Regent Physics Review

Physics Units

I. Physics Skills II. Mechanics III. Energy IV. Electricity and

Magnetism V. Waves VI. Modern Physics

I. Physics Skills

A. Scientific Notation B. Graphing C. Significant Figures D. Units E. Prefixes F. Estimation

A. Scientific Notation

Use for very large or very small numbers

Write number with one digit to the left of the decimal followed by an exponent (1.5 x 105)

Examples: 2.1 x 103 represents 2100 and 3.6 x 10-4 represents 0.00036

Scientific Notation Problems

1. Write 365,000,000 in scientific notation

2. Write 0.000087 in scientific notation

Answers

1.) 3.65 x 108

2.) 8.7 x 10-5

B. Graphing

Use graphs to make a “picture” of scientific data

“independent variable”, the one you change in your experiment is graphed on the “x” axis and listed first in a table

“dependent variable”, the one changed by your experiment is graphed on the “y” axis and listed second in a table

Best fit “line” or “curve” is drawn once points are plotted. Does not have to go through all points. Just gives you the “trend” of the points

The “slope” of the line is given as the change in the “y” value divided by the change in the “x” value

Types of Graphs

1. Direct Relationship means an increase/decrease in one variable causes an increase/decrease in the other

Example below

2. Inverse(indirect) relationship means that an increase in one variable causes a decrease in the other variable and vice versa

Examples

3. Constant proportion means that a change in one variable doesn’t affect the other variable

Example;

4. If either variable is squared(whether the relationship is direct or indirect), the graph will curve more steeply.

C. Significant figures

Uncertainty in measurements is expressed by using significant figures

The more accurate or precise a measurement is, the more digits will be significant

Significant Figure Rules

1. Zeros that appear before a nonzero digit are not significant (examples: 0.002 has 1 significant figure and 0.13 has 2 significant figures)

2. Zeros that appear between nonzero digits are significant (examples: 1002 has 4 significant figures and 0.405 has 3 significant figures)

Significant figures rules(cont.)

3. zeros that appear after a nonzero digit are significant only if they are followed by a decimal point (20. has 2 sig figs) or if they appear to the right of the decimal point (35.0 has 3 sig figs)

Sig Fig problems

1. How many significant figures does 0.050900 contain?

2. How many significant figures does 4800 contain?

Answers

1. 5 sig figs

2. 2 sig figs

D. Units

1. Fundamental units are units that can’t be broken down

2. Derived units are made up of other units and then renamed

3. SI units are standardized units used by scientists worldwide

Fundamental Units

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

Kilogram (kg)– mass Second (s)– time, period Ampere (A)– electric

current Degree (o)– angle

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

Derived Units (cont.)

Joule (J)– work, all types of energy

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)

Derived Units (cont.)

Ohm (Ω)– resistance Ohm times meter (Ω.m)–

resistivity Weber (Wb)– number of

magnetic field (flux) lines Tesla (T)– magnetic field

(flux) density Hertz (Hz)-- frequency

E. Prefixes

Adding prefixes to base units makes them smaller or larger by powers of ten

The prefixes used in Regents Physics are tera, giga, mega, kilo, deci, centi, milli, micro, nano and pico

Prefix Examples

A terameter is 1012 meters, so… 4 Tm would be 4 000 000 000 000 meters

A gigagram is 109 grams, so… 9 Gg would be 9 000 000 000 grams

A megawatt is 106 watts, so… 100 MW would be 100 000 000 watts

A kilometer is 103 meters, so… 45 km would be 45 000 meters

Prefix examples (cont.)

A decigram is 10-1 gram, so… 15 dg would be 1.5 grams

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

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

Prefix examples (cont.)

A microvolt is 10-6 volt, so… 8 µV would be 0.000 008 volt

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

A picometer is 10-12 meter, so… 677 pm would be 0.000 000 000 677 meter

Prefix Problems

1.) 16 terameters would be how many meters?

2.) 2500 milligrams would be how many grams?

3.) 1596 volts would be how many gigavolts?

4.) 687 amperes would be how many nanoamperes?

Answers

1.) 16 000 000 000 000 meters

2.) 2.500 grams

3.) 1596 000 000 000 gigavolts

4.) 0.000 000 687 amperes

F. Estimation

You can estimate an answer to a problem by rounding the known information

You also should have an idea of how large common units are

Estimation (cont.)

2 cans of Progresso soup are just about the mass of 1 kilogram

1 medium apple weighs 1 newton

The length of an average Physics student’s leg is 1 meter

Estimation Problems

1.) Which object weighs approximately one newton? Dime, paper clip, student, golf ball

2.) How high is an average doorknob from the floor? 101m, 100m, 101m, 10-2m

Answers

1.) golf ball

2.) 100m

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

A. Kinematics; vectors, velocity, acceleration

In physics, quantities can be vector or scalar

VECTOR quantities have a magnitude (a number), a unit and a direction

Example; 22m(south)

SCALAR quantities only have a magnitude and a unit

Example; 22m

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

SCALAR quantities; distance, mass, time, speed, work(energy), power

Distance vs. Displacement

Distance is the entire pathway an object travels

Displacement is the “shortest” pathway from the beginning to the end

Distance/Displacement Problems

1.) A student walks 12m due north and then 5m due east. What is the student’s resultant displacement? Distance?

2.) A student walks 50m due north and then walks 30m due south. What is the student’s resultant displacement? Distance?

Answers

1.) 13m (NE) for displacement17 m for distance

2.) 20m (N) for displacement 80 m for distance

Speed vs. Velocity

Speed is the distance an object moves in a unit of time

Velocity is the displacement of an object in a unit of time

Average Speed/Velocity Equations

t

dv

2if vv

v

Symbols

timet

distanceor nt displacemed

speedor velocity initialv

speedor velocity finalv

speedor velocity averagev

i

f

Speed/Velocity Problems

1.) A boy is coasting down a hill on a skateboard. At 1.0s he is traveling at 4.0m/s and at 4.0s he is traveling at 10.0m/s. What distance did he travel during that time period? (In all problems given in Regents Physics, assume acceleration is constant)

Answers

1.) You must first find the boy’s average speed before you are able to find the distance

Answers (cont.)

21m wasd travellehe distance eso......th 3.0st

v use

/0.7vso.... /0.4

/0.10v

2

use

f

t

dthen

smsmv

sm

vvv

i

if

Acceleration

The time rate change of velocity is acceleration (how much you speed up or slow down in a unit of time)

We will only be dealing with constant (uniform) acceleration

Symbols (cont.)

)(vy in velocit change f ivthev

onacceleratia

Constant Acceleration Equations

adv

att

atvv

t

va

i

if

2v

2

1vd

22f

2i

Constant Acceleration Problems

1.) A car initially travels at 20m/s on a straight, horizontal road. The driver applies the brakes, causing the car to slow down at a constant rate of 2m/s2 until it comes to a stop. What was the car’s stopping distance? (Use two different methods to solve the problem)

Answers

First Methodvi=20m/s

vf=0m/s

a=2m/s2

Use vf2=vi

2+2ad to find “d”

d=100m

Answer (cont.)

