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Honors Physics Review Notes
2008–2009
Tom StrongScience Department
Mt Lebanon High Schooltomstrong@gmail.com
The most recent version of this can be found at http://www.tomstrong.org/physics/
Chapter 1 — The Science of Physics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1Chapter 2 — Motion in One Dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3Chapter 3 — Two-Dimensional Motion and Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4Chapter 4 — Forces and the Laws of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5Chapter 5 — Work and Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7Chapter 6 — Momentum and Collisions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8Chapter 7 — Rotational Motion and the Law of Gravity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9Chapter 8 — Rotational Equilibrium and Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10Chapter 12 — Vibrations and Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12Chapter 13 — Sound . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14Chapter 14 — Light and Reflection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16Chapter 15 — Refraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20Chapter 16 — Interference and Diffraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22Chapter 17 — Electric Forces and Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24Chapter 18 — Electrical Energy and Capacitance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26Chapter 19 — Current and Resistance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27Chapter 20 — Circuits and Circuit Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29Chapter 21 — Magnetism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31Chapter 22 — Induction and Alternating Current . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32Variables and Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33Mathematics Review for Physics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36Physics Using Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41Data Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43Final Exam Description . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44Final Exam Equation Sheet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45Mixed Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47Mixed Review Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65
ii
These notes are meant to be a summary of important points covered in the Honors Physics class at Mt. Lebanon HighSchool. They are not meant to be a replacement for your own notes that you take in class, nor are they a replacement foryour textbook. Much of the material in here is taken from the textbook without specifically acknowledging each case, inparticular the organization and overall structure exactly match the 2002 edition of Holt Physics by Serway and Faughnand many of the expressions of the ideas come from there as well.
The mixed review exercises were taken from the supplementary materials provided with the textbook. They are arepresentative sampling of the type of mathematical problems you may see on the final exam for the course. There willalso be conceptual questions on the exam that may not be covered by the exercises included here. These exercises areprovided to help you to review material that has not been seen in some time, they are not meant to be your only resourcefor studying. Exercises are included from chapters 1–8, 12–17, and 19–22.
The answers at the end of the review are taken from the textbook, often without verifying that they are correct. Usethem to help you to solve the problems but do not accept them as correct without verifying them yourself.
This is a work in progress and will be changing and expanding over time. I have attempted to verify the correctnessof the information presented here, but the final responsibility there is yours. Before relying on the information in thesenotes please verify it against other sources.
Honors Physics 2008–2009 Mr. Strong
1
Chapter 1 — The Science of Physics
1.1 What is Physics?Some major areas of Physics:Mechanics — motion and its causes — falling objects,
friction, weightThermodynamics — heat and temperature — melting
and freezing processes, engines, refrigeratorsVibrations and Waves — specific types of repeating
motions — springs, pendulums, soundOptics — light — mirrors, lenses, colorElectromagnetism — electricity, magnetism, and light
— electrical charge, circuitry, magnetsRelativity — particles moving at very high speeds — par-
ticle accelerators, particle collisions, nuclear energyQuantum Mechanics — behavior of sub-microscopic
particles — the atom and its parts
The steps of the Scientific Method1. Make observations and collect data that lead to a
question2. Formulate and objectively test hypotheses by experi-
ments (sometimes listed as 2 steps)3. Interpret results and revise the hypotheses if neces-
sary4. State conclusions in a form that can be evaluated by
others
1.2 Measurements in ExperimentsMeasurementsThere are 7 basic dimensions in SI (Systeme International),the 3 we will use most often are:
• Length — meter (m) — was 1/10,000,000 of the dis-tance from the equator to the North Pole — now thedistance traveled by light in 3.3× 10−9 s• Mass — kilogram (kg) — was the mass of 0.001 cubic
meters of water, now the mass of a specific platinum-iridium cylinder
• Time — second (s) — was a fraction of a mean so-lar day, now 9,162,631,700 times the period of a radiowave emitted by a Cesium-133 atom
Common SI Prefixes
Prefix Multiple Abbrev.
nano- 10−9 n deci- 10−1 dmicro- 10−6 µ kilo- 103 kmilli- 10−3 m mega- 106 Mcenti- 10−2 c giga- 109 G
Accuracy vs. Precision
• Accuracy describes how close a measured value is tothe true value of the quantity being measured
Problems with accuracy are due to error. To avoiderror:
– Take repeated measurements to be certain thatthey are consistent (avoid human error)
– Take each measurement in the same way (avoidmethod error)
– Be sure to use measuring equipment in goodworking order (avoid instrument error)
• Precision refers to the degree of exactness with whicha measurement is made and stated.
– 1.325 m is more precise than 1.3 m
– lack of precision is usually a result of the limi-tations of the measuring instrument, not humanerror or lack of calibration
– You can estimate where divisions would fall be-tween the marked divisions to increase the pre-cision of the measurement
1.3 The Language of Physics
There are many symbols that will be used in this class, someof the more common will be:
Symbol Meaning
∆x Change in xxi, xf Initial, final values of x∑
F Sum of all F
Dimensional analysis provides a way of checking tosee if an equation has been set up correctly. If the units re-sulting from the calculation are not those that are expectedthen it’s very unlikely that the numbers will be correct ei-ther. Order of magnitude estimates provide a quickway to evaluate the appropriateness of an answer — if theestimate doesn’t match the answer then there’s an errorsomewhere.
Honors Physics 2008–2009 Mr. Strong
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Counting Significant Figures in a Number
Rule Example
All counted numbers have an infinite numberof significant figures
10 items, 3 measurements
All mathematical constants have an infinitenumber of significant figures
1/2, π, e
All nonzero digits are significant 42 has two significant figures; 5.236 has four
Always count zeros between nonzero digits 20.08 has four significant figures; 0.00100409 has six
Never count leading zeros 042 and 0.042 both have two significant figures
Only count trailing zeros if the number con-tains a decimal point
4200 and 420000 both have two significant figures; 420.has three; 420.00 has five
For numbers in scientific notation apply theabove rules to the mantissa (ignore the expo-nent)
4.2010× 1028 has five significant figures
Counting Significant Figures in a Calculation
Rule Example
When adding or subtracting numbers, find the number which is knownto the fewest decimal places, then round the result to that decimalplace.
21.398 + 405 − 2.9 = 423 (3significant figures, roundedto the ones position)
When multiplying or dividing numbers, find the number with the fewestsignificant figures, then round the result to that many significant fig-ures.
0.049623 × 32.0/478.8 =0.00332 (3 significant figures)
When raising a number to some power count the number’s significantfigures, then round the result to that many significant figures.
5.82 = 34 (2 significantfigures)
Mathematical constants do not influence the precision of any compu-tation.
2× π × 4.00 = 25.1 (3 signif-icant figures)
In order to avoid introducing errors during multi-step calculations, keepextra significant figures for intermediate results then round properlywhen you reach the final result.
Rules for Rounding
Rule Example
If the hundredths digit is 0 through 4 drop it and all following digits. 1.334 becomes 1.3
If the hundredths digit is 6 though 9 round the tenths digit up to thenext higher value.
1.374 becomes 1.4
If the hundredths digit is a 5 followed by other non-zero digits thenround the tenths digit up to the next higher value.
1.351 becomes 1.4
If the hundredths digit is a 5 not followed by any non-zero digits thenif the tenths digit is even round down, if it is odd then round up.
1.350 becomes 1.4, 1.250 be-comes 1.2
(assume that the result was to be rounded to the nearest 0.1, for other precisions adjust accordingly)
Honors Physics 2008–2009 Mr. Strong
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Chapter 2 — Motion in One Dimension
2.1 Displacement and Velocity
The displacement of an object is the straight line (vector)drawn from the object’s initial position to its new position.Displacement is independent of the path taken and is notnecessarily the same as the distance traveled. Mathemati-cally, displacement is:
∆x = xf − xi
The average velocity, equal to the constant velocitynecessary to cover the given displacement in a certain timeinterval, is the displacement divided by the time intervalduring which the displacement occurred, measured in m
s
vavg =∆x∆t
=xf − xi
tf − ti
The instantaneous velocity of an object is equivalentto the slope of a tangent line to a graph of x vs. t at thetime of interest.
The area under a graph of instantaneous velocity vs.time (v vs. t) is the displacement of the object during thattime interval.
2.2 Acceleration
The average acceleration of an object is the rate ofchange of its velocity, measured in m
s2 . Mathematically, itis:
aavg =∆v∆t
=vf − vi
tf − ti
Like velocity, acceleration has both magnitude and di-rection. The speed of an object can increase or decreasewith either positive or negative acceleration, depending onthe direction of the velocity — negative acceleration doesnot always mean decrease in speed.
The instantaneous acceleration of an object is equiv-alent to the slope of a tangent line to the v vs. t graph atthe time of interest, while the area under a graph of instan-taneous acceleration vs. time (a vs. t) is the velocity ofthe object. In this class acceleration will almost always beconstant in any problem.
Displacement and velocity with constant uniform accel-eration can be expressed mathematically as any of:
vf = vi + a∆t
∆x = vi∆t+12a∆t2
∆x =12(vi + vf )∆t
v2f = v2
i + 2a∆x
2.3 Falling ObjectsIn the absence of air resistance all objects dropped near thesurface of a planet fall with effectively the same constantacceleration — called free fall. That acceleration is alwaysdirected downward, so in the customary frame of referenceit is negative, so:
ag = g = −9.81ms2
Honors Physics 2008–2009 Mr. Strong
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Chapter 3 — Two-Dimensional Motion and Vectors
3.1 Introduction to VectorsVectors can be added graphically.
Vectors can be added in any order:
~V1 + ~V2 = ~V2 + ~V1
To subtract a vector you add its opposite:
~V1 − ~V2 = ~V1 + (−~V2)
Multiplying a vector by a scalar results in a vector insame direction as the original vector with a magnitude equalto the original magnitude multiplied by the scalar.
3.2 Vector OperationsTo add two perpendicular vectors use the Pythagorean The-orem to find the resultant magnitude and the inverse of thetangent function to find the direction: Vr =
√V 2
x + V 2y and
the direction of Vr, θr, is tan−1 Vy
Vx
Just as 2 perpendicular vectors can be added, any vectorcan be broken into two perpendicular component vectors:~V = ~Vx + ~Vy where ~Vx = V cos θı and ~Vy = V sin θ;
Two vectors with the same direction can be added byadding their magnitudes, the resultant vector will have thesame direction as the vectors that were added.
Any two (or more) vectors can be added by first decom-posing them into component vectors, adding all of the x andy components together, and then adding the two remainingperpendicular vectors as described above.
3.3 Projectile MotionThe equations of motion introduced in chapter 2 are ac-tually vector equations. Once the new symbol for displace-ment, ~d = ∆xı+∆y has been introduced the most commonones can be rewritten as
~d = ~vi∆t+12~a∆t2
~vf = ~vi + ~a∆t
Since motion in each direction is independent of motion inthe other, objects moving in two dimensions are easier toanalyze if each dimension is considered separately.
