Post on 07-Apr-2018
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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
Quantities Horizontal Vertical
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.)
Quantities Horizontal Vertical
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 ( )
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.)
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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)
Uniform Circular Motion
Diagram
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Uniform Circular Motion
Equations
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r
mv
r
v
2
c
2
ca
!
!
Uniform Circular Motion
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
Coefficient of Friction
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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
Coefficient of Friction
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
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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
Momentum and Impulse
Equations
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p=mv
J=Ft
J=p
pbefore=pafter
Momentum and Impulse
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)
Conservation of Energy
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..
Conservation of Energy
Examples (cont.)
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Conservation of Energy
Examples (cont.)
Position #1
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Position #2
Position #3
Conservation of Energy
(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
Conservation of Energy
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
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Field lines
Electrostatics Problems
1.) What is the magnitude ofthe electric field strength when
an electron experiences a 5.0Nforce?
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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!!
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Resistivity
Resistivity is an propertyinherent to a material (and its
temperature) that is directlyproportional to the resistanceof the material
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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?
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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
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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
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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
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Answers
1.)
;!;
!;
;
;
!
!
14
4
4
1
4
1
2
11
1111se
321
eq
eq
R
RRRR
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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
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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)
Doppler Effect (cont.)
Joe Shmoe
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more.
Doppler Effect (cont.)
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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Doppler Effect (cont.)
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