Second Method

100md

d"" find to2

1vitd then...use

t"" find toa Use

2

at

t

v

B. Kinematics: freefall

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

The rate at which objects fall is known as “g”, the acceleration due to gravity

On earth, the “g” is 9.81m/s2

Solving Freefall Problems

To solve freefall problems use the constant acceleration equations

Assume a freely falling object has a vi=0m/s

Assume a freely falling object has an a=9.81m/s2

Freefall Problems

1.) How far will an object near Earth’s surface fall in 5s?

2.) How long does it take for a rock to fall 60m? How fast will it be going when it hits the ground

3.) In a vacuum, which will hit the ground first if dropped from 10m, a ball or a feather?

Answers

1.)

mdso

st

sma

sm

attvi

6.122.....

5

/81.9

/0v

.. with....2

1d Use

2

i

2

Answers (cont.)

2.)

smvso

atv

stso

md

sma

sm

att

f

i

/3.34....

v

use......en th5.3....

60

/81.9

/0v

. with....2

1vd Use

f

2

i

2i

Answers (cont.)

3.) Both hit at the same time because “g” the acceleration due to gravity is constant. It doesn’t depend on mass of object because it is a ratio

m

Fg g

Solving Anti??? Freefall Problems

If you toss an object straight up, that is the opposite of freefall.

So… vf is now 0m/s a is -9.8lm/s2

Because the object is slowing down not speeding up

Antifreefall Problems

1.) How fast do you have to toss a ball straight up if you want it reach a height of 20m?

2.) How long will the ball in problem #1 take to reach the 20m height?

Answers

1.)

sm

md

sma

sm

adv

f

i

/8.19v

so........ 20

/81.9

/0v

... when....2 vUse

i

2

22f

Answers (cont.)

2.)

stso

md

sma

smv

smv

attatv

f

i

i

0.2.......

20

/81.9

/0

/8.192

1vdor vUse

2

2if

C. Statics

The study of the effect of forces on objects at rest

Force is a push or pull The unit of force is the

newton(N) (a derived vector quantity)

Adding forces

When adding concurrent (acting on the same object at the same time) forces follow three rules to find the resultant (the combined effect of the forces)

1.) forces at 00, add them2.) forces at 1800, subtract

them3.) forces at 900, use

Pythagorean Theorem

Force Diagrams

Forces at 00

Forces at 1800

Forces at 900

Composition of Forces Problems

1.) Find the resultant of two 5.0N forces at 00? 1800? And 900?

Answers

1.) 00

5N 5N = 10N

Answers (cont.)

1.) 1800

5N 5N = 0N

Answers (cont.)

1.) 900

5N =

5N

7.1N

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 Graphical 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

of the force to each of the axes

3.) From the tail of the resultant force to where the lines intersect the axes, are the lengths of the component forces

Resolution Diagram

Black arrow=resultant forceOrange lines=reference linesGreen arrows=component

forces

y

x

Resolving Forces Using Algebraic Method

A and Abetween angle

forceresultant theA

component A

component A

component verticalfind tosinA

component horizontal find tocosA

use... lly,algebraica components find To

x

y

x

y

x

vertical

horizontal

A

A

Equilibrium

Equilibrium occurs when the net force acting on an object is zero

Zero net force means that when you take into account all the forces acting on an object, they cancel each other out

Equilibrium (cont.)

An object in equilibrium can either be at rest or can be moving with constant (unchanging) velocity

An “equilibrant” is a force equal and opposite to the resultant force that keeps an object in equilibrium

Equilibrium Diagram

Black arrows=componentsBlue arrow=resultantRed arrow=equilibrant

= +

Problems

1.) 10N, 8N and 6N forces act concurrently on an object that is in equilibrium. What is the equilibrant of the 10N and 6N forces? Explain.

2.) A person pushes a lawnmower with a force of 300N at an angle of 600 to the ground. What are the vertical and horizontal components of the 300N force?

Answers

1.) The 8N force is the equilibrant (which is also equal to and opposite the resultant) The 3 forces keep the object in equilibrium, so the third force is always the equilibrant of the other two forces.

Answers (cont.)

2.)

NNAand

NNAso

A

A

y

x

y

26060sin300....

15060cos300......

60

300NA

component verticalfind tosinA and

component horizontal find tocosA Use

0

0

0

x

C. Dynamics

The study of how forces affect the motion of an object

Use Newton’s Three Laws of Motion to describe Dynamics

Newton’s 1st Law of Motion

Also called the law of inertia Inertia is the property of an

object to resist change. Inertia is directly proportional to the object’s mass

“An object will remain in equilibrium (at rest or moving with constant speed) unless acted upon by an unbalanced force”

Newton’s Second Law of Motion

“When an unbalanced (net) force acts on an object, that object accelerates in the direction of the force”

How much an object accelerates depends on the force exerted on it and the object’s mass (See equation)

m

Fneta

Symbols

Fnet=the net force exerted on an object (the resultant of all forces on an object) in newtons (N)

m=mass in kilograms (kg)

a=acceleration in m/s2

Newton’s Third Law of Motion

Also called law of action-reaction

“When an object exerts a force on another object, the second object exerts a force equal and opposite to the first force”

Masses of each of the objects don’t affect the size of the forces (will affect the results of the forces)

Free-Body Diagrams

A drawing (can be to scale) that shows all concurrent forces acting on an object

Typical forces are the force of gravity, the normal force, the force of friction, the force of acceleration, the force of tension, etc.

Free-body Forces

Fg is the force of gravity or weight of an object (always straight down)

FN is the “normal” force (the force of a surface pushing up against an object)

Ff is the force of friction which is always opposite the motion

Fa is the force of acceleration caused by a push or pull

Free-Body Diagrams

If object is moving with constant speed to the right….

Black arrow=Ff

Green arrow=Fg

Yellow arrow=FN

Blue arrow=Fa

Free-Body Diagrams on a Slope

When an object is at rest or moving with constant speed on a slope, some things about the forces change and some don’t

1.) Fg is still straight down 2.) Ff is still opposite motion 3.) FN is no longer equal and

opposite to Fg

4.) Ff is still opposite motion More……

Free-Body Diagrams on a Slope

Fa=Ff=Ax=Acosθ

FN=Ay=Asinθ

Fg=mg (still straight down)

On a horizontal surface, force of gravity and normal force are equal and opposite

On a slope, the normal force is equal and opposite to the “y” component of the force of gravity

Free-Body Diagram on a Slope

Green arrow=FN

Red arrow=Fg

Black arrows=Fa and Ff

Orange dashes=Ay

Purple dashes=Ax

Dynamics Problems

1.) Which has more inertia a 0.75kg pile of feathers or a 0.50kg pile of lead marbles?

2.) An unbalanced force of 10.0N acts on a 20.0kg mass for 5.0s. What is the acceleration of the mass?

Answers

1.) The 0.75kg pile of feathers has more inertia because it has more mass. Inertia is dependent on the “mass” of the object

Answers (cont.)

2.)

