifference from the battery

(not necessarily equally) The sum of the resistances of

all resistors is equal to theequivalent resistance of the

entire circuit

Series Circuit Equations

.....

......

.....

321

321

321

!

!

!!!!

RRRR

VVVV

IIII

eq


218/343

same.thebeuldcircuit otheandresistor

"equivalent"singleathcircuit iainresistorsthereplacecouldyouthatmeans

Resistancequivalent

Meters

Ammeter measures the currentin a circuit

Voltmeter measures thepotential difference across aresistor or battery


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resistor or battery

Ohmmeter measures theresistance of a resistor

Circuit Diagram Symbols

Cell

Battery


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Switch

Voltmeter

ammeter

More Circuit DiagramSymbols

Resistor

Variable Resistor


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Lamp

Series Circuit DiagramExample


222/343

Series Circuit Problems

1.) Two resistors withresistances of 4 and 6 are

connected in series to a 25Vbattery. Determine thepotential drop through each


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p p gresistor, the current through

each resistor and theequivalent resistance.

Answers

1.) You can use R=V/I as soonas you have 2 pieces of

information at each resistor.You can also use the seriescircuit laws.


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First, use Req=R1+R2 to find Req4+6=10=Req (use thisas

the Ratthe battery)

More

Answers (cont.)

Then use R=V/I at the battery tofind I at the battery

So10=25V/I and I=2.5Aand all currents are equal in aseries circuit, so I1and


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, 1I2=2.5A

More.

Answers (cont.)

Now that you have 2 pieces ofinformation at each resistor,

you can use R=V/I to findpotential differences

At the 4 resistor, 4=V/2.5A


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soV1=10

V

At the 6 resistor, 6=V/2.5A

SoV2=15V

D. Parallel Circuits

Parallel circuits have more than onepathway that the current can passthrough

Current is shared (not necessarilyequally) among the pathways

Voltage is the same in each


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Voltage is the same in each

pathway as across the battery Equivalent resistance is the sum of

the reciprocal resistances of thepathways (Req is always less than R

of any individual pathway)

Parallel CircuitEquations

....

.....

321

321

!!!!

!

VVVV

IIII


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.....1111

321! RRRReq

same.thebeuldcircuit wotheandresistor

"equivalent"singleathcircuit wiainresistorsthe

replacecouldyouthatmeansResistanceEquivalent

Parallel Circuit DiagramExample


229/343


230/343

Answers

1.)

;!;

!;

;

;

!

!

14

4

4

1

4

1

2

11

1111se

321

eq

eq

R

RRRR


231/343

So equivalent resistance=1

More..

Answers (cont.)

Use V=V1=V2=V3

So since V=20V,V1=V2=V3=20V


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More

Answers (cont.)

Using R=V/I, substitute thepotential differences and

resistances for each resistor tofind the current for eachresistor

I 20A


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I=20A

I1=10A

I2=5A

I3=5A

E. Electric Power andEnergy

Electric power is the product ofpotential difference and current

measured in watts just likemechanical power

Electric energy is the productf d ti d i


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of power and time measured in

joules just like other forms ofenergy

Electric Power andEnergy Equations

R

VRIVIP

22 !!!


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R

tVRtIVItPtW

22 !!!!

Electric Energy andPower Problems

1.) How much time does it take a60W light bulb to dissipate 100J ofenergy?

2.) What is the power of an electricmixer while operating at 120V, if ithas a resistance of 10?


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3.) A washing machine operates at220V for 10 minutes, consuming3.0x106J of energy. How muchcurrent does it draw during thistime?

Answers

1.) Use W=Pt

So100J=60Wxt

Sot=0.6s

2.) Use


237/343

SoP=1440W

;!!

10(120V)so...P

R

VP22

Answers (cont.)