~dx = ~vxi∆t+12~ax∆t2
~dy = ~vyi∆t+12~ay∆t2
In the specific case of projectile motion there is noacceleration in the horizontal (ı) direction (~ax = 0) andthe acceleration in the vertical () direction is constant(~ay = ~ag = ~g = −9.81 m
s2 ). While ~vy will change over time,~vx remains constant. Neglecting air resistance, the path fol-lowed by an object in projectile motion is a parabola. Thefollowing equations use these simplifications to describe themotion of a projectile launched with speed ~vi and directionθ up from horizontal:
~vx = vi cos θı = constant
~vyi = vi sin θ
~dx = vi cos θ∆tı
~dy = vi sin θ∆t+12~g∆t2
To solve a projectile motion problem use one part of theproblem to find ∆t, then once you know ∆t you can usethat to solve the rest of the problem. In almost every casethis will involve using the vertical part of the problem tofind ∆t which will then let you solve the horizontal partbut there may be some problems where the opposite ap-proach will be necessary. For the special case of a projectilelaunched on a level surface the range can be found with
dx =v2 sin 2θag
Honors Physics 2008–2009 Mr. Strong
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Chapter 4 — Forces and the Laws of Motion
4.1 Changes in MotionForce causes change in velocity. It can cause a station-ary object to move or a moving object to stop or otherwisechange its motion.
The unit of force is the newton (N), equivalent to kg·ms2 ,
which is defined as the amount of force that, when actingon a 1 kg mass, produces an acceleration of 1m
s2 .Contact forces act between any objects that are in
physical contact with each other, while field forces actover a distance.
A force diagram is a diagram showing all of the forcesacting on the objects in a system.
A free-body diagram is a diagram showing all of theforces acting on a single object isolated from its surround-ings.
4.2 Newton’s First LawNewton’s first law states that
An object at rest remains at rest, and an ob-ject in motion continues in motion with constantvelocity (that is, constant speed in a straightline) unless the object experiences a net exter-nal force.
This tendency of an object to not accelerate is calledinertia. Another way of stating the first law is that if thenet external force on an object is zero, then the accelerationof that object is also zero. Mathematically this is∑
~F = 0→ ~a = 0
An object experiencing no net external force is said tobe in equilibrium, if it is also at rest then it is in staticequilibrium
4.3 Newton’s Second and Third LawsNewton’s second law states that
The acceleration of an object is directly pro-portional to the net external force acting on theobject and inversely proportional to the object’smass.
Mathematically this can be stated as:∑~F = m~a
Which in the case of no net external force (∑ ~F = 0) also
illustrates the first law:∑~F = 0→ m~a = 0→ ~a = 0
Newton’s third law states that
If two objects interact, the magnitude of theforce exerted on object 1 by object 2 is equal tothe magnitude of the force simultaneously ex-erted on object 2 by object 1, and these twoforces are opposite in direction.
Mathematically, this can be stated as:
~F1,2 = −~F2,1
The two equal but opposite forces form an action-reactionpair.
4.4 Everyday Forces
The weight of an object (~Fg) is the gravitational force ex-erted on the object by the Earth. Mathematically:
~Fg = m~g where ~g = −9.81ms2j
The normal force (~Fn) is the force exerted on an objectby the surface upon which the object rests. This force isalways perpendicular to the surface at the point of contact.
The force of static friction (~Fs) is the force that op-poses motion before an object begins to move, it will pre-vent motion so long as it has a magnitude greater than theapplied force in the direction of motion. The maximummagnitude of the force of static friction is the product ofthe magnitude of the normal force times the coefficientof static friction (µs) and its direction is opposite thedirection of motion:
µs =Fs,max
Fn→ Fs,max = µsFn
The force of kinetic friction (~Fk) is the force that op-poses the motion of an object that is sliding against a sur-face. The magnitude of the force of kinetic friction is theproduct of the magnitude of the normal force times thecoefficient of kinetic friction (µk) and its direction isopposite the direction of motion:
µk =Fk
Fn→ Fk = µkFn
The force of friction between two solid objects dependsonly on the normal force and the coefficient of friction, itis independent of the the surface area in contact betweenthem.
For an object at rest with a force applied to it the fric-tional force will vary as the applied force is increased.
Honors Physics 2008–2009 Mr. Strong
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Fapp
Ffriction
µkFn
µsFn
µsFnstatic kinetic
Fs = Fapp
Fk = µkFn
Air resistance (~FR) is the force that opposes the mo-tion of an object though a fluid. For small speeds FR ∝ vwhile for large speeds FR ∝ v2. (Exactly what is small orlarge depends on things that are outside the scope of thisclass.) Air resistance is what will cause falling objects toeventually reach a terminal speed where FR = Fg.
Solving Friction Problems
When solving friction problems start by drawing a diagramof the system being sure to include the gravitational force(Fg), normal force (Fn), frictional force (Fk, Fs, or Fs,max)and any applied force(s).
Determine an appropriate frame of reference, rotatingthe x and y coordinate axes if that is convenient for theproblem (such as an object on an inclined plane). Resolveany vectors not lying along the coordinate axes into com-ponents.
Use Newton’s second law (∑ ~F = m~a) to find the rela-
tionship between the acceleration of the object (often zero)and the applied forces, setting up equations in both the xand y directions (
∑ ~Fx = m~ax and∑ ~Fy = m~ay) to solve
for any unknown quantities.
Honors Physics 2008–2009 Mr. Strong
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Chapter 5 — Work and Energy
5.1 WorkA force that causes a displacement of an object does work(W ) on the object. The work is equal to the product of thedistance that the object is displaced times the component ofthe force in the direction of the displacement, or if θ is theangle between the displacement vector and the (constant)net applied force vector, then:
W = Fnetd cos θ
The units of work are Joules (J) which are equivalentto N ·m or kg·m2
s2 . The sign of the work being done is sig-nificant, it is possible to do a negative amount of work.
5.2 EnergyWork done to change the speed of an object will accumu-late as the kinetic energy (KE , or sometimes K) of theobject. Kinetic energy is:
KE =12mv2
If work is being done on an object, the work-kineticenergy theorem shows that:
Wnet = ∆KE
In addition to kinetic energy, there is also potentialenergy (PE , or sometimes U) which is the energy storedin an object because of its position. If the energy is storedby lifting the object to some height h, then the equation forgravitational potential energy is:
PE g = mgh
Potential energy can also be stored in a compressedor stretched spring, if x is the distance that a spring isstretched from its rest position and k is the spring con-stant measuring the stiffness of the spring (in newtons permeter) then the elastic potential energy that is storedis:
PE elastic =12kx2
The units for all types of energy are the same as thosefor work, Joules.
5.3 Conservation of EnergyThe sum an objects kinetic energy and potential energy isthe object’s mechanical energy (ME ). In the absence offriction, the total mechanical energy of a system will remainthe same. Mathematically, this can be expressed as:
ME i = ME f
In the case of a single object in motion, this becomes:
12mv2
i +mghi =12mv2
f +mghf
5.4 PowerThe rate at which energy is transferred is called power (P ).The mathematical expression for power is:
P =W
∆t=Fd
∆t= F
d
∆t= Fv
The units for power are Watts (W) which are equivalentto J
s or kg·m2
s3
Honors Physics 2008–2009 Mr. Strong
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Chapter 6 — Momentum and Collisions
6.1 Momentum and ImpulseMomentum is a vector quantity described by the productof an object’s mass times its velocity:
~p = m~v
If an object’s momentum is known its kinetic energy canbe found as follows:
KE =p2
2m
The change in the momentum of an object is equal tothe impulse delivered to the object. The impulse is equalto the constant net external force acting on the object timesthe time over which the force acts:
∆~p = ~F∆t
Any force acting on on object will cause an impulse,including frictional and gravitational forces.
In one dimension the slope of a graph of the momen-tum of an object vs. time is the net external force actingon the object. The area under a graph of the net externalforce acting on an object vs. time is the total change inmomentum of that object.
Momentum and impulse are measured in kg·ms — there
is no special name for that unit.
6.2 Conservation of MomentumMomemtum is always conserved in any closed system:
Σ~pi = Σ~pf
For two objects, this becomes:
m1~v1i +m2~v2i = m1~v1f +m2~v2f
Any time two objects interact, the change in momen-tum of one object is equal in magnitude and opposite indirection to the change in momentum of the other object:
∆~p1 = −∆~p2
6.3 Elastic and Inelastic CollisionsThere are three types of collisions:
• Elastic — momentum and kinetic energy are con-served. Both objects return to their original shapeand move away separately. Generally:
Σ~pi = Σ~pf
ΣKEi = ΣKEf
For two objects:
m1~v1i +m2~v2i = m1~v1f +m2~v2f
12m1v
21i +
12m2v
22i =
12m1v
21f +
12m2v
22f
• Inelastic — momentum is conserved, kinetic energyis lost. One or more of the objects is deformed in thecollision. Generally:
Σ~pi = Σ~pf
ΣKEi > ΣKEf
For two objects:
m1~v1i +m2~v2i = m1~v1f +m2~v2f
12m1v
21i +
12m2v
22i >
12m1v
21f +
12m2v
22f
• Perfectly inelastic — momentum is conserved, ki-netic energy is lost. One or more objects may bedeformed and the objects stick together after the col-lision. Generally:
Σ~pi = Σ~pf
ΣKEi > ΣKEf
For two objects:
m1~v1i +m2~v2i = (m1 +m2)~vf
12m1v
21i +
12m2v
22i >
12(m1 +m2)v2
f
True elastic or perfectly inelastic collisions are very rarein the real world. If we ignore friction and other small en-ergy losses many collisions may be modeled by them.
Newton’s Laws in terms of Momentum1. Inertia:
Σ~F = 0→ ~p = constant
2. ~F = m~a:∆~p = ~F∆t
3. Every action has an equal and opposite reaction:
∆~p1 = −∆~p2
Honors Physics 2008–2009 Mr. Strong
9
Chapter 7 — Rotational Motion and the Law of Gravity
7.1 Measuring Ratational Motion
For an object moving in a circle with radius r through anarc length of s, the angle θ (in radians) swept by the objectis:
θ =s
r
The conversion between radians and degrees is:
θ(rad) =π
180oθ(deg)
The angular displacement (∆θ) through which anobject moves from θi to θf is, in rad:
∆θ = θf − θi =sf − si
r=
∆sr
The average angular speed (ω) of an object is, in rads ,
the ratio between the angular displacement and the timeinterval required for that displacement:
ωavg =θf − θi
∆t=
∆θ∆t
The average angular acceleration (α) is, in rads2 :
αavg =ωf − ωi
∆t=
∆ω∆t
For each quantity or relationship in linear motion thereis a corresponding quantity or relationship in angular mo-tion:
Linear Angular
x θv ωa α
vf = vi + a∆t ωf = ωi + α∆t
∆x = vi∆t+ 12a∆t
2 ∆θ = ωi∆t+ 12α∆t2
v2f = v2
i + 2a∆x ω2f = ω2
i + 2α∆θ
∆x = 12 (vi + vf )∆t ∆θ = 1
2 (ωi + ωf )∆t
7.2 Tangential and Centripetal AccelerationIn addition to the angular speed of an object moving withcircular motion it is also possible to measure the object’stangential speed (vt) or instantaneous linear speed whichis measured in m
s as follows:
vt = rω
An object’s tangential acceleration (at) can also bemeasured, in m
s2 , as follows:
at = rα
The centripetal acceleration (ac) of an object is di-rected toward the center of the object’s rotation and hasthe following magnitude:
ac =v2
t
r= rω2
7.3 Causes of Circular MotionThe centripetal force (Fc) causing a centripetal accel-eration is also directed toward the center of the object’srotation, and has the following magnitude:
Fc = mac =mv2
t
r= mrω2
The centripetal force keeping planets in orbit is a grav-itational force (Fg) and it is found with Newton’s Univer-sal Law of Gravitation:
Fg = Gm1m2
r2
where G is the constant of universal gravitation whichhas been determined experimentally to be
G = 6.673× 10−11 N ·m2
kg2
An object in a circular orbit around the Earth will sat-isfy the equation
v =
√GME
r
where ME is the mass of the Earth.