2

net

net

0.50m/sso......a

this)needt (don' 0.5

0.20

0.10F

h....... witm

Fa Use

st

kgm

N

More Problems

3.) A 10N book rests on a horizontal tabletop. What is the force of the tabletop on the book?

4.) How much force would it take to accelerate a 2.0kg object 5m/s2? How much would that same force accelerate a 1.0kg object?

Answers

1.) The force of the tabletop on the book is also 10N (action/reaction)

Answers (cont.)

2.)

2

net

net

2

net

10m/sso.....a

part)first in as (same N10F

1.0kgm

:part Second

10so....F 2.0kgm

5m/sa

:partFirst

question theof partsboth for F

a Use

N

m

E. 2-Dimensional Motion

To describe an object moving 2-dimensionally, the motion must be separated into a “horizontal” component and a “vertical” component (neither has an effect on the other)

Assume the motion occurs in a “perfect physics world”; a vacuum with no friction

Types of 2-D Motion

1.) Projectiles fired horizontally

an example would be a baseball tossed straight horizontally away from you

Projectile Fired Horizontally

Use the table below to solve these type of 2-D problems

Quantities Horizontal Vertical

vi Same as vf 0m/s

vf Same as vi

a 0m/s2 9.81m/s2

d

t Same as vertical time

Same as horizontal time

Types of 2-D Motion

2.) Projectiles fired at an angle

an example would be a soccer ball lofted into the air and then falling back onto the ground

Projectile Fired at an Angle

Use the table below to solve these type of 2-D problems

Ax=Acosθ and Ay=Asinθ

Quantities Horizontal Vertical

vi Ax Ay

vf Ax 0m/s

a 0m/s2 - 9.81m/s2

d

t twice vertical time

2-D Motion Problems

1.) A girl tossed a ball horizontally with a speed of 10.0m/s from a bridge 7.0m above a river. How long did the ball take to hit the river? How far from the bottom of the bridge did the ball hit the river?

Answers

1.) In this problem you are asked to find time and horizontal distance (see table on the next page)

Answers (cont.)

Quantities Horizontal Vertical

vi 10.0m/s 0.0m/s

vf 10.0m/s Don’t need

a 0.0m/s 9.81m/s2

d ? 7.0m

t ? ?

Answers (cont.)

Use

14.3mhorizontal so......d

vertical)as same(43.1

/0.10v

d"" horizontal find n toinformatio horizontal with v Use

43.1......

/81.9

/0.0v

7.0md

t"" find n toinformatio tical with ver2

1vd Use

2

i

2i

st

sm

t

d

stso

sma

sm

att

More 2-D Motion Problems

2.) A soccer ball is kicked at an angle of 600 from the ground with an angular velocity of 10.0m/s. How high does the soccer ball go? How far away from where it was kicked does it land? How long does its flight take?

Answers

2.) In this problem you are asked to find vertical distance, horizontal distance and horizontal time. Finding vertical time is usually the best way to start. (See table on next page)

Answers (cont.)

Quantities Horizontal

Vertical

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.)

Find vertical “t” first usingvf=vi+at with….

vf=0.0m/s

vi=8.7m/s

a=-9.81m/s2

So…vertical “t”=0.89s and horizontal “t” is twice that

and equals 1.77s

Answers (cont.)

Find horizontal “d” using

8.87m horizontal dso........

77.1

/0.5v

... when.v

st

sm

t

d

Answers (cont.)

Find vertical “d” by using

mdso

sma

st

smv

att

i

85.3.......

/81.9

89.0

/7.8

... when...2

1vd

2

2i

F. Uniform Circular Motion

When an object moves with constant speed in a circular path

The force (centripetal) will be constant towards the center

Acceleration (centripetal) will only be a direction change towards the center

Velocity will be tangent to the circle in the direction of movement

Uniform Circular Motion Symbols

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)

Uniform Circular Motion Diagram

Uniform Circular Motion Equations

r

mv

r

v

2

c

2

c

F

a

Uniform Circular Motion Problems

1.) A 5kg cart travels in a circle of radius 2m with a constant velocity of 4m/s. What is the centripetal force exerted on the cart that keeps it on its circular pathway?

Answers

1.)

N40so.....F

force lcentripeta find toF useThen

8m/sso.....a

onaccelerati lcentripeta find toa useFirst

c

c

2c

2

c

cma

r

v

G. Mass, Weight and Gravity

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 attraction between objects

Mass

Mass is measured in kilograms (kg)

Mass doesn’t change with location (for example, if you travel to the moon your mass doesn’t change)

Weight

Weight is measured in newtons (N)

Weight “does” change with location because it is dependent on the pull of gravity

Weight is equal to mass times the acceleration due to gravity

Weight/Force of Gravity Equations

earth)on m/s (9.81gravity todueon accelerati the timesmass its to

equal isobject an of )(or weightgravity of force themeanswhich

m

Fg

objects 2between gravity of

force the torelated (squared) indirectly is anceand...dist

objects 2between gravity of force

the torelateddirectly is mass that meanswhich

F

2

g

221

g

r

mmG

Gravitational Field Strength g=acceleration due to

gravity but it 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

1.) If the distance between two masses is doubled, what happens to the gravitational force between them?

2.) If the distance between two objects is halved and the mass of one of the objects is doubled, what happens to the gravitational force between them?

Answers

1.) Distance has an inverse squared relationship with the force of gravity.

So…since r is multiplied by 2 in the problem, square 2……so….22=4, then take the inverse of that square which equals ¼….so….the answer is “¼ the original Fg”

221

r

mmGFg

Answers (cont.)

2.) Mass has a direct relationship with Fg and distance has an inverse squared relationship with Fg.

First…since m is doubled so is Fg and since r is halved, square ½ , which is ¼ and then take the inverse which is 4.

Then…combine 2x4=8 So…answer is “8 times Fg”

More Problems

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

4.) An object with a mass of 10kg has a weight of 4N on Planet X. What is the acceleration due to gravity on Planet X? What is the gravitational field strength on Planet X?

Answers

3.)

4.)

strength field nalgravitatiofor 0.4N/kgg

gravity todueon acceleratifor 0.4m/sg

m

Fg Using

106.1

/6.67x10G

. with...F Using

2

g

11

2211-

2

1g

2

NxF

kgmN

r

mmG

g

H. Friction

The force that opposes motion measured in newtons (N)

Always opposite direction of motion

“Static Friction” is the force that opposes the “start of motion”

“Kinetic Friction” is the force of friction between objects in contact that are in motion

Coefficient of Friction

The ratio of the force of friction to the normal force (no unit, since newtons cancel out)

EquationFf= μFN

μ=coefficient of frictionFf=force of friction

FN=normal force

Coefficient of Friction

The smaller the coefficient, the easier the surfaces slide over one another

The larger the coefficient, the harder it is to slide the surfaces over one another

Use the table in the reference tables

Coefficient of Friction Problems

1.) A horizontal force is used to pull a 2.0kg cart at constant speed of 5.0m/s across a tabletop. The force of friction between the cart and the tabletop is 10N. What is the coefficient of friction between the cart and the tabletop? Is the friction force kinetic or static? Why?

Answers

1.)