3.) Use W=VIt but first needto change the 10 minutes to

600 seconds. So3.0x106J=220V(I)600s

SoI=22.7A


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F. Magnetism andElectromagnetism

Magnetism is a force ofattraction or repulsion

occurring when spinningelectrons align

A magnet has two ends calledpoles


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poles

The N pole is north seeking

The S pole is south seeking

Like poles repel

Unlike poles attract

If you break a magnet, thepieces still have N and S poles

The earth has an S pole atthe North Pole

The earth has a N pole at theSouth Pole


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South Pole

The N pole of a compass willpoint towards earths S polewhich is the geographic North

pole

There is a field around eachmagnet

Imaginary magnetic field orflux lines show where amagnetic field is

You can use a compass to map


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You can use a compass to mapa magnetic field

Units for Magnetic Field

The weber (Wb) is the unit formeasuring the number of field

lines

The tesla (T) is the unit formagnetic field or flux density


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magnetic field or flux density

1T=1Wb/m2

Magnetic Field Diagrams


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Magnetic Field Diagrams(cont.)


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Magnetic Field Diagrams(cont.)


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Electromagnetism

Moving a conductor through amagnetic field will induce apotential difference which may

cause a current to flow (conductormust break the field lines for thisto occur)

A wire with a current flowing


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g

through it creates a magnetic field

Somagnetism creates electricityand electricity creates magnetism

ElectromagnetismDiagram

Must move wire into and out ofpage to induce potential

current by breaking field lines


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v. Waves

A.) Wave Characteristics

B.) Periodic Wave Phenomena

C.) Light D.) Reflection and Refraction

E.) Electromagnetic Spectrum


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A. Wave Characteristics

A wave is a vibration in amedium or in a field

Sound waves must travelthrough a medium (material)

Light waves may travel ineither a medium or in an


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electromagnetic field

Waves transfer energy only,not matter

A pulse is a single disturbanceor vibration

A periodic wave is a series ofregular vibrations


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Types of Waves

1.) Longitudinal Waves arewaves that vibrate parallel to

the direction of energytransfer. Sound andearthquake p waves areexamples.


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Types of Waves

2.) Transverse waves arewaves that vibrate

perpendicularly to the directionof energy transfer. Light wavesand other electromagneticwaves are examples.


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Frequency

Frequency is how many wavecycles per second

The symbol for frequency is f The unit for frequency is hertz

(Hz)

1Hz=1/s


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Frequency is pitch in sound

Frequency is color in visiblelight

Frequency (cont.)

High frequency wave


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Low frequency wave

Period

Period is the time required forone wave cycle

The symbol for period is T The unit for period is the

second (s)

The equation for period is


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T=1/f

Period and frequency areinversely proportional

Period (cont.)

Short period wave


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Long period wave

Amplitude

The amplitude of a wave is theamount of displacement fromthe equilibrium line for thewave (how far crest or troughis from the equilibrium line)

Symbol for amplitude is A

U it f lit d i t ( )


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Unit for amplitude is meter (m)

Amplitude is loudness in sound

Amplitude is brightness in light

Amplitude (cont.)

High amplitude wave

L lit d


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Low amplitude wave

Wavelength

Wavelength is the distancebetween points of a completewave cycle

Symbol for wavelength is

Unit for wavelength is meter(m)

A ith hi h f


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A wave with a high frequencywill have short wavelengthsand short period

Wavelength (cont.)

Short wavelength

Long wavelength


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Long wavelength

Phase

Points that are at the sametype of position on a wavecycle (including same directionfrom equilibrium and moving insame direction) are in phase


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Phase (cont.)

Points that are in phase arewhole wavelengths apart (i.e.,1 apart, 2 apart, 3 apart,

etc.)

Points that are in phase aremultiples of 360 degrees apart

(i e 360 degrees apart 720


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(i.e., 360 degrees apart, 720degrees apart, 1080 degreesapart, etc.)

Phase (cont.)

A B C

D E F

Points A, B and C are in phase

Points D, E and F are in phase

Points A and B are 360 degrees apart


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Points A and B are 360 degrees apartand 1 apart

Points D and F are 720 degrees apart

and 2 apart

Phase (cont.)