Honors Physics 2008–2009 Mr. Strong
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Chapter 8 — Rotational Equilibrium and Dynamics
8.1 Torque
Just as a net external force acting on an object causes linearacceleration, a net torque (τ) causes angular acceleration.The torque caused by a force acting on an object is:
τ = rF sin θ
where F is the force causing the torque, r is the distancefrom the center to the point where the force acts on theobject, and θ is the angle between the force and a radialline from the object’s center through the point where theforce is acting. Torque is measured in N ·m.
The convention used by the book is that torque in acounterclockwise direction is positive and torque in a clock-wise direction is negative (this corresponds to the right-hand rule). If more that one force is acting on an objectthe torques from each force can be added to find the nettorque:
τnet =∑
τ
8.2 Rotation and Inertia
The center of mass of an object is the point at which allthe mass of the object can be said to be concentrated. Ifthe object rotates freely it will rotate about the center ofmass.
The center of gravity of an object is the point throughwhich a gravitational force acts on the object. For mostobjects the center of mass and center of gravity will be thesame point.
The moment of inertia (I) of an object is the object’sresistance to changes in rotational motion about some axis.Moment of inertia in rotational motion is analogous to massin translational motion.
Some moments of inertia for various common shapes are:
Shape I
Point mass at a distance r from the axis mr2
Solid disk or cylinder of radius r aboutthe axis
12mr
2
Solid sphere of radius r about its diameter 25mr
2
Thin spherical shell of radius r about itsdiameter
23mr
2
Thin hoop of radius r about the axis mr2
Thin hoop of radius r about the diameter 12mr
2
Thin rod of length l about its center 112ml
2
Thin rod of length l about its end 13ml
2
An object is said to be in rotational equilibrium whenthere is no net torque acting on the object. If there is also
no net force acting on the object (translational equilib-rium) then the object is in equilibrium (without any qual-ifying terms).
Type Equation Meaning
TranslationalEquilibrium
∑ ~F = 0 The net force on the ob-ject is zero
Rotational Equi-librium
∑τ = 0 The net torque on the
object is zero
8.3 Rotational DynamicsNewton’s second law can be restated for angular motion as:
τnet = Iα
This is parallel to the equation for translational motionas follows:
Type of Motion Equation
Translational ~F = m~aRotational τ = Iα
Just as moment of inertia was analogous to mass, theangular momentum (L) in rotational motion is analogousto the momentum of an object in translational motion.
This is parallel to the equation for translational motionas follows:
Type of Motion Equation
Translational ~p = m~vRotational L = Iω
The angular momemtum of an object is conserved in theabsence of an external force or torque.
Rotating objects have rotational kinetic energy ac-cording to the following equation:
KE r =12Iω2 =
L2
2I
Just as other types of mechanical energy may be con-served, rotational kinetic energy is also conserved in theabsence of friction.
8.4 Simple MachinesThere are six fundamental types of machines, called sim-ple machines, with which any other machine can be con-structed. They are levers, inclined planes, wheels, wedges,pulleys, and screws.
The main purpose of a machine is to magnify the outputforce of the machine compared to the input force, the ratioof these forces is called the mechanical advantage (MA)
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of the machine. It is a unitless number according to thefollowing equation:
MA =output forceinput force
=Fout
Fin
When frictional forces are accounted for, some of theoutput force is lost, causing less work to be done by themachine than by the original force. The ratio of work doneby the machine to work put in to the machine is called theefficiency (eff ) of the machine:
eff =Wout
Win
If a machine is perfectly efficient (eff = 1) then the idealmechanical advantage (IMA) can be found by comparingthe input and output distances:
IMA =din
dout
This leads to another way to find the efficiency of themachine as well:
eff =MAIMA
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Chapter 12 — Vibrations and Waves
12.1 Simple Harmonic MotionHooke’s law states that for a given spring the force thespring exerts (~F ) is proportional to the negative of the dis-placement that the spring is stretched from rest (~x), or
~F = −k~x
Simple harmonic motion is the repetitive, back-and-forth motion of an object such as a pendulum or a massoscillating on the end of a spring. The motion is over thesame path each time and passes through the equilibriumposition twice each cycle under the influence of a restor-ing force (a force that pushes the object back toward theequilibrium position).
The displacement (x), velocity (v), and acceleration (a)of an object in simple harmonic motion when graphed withrespect to time are all sinusoids, if they are all plotted onthe same graph then their relationship can be seen
t
xva
12.2 Measuring Simple Harmonic MotionThe amplitude of harmonic motion measures how far theobject moves.
The period (T ) is the time that it takes for one com-plete cycle of motion.
The frequency (f) is the number of complete cycles ofmotion occurring in one second. Frequency is measured ininverse seconds (s−1) or Hertz (Hz). Frequency and periodare inverses of each other, so:
f =1T
T =1f
For a simple pendulum of length L, the period of smalloscillations is given by:
Tpendulum = 2π
√L
g
For a mass-spring system with mass m and spring constantk, the period of oscillation is:
Tspring = 2π√m
k
12.3 Properties of WavesMechanical waves require the presence of a mediumthrough which to move, such as a ripple traveling on thesurface of a pond, or a pulse traveling along a stretchedspring. If the vibrations of the wave are perpendicular to
the direction of motion of the wave then it is a transversewave, if the vibrations are parallel to the direction of mo-tion then it is a longitudinal wave.
The displacement of a wave moving with simple har-monic motion can be modeled by a sinusoidal wave.
rCrest
rTrough Wavelength (λ)
Amplitude
The wavelength (λ) of a wave is the shortest distancebetween corresponding parts of two waves, such as from onecrest or trough to the next one.
The wave speed (v) is the speed with which a wavepropagates, it is equal to the frequency of the wave timesits wavelength:
v = fλ =λ
T
The energy transferred by a wave is proportional to thesquare of its amplitude.
12.4 Wave Interactions
When two (or more) waves come together they pass througheach other, and through the principle of superposition thetotal displacement of the medium is equal to the sum of thedisplacements of the overlapping waves at each point.
When two waves overlap with displacements in the samedirection the resulting wave has an amplitude greater thaneither of the two overlapping waves, this is called construc-tive interference. In the other case, when two overlap-ping waves have displacements in opposite directions theresulting wave will have a displacement less than the dis-placement of the wave with larger amplitude, possibly evenno displacement at all. This is called destructive inter-ference.
→
←
→Destructive
→ ←→
Constructive
When a wave traveling through some medium reaches afixed boundary (a transition to a medium with a smaller,possibly zero, wave speed) the wave will reflect back in-verted with respect to the initial wave. If the wave insteadreflects off of a free boundary (a transition to a mediumwith a greater wave speed) then the wave will reflect off theboundary in the same orientation that it arrived.
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If a wave is traveling in a confined space with just theright frequency then a standing wave may occur. A stand-ing wave results from a wave constructively and destruc-tively interfering with its own reflection. There will be somepoints (at least two, the ends) called nodes where completedestructive interference occurs, and between every pair ofnodes there will be an antinode where constructive inter-ference causes the oscillation to reach a relative maximumamplitude.s s
s s ss s s s
n = 1, λ = 2L
n = 2, λ = L
n = 3, λ = 23L
Node Node Node NodeAntinode Antinode Antinode
For a vibrating spring or string of length L the wave-
lengths of possible standing waves are:
λ = 2L, L,23L,
12L,
25L . . . λ =
2nL
where n is the number of the harmonic that correspondsto that standing wave. The nth harmonic will have n + 1nodes and n antinodes.
n Nodes Antinodes λ
1 2 1 λ = 2L2 3 2 λ = L3 4 3 λ = 2
3L4 5 4 λ = 1
2L5 6 5 λ = 2
5L· · ·
n n+ 1 n λ = 2nL
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Chapter 13 — Sound
13.1 Sound WavesSound waves are longitudinal waves consisting of successivecompressions (areas of high pressure) and rarefactions(areas of low pressure) caused by and radiating outward inspherical wavefronts from a vibrating object.
The frequency a sound wave is perceived to have is calledthe pitch.
Sound waves travel outward in 3 dimensions along spher-ical wavefronts. If a ray is drawn from the source outthrough the wavefronts a graph of pressure vs. distancewill be a sinusoid. When r λ the wave fronts begin tolook like and can be approximated by plane waves.
Humans can hear sounds with frequencies from about20 Hz to about 20,000 Hz with the greatest sensitivity inthe range of 500 Hz to 5000 Hz. Frequencies outside thatrange are called either infrasonic (below 20 Hz) or ultra-sonic (above 20,000 Hz).
The speed at which a sound wave propagates dependson the medium. In air at sea level and room temperaturethe speed of sound is approximately 343 m
s .
Doppler EffectIf a sound source and the detector are moving relative toone another then the Doppler effect can be observed.
r rvs = 0 vs < v
r rvs = v vs > v
This is a change in the pitch (f ′) relative to the frequencyas emitted by the source (f). For a wave traveling at speedv, a detector moving at velocity vd (positive if toward thesource), and a source moving at velocity vs (positive if to-ward the detector), the pitch will be
f ′ = f
(v + vd
v − vs
)13.2 Sound Intensity and ResonanceIntensityAs a sound wave travels through a medium energy is trans-ferred from one molecule to the next. The rate at which
this energy is transferred is called the intensity and canbe described as power per unit area and measured in wattsper square meter.
In a spherical wave the energy is spread across the sur-face of a sphere, so the expression for intensity becomes
intensity =P
4πr2
The quietest sound audible to the human ear (thethreshold of hearing) is about 1.0×10−12 W
m2 . The loud-est sound a typical person can tolerate (the threshold ofpain) is about 1.0 W
m2 .
Perceived IntensityThe perceived loudness of a sound is dependent on the log-arithm of the intensity, so it is measured in decibels (dB).The sound level measured in decibels is:
dB = 10 log10
intensityintensityr
where intensity is the intensity of the sound being measuredand intensityr is the reference intensity. If nothing else isspecified the reference intensity is set equal to the thresholdof hearing or intensityr = 1.0× 10−12 W
m2 .An increase of 10 decibels will be perceived as being
twice as loud but will cause the intensity to increase by afactor of ten.