The friction force is kinetic because the cart is moving over the tabletop

51.0

FF singso.......u 10

6.19/81.90.2

Nf

2N

NF

NsmkgmgFF

f

g

I. Momentum and Impulse

Momentum is a vector quantity that is the product of mass and velocity (unit is kg.m/s)

Impulse is the product of the force applied to an object and time (unit is N.s)

Momentum and Impulse Symbols

p=momentum

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

J=impulse

Momentum and Impulse Equations

p=mv

J=Ft

J=Δp

pbefore=pafter

Momentum and Impulse Problems

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

2.) A 1.0kg cart at rest is hit by a 0.2kg cart moving to the right at 10.0m/s. The collision causes the 1.0kg cart to move to the right at 3.0m/s. What is the velocity of the 0.2kg cart after the collision?

Answer

1.) Use J=Δp so……J=10N.s(E)=Δp=10kg.m/s(E) and since original p was

0kg.m/s and Δp=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 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 or 1kg.m/s(left)

more…..

Answers (cont.)

So if p after collision of 0.2kg cart is 1kg.m/s(left) and

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

And v=5m/s(left)

III. Energy

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 an object a distance

Work is scalar The unit of work is the Joule

(J) Work and energy are

manifestations of the same thing, that is why they have the same unit of Joules

Work and Power (cont.)

Power is the rate at which work is done so there is a “time” factor in power but not in work

Power and time are inversely proportional; the less time it takes to do work the more power is developed

The unit of power is the watt (W)

Power is scalar

Work and Power Symbols

W=work in Joules (J) F=force in newtons

(N) d=distance in

meters (m) ΔET=change in total

energy in Joules (J) P=power in watts

(W) t=time in seconds

(s)

(m/s)in speed averagev

Work and Power Equations

W=Fd=ΔET

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

vFt

Fd

t

WP

Work and Power Problems

1.) A 2.5kg object is moved 2.0m in 2.0s after receiving a horizontal push of 10.0N. How much work is done on the object? How much power is developed? How much would the object’s total energy change?

2.) A horizontal 40.0N force causes a box to move at a constant rate of 5.0m/s. How much power is developed? How much work is done against friction in 4.0s?

Answers

1.) to find work use W=Fd So…W=10.0N x 2.0m=20.0J

To find power use P=W/t So…P=20.0J/2.0s=10.0W

To find total energy change it’s the same as work done so……

ΔET=W=20.0J

Answers (cont.)

2.) To find power use

So… P=40.0N x 5.0m/s=200W

To find work use P=W/t so…200W=W/4.0s

So….W=800J

vFP

More problems

3.) A 2.0kg object is raised vertically 0.25m. What is the work done raising it?

4.) A lift hoists a 5000N object vertically, 5.0 meters in the air. How much work was done lifting it?

Answers

3.) to find work use W=Fd with F equal to the weight of the object

So…..W=mg x d So...W=2.0kgx9.81m/

s2x0.25m So…W=4.9J

Answers (cont.)

4.) to find work use W=Fd Even though it is vertical

motion, you don’t have to multiply by “g” because weight is already given in newtons

So…W=Fd=5000N x 5.0m And W=25000J

B. Potential and Kinetic Energy

Gravitational Potential Energy is energy of position above the earth

Elastic Potential Energy is energy due to compression or elongation of a spring

Kinetic Energy is energy due to motion

The unit for all types of energy is the same as for work the Joule (J). All energy is scalar

Gravitational Potential Energy Symbols and Equation

ΔPE=change in gravitational potential energy in Joules (J)

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

in (m/s2) Δh=change in height in meters

(m)

Equation ΔPE=mgΔh ***Gravitational PE only

changes if there is a change in vertical position

Gravitational PE Problems

1.) How much potential energy is gained by a 5.2kg object lifted from the floor to the top of a 0.80m high table?

2.) How much work is done in the example above?

Answers

1.) Use ΔPE=mgΔh to find potential energy gained so ΔPE=5.2kgx9.81m/s2x0.80m

So…ΔPE=40.81J

2.) W=ΔET so..W is also 40.81J

Kinetic Energy Symbols and Equation

KE=kinetic energy in Joules (J)

m=mass in kilograms (kg)

v=velocity or speed in (m/s)

2

2

1KE mv

Kinetic Energy Problems

1.) If the speed of a car is doubled, what happens to its kinetic energy?

2.) A 6.0kg cart possesses 75J of kinetic energy. How fast is it going?

Answers

1.) Using KE=1/2mv2 if v is doubled, because v if squared KE 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 total amount of energy is conserved

Total energy includes potential energy, kinetic energy and internal energy

Energy within a system can be transferred among different types of energy but it can’t be destroyed

Conservation of Energy in a Perfect Physics World

In a perfect physics world since there is no friction there will be no change in internal energy so you don’t have to take that into account

In a perfect physics world energy will transfer between PE and KE

In the “Real” World

In the real world there is friction so the internal energy of an object will be affected by the friction (such as air resistance)

Conservation of Energy Symbols

ET=total energy of a system

PE=potential energy KE=kinetic energy Q=internal energy ***all units are Joules (J)

Conservation of Energy Equations

In a real world situation, ET=KE+PE+Q because friction exists and may cause an increase in the internal energy of an object

In a “perfect physics world” ET=KE+PE with KE+PE equal to the total “mechanical energy of the system object

Conservation of Energy Examples (perfect physics world)

position #1

position #2

position #3 more..

Conservation of Energy Examples (cont.)

Conservation of Energy Examples (cont.)

Position #1

Position #2

Position #3

Conservation of Energy (perfect physics world)

Position #1 the ball/bob has not starting falling yet so the total energy is all in gravitational potential energy

Position #2 the ball/bob is halfway down, so total energy is split evenly between PE and KE

Position #3 the ball/bob is at the end of its fall so total energy is all in KE

Conservation of Energy Problems

1.) A 2.0kg block starts at rest and then slides along a frictionless track. What is the speed of the block at point B?

A

7.0m B

Answer

Since there is no friction, Q does not need to be included

So…use ET=PE+KE At position B, the total energy

is entirely KE Since you cannot find KE

directly, instead find PE at the beginning of the slide and that will be equal to KE at the end of the slide more…..

Answer (cont.)

PE (at position A) =mgΔh=2.0kgx9.81m/s2x7.0m =137.3J

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

So ET (at position A)=137.3J At position B there is no

height so the PE is 0JMore….

Answer (cont.)

At position B the total energy still has to be 137.3J because energy is conserved and because there is no friction no energy was “lost” along the slide

So….ET(position B)=137.3J=0J+KE So…KE also equals 137.3J at

position B More…

Answer (cont.)

Use KE=1/2mv2

So…KE=137.3J=1/2x2.0kgxv2

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

More Conservation of Energy Problems

position #1

position #2

position #3

1.) From what height must you drop the 0.5kg ball so that the it will be traveling at 25m/s at position #3(bottom of the fall)?

2.)How fast will it be traveling at position #2 (halfway down)?

*Assume no friction

Answers

1.) At position #3, total energy will be all in KE because there is no height and no friction

So…use ET=KE=1/2mv2

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

So…KE=156.25J=PE (at position#1)

So…ΔPE=156.25J=mgΔh And Δh=31.86m

Answers (cont.)