A C

D

B

Points A and B are 180 degrees apartand apart (not in phase)

Points C and D are 90 degrees apart


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Points C and D are 90 degrees apartand apart (not in phase)

Speed

Speed of a wave is equal to theproduct of wavelength andfrequency

Symbol for speed is v

Unit for speed is m/s

Equation for speed is v=f

Can also use basic speed


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Can also use basic speedequation v=d/t if needed

Speed (cont.)

The speed of a wave dependson its type and the medium itis traveling through

The speed of light (and allelectromagnetic waves) in avacuum and in air is

3 00x108m/s


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3.00x10 m/s The speed of sound in air at

STP (standard temperature and

pressure) is 331m/s

Wave CharacteristicsProblems

1.) A wave has a speed of7.5m/s and a period of 0.5s.What is the wavelength of thewave?

2.) If the period of a wave isdoubled, what happens to its

frequency?


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frequency? 3.) Points on a wave are 0.5

apart, 7200 apart, 4 apart.

Which points are in phase?

Answers

1.) First use T=1/f to find f,then use v=f to find

So0.5s=1/f and f=2.0Hz Then7.5m/s=2.0Hz() and=3.75m

2 ) T and f are inversely related


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2.) T and f are inversely relatedso if T is doubled fis halved.

Answers (cont.)

3.) Points are in phase only ifthey are whole wavelengthsapart or multiples of 360degrees apart. So..the pointsthat are 720degreesapartare inphase and the points

that are 4 apart are inphase The points that are


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that are 4 apartare inphase. The points that are0.5 apart are not in phase.

More problems

5.0m

4 ) What is the wavelength


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4.) What is the wavelengthof the wave shown above?

Answer

4.) The wave is 5.0m long andcontains 2.5 cycles. You wantto know the length of 1 cycle.

So5.0m/2.5cycles=2.0m/cycle

So=2.0m


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B. Periodic WavePhenomena

When waves interact with oneanother many different

phenomena result

Those phenomena are thedoppler effect, interference,standing waves, resonance and

diffraction


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Circular Waves

Some phenomena are easier to explainusing circular waves

A is the from

crest to crest

is the

wavefront (all

points on the

circle are in A A

phase)


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Doppler Effect

If the source of waves ismoving relative to an observer,the observed frequency of thewave will change (actual fproduced by the source doesntchange and movement must be

fast for observer to notice)


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Doppler Effect (cont.)

If source is moving towards theobserver, f will increase (higherpitch if sound wave, shifttowards blue if light wave)

If source is moving away fromthe observer, f will decrease

(lower pitch if sound wave,shift towards red if light wave)


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shift towards red if light wave)


Joe Shmoe


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more.


In the diagram on the previouspage, the wave source ismoving towards Joe

In the diagram on the previouspage, the wave source ismoving away from Shmoe


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The effect on Joe is that the fof the wave will be higher forhim (shorter and T, higher

pitch or bluer light)

The effect on Shmoe is thatthe f of the wave will be lower

for him (longer and T, lowerpitch or redder light)


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pitch or redder light)


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Interference (cont.)

Maximum constructiveinterference occurs whenwaves are in phase

Maximum destructive interenceoccurs when waves are 180degrees out of phase


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Maximum ConstructiveInterference

Above results in


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Maximum DestructiveInterference

Above results in


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Standing Waves

Standing waves occur whenwaves having same A and

same f travel in oppositedirections. Vibrates so that


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it looks like waves arestationary


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Resonance (cont.)

Examples of resonance:

Find the Tacoma NarrowsBridge video on the internet

A singer being able to break aglass with her voice


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Diffraction

When waves bend behind abarrier instead of going straightthrough that is called

diffraction


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Wave

fronts

C. Light

Light is the part of theelectromagnetic spectrum thatis visible to humans

The speed of light symbol is c

The speed of light is constant,3.00x108m/s in air or a vacuum

Instead of using v=f, whenits light you can substitute


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it s light you can substitutec=f

Light (cont.)

No object can travel faster thanthe speed of light (one ofEinsteins ideas)

The speed of light in a materialis always less than c


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Light Problems

1.) Determine the wavelengthin a vacuum of a light wavehaving a frequency of

6.4x1014Hz.