ResonanceJust as was observed with standing waves in a spring, cer-tain physical systems show a particular response at certainfrequencies called resonance. When a system is driven at(or very close to) its resonant frequency the amplitudeof its vibration will continue to increase.
13.3 HarmonicsStanding WavesStrings and air-filled pipes can exhibit standing waves atcertain frequencies. The lowest frequency able to form astanding wave is called the fundamental frequency (f1),those above it are harmonics of that frequency. The fre-quency of each harmonic frequency is equal to the funda-mental frequency multiplied by the harmonic number (n),this forms a harmonic series. The fundamental frequencyis also referred to as the first harmonic frequency.
Depending on the medium in which the standing wavesare formed the characteristics of the waves will vary. Apipe open at both ends has the same harmonic series asa vibrating string (producing all of the harmonics), whilea pipe closed at one end will only produce odd-numberedharmonics.
The harmonics that are present form a sound source willaffect the timbre or sound quality of the resulting sound.
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Standing Waves on a String or Open Pipe of Length L
n Nodes Antinodes String Diagram Pipe Diagram Wavelength Frequency
1 2 1 λ1 = 2L f1 =v
2L
2 3 2 λ2 = L f2 =v
L= 2f1
3 4 3 λ3 =23L f3 =
3v2L
= 3f1
n n+ 1 n n = 1, 2, 3, . . . λn =2nL fn = n
v
2L= nf1
Standing Waves in a Pipe of Length L Closed at One End
n Nodes Antinodes Diagram Wavelength Frequency
1 1 1 λ1 = 4L f1 =v
4L
3 2 2 λ3 =43L f3 =
3v4L
= 3f1
5 3 3 λ5 =45L f5 =
5v4L
= 5f1
nn+ 1
2n+ 1
2n = 1, 3, 5, . . . λn =
4nL fn = n
v
4L= nf1
Beat Frequency
If two sources emit sound waves that are close to one an-other in frequency but not exactly the same then a beatwill be observed. When the amplitudes of the waves areadded together the resulting sound oscillates between loudand soft. The number of beats heard per second is equal tothe absolute value of the difference between the frequenciesof the two sounds. The diagram below shows one second of
the sum of a 34 Hz wave and a 40 Hz wave yielding a 6 Hzbeat frequency.
f = 40 Hz
f = 34 Hz
Sum
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Chapter 14 — Light and Reflection
14.1 Characteristics of LightLight is an electromagnetic wave as opposed to a me-chanical wave, so it does not need a medium through whichto travel although it can pass through many media. Beingan electromagnetic wave it is composed of a vibrating elec-tric field and a vibrating magnetic field, the two fields areat right angles to each other.
The speed of light depends on the medium throughwhich it travels, it is fastest in a vacuum and slower indenser media. The speed of light in a vacuum is exactly299 792 458 m
s (the definition of a meter comes from thespeed of light so this is defined as the exact value). Formost purposes, a value of 3.00×108 m
s is sufficiently precise.The speed of light in a vacuum is a fundamental constantin many calculations and has been assigned the symbol c,thus for light and other electromagnetic waves the wavespeed equation v = fλ becomes:
c = fλ
Visible light is only one part of the electromagneticspectrum corresponding to wavelengths from 400 nm (vi-olet light) to 700 nm (red light).
Huygens’ principle states that all of the points on awave front can be treated as point sources and each of thosepoint sources will produce a circular or spherical wavelet.The wavelets will constructively interfere on a line tangentto the fronts of all of the wavelets, this will create the nextwave front. This principle will become more important laterwhen we discuss diffraction, for now we can use the ray ap-proximation which states that light can be considered tobe straight lines traveling outward from the source perpen-dicular to the wave fronts.
The brightness of a surface lit by a single light is in-versely proportional to the square of the distance betweenthe source and the surface, for a source twice as far awaythe surface is one-fourth as bright, etc.
14.2 Flat Mirrors
ReflectionLight traveling through a uniform medium will move in astraight line regardless of the medium.
When light encounters a boundary between media partof the light will pass through, part will be absorbed, andpart will be bounced off of the boundary, or reflected. Ifthe medium is one that is opaque to the light then noneof it will pass through, the light will either be absorbed orreflected instead.
If the surface is rough the result is diffuse reflection,if it’s smooth then it’s specular reflection. Smooth vs.rough depends on the relative sizes of the surface variationsand λ.
The light ray coming to the surface is the incident ray,the ray bouncing off of it is the reflected ray, and a lineperpendicular to the surface at the point where the incidentray strikes it is the normal line. The angle between theincident ray and the normal line is the angle of incidence(θi) and the angle between the normal line and the reflectedray is the angle of reflection (θ′). The angle of incidenceis always equal to the angle of reflection (θi = θ′).
@@
@R
Reflecting Surface
Incident Ray Reflected Ray
Normal Line
θi θ′
Flat MirrorsIf you put an object at some distance in front of a flat mirror(the object distance, p) light from the object will reflectoff of the mirror and appear to come from a location behindthe mirror’s surface. As a convention, the object’s image issaid to be at that location (the image distance, q) behindthe mirror. For a flat mirror the object is the same dis-tance in front of the mirror as the image is behind it andthe height of the image and object are equal (h′ = h).
Ray DiagramsRay diagrams are drawn to locate the image formed by amirror. To simplify matters we just consider a single pointon an object resting on the principal axis. Rays are drawnfrom the tip of the object to the mirror and allowed to reflectback. In the diagram below ray 1 is drawn perpendicularto the mirror, it reflects back along itself and away. Ray 2is drawn from the object to the mirror at the principal axiswhere it reflects back out at the same angle. The two raysas drawn do not meet to form an image so the need to be ex-tended backward behind the mirror to where the image willbe formed. The rays extended backward are not actuallythere but they seem to originate at the source of the image.Rays that are not there but seem to be are called virtualrays (real rays are the rays that are actually there) andthe image that they form is a virtual image as opposedto a real image that is formed by real rays. A real imagecould be projected onto a screen, there is no way to see avirtual image other than to observe it in the mirror.
HHHH
HHHHHj
-
h h′
p q1
2
Object Image
MirrorPrincipal Axis@
@@I
Ray diagram of a virtual image formed by a flat mirror
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14.3 Curved Mirrors
Curved mirrors can be either concave (curving inward inthe middle) or convex (curving outward in the middle).They can be approximated as a section of a sphere withradius of curvature R, and will have a focal length (thedistance from the center of the mirror to the focal point,F ) of f = R/2. By convention f and R are positive for aconcave mirror and negative for a convex mirror. Similarly,p and q are considered to be positive if they are measuredto a point in front of the mirror and negative if the point isbehind the mirror. For a flat mirror f = R =∞.
For a given mirror of focal length f , the relationship be-tween focal length, object distance (p), and image distance(q) is:
1p
+1q
=1f
For an object of height h and an image of height h′ theirratio is the magnification, M and is given as:
M =h′
h= −q
p
Sign conventions for mirrors
Symbol Quantity Situation Sign
p Objectdistance
Object is in front of themirror
+
q Imagedistance
Image is in front of themirror (real image)
+
Image is behind the mirror(virtual image)
−
h Objectheight
Object is above the axis +
h′ Imageheight
Image is above the axis +
Image is below the axis −f Focal
lengthFocal point is in front ofthe mirror (concave mir-ror)
+
Focal point is behind themirror (convex mirror)
−
No focal point (flat mirror) ∞R Radius
of Cur-vature
Center of curvature is infront of the mirror (con-cave mirror)
+
Center of curvature is be-hind the mirror (convexmirror)
−
No center of curvature(flat mirror)
∞
Images can be located through ray diagrams as well.The general principle is the same as the ray diagram forthe flat mirror, the usual rays to be drawn are:
1. A ray that comes in parallel to the principal axis andgoes out through the focal point.
2. A ray that comes in through the focal point and goesout parallel to the principal axis.
3. A ray that goes in through the center of curvature, re-flects perpendicularly off of the mirror, and goes backout along the same line where it came in.
Each of these rays is labeled with its corresponding num-ber in the ray diagrams below. For more precise results therays are drawn to a line perpendicular to the principal axisrather than to the mirror itself.
sC
sFObject Image
1
2
1)
3
Concave mirror, p < f , virtual upright image, M > 1
sC
sFObject
:9
- 91
3
No Image
Concave mirror, p = f , no image
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sC
sFObject
Image
HHHHH
HHHH
HHHH
HHj
1
19
-
22
*
3
3
Concave mirror, f < p < R, real inverted image, |M | > 1
sC
sFObject
Image
-
1
1
HHHHHH
HHHjH
HH
22
Concave mirror, f = R, real inverted image, |M | = 1
sC
sFObject
Image
-
1
1
XXXXXXXXXXXXXXXzXXX
2
2
HHHHHHH
HHHHHH
HHjHHH
HHY3
3
Concave mirror, f > R, real inverted image, |M | < 1
sC
sFObject Image
-HHHHHY1
XXXXXzXXXXXX2
-3 H
HHHHH
Convex mirror, virtual upright image, M < 1
Parabolic Mirrors
Spherical mirrors come close to focusing incoming parallelrays to a single point, particularly if all of the rays strikeclose to the center of the mirror. The farther the rays strikefrom the center the greater the error in focusing at a singlepoint, this is called spherical aberration. If a parabolicmirror is used instead of a spherical mirror the sphericalaberration is eliminated. High-quality optical instrumentsuse parabolic mirrors for their improved optical properties.
14.4 Color and Polarization
Color
The primary colors of light are red, green, and blue. Colorsof light mix through additive mixing, when all three primarycolors are mixed white light is produced. The primary col-ors of pigment are cyan, magenta, and yellow. Colors ofpigment mix through subtractive mixing, when all threeprimary colors are mixed black is produced. The primarycolors for additive mixing are the secondary colors for sub-tractive mixing and vice versa.
Pigments appear to be the color that they are becausethat color of light is reflected, all other colors are absorbed.When two pigments are mixed the resulting mix will absorbthe colors absorbed by both pigments that were mixed andonly reflect the colors that neither pigment will absorb.Primary Colors
In the diagrams below the circles represent the primary col-ors and the overlapping areas the result of mixing thosecolors.
Red
Green BlueCyan
White
Yellow Magenta
(Light)Additive
Yellow
Cyan MagentaBlue
Black
Green Red
(Pigment)Subtractive
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Additive and subtractive primary colors
Color Additive (mixinglight)
Subtractive (mixingpigment)
red primary complimentary tocyan
green primary complimentary tomagenta
blue primary complimentary toyellow
cyan(bluegreen)
complimentary tored
primary
magenta(redblue)
complimentary togreen
primary
yellow complimentary toblue
primary
PolarizationLight can be polarized, or set to vibrate in the same planes(one for electric field, one for magnetic field), by passing itthrough certain materials.