2.) Since position #2 is half way down total energy will be half in PE and half in KE

So…KE at position #2 will be half that at position #3

So…KE at position #2 is 78.125J

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”

Energy is stored in a spring when the spring is stretched or compressed

The work done to compress or stretch a spring becomes its elastic potential energy

Spring Symbols

Fs=force applied to stretch or compress the spring in newtons (N)

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

x=the change in length in the spring from the equilibrium position in meters (m)

Spring Equations

Fs=kx

PEs=1/2kx2

Spring Diagrams

Spring Problems

1.) What is the potential energy stored in a spring that stretches 0.25m from equilibrium when a 2kg mass is hung from it?

2.) 100J of energy are stored when a spring is compressed 0.1m from equilibrium. What force was needed to compress the spring?

Answers

1.) Using PEs=1/2kx2 you know “x” but not “k”

You can find “k” using Fs=kx With Fs equal to the weight of

the 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=1/2kx2

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

So 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

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

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

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

4.) Electromagnetic Energy is the energy associated with electric and magnetic fields

IV. Electricity and Magnetism

A. Electrostatics/Electric Fields

B. Current Electricity C. Series Circuits D. Parallel Circuits E. Electric Power and

Energy F. Magnetism and

Electromagnetism

A. Electrostatics

Atomic Structure—the atom consists of proton(s) and neutron(s) in the nucleus and electrons outside the nucleus.

The proton and neutron have similar mass (listed in reference table)

The electron has very little mass (also in reference table)

A proton has a positive charge that is equal in magnitude but opposite in sign to the electron’s negative charge

A neutron has no charge so it is neutral

The unit of charge is the coulomb (C)

Each proton or each electron has an “elementary charge” (e) of 1.60x10-19C

The magnitude of the charge on both an electron and a proton are the same, only the signs are different

An object has a neutral charge or no net charge if there are equal numbers of protons and neutrons

An object will have a net “negative” charge if there are more electrons than protons

An object will have a net “positive” charge if there are less electrons than protons

Transfer of charge occurs only through movement of electrons

If an object loses electrons, it will have a net positive charge

If an object gains electrons, it will have a net negative charge

When the 2 spheres touch

Next….

And then are pulled apart. This is what happens

On the previous page, only charge is transferred

Total charge stays the same

So charge is always conserved

There is “CONSERVATION OF CHARGE” just like with energy and momentum

Transfer of Charge Problem

1.) Sphere #1 touches sphere #2 and then is separated. Then sphere #2 touches #3 and is separated. What are the final charges on each sphere?

Answer

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 repulsive

k=electrostatic constant 8.99x109N.m2/C2

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

meters (m) E=electric field strength in

(N/C)

Electrostatic Equations

221

r

qkqFe

q

FE e

Electric Fields

An electric field is an area around a charged particle in which electric force can be detected

Electric fields are detected and mapped using “positive” test charges

Field lines are the imaginary lines along which a positive test charge would move (arrows show the direction)

Electric Field Line Diagrams

Negative charge Positive charge

Parallel Plates +

-

Electric Field Line Diagrams

Negative and positive charges

Field lines

Electric Field Line Diagrams

Two positive charges

Field lines

Electric Field Line Diagrams

Two negative charges

Field lines

Electrostatics Problems

1.) What is the magnitude of the electric field strength when an electron experiences a 5.0N force?

2.) Is a charge of 4.8x10-

19C? possible? A charge of 5.0x10-19C? 3.) What is the electrostatic

force between two 5.0C charges that are 1.0x10-4m apart?

Answers

1.) Use

CxC

Cxq

N

1919-

19

e

e

101.31.60x10

5.0Nso...E

electron oneon charge

elementary is that because 1060.1

0.5F

withq

FE

Answers (cont.)

2.) Only whole number multiples of the elementary charge are possible. To find out if charge is possible, divide by 1.60x10-19C.

4.8x10-19C is possible because when it is divided you get 3, which is a whole number.

5.0x10-19C is not possible because when it is divided you get 3.125 which is not a whole number.

Answers (cont.)

3.) Use

CxF

mx

CCCmmNx

r

qkq

e19

24

229

e

221

e

1025.2

)100.1(

)5)(5(/.1099.8F

F

B. Current Electricity

Current is the rate at which charge passes through a closed pathway (a circuit)

The unit of current is ampere (A)

Conditions needed for a Circuit

Must have a “potential difference” supplied by a battery or power source

Must have a “pathway” supplied by wires

Can have a resistor(s) Can have meters

(ammeters, voltmeters, ohmmeters)

Current Electricity Quantities

Current is the flow of charge past a point in a circuit in a unit of time. Measured in amperes (A)

Potential Difference is the work done to move a charge between two points. Measured in volts (V)

Resistance is the opposition to the flow of current. Measured in ohms (Ω)

Current Electricity Symbols

I=current in amperes (A) Δq=charge in coulombs (C) t=time in seconds (s) V=potential difference in volts

(V) W=work in Joules (J) R=resistance in ohms (Ω) ρ=resistivity in (Ω.m) L=length of wire in meters (m) A=cross-sectional area of wire

in square meters (m2)

Current Electricity Equations

A

LR

I

VR

q

WV

t

qI

Conductivity

A material is a “good” conductor if electric current flows easily through it

Metals are good conductors because they have many electrons available and they are not tightly bound to the atom

If a material is an extremely poor conductor, it is called an “insulator

Resistivity

Resistivity is an property inherent to a material (and its temperature) that is directly proportional to the resistance of the material

Use the “short, thick, cold” rule to remember what affects resistance

Short, thick, cold wires have the lowest resistance (best conduction)

Current Electricity Problems

1.) How many electrons pass a point in a wire in 2.0s if the wire carries a current of 2.5A?

2.) A 10 ohm resistor has 20C of charge passing through it in 5s. What it the potential difference across the resistor?

Answers

1.)

electrons 3.13x10 electrons ofnumber ....

1.60x101e use tablesreference thefrom

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

2.5A witht

qI Use

19

19-

so

C

s

q

Answers (cont.)

2.)

V40....

20C/5s

V10....

/

VRget to

togethert

qI and

VR Use

Vso

so

tq

I

Resistivity Problem

3.) What is the resistance of a 5.0m long aluminum wire with a cross-sectional area of 2.0x10-6m2 at 200C?

Answer

3.)

07.0...

100.2

)0.5)((2.82x10so....R

tablesreference from ""h witR Use

26

8-

Rso

mx

mm

A

L

C. Series Circuits

Series circuit is a circuit with a single pathway for the current

Current stays the same throughout the circuit

Resistors share the potential difference from the battery (not necessarily equally)

The sum of the resistances of all resistors is equal to the equivalent resistance of the entire circuit

Series Circuit Equations

same. thebe uldcircuit wo theandresistor

"equivalent" single ath circuit wi ain resistors the

replace couldyou that means Resistance Equivalent

.....

......

.....