2.) What is the frequency of a

light wave with a wavelengthof 5.6x10-7m?


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Answers

1.) Use c=f

3.00x108m/s=6.4x1014Hz()

So=4.7x10

-7

m

2.) Use c=f

3.00x108

m/s=f(5.6x10-7

m) Sof=5.4x1014Hz


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D. Reflection andRefraction

Reflection occurs when a ray oflight hits a boundary andbounces back into the same

medium

Refraction occurs when a rayof light enters a new medium

and changes direction becauseof a change in speed


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More Reflection

The law of reflection

i r

reflectionofangleincidenceofangle

!!

!

r

i

ri

UU

UU


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Dotted line is the normal(reference line)

More Reflection

In a mirror the image of anobject is flipped laterally (yourright hand is your left hand in a

mirror image)

To view an object in a planemirror, you need a minimum of

the height of the object forthe height of the mirror


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More Refraction

When light enters a differentmedium, the change indirection depends on the

density of the new medium

If new medium is denser, thelight will slow down and bend

towards the normal If new medium is less dense,

th li ht ill d d


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the light will speed up andbend away from the normal

More Refraction

Refraction Equations

2211

RefractionofIndex

sinsin

Lawsnell'

UU

!

!

v

cn

nn


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2

1

2

1

1

2

PP!!

vv

nn

Other

More Refraction

Symbols n=index of refraction (no units)

v=speed of the wave in m/s

=wavelength in m c=speed of light in a vacuum

(3.00x108m/s)

1=angle of incidence

2=angle of refraction


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Refraction Diagram

1air

glass 2

Light travels from air into glass which


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Light travels from air into glass whichis more dense so it slows down andbends towards the normal

Refraction Diagram

1

air

glass 2

Light travels from glass into air which


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g gis less dense so it speeds up andbends away from the normal

Absolute Index ofRefraction

Absolute index of refraction isthe ratio of the speed of lightin a vacuum to the speed of

light in a specific medium The symbol is n

The higher the n number the

denser the medium The lower the n number the

l d th di


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less dense the medium

Absolute Index ofRefraction (cont.)

Higher n means slower v oflight in the medium

Lower n means higher v oflight in the medium

Use reference tables to find n

numbers


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Problems

1.) What is the speed of lightin diamond?

2.) What is the ratio of the

speed of light in corn oil to thespeed of light in water?


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Problems (cont.)

3.) A ray of monochromaticlight having a frequency of5.09x1014Hz is traveling

through water. The ray isincident on corn oil at an angleof 600 to the normal. What isthe angle of refraction in cornoil?


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Answers

1.)

smxvso

vsm

/1024.1....

/3.00x10so....2.42

2.42diamondofnwithv

cnse

8

8

!

!

!!


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Answers (cont.)

2.)

2

1

1

2

1

2

2

1

1.47

1.33...

1.33isaterofn1.47,isoilcornofn

vUse

v

v

n

nso

n

n

v

!!

!


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Answers (cont.)

3.) Use SnellsLaw

0

2

2

0

2211

52...

sin47.1sin60so....1.33

secondtheoilcornmedium,firsttheisaterThe

sinsin

!

!

!

U

U

UU

so

nn


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E. ElectromagneticSpectrum

Electromagnetic waves includegamma rays, x rays, ultravioletrays, infrared rays,

microwaves, t.v. and radiowaves, long waves

Electromagnetic waves are

produced by acceleratingcharges (produce alternatingelectric and magnetic fields)


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ElectromagneticSpectrum (cont.)

High energy electromagneticwaves have high f and small

Low energy electromagnetic

waves have low f and long

Energy in a wave can beincreased by increasing f,

decreasing and T andincreasing the duration of thewave


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ElectromagneticSpectrum (cont.)

In the visible light spectrum,the violet end is high f, shortand high energy

In the visible light spectrum,the red end is low f, long and

low energy


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VI. Modern Physics

A. Quantum Theory

B. Atomic Models

C. Nucleus

D. Standard Model

E. Mass-E

Regent Physics by Abhishek Jaguessar

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Transcript of Regent Physics by Abhishek Jaguessar