If a polarized beam of light is passed through a secondpolarizing filter only the component of that light match-ing the polarization of the new filter will pass through, sothe overall intensity of the light will be proportional to thecosine of the angle between the polarizing filters.
Light can be polarized by reflection at certain angles(about 34 up from the surface of a glass plate works well,the best angle varies depending on the material). Polar-ization is perpendicular to the plane containing the rays orhorizontal for reflection off of a level surface. A pair of sun-glasses with vertical polarization will remove the polarizedglare from reflecting surfaces on a sunny day.
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Chapter 15 — Refraction
15.1 RefractionAs light travels from one medium to another the speed oflight also changes depending upon the densities of the me-dia.
If a ray of light reaches a boundary between two me-dia part of the ray will be reflected and part may also berefracted or bent while passing across the boundary.
@@
@@@R
AAAAAU
AirGlass
θi θ′
θr
AAAAAU@
@@
@@R
GlassAir
θi θ′
θr
The index of refraction (n) of a medium is a measureof how much the speed of light changes in that mediumcompared to in a vacuum. For a medium with speed oflight v, the index of refraction is:
n =c
v
The amount that light will be refracted at a boundarydepends on the respective indices of refraction of the media.All angles are measured from a line normal to the boundaryat the point that the light intersects the boundary. For anangle of incidence θi in a medium with index of refractionni and an angle of refraction θr in a medium with indexof refraction nr the relationship is given by Snell’s law:
ni sin θi = nr sin θr
15.2 Thin LensesLenses operate similarly to curved mirrors with a few im-portant differences: lenses have a focal point on both sidesand some of the sign conventions are changed.
Sign conventions for lenses
Symbol Positive (+) Negative (−)
p Object is in front ofthe lens
Object is in back ofthe lens
q Image is in back ofthe lens
Image is in front ofthe lens
f Converging lens Diverging lens
With those sign conventions in place the following equa-tions still hold for focal length and magnification:
1p
+1q
=1f
M =h′
h= −q
p
Ray diagrams can also be drawn to locate and describethe image that will be formed by a lens. The usual rays tobe drawn are:
1. A ray that comes in parallel to the principal axis andgoes out through the focal point.
2. A ray that comes in through the focal point and goesout parallel to the principal axis.
3. A ray that goes straight through the center of the lens
Each of these rays is labeled with its corresponding numberin the ray diagrams below.
sF sF-
-
-
HHHH
HHHHHH
HHH
-*H
HHj
Object at ∞
Image
Converging lens, p =∞, point image, q = f
1
3
1
sF
sFhhhhhhhhhhhhhhhhhhhhhh
XXXXXXXXXXXXXXX
-XXXXXXXXz
HHHHHH
HH
HHHj
-
Object
Image
Converging lens, p > 2f , real image, f < q < 2f
1
2
3
sF
sF-XXXzXXXXXXXXXXXXXXXXXXXX
HHHjHHH
HHHHHHH
HHHHHH
HHHH
HHHj-
XXXXXXXXz
Object
ImageConverging lens, p = 2f , real image, q = 2f
1
2
3
sF
sFhhhhhhhhhhhhhhhhhhhhhh
HHHHH
HHH
-HHHHHj
-
XXXXXXXXXXXXXXX
XXXXXXXXz
Object
ImageConverging lens, f < p < 2f , real image, q > 2f
1
23
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sF
sF-H
HHjH
HHHHHH
HHHHHHj
HHHHHHH
HHHHHH
HH
HHHHHH
HHHHHHH
HHHj
Object
No image
Converging lens, p = f , no image
1
3
sF
sF
HHHHH
HHHHHH
HH
HHHHH
HHHHHj
XXXXXXXXXXXXX
XXXXXXXXz
-HHHXXXXX
Object
Image
Converging lens, p < f , virtual image
1
23
sF sF
HHHHHH
HHHHj
PPPPP -
Image
Object
Diverging lens, virtual image
1
2
3
Lenses can be used for the correction of defects in hu-man vision. If a person has hyperopia (farsightedness)light rays tend to focus behind the retina, a converging lenswill cause them to focus on the retina. Similarly, if a personhas myopia (nearsightedness) a diverging lens will causethe light rays to focus on the retina instead of in front of it.
Lenses can be used in combination, the image formedby the first lens is treated as the object for the second lens.The overall magnification of the system is the product ofthe magnification of the individual lenses.
15.3 Optical Phenomena
*
HH
HHHHj
-
7
12
3 3
2
1
refra
cted
ray
reflected ray
θr
θi
AirGlass
When light passes from a medium with a higher indexof refraction to one with a lower index of refraction an ef-fect called total internal reflection can sometimes beobserved. If the angle of incidence is greater than somecritical angle θc then there will be no refracted ray. Theexpression for that critical angle is:
sin θc =nr
nifor ni > nr
The diagram above shows incident rays at less than the crit-ical angle (1), equal to the critical angle (2), and greaterthan the critical angle (3). All three incident rays wouldproduce reflected rays but they have been omitted from 1and 2 for clarity.
Refraction can occur even if there is not an abruptboundary between the two media, if for example warm (lessdense) air is trapped below cooler (more dense) air thenlight rays passing through the air will be refracted slightly.It is very difficult to predict the exact amount of refractionthat will be observed.
When light travels through the atmosphere it is scat-tered with the most scattering happening to the shorterwavelengths of light, this causes the sky to appear to beblue and sunsets, after the sunlight has had most of theblue scattered when traveling through the atmosphere, toappear orange or red. The name for this phenomenon isRayliegh scattering.
The index of refraction in a medium is dependent onthe wavelength of light involved, because of this a boundarybetween media may refract different wavelengths by differ-ent amounts, a phenomenon known as dispersion. Thiscauses the display of a spectrum by a prism, rainbows, andchromatic aberration or the focusing of different colorsof light by a lens at slightly different distances behind thelens.
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Chapter 16 — Interference and Diffraction
16.1 InterferenceInterference can only happen between waves of the samewavelength. If the crests of two waves arrive at the samepoint then they have a phase difference of 0 and they aresaid to be in phase, if the crest of one wave matches atrough of the other then they have a phase difference of180 and they are said to be out of phase.
If monochromatic light from a single source is passedthrough two narrow parallel slits they will serve as coher-ent sources and will produce an interference pattern witha bright fringe where constructive interference occurs anddark ones in between as shown below. s
sd
1st bright fringe
0th dark fringe
The fringes are numbered with an order number, thecenter one being assigned the number zero (the zeroth-order maximum or central maximum) and others num-bering outward from there. The dark fringes are also num-bered from zero although there are two zeroth-order darkfringes and only one bright one. The diagram below showshow the fringes produced by two slits a distance d apartare numbered (one side of the diagram is omitted to savespace), the angle θ shown below corresponds to the 2ndbright fringe.
dθ
m = 2 (bright)
m = 1 (bright)
m = 0 (bright)m = 0 (dark)
m = 0 (dark)
m = 1 (dark)
m = 2 (dark)
For two slits separated by a distance d the mth ordermaximum is located at some angle θ from a normal linedrawn between the slits, that angle also depends on thewavelength of the light and can be found from the equation
d sin θ = mλ m = 0,±1,±2,±3 . . .
with the corresponding dark fringes being found from thesimilar equation
d sin θ =(m+
12
)λ m = 0,±1,±2,±3 . . .
If the difference in distances is an integral multiple ofthe wavelength then a bright fringe will be seen, if the dif-ference is half a wavelength from an integral multiple of thewavelength then it will be a dark fringe.
When a white light source is used, instead of simplefringes a very different pattern is observed since each colorof light refracts differently and the resulting fringes mix toform other colors.16.2 Diffraction
Since each point on a wave front is a source of Huygenswavelets interference patterns can be observed from a sin-gle slit. Light deviates from a straight-line path and entersthe area that would normally be in shadow (diffraction)and produces a diffraction pattern when the wavelets in-terfere with each other. This can occur when light passesthrough a slit or when it passes by the edge of an object.
When light passes through a diffraction grating each ad-jacent pair of grooves in the grating will behave like the twoslits described previously, the same equations will work todescribe the location of the bright and dark fringes.
Diffraction plays a mayor part in determining the re-solving power of an optical instrument. The resolvingpower is the instrument’s ability to distinguish two objectsthat appear to be close to each other as separate imagesand it is measured as the minimum angle between the twoobjects (with the instrument at the vertex of the angle) thatcan be discerned. This angle is proportional to the wave-length of the light and inversely proportional to the size ofthe aperture through which the light must pass.16.3 Lasers
In a typical light source the individual rays of light areemitted haphazardly, the waves do not share a commonwavelength or phase.
-
6 AAAU
-
Ground energy level
Metastable energy level
Excited energy level
Pumping energy
Spontaneous energy decay
Stimulated emission
Q0
Qe
Qm
As shown above, lasers take advantage of properties ofthe energy levels of an electron in certain types of atoms.They do so by having a set of energy levels where some
Honors Physics 2008–2009 Mr. Strong
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initial energy (the pumping energy) will raise an electronto an unstable, excited state. The electron will then spon-taneously decay to a lower, metastable state (a state theelectron can hold for some time) where it will remain untilthe electron is struck by a photon from another decayingelectron, this will trigger the electron to drop back to theground state and release an additional photon. Since all ofthe stimulated emission photons come from the same en-ergy transition in the atoms they all have the same energyand thus frequency and wavelength. The waves are alsoproduced in phase with each other, yielding what is calledcoherent light. The beam of a laser appears much moreintense than a similar light source because of the coherenceand the tendency of lasers to emit a nearly parallel beamof light (it does not spread very much from its source untilit strikes an object).
Lasers are often used in information storage and re-trieval (reading and writing CDs and DVDs), in measure-ment (finding the time for a laser to reflect from a distantobject) and in medicine (precisely delivering energy to spe-cific targets in the body that might not otherwise be reach-able).
Thin-Film InterferenceWhen light bounces off of a soap bubble it is reflected offof both the outside and inside surfaces. Depending on theexact thickness of the soap bubble the light reflecting off ofone of the two surfaces will be either in or out of phase withthe light reflecting from the other one. If they are in phasethen a bright band will be observed, if not then it will bedark. To see this sort of interference the bubble must be atleast 1
4λ thick, any thinner and the waves will be less than180 out of phase.
R@@
@
AAsn = 1
n = 1.33n = 1
airsoap filmair
The black dot on the diagram above indicates that thephase of the reflected light at that point is shifted by 180
because the reflecting surface has a higher index of refrac-tion. Since all other reflections are from a reflecting surfacewith a lower index of refraction they will not change thephase of the incident rays. If the thickness of the soap filmis 1
4λ then there will be constructive interference, if it is 12λ
then there will be descructive interference, and the patternwill repeat with each 1
4λ added to the thickness.Since the bubbles vary in thickness from top to bottom
the interference pattern observed will also vary. The pat-tern will generally consist of horizontal bands with someirregularities as the fluid in the bubble moves around. Ifthe bubble is lit with only one color of light then the bandswill be light and dark, if the bubble is lit with white lightthen a rainbow pattern will be observed.