321

321

321

RRRR

VVVV

IIII

eq

Meters

Ammeter measures the current in a circuit

Voltmeter measures the potential difference across a resistor or battery

Ohmmeter measures the resistance of a resistor

Circuit Diagram Symbols

Cell

Battery

Switch

Voltmeter ammeter

More Circuit Diagram Symbols

Resistor

Variable Resistor

Lamp

Series Circuit Diagram Example

Series Circuit Problems

1.) Two resistors with resistances of 4Ω and 6Ω are connected in series to a 25V battery. Determine the potential drop through each resistor, the current through each resistor and the equivalent resistance.

Answers

1.) You can use R=V/I as soon as you have 2 pieces of information at each resistor. You can also use the series circuit laws.

First, use Req=R1+R2 to find Req

4Ω+6Ω=10Ω=Req (use this as the R at the battery)

More…

Answers (cont.)

Then use R=V/I at the battery to find I at the battery

So…10Ω=25V/I and I=2.5A and all currents are equal in a series circuit, so I1and I2=2.5A

More….

Answers (cont.)

Now that you have 2 pieces of information at each resistor, you can use R=V/I to find potential differences

At the 4Ω resistor, 4Ω=V/2.5A

so V1=10V At the 6Ω resistor,

6Ω=V/2.5A So…V2=15V

D. Parallel Circuits

Parallel circuits have more than one pathway that the current can pass through

Current is shared (not necessarily equally) among the pathways

Voltage is the same in each pathway as across the battery

Equivalent resistance is the sum of the reciprocal resistances of the pathways (Req is always less than R of any individual pathway)

Parallel Circuit Equations

.....1111

....

.....

321

321

321

RRRR

VVVV

IIII

eq

same. thebe uldcircuit wo theandresistor

"equivalent" single ath circuit wi ain resistors the

replace couldyou that means Resistance Equivalent

Parallel Circuit Diagram Example

Parallel Circuit Problems

1.) Three resistors of 2Ω, 4Ω and 4Ω are connected in parallel to an applied potential difference of 20V. Determine equivalent resistance for the circuit, the potential difference across each resistor and the current through each resistor.

Answers

1.)

So equivalent resistance=1Ω

More…..

14

4

4

1

4

1

2

11

1111 Use

321

eq

eq

R

RRRR

Answers (cont.)

Use V=V1=V2=V3

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

More……

Answers (cont.)

Using R=V/I, substitute the potential differences and resistances for each resistor to find the current for each resistor

I=20A I1=10A I2=5A I3=5A

E. Electric Power and Energy

Electric power is the product of potential difference and current measured in watts just like mechanical power

Electric energy is the product of power and time measured in joules just like other forms of energy

Electric Power and Energy Equations

R

tVRtIVItPtW

R

VRIVIP

22

22

Electric Energy and Power Problems

1.) How much time does it take a 60W light bulb to dissipate 100J of energy?

2.) What is the power of an electric mixer while operating at 120V, if it has a resistance of 10Ω?

3.) A washing machine operates at 220V for 10 minutes, consuming 3.0x106J of energy. How much current does it draw during this time?

Answers

1.) Use W=Pt So…100J=60Wxt So…t=0.6s

2.) Use

So…P=1440W

10

(120V)so...P

R

VP

22

Answers (cont.)

3.) Use W=VIt but first need to change the 10 minutes to 600 seconds.

So…3.0x106J=220V(I)600s So…I=22.7A

F. Magnetism and Electromagnetism

Magnetism is a force of attraction or repulsion occurring when spinning electrons align

A magnet has two ends called “poles”

The “N” pole is “north seeking” The “S” pole is “south seeking” Like “poles” repel Unlike “poles” attract

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

The earth has an “S” pole at the North Pole

The earth has a “N” pole at the South Pole

The “N” pole of a compass will point towards earth’s “S” pole which is the geographic North pole

There is a “field” around each magnet

Imaginary magnetic field or flux lines show where a magnetic field is

You can use a compass to map a magnetic field

Units for Magnetic Field

The weber (Wb) is the unit for measuring the number of field lines

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

1T=1Wb/m2

Magnetic Field Diagrams

Magnetic Field Diagrams (cont.)

Magnetic Field Diagrams (cont.)

Electromagnetism

Moving a conductor through a magnetic field will induce a potential difference which may cause a current to flow (conductor must “break” the field lines for this to occur)

A wire with a current flowing through it creates a magnetic field

So…magnetism creates electricity and electricity creates magnetism

Electromagnetism Diagram

Must move wire into and out of page to induce potential current by breaking field lines

v. Waves

A.) Wave Characteristics B.) Periodic Wave

Phenomena C.) Light D.) Reflection and

Refraction E.) Electromagnetic

Spectrum

A. Wave Characteristics

A wave is a vibration in a “medium” or in a “field”

Sound waves must travel through a medium (material)

Light waves may travel in either a medium or in an electromagnetic field

Waves transfer energy only, not matter

A pulse is a single disturbance or vibration

A periodic wave is a series of regular vibrations

Types of Waves

1.) Longitudinal Waves are waves that vibrate parallel to the direction of energy transfer. Sound and earthquake p waves are examples.

Types of Waves

2.) Transverse waves are waves that vibrate perpendicularly to the direction of energy transfer. Light waves and other electromagnetic waves are examples.

Frequency

Frequency is how many wave cycles per second

The symbol for frequency is f

The unit for frequency is hertz (Hz)

1Hz=1/s Frequency is pitch in sound Frequency is color in visible

light

Frequency (cont.)

High frequency wave

Low frequency wave

Period

Period is the time required for one wave cycle

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

second (s) The equation for period is

T=1/f Period and frequency are

inversely proportional

Period (cont.)

Short period wave

Long period wave

Amplitude

The amplitude of a wave is the amount of displacement from the equilibrium line for the wave (how far crest or trough is from the equilibrium line)

Symbol for amplitude is A Unit for amplitude is meter (m) Amplitude is loudness in sound Amplitude is brightness in light

Amplitude (cont.)

High amplitude wave

Low amplitude wave

Wavelength

Wavelength is the distance between points of a complete wave cycle

Symbol for wavelength is λ Unit for wavelength is

meter (m) A wave with a high

frequency will have short wavelengths and short period

Wavelength (cont.)

Short wavelength λ

λ Long wavelength

Phase

Points that are at the same type of position on a wave cycle (including same direction from equilibrium and moving in same direction) are “in phase”

Phase (cont.)

Points that are in phase are whole wavelengths apart (i.e., 1λ apart, 2λ apart, 3λ apart, etc.)

Points that are in phase are multiples of 360 degrees apart (i.e., 360 degrees apart, 720 degrees apart, 1080 degrees apart, etc.)

Phase (cont.)

A B C

D E F

Points A, B and C are in phasePoints D, E and F are in phasePoints A and B are 360 degrees

apart and 1λ apartPoints D and F are 720 degrees

apart and 2λ apart

Phase (cont.)

A C

D

BPoints A and B are 180 degrees

apart and ½ λ apart (not in phase)Points C and D are 90 degrees apart

and ¼ λ apart (not in phase)

Speed

Speed of a wave is equal to the product of wavelength and frequency

Symbol for speed is v Unit for speed is m/s Equation for speed is v=fλ Can also use basic speed

equation v=d/t if needed

Speed (cont.)