Thin films of oil floating on water (as you may see onyour driveway after it rains) exhibit similar properties, in-stead of seeing horizontal bands around a bubble you seeconcentric irregular rings of color as the thickness of the oildecreases toward the edges.
R@@
@
AAsn = 1
n = 1.4n = 1.33
airoilwater
A similar principle controls the function of anti-reflection coatings on lenses, a very thin coating of mag-nesium fluoride is deposited on the surface of a glass lensto provide destructive interference for most wavelengths ofvisible light.
R@@
@
AAs sn = 1
n = 1.38n = 1.72
airMgF2
glass
The main difference between the anti-reflection coatingsand the soap or oil films is that there is an extra phase in-version at the coating-glass boundary. Because of this theinterference at 1
4λ will be destructive and at 12λ it will be
constructive.
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Chapter 17 — Electric Forces and Fields
17.1 Electric Charge
Electric charge comes in two varieties, negative (an excessof electrons) and positive (a lack of electrons). Positivecharges attract negative charges and vice versa, two likecharges will repel each other.
As shown by Robert Millikan in 1909 charge must comein multiples of a fundamental unit. This fundamental unitof charge, e, is the charge on a single electron or proton. Theunit by which charge is usually measured is the coulomb(C), one coulomb is 6.2× 1018 e and the charge on a singleelectron is −1.60× 10−19 C.
Electric charge is conserved. It can me transferred fromone object to another by moving electrons but it can notbe created or destroyed.
Materials can be either conductors (materials throughwhich electrons flow freely), insulators (materials that donot permit the free flow of electrons) and semiconductors(materials that will either conduct or not depending on thecircumstances).
Both conductors and insulators can be charged by con-tact (by placing two dissimilar objects together, the effectcan be intensified by rubbing them together) although aconductor must not be connected to a ground for the effectto accumulate.
A conductor can also be charged by induction by bring-ing another charged object such as a rubber rod near it (1,below), allowing the displaced charge to drain off on theother side (2) then removing the drain while the charged ob-ject remains (3) then finally removing the original chargedobject leaving the induced charge (4).
h− h− h− h+ h+h+h+ h+h+h−h−h− h−h−h−
1
h− h− h− h+ h+h+h+ h+h+2
h− h− h− h+ h+h+h+ h+h+3
h+h+h+ h+h+h+
4A surface charge can be induced on an insulator by po-
larization (by bringing a charged object close to the sur-face, this will cause an opposite charge on the near surfaceand a charge the same as the charged object on the far side).
h+h+h+ h+h+h+ h−h−
h−h−h−
h+h+h+h+h+
17.2 Electric ForceAccording to Coulomb’s law, the electric force between twocharges is proportional to the product of the charges andinversely proportional to the square of the distance betweenthem.
Felectric = kCq1q2r2
This is dependent on the Coulomb constant (kC):
kC = 8.987 551 788× 109 N ·m2
C2≈ 8.99× 109 N ·m2
C2
This is very similar to the equation for universal gravita-tion, substituting charge for mass and making correspond-ing changes in the constant.
Like gravity, the electric force is a field force so it doesnot require contact to act. The resultant electric force onany charge is the vector sum of the individual vector forceson that charge (the superposition principle). An timemultiple forces act on a single charged object and the sumof the forces is zero the object is said to be in equilibrium.
17.3 The Electric Field
Electric Field StrengthA charged object sets up an electric field around it, whena second charge is present in this field an electric force willresult. The strength of the electric field (E) is defined to bethe force that would be experienced by a small positive testcharge (q0) divided by the magnitude of the test charge.Since the field strength is a force divided by a charge theunits are newtons per coulomb (N
C ). Mathematically, thiscan be expressed as
E =Felectric
q0
Combining the above equation with Coulomb’s law gives anequation that will give the electric field strength around acharge
E =kCq
r2
Since the test charge is always positive a positive charge willproduce a field directed away from the charge, a negativecharge will produce a field directed inward. The directionof the electric field vector, ~E, is the direction in which anelectric force would act on a positive test charge. The equa-tion will provide the magnitude of the vector, the directionis found from the sign of the charge.
Electric Field LinesA convenient aid for visualizing electric field patterns is todraw electric field lines. They consist of lines drawn tan-gent to the electric field vector at any point, the numberof lines drawn being proportional to the magnitude of the
Honors Physics 2008–2009 Mr. Strong
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field strength. It is important to remember that electricfield lines do not actually exist as drawn since the field iscontinuous at every point outside of a conductor; the fieldlines are only drawn as a representation of the field.Rules for drawing field lines
• Field lines must begin on positive charges or at in-finity and must terminate on negative charges or atinfinity.
• The number of lines drawn leaving a positive chargeor approaching a negative charge is proportional tothe magnitude of the charge.
• No two field lines from the same field can cross eachother.
6
6
66
6
66
6
66
6
6
A single positive charge
????
??????
??
A single negative charge
Equal positive and negative charges
Equal positive charges
+2 −1
A postive charge twice the size of a negative charge
Conductors in Electrostatic EquilibriumAn electrical conductor such as a metallic object containsexcess electrons (or a lack of them) that are free to movearound the conductor. When no net motion of charge isoccurring within the conductor it is said to be in electro-static equilibrium. Any conductor in electrostatic equi-librium has the following properties:
• The electric field is zero everywhere inside the con-ductor.
• Any excess charge on an isolated conductor is entirelyon the surface of the conductor.
• The electric field immediately outside of a chargedconductor is perpendicular to the surface of the con-ductor.
• On an irregularly shaped conductor the charge willtend to accumulate where the radius of outward cur-vature is smallest. (This will produce the strongestelectric fields just outside of those points.)
For example, consider this irregular conductor with an ex-cess of electrons:
−−
−
−
−
−−
−
−
−
−−
−
−−−
−
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Chapter 18 — Electrical Energy and Capacitance
18.1 Electrical Potential EnergyA uniform electric field is an electric field that has all ofthe field lines parallel to and equidistant from each other.An electric field caused by a charge that is very far away iseffectively uniform over a small area since the field lines arenearly parallel and the difference in strength over a smalldistance is negligible. Such a field could also be caused bya pair of oppositely charged plates, the field will be uniformin the region between the plates.
A uniform electric field will always exert a constant forceon a charged object, if that object is displaced in the fieldthen a force will be required causing work to be done andenergy to be accumulated.
-
-
-
-
-
-E
-q F = qE
q q
∆d
Just as gravitational potential energy was determinedby taking the product of the weight of an object and itsheight from some reference height the same can be donefor electrical potential energy (PEelectric) by taking theproduct of the force generated by the field and its positionin the field. Mathematically, in a uniform electric field ofstrength E with a displacement d measured in the directionof E
PEelectric = −qEd
The negative sign is needed because the displacementof a positive charge to increase the potential energy shouldbe in the direction opposite the direction of the field. Ifthe initial and final positions are taken before and after adisplacement the change in potential energy can be foundthusly
∆PEelectric = −qE∆d
18.2 Potential Difference
Electric potential (V ) is defined as the electrical potentialenergy per unit charge, so
V =PEelectric
q=−qEd
q→ V = −Ed
The relative difference in electric potential between twopoints is the potential difference (∆V ) which, like elec-tric potential, is measured in J
C which are given the namevolts (V).
∆V =∆PEelectric
q= −E∆d
Note that the electric potential at a given point is inde-pendent of the charge at that point and that it’s only theelectric potential difference between two points that is use-ful, the electric potential at a single point is based on anarbitrary reference similar to position.
18.3 Capacitance
Because of time constraints the topic of capacitance wasomitted this year.
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Chapter 19 — Current and Resistance
19.1 Electric CurrentCurrentElectric current (I) is the rate at which positive chargesmove through a conductor past a fixed point. For a charge∆Q moving during a time ∆t the current is
I =∆Q∆t
Current is measured in amperes (A) which are a fun-damental SI unit from which coulombs are derived.(1 A = 1 C
s or more properly 1 C = 1 A · s)
Charge CarriersElectric charge can be either positive or negative, so eitherpositive or negative charges could be carried by the parti-cles moving to make up the current. The convention is toregard current as being the movement of positive charges,but in common conductors such as metals it is actually thenegatively charged electrons that are moving. In such a casethe current is considered to be positive charges moving inthe opposite direction as shown below.
i+ -
-
i+ - i+ -
-i++i+ i− i -
i−
EquivalentConventional
Current
Motion of ChargeCarriers
When studying current on a macroscopic scale there isno discernible difference between the actual motion of thecharge carriers and the conventional current, it is whenstudying the charge carriers themselves that a distinctionmust be made.
Sources and Types of CurrentThe charge carriers (electrons, protons) are always present,all source of current does is move them to points with apotential difference between them and then allow them toflow back to a lower energy state through some device.
Batteries do this by converting chemical energy to elec-trical energy, generators do this by converting mechanicalenergy to electrical energy. Batteries always produce a fixedpotential difference between their terminals (direct cur-rent, limited of course as the battery runs out of storedchemical energy), generators can either produce the sametype of potential difference or one that reverses itself manytimes a second (alternating current). Alternating cur-rent has several practical advantages for energy transmis-sion, we will address them later.
19.2 ResistanceThe current through some circuit depends on the potentialdifference in the current source, a larger potential differ-
ence produces a larger current. If the circuit is replacedwith a different one that can also change the current thatwill flow, some materials allow current to flow more easilythan others. The opposition to the flow of current througha conductor is called resistance (R) which is defined to be
R =∆VI
Resistance is measured in VA which are called ohms (Ω).
Ohm’s LawOhm’s law states that for many materials the resistance isconstant over a wide range of applied potential differences.Mathematically
∆VI
= constant
Materials that follow this relationship are said to be ohmic,others (such as semiconductors) are non-ohmic. Materi-als may be ohmic at one temperature and non-ohmic atanother.
An easy way to remember Ohm’s law as well as toquickly derive it for any of the values is to show it thisway:
∆V
I R
If you put our finger on top of any one of the three quan-tities (∆V , I, or R) the remaining two will be in the rightposition to tell you what to do with them to find the oneyou have covered. For example, to find current cover the Iwith your finger, what you then see is ∆V
R which is equal tothe current.
R depends on length, cross-section, material, tempera-ture. Longer conductors, smaller cross-sections, and highertemperatures all increase the resistance of a conductor andvice versa, replacing the material making up the conductorwith another material will also change the resistance (forexample, copper has lower resistance than aluminum).
Human Body ResistanceThe resistance of the human body is about 500 000 Ω if theskin is dry, if the skin is soaked with salt water (or sweat)the resistance can drop to 100 Ω or so. A current throughyour body less than about 0.01 A is either imperceptibleor felt as a slight tingling, greater currents are painful, anda current of about 0.15 A through the chest cavity can befatal.
SuperconductorsSome materials have zero resistance below a certain crit-ical temperature, these materials are known as super-
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conductors. The critical temperatures are generally verylow (less than 10 K for most materials that superconduct)although some have been found that superconduct up toabout 125 K.