The speed of a wave depends on its “type” and the medium it is traveling through

The speed of light (and all electromagnetic waves) in a vacuum and in air is 3.00x108m/s

The speed of sound in air at STP (standard temperature and pressure) is 331m/s

Wave Characteristics Problems

1.) A wave has a speed of 7.5m/s and a period of 0.5s. What is the wavelength of the wave?

2.) If the period of a wave is doubled, what happens to its frequency?

3.) Points on a wave are 0.5λ apart, 7200 apart, 4λ apart. Which points are in phase?

Answers

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

So…0.5s=1/f and f=2.0Hz Then…7.5m/s=2.0Hz(λ)

and λ=3.75m

2.) T and f are inversely related so if T is doubled f is halved.

Answers (cont.)

3.) Points are in phase only if they are whole wavelengths apart or multiples of 360 degrees apart. So..the points that are 720 degrees apart are in phase and the points that are 4λ apart are in phase. The points that are 0.5λ apart are not in phase.

More problems

5.0m

4.) What is the wavelength of the wave shown above?

Answer

4.) The wave is 5.0m long and contains 2.5 cycles. You want to know the length of 1 cycle.

So…5.0m/2.5 cycles=2.0m/cycle

So…λ=2.0m

B. Periodic Wave Phenomena

When waves interact with one another many different “phenomena” result

Those phenomena are the doppler effect, interference, standing waves, resonance and diffraction

Circular Waves

Some phenomena are easier to explain using circular waves

A is the λ from crest to crest is the wavefront (all points on the circle are in A A phase)

Doppler Effect

If the source of waves is moving relative to an observer, the observed frequency of the wave will change (actual f produced by the source doesn’t change and movement must be fast for observer to notice)

Doppler Effect (cont.)

If source is moving towards the observer, f will increase (higher pitch if sound wave, shift towards blue if light wave)

If source is moving away from the observer, f will decrease (lower pitch if sound wave, shift towards red if light wave)

Doppler Effect (cont.)

Joe Shmoe

more….

Doppler Effect (cont.)

In the diagram on the previous page, the wave source is moving towards Joe

In the diagram on the previous page, the wave source is moving away from Shmoe

Doppler Effect (cont.)

The effect on Joe is that the f of the wave will be higher for him (shorter λ and T, higher pitch or bluer light)

The effect on Shmoe is that the f of the wave will be lower for him (longer λ and T, lower pitch or redder light)

Interference

When waves travel in the same region and in the same plane at the same time (superposition) they can interfere with each other

In “constructive” interference, waves build on one another to increase amplitude

In “destructive” interference, waves destroy on another to decrease amplitude

Interference (cont.)

Maximum constructive interference occurs when waves are in phase

Maximum destructive interence occurs when waves are 180 degrees out of phase

Maximum Constructive Interference

Above results in

Maximum Destructive Interference

Above results in

Standing Waves

Standing waves occur when waves having same A and same f travel in opposite directions. Vibrates so that it looks like waves are stationary

Resonance

A body with an ability to vibrate has a natural frequency at which it will vibrate

If you put an object that is vibrating with that same natural frequency next to that body, the body will also start vibrating (don’t need to touch)

This is called “resonance”

Resonance (cont.)

Examples of resonance: Find the “Tacoma Narrows

Bridge” video on the internet

A singer being able to break a glass with her voice

Diffraction

When waves bend behind a barrier instead of going straight through that is called diffraction

Wave fronts

C. Light

Light is the part of the electromagnetic spectrum that is 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=fλ, when it’s light you can substitute c=fλ

Light (cont.)

No object can travel faster than the speed of light (one of Einstein’s ideas)

The speed of light in a material is always less than c

Light Problems

1.) Determine the wavelength in a vacuum of a light wave having a frequency of 6.4x1014Hz.

2.) What is the frequency of a light wave with a wavelength of 5.6x10-7m?

Answers

1.) Use c=fλ 3.00x108m/s=6.4x1014Hz(λ) So… λ=4.7x10-7m

2.) Use c=fλ 3.00x108m/s=f(5.6x10-7m) So…f=5.4x1014Hz

D. Reflection and Refraction

Reflection occurs when a ray of light hits a boundary and bounces back into the same medium

Refraction occurs when a ray of light enters a new medium and changes direction because of a change in speed

More Reflection

The law of reflection

θi θr

Dotted line is the “normal”(reference line)

reflection of angle

incidence of angle

r

i

ri

More Reflection

In a mirror the image of an object is flipped laterally (your right hand is your left hand in a mirror image)

To view an object in a plane mirror, you need a minimum of ½ the height of the object for the height of the mirror

More Refraction

When light enters a different medium, the change in direction depends on the density of the new medium

If new medium is denser, the light will slow down and bend towards the normal

If new medium is less dense, the light will speed up and bend away from the normal

More Refraction

Refraction Equations

2

1

2

1

1

2

2211

Refraction ofIndex

sinsin

Law sSnell'

v

v

n

n

Other

v

cn

nn

More Refraction

Symbols n=index of refraction (no units) v=speed of the wave in m/s λ=wavelength in m c=speed of light in a vacuum

(3.00x108m/s) θ1=angle of incidence θ2=angle of refraction

Refraction Diagram

θ1

air

glass θ2

Light travels from air into glass which is more dense so it slows down and bends towards the normal

Refraction Diagram

θ1

air

glass θ2

Light travels from glass into air which is less dense so it speeds up and bends away from the normal

Absolute Index of Refraction

Absolute index of refraction is the ratio of the speed of light in 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 less dense the medium

Absolute Index of Refraction (cont.)

Higher “n” means slower v of light in the medium

Lower “n” means higher v of light in the medium

Use reference tables to find “n” numbers

Problems

1.) What is the speed of light in diamond?

2.) What is the ratio of the speed of light in corn oil to the speed of light in water?

Problems (cont.)

3.) A ray of monochromatic light having a frequency of 5.09x1014Hz is traveling through water. The ray is incident on corn oil at an angle of 600 to the normal. What is the angle of refraction in corn oil?

Answers

1.)

smxvso

v

sm

/1024.1....

/3.00x10so....2.42

2.42 diamond ofn with v

cn Use

8

8

Answers (cont.)

2.)

2

1

1

2

1

2

2

1

1.47

1.33...

1.33 is water ofn 1.47, is oilcorn ofn

v

Use

v

v

n

nso

n

n

v

Answers (cont.)

3.) Use Snell’s Law

02

20

2211

52...

sin47.1sin60so....1.33

second theoilcorn medium,first theis water The

sinsin

so

nn

E. Electromagnetic Spectrum

Electromagnetic waves include gamma rays, x rays, ultraviolet rays, infrared rays, microwaves, t.v. and radio waves, long waves

Electromagnetic waves are produced by accelerating charges (produce alternating electric and magnetic fields)

Electromagnetic Spectrum (cont.)