19.3 Electric PowerJust as power was studied previously as the rate of con-version of mechanical energy it is also the rate at whichelectrical energy is transferred. Since
P =W
∆t=
∆PE∆t
=q∆V∆t
and I =q
∆t
it follows that power is then
P = I∆V
or that power is the product of the current and the potentialdifference. Combining the equation for power with Ohm’s
law (∆V = IR) yields the relationships
P = I2R and P =∆V 2
R
When electric utilities sell energy they use the derivedunit kilowatt-hours (kW · h), one of which is defined tobe the energy delivered at a constant rate of 1 kW in1 h. A brief exercise in algebra will show that 1 kW · h =3.6× 106 J.
Since power converted by a resistive load is P = I2R,the power lost to resistance in a wire is proportional to thesquare of the current. Reducing the current provides a con-siderable reduction in lost power so electrical transmissionlines operate at very high potential differences to keep thecurrent low, up to 765 000 V in the highest potential differ-ence lines. Transformers are used to reduce that potentialdifference to 120 V close to the point where it will be used.
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Chapter 20 — Circuits and Circuit Elements
20.1 Schematic Diagrams and Circuits
Electric circuits are generally drawn as schematic dia-grams that show the electrical components and their con-nections but leave out unnecessary detail. There are severalversions of the symbols used, the ones most commonly usedby your textbook are:
Component Symbol Comments
Wire Wires are representedby drawing lines be-tween the elements to beconnected
Resistor orload
All resistive loads sharethis symbol. If a distinc-tion needs to be madeit is generally made byadding a label.
Lamp A lamp (or bulb) isshown as a resistor witha circle around it to rep-resent the glass
Battery The longer line rep-resents the positiveterminal.
Switch An open switch is shown.A closed switch wouldbring the arm down tomake contact on bothsides.
A schematic diagram of a simple flashlight could look likethis
−∆V+
The battery is shown on the right, the lamp on the left, anda switch is at the top to allow the circuit to be opened orclosed as desired. No details such as the flashlight housing,lens, or reflector are shown, they do not affect the electricalfunction of the flashlight so they are omitted.
The battery provides a potential difference at two pointsin the circuit, when the switch is open nothing will happen.When the switch is closed then the circuit is completed andthe current can flow through the lamp, the amount of cur-rent flowing will depend on the potential difference providedby the battery and the resistance of the lamp.
20.2 Resistors In Series or In ParallelFinding the current in a circuit that only contains a singleresistor is simple, but finding the current in circuits withmore than one resistor becomes more complex.
Resistors In Series
−∆V+
R1
R2
In the case where multiple resistors are connected in se-ries the fact that the same current must flow through all ofthem leads to the equations
Req = R1 +R2 +R3 + . . .
∆Vt = ∆V1 + ∆V2 + ∆V3 + . . .
It = I1 = I2 = I3 . . .
which indicates that in a series circuit the equivalent re-sistance is equal to the sum of the individual resistances.All of the resistances can be replaced with a single resistorequal to their sum and the current will be the same
−∆V+
Req
Once the current has been found through the equivalent re-sistance the potential difference across each resistor (∆Vi)can be found by once again taking advantage of the fact thethe current is the same through resistors in series and thesum of the individual potential differences is equal to thetotal potential difference, so
∆Vi = ItRi
If any element in a series circuit stops conducting then theflow of current through the entire circuit will stop as if aswitch had been opened at that point.
Resistors In Parallel
−∆V+
R1 R2
For the case where multiple resistors are connected in seriesthe fact that the potential difference across all of the resis-tors is the same and the total current must be the sum ofthe individual currents through each resistor leads to
1Req
=1R1
+1R2
+1R3
. . .
∆Vt = ∆V1 = ∆V2 = ∆V3 . . .
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It = I1 + I2 + I3 . . .
The current through each resistor (Ii) can be found with
Ii =∆Vt
Ri
Because of the reciprocal relationship used in findingthe equivalent resistance for a parallel circuit the equiva-lent resistance will always be smaller than the smallest ofthe individual resistances.
If any element in a parallel circuit stops conducting thenthe flow of electricity will continue through the remainingparallel elements.
20.3 Complex Resistor Combinations
−∆V+
R1
R2 R3
In a circuit like the one above with both parallel and serieselements the methods to find each type of equivalent resis-tance must be applied repetitively until only one equivalentresistance is left. In this circuit R1 is in series with the com-bination of R2 and R3 in parallel, so first the two parallelresistors R2 and R3 combine in parallel to give the equiva-lent resistance R2,3, then R2,3 and R1 combine in series togive the overall equivalent resistance for the entire circuit.
Once the equivalent resistance has been found you workbackward from there to find the current in and potentialdifference across each resistor.
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Chapter 21 — Magnetism
21.1 Magnets and Magnetic Fields
Every magnet has two magnetic poles, one designated asnorth, one as south. A single magnet can be cut or brokeninto two pieces, each piece of a magnet will also be a magnetwith two poles. (It is impossible to have a single magneticpole without the corresponding opposite pole.) Like mag-netic poles repel each other, unlike poles attract.
Just as an electric charge creates an electric field, amagnetic dipole (a pair of magnetic poles) creates a mag-netic field around it and interacting with other magneticdipoles in the vicinity.
The magnetic field strength (B) is measured in tesla(T) which are N
A·m (this is related to the force exerted on amoving charge).
The direction of a magnetic field is defined as the direc-tion that the north pole of a magnet would point if placedin the field. The magnetic field of a magnet points from thenorth pole of the magnet to the south pole. By convention,when magnetic field lines are drawn they are representedas arrows when laying in the plane of the page, if they arecoming up out of the page they are represented by a solidcircle, and if they are going down into the page they arerepresented by an ‘x’. It may help to think of an arrow likewould be shot from a bow, if it is coming toward you thenyou will see the point, if it is moving away then you will seethe feathers on the end.
The magnetic north pole of the earth corresponds to thegeographic south pole and vice versa.
21.2 Electromagnetism and Magnetic Do-mains
A magnetic field exists around any current-carrying wire,the direction of the field around the wire follows a circularpath around the wire according to the right-hand rule — ifyou point the thumb of your right hand in the direction ofthe current in the wire and curl your fingers the magneticfield will follow your fingers.
The magnetic field created by a solenoid or coil is similarto the magnetic field of a permanent magnet, the directionis once again determined by the right-hand rule — this timeif you wrap the fingers of your right hand in the directionthat the current flows in the coil then the magnetic fieldinside the coil will be in the direction of your thumb andwill be nearly uniform within the coil.
A magnetic domain is a group of atoms whose mag-netic fields are aligned. In some materials (wood, plastic,etc.) the magnetic fields of the individual atoms are ar-ranged randomly and can not be aligned, others such asiron, cobalt, and nickel the magnetic fields do not com-
pletely cancel. These are known as ferromagnetic materi-als. When a substance becomes magnetized by an externalmagnetic field the individual domains are oriented to pointin the same direction. If a material is easily magnetized thenthe domains will return to a random organization when theexternal field is removed, if it is not easily magnetized thenthe organization of the domains will remain when the fieldis removed and the substance will remain magnetized.
21.3 Magnetic Force
Charged Particles in a Magnetic FieldWhen a charge moves through a magnetic field it will expe-rience a force caused by the magnetic field. This force hasa maximum when the charge moves perpendicularly to themagnetic field, decreases at other angles, and becomes zerowhen the charge moves along the field lines.
The magnitude of the magnetic field (B) is given by
B =Fmagnetic
qv
where q is the moving charge and v is its velocity.The direction of the force on a positive charge moving
through a magnetic field can be found using the right-handrule. If you point the fingers in the direction of the mag-netic field and your thumb in the direction of the motion ofthe charge and then bend your fingers at an angle the forcewill then be in the direction of your fingers. If the chargeis negative then find the direction for a positive charge andthen reverse it.
A charged particle moving perpendicularly to a uniformmagnetic field will move in a circular path. If the particle isnot moving perpendicularly to the magnetic field then thecomponent of the particle’s velocity in the direction of thefield will remain unchanged and the particle will move in ahelical path.
Magnetic Force on a Current-Carrying ConductorA length of wire, l, within an external magnetic field, B,carrying a current, I, undergoes a magnetic force of
Fmagnetic = BIl
Two parallel current-carrying wires exert on one anotherforces that are equal in magnitude and opposite in direction.If the currents are in the same direction the two wires at-tract one another, if they are in opposite directions thenthey repel each other.
The magnetic fields developed by conductors in an elec-tric field are used to produce sound with loudspeakers andare also used to cause the needle to move in a galvanometer.
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Chapter 22 — Induction and Alternating Current
22.1 Induced CurrentMoving a conductor in an electric field or changing the mag-netic field around a conductor will induce an emf (alsoknown as an electromotive force, but it is actually a po-tential difference and not a force) in the conductor throughelectromagnetic induction. The charges already presentin the conductor will be caused to move in opposite di-rections by the motion of the conductor through the field,thus a current can be induced without a battery or othersource present. This effect is greatest when the motion ofthe conductor is perpendicular to both the conductor andthe magnetic field.
If a loop (or coil) of wire is moved linearly through amagnetic field the orientation of the loop will affect the cur-rent induced, the current will be greatest when the planeof the loop is perpendicular to the magnetic field and zerowhen it is parallel to the field.
Induced emf for a moving loop can be calculated usingFaraday’s law of magnetic induction
emf = −N∆(AB cos θ)∆t
where A is the area of the loop, B is the strength of themagnetic field, θ is the angle the field lines make with anormal vector to the plane of the loop, and N is the num-ber of turns in the loop. The effect of this equation is thatif more magnetic field lines are cut by the loop of wire theemf will be increased, whether by moving the loop or bychanging the field itself.
Lenz’s law states that the magnetic field of the inducedcurrent opposes the change in the applied magnetic field.
22.2 Alternating Current, Generators, andMotorsGenerators use induction to convert mechanical energy toelectrical energy. The produce a continuously changing emfby rotating a coil of wire inside a uniform magnetic field.When the loop is perpendicular to the magnetic field everyportion of the loop is moving parallel to the magnetic fieldso there is no current induced, when the loop is parallelto the magnetic field then the sides of the loop are cross-ing the largest number of magnetic field lines so the largest
emf is then produced. The emf will vary from zero to themaximum twice every rotation of the coil and will follow asinusoidal path, this is called alternating current.
The maximum potential difference is only reached dur-ing a small part of each cycle, the effective potential dif-ference is the square root of the average of the square ofthe potential difference over time, known as the root-mean-square potential difference (∆Vrms). This and the similarexpression for current are
∆Vrms =∆Vmax√
2= 0.707∆Vmax
Irms =Imax√
2= 0.707Imax
The rms potential difference can be used as the potentialdifference term and the rms current can be used as the cur-rent term in all of the electrical equations from previouschapters. Electrical meters also usually read rms values foralternating current.
Motors use an arrangement similar to that of generatorsto convert electrical energy into mechanical energy, the dif-ference is that in a motor the electrical energy is convertedto mechanical energy, a generator does the opposite.