High energy electromagnetic waves have high f and small λ

Low energy electromagnetic waves have low f and long λ

Energy in a wave can be increased by increasing f, decreasing λ and T and increasing the duration of the wave

Electromagnetic Spectrum (cont.)

In the visible light spectrum, the violet end is high f, short λ and high energy

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

VI. Modern Physics

A. Quantum Theory B. Atomic Models C. Nucleus D. Standard Model E. Mass-Energy

A. Quantum Theory

E-m energy is absorbed and emitted in specific amounts called “quanta” according to the quantum theory

A photon is the basic unit of e-m energy (particle of light) In wave theory E-m energy

is emitted continuously Energy of a photon is

directly related to its frequency

Quantum Theory (cont.)

Light acts like a wave during diffraction, interference and the doppler effect

Light acts like a particle (quantum idea) during the photoelectric effect and atomic spectra

Quantum Theory (cont.)

Equation

Ephoton=energy of a photon in joules (J)

h=Planck’s constant (6.63x10-34J.s) f=frequency in hertz (Hz) λ=wavelength in meters (m) c=speed of light in vacuum

3.00x108m/s

hc

hfEphoton

Quantum Theory (cont.)

Photoelectric Effect—Einstein used the particle idea of light (quantum theory) to explain why light incident on a photosensitive metal would only cause electrons to be emitted if its frequency was high enough

Quantum Theory (cont.)

When light acts like a particle, it has momentum just like a particle

Compton Effect—when an e-m photon hits an electron, the photon transfers some of its energy and momentum to the electron

Particles can also act like waves by having a wavelength (usually too small to detect)

B. Atomic Models

Thomson Model of the atom introduced the idea of an atom containing both positive and negative charges (balancing each other because atom is neutral)

negative charge positive charge

Atomic Models (cont.)

Rutherford Model of the atom introduced “nucleus” idea with small positive core at center surrounded by orbiting negative electrons

positive charge negative charge

Atomic Models (cont.)

Bohr Model of the atom kept Thomson and Rutherford ideas but added “energy levels” to explain why electrons don’t crash into the nucleus

Positive chargeNegative charge

Atomic Models (cont.)

Bohr Model Ideas Electrons can only gain or

lose energy in specific amounts (quanta)

Electrons can occupy only certain orbits that have fixed radii

Orbits closer to the nucleus have lower energy, those farther away, higher energy

Atomic Models (cont.)

Bohr Model Ideas (cont.) Electrons can jump up to

higher orbits by “absorbing” quanta of energy

Electrons will emit quanta of energy when they fall down to lower orbits

The orbit closest to the nucleus is called “ground state”

Atomic Models (cont.)

Bohr Model Ideas (cont.) Any orbit (level) above

ground state is called an “excited state”

An atom is “ionized” when an electron has been removed

“The ionization potential” is the amount of energy needed to remove an electron from the atom

Use energy level diagrams in the reference tables

Atomic Models (cont.)

Cloud Model was introduced because the Bohr model couldn’t explain atoms with many electrons

Cloud model has same basic setup, nucleus and electrons, but there are no fixed orbits

Electrons are spread out in electron clouds instead

Atomic Spectra

Each element has a specific spectrum (lines of specific frequencies of e-m energy that are either emitted or absorbed)

Atomic Spectra (cont.)

Emission Spectra (Bright-line) occur when electrons fall to lower energy levels and “emit” quanta of energy

Absorption Spectra (Dark line) occur when electrons absorb energy in order to jump up energy levels leaving behind dark lines

C. Nucleus

The nucleus is the core of the atom made up of protons and neutrons

The nucleus contains almost all of the mass of the atom

The nucleus is positively charged because of the protons

Nucleus (cont.)

Each proton has a charge of +1e which is equal to +1.60x10-19C

Each neutron has a charge of 0e so it is neutral

The mass of a proton or a neutron is 1.67x10-27kg

Nucleus (cont.)

The nuclear force (strong force) is the strongest force known, that is why nuclear reactions (fission and fusion) create such large amounts of energy

D. Standard Model

The particles of an atom; the electrons, neutrons and protons, are not the whole story

An electron is a fundamental particle, but a neutron and proton are not

A fundamental particle can’t be broken down into smaller particles (at least we think so)

Standard Model (cont.)

There are 3 categories of fundamental particles; hadrons, leptons and force carrier particles

Hadrons can interact with all 4 fundamental forces of nature. Protons and neutrons are hadrons

Leptons can interact with all the forces except the strong nuclear force. Electrons are leptons

Standard Model (cont.)

matter

hadrons leptons

Standard Model (cont.)

A Baryon is a particle that can be changed into a proton or a neutron

A Meson is a particle of intermediate mass

Standard Model (cont.)

hadrons

Baryons Mesons

Standard Model (cont.)

A quark is a particle that makes up baryons and mesons

Quarks have partial charges, +1/3e, -1/3e, +2/3e or -2/3e

See reference tables for lists of quarks

Each particle has an antiparticle having all same characteristics except opposite charge and magnetic moment

Standard Model (cont.)

Baryon

quark quark quark

Standard Model (cont.)

Meson

quark antiquark

Four Fundamental Forces of Nature

1.) Strong nuclear force is by far the strongest, very close ranged (nuclear distances)

2.) Electromagnetic force is the next strongest and closed ranged

3.) Weak nuclear force is the third strongest, very close ranged

4.) Gravitational force is weakest and very long ranged

E. Mass-Energy

According to Albert Einstein, mass and energy are different manifestations of the same thing E=mc2

E=energy in joules (J) m=mass in kilograms (kg) c=3.00x108m/s

Mass-Energy (cont.)

The mass-energy relationship can also be seen in the following relationship

1u=931Mev u=universal mass unit Mev=megaelectronvolt

(106eV)

****eV is an energy unit for very small amounts of energy

1eV also equals 1.60x10-19J

Mass-Energy (cont.)

Just like energy has to be conserved, so does mass energy

Mass can be converted to energy and energy can be converted to mass

But mass cannot be created or destroyed

Modern Physics Problems

1.) How much energy would be produced if a 1.0kg mass was completely converted to energy?

2.) Determine the energy of a photon with a frequency of 3.5x1014Hz.

Modern Physics Problems (cont.)

3.) The rest mass of a neutron is 1.0087u. Determine the energy equivalent in megaelectronvolts.

4.) What type of e-m energy will be either absorbed or emitted when an electron falls from the n=3 to the n=1 level in a hydrogen atom?

Answers

1.) Use E=mc2

E=1.0kg(3.0x108m/s)2

E=9x1016J

2.) Use Ephoton=hf

Ephoton=6.63x10-

34J.s(3.5x1015Hz) Ephoton=2.3x10-18J

Answers (cont.)

3.) Use 1u=931MeV So…1.0087u x

931MeV/1u=939MeV for the energy

equivalent

Answers (cont.)

4.) First subtract energy levels so…13.6eV-1.51eV=12.09eV

Then change to joules using 1eV=1.60x10-19J

So…12.09eV=1.93x10-18J Then use E=hf1.93x10-18J=6.63x10-34J.s(f)So….f=2.92x1015Hz which is

ultraviolet light(find that in reference table)

The End

You made it!!!!!