22.3 Inductance
Mutual inductance involves the induction of a current inone circuit by means of a changing current in another cir-cuit. The example seen most often is a transformer, twocoils of wire wound on the same iron core, this will changethe potential difference of an alternating current source. Fora transformer with primary and secondary windings of N1
and N2 turns, an input potential difference of ∆V1 will bechanged into an output potential difference ∆V2
∆V2 =N2
N1∆V1
The power produced by the output of a transformer isthe same as the power consumed by the transformer, soP1 = P2 and combining the two equations eventually yields∆V1I1 = ∆V2I2.
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Variables and Notation
SI Prefixes
Prefix Mult. Abb. Prefix Mult. Abb.
yocto- 10−24 y yotta- 1024 Yzepto- 10−21 z zetta- 1021 Zatto- 10−18 a exa- 1018 Efemto- 10−15 f peta- 1015 Ppico- 10−12 p tera- 1012 Tnano- 10−9 n giga- 109 Gmicro- 10−6 µ mega- 106 Mmilli- 10−3 m kilo- 103 kcenti- 10−2 c hecto- 102 hdeci- 10−1 d deka- 101 da
Units
Symbol Unit Quantity Composition
kg kilogram Mass SI base unitm meter Length SI base units second Time SI base unitA ampere Electric current SI base unitcd candela Luminous intensity SI base unitK kelvin Temperature SI base unit
mol mole Amount SI base unit
Ω ohm Resistance VA or m2·kg
s3·A2
C coulomb Charge A · sF farad Capacitance C
V or s4·A2
m2·kg
H henry Inductance V·sA or m2·kg
A2·s2
Hz hertz Frequency s−1
J joule Energy N ·m or kg·m2
s2
N newton Force kg·ms2
rad radian Angle mm or 1
T tesla Magnetic field NA·m
V volt Electric potential JC or m2·kg
s3·A
W watt Power Js or kg·m2
s3
Wb weber Magnetic flux V · s or kg·ms2·A
Notation
Notation Description
~x Vectorx Scalar, or the magnitude of ~x|~x| The absolute value or magnitude of ~x∆x Change in x∑x Sum of all x
Πx Product of all xxi Initial value of xxf Final value of xx Unit vector in the direction of x
A→ B A implies BA ∝ B A is proportional to BA B A is much larger than B
Greek Alphabet
Name Maj. Min. Name Maj. Min.
Alpha A α Nu N νBeta B β Xi Ξ ξ
Gamma Γ γ Omicron O oDelta ∆ δ Pi Π π or $
Epsilon E ε or ε Rho P ρ or %Zeta Z ζ Sigma Σ σ or ςEta H η Tau T τ
Theta Θ θ or ϑ Upsilon Υ υIota I ι Phi Φ φ or ϕ
Kappa K κ Chi X χLambda Λ λ Psi Ψ ψ
Mu M µ Omega Ω ω
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Variables
Variable Description Units
α Angular acceleration rads2
θ Angular position radθc Critical angle o (degrees)θi Incident angle o (degrees)θr Refracted angle o (degrees)θ′ Reflected angle o (degrees)∆θ Angular displacement radτ Torque N ·mω Angular speed rad
sµ Coefficient of friction (unitless)µk Coefficient of kinetic friction (unitless)µs Coefficient of static friction (unitless)~a Acceleration m
s2
~ac Centripetal acceleration ms2
~ag Gravitational acceleration ms2
~at Tangential acceleration ms2
~ax Acceleration in the x direction ms2
~ay Acceleration in the y direction ms2
A Area m2
B Magnetic field strength TC Capacitance F~d Displacement m
d sin θ lever arm m~dx or ∆x Displacement in the x direction m~dy or ∆y Displacement in the y direction m
E Electric field strength NC
f Focal length m~F Force N~Fc Centripetal force N
~Felectric Electrical force N~Fg Gravitational force N~Fk Kinetic frictional force N
~Fmagnetic Magnetic force N~Fn Normal force N~Fs Static frictional force N~F∆t Impulse N · s or kg·m
s~g Gravitational acceleration m
s2
Variable Description Units
h Object height mh′ Image height mI Current AI Moment of inertia kg ·m2
KE or K Kinetic energy JKE rot Rotational kinetic energy JL Angular Momentum kg·m2
sL Self-inductance Hm Mass kgM Magnification (unitless)M Mutual inductance HMA Mechanical Advantage (unitless)ME Mechanical Energy Jn Index of refraction (unitless)p Object distance m~p Momentum kg·m
s
P Power WPE or U Potential Energy JPE elastic Elastic potential energy JPE electric Electrical potential energy J
PE g Gravitational potential energy Jq Image distance m
q or Q Charge CQ Heat, Entropy JR Radius of curvature mR Resistance Ωs Arc length mt Time s
∆t Time interval s~v Velocity m
s
vt Tangential speed ms
~vx Velocity in the x direction ms
~vy Velocity in the y direction ms
V Electric potential V∆V Electric potential difference VV Volume m3
W Work Jx or y Position m
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Constants
Symbol Name Established Value Value Used
ε0 Permittivity of a vacuum 8.854 187 817× 10−12 C2
N·m2 8.85× 10−12 C2
N·m2
φ Golden ratio 1.618 033 988 749 894 848 20
π Archimedes’ constant 3.141 592 653 589 793 238 46
g, ag Gravitational acceleration constant 9.79171 ms2 (varies by location) 9.81 m
s2
c Speed of light in a vacuum 2.997 924 58× 108 ms (exact) 3.00× 108 m
s
e Natural logarithmic base 2.718 281 828 459 045 235 36
e− Elementary charge 1.602 177 33× 1019 C 1.60× 1019 C
G Gravitational constant 6.672 59× 10−11 N·m2
kg2 6.67× 10−11 N·m2
kg2
kC Coulomb’s constant 8.987 551 788× 109 N·m2
C2 8.99× 109 N·m2
C2
NA Avogadro’s constant 6.022 141 5× 1023 mol−1
Astronomical Data
Symbol Object Mean Radius Mass Mean Orbit Radius Orbital Period
Á Moon 1.74× 106 m 7.36× 1022 kg 3.84× 108 m 2.36× 106 s
À Sun 6.96× 108 m 1.99× 1030 kg — —
 Mercury 2.43× 106 m 3.18× 1023 kg 5.79× 1010 m 7.60× 106 s
à Venus 6.06× 106 m 4.88× 1024 kg 1.08× 1011 m 1.94× 107 s
Ê Earth 6.37× 106 m 5.98× 1024 kg 1.496× 1011 m 3.156× 107 s
Ä Mars 3.37× 106 m 6.42× 1023 kg 2.28× 1011 m 5.94× 107 s
Ceres1 4.71× 105 m 9.5× 1020 kg 4.14× 1011 m 1.45× 108 s
Å Jupiter 6.99× 107 m 1.90× 1027 kg 7.78× 1011 m 3.74× 108 s
Æ Saturn 5.85× 107 m 5.68× 1026 kg 1.43× 1012 m 9.35× 108 s
Ç Uranus 2.33× 107 m 8.68× 1025 kg 2.87× 1012 m 2.64× 109 s
È Neptune 2.21× 107 m 1.03× 1026 kg 4.50× 1012 m 5.22× 109 s
É Pluto1 1.15× 106 m 1.31× 1022 kg 5.91× 1012 m 7.82× 109 s
Eris21 2.4× 106 m 1.5× 1022 kg 1.01× 1013 m 1.75× 1010 s
1Ceres, Pluto, and Eris are classified as “Dwarf Planets” by the IAU2Eris was formerly known as 2003 UB313
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Mathematics Review for Physics
This is a summary of the most important parts of math-ematics as we will use them in a physics class. There arenumerous parts that are completely omitted, others aregreatly abridged. Do not assume that this is a completecoverage of any of these topics.
AlgebraFundamental properties of algebra
a+ b = b+ a Commutative law for ad-dition
(a+ b) + c = a+ (b+ c) Associative law for addi-tion
a+ 0 = 0 + a = a Identity law for additiona+ (−a) = (−a) + a = 0 Inverse law for addition
ab = ba Commutative law formultiplication
(ab)c = a(bc) Associative law for mul-tiplication
(a)(1) = (1)(a) = a Identity law for multipli-cation
a 1a = 1
aa = 1 Inverse law for multipli-cation
a(b+ c) = ab+ ac Distributive law
Exponents(ab)n = anbn (a/b)n = an/bn
anam = an+m 0n = 0an/am = an−m a0 = 1(an)m = a(mn) 00 = 1 (by definition)
Logarithms
x = ay → y = loga x
loga(xy) = loga x+ loga y
loga
(x
y
)= loga x− loga y
loga(xn) = n loga x
loga
(1x
)= − loga x
loga x =logb x
logb a= (logb x)(loga b)
Binomial Expansions
(a+ b)2 = a2 + 2ab+ b2
(a+ b)3 = a3 + 3a2b+ 3ab2 + b3
(a+ b)n =n∑
i=0
n!i!(n− i)!
aibn−i
Quadratic formulaFor equations of the form ax2 +bx+c = 0 the solutions are:
x =−b±
√b2 − 4ac
2a
Geometry
Shape Area Volume
Triangle A = 12bh —
Rectangle A = lw —
Circle A = πr2 —
Rectangularprism
A = 2(lw + lh+ hw) V = lwh
Sphere A = 4πr2 V = 43πr
3
Cylinder A = 2πrh+ 2πr2 V = πr2h
Cone A = πr√r2 + h2 + πr2 V = 1
3πr2h
TrigonometryIn physics only a small subset of what is covered in atrigonometry class is likely to be used, in particular sine, co-sine, and tangent are useful, as are their inverse functions.As a reminder, the relationships between those functionsand the sides of a right triangle are summarized as follows:
θ
opp
adj
hypsin θ =
opp
hypcos θ =
adj
hyp
tan θ =opp
adj=
sin θcos θ
The inverse functions are only defined over a limited range.The tan−1 x function will yield a value in the range −90 <θ < 90, sin−1 x will be in −90 ≤ θ ≤ 90, and cos−1 xwill yield one in 0 ≤ θ ≤ 180. Care must be taken toensure that the result given by a calculator is in the correctquadrant, if it is not then an appropriate correction mustbe made.
Degrees 0o 30o 45o 60o 90o 120o 135o 150o 180o
Radians 0 π6
π4
π3
π2
2π3
3π4
5π6 π
sin 0 12
√2
2
√3
2 1√
32
√2
212 0
cos 1√
32
√2
212 0 − 1
2 −√
22 −
√3
2 −1
tan 0√
33 1
√3 ∞ −
√3 −1 −
√3
3 0
sin2 θ + cos2 θ = 1 2 sin θ cos θ = sin(2θ)
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Trigonometric functions in terms of each other
sin θ = sin θ√
1− cos2 θ tan θ√1+tan2 θ
cos θ =√
1− sin2 θ cos θ 1√1+tan2 θ
tan θ = sin θ√1−sin2 θ
√1−cos2 θcos θ tan θ
csc θ = 1sin θ
1√1−cos2 θ
√1+tan2 θtan θ
sec θ = 1√1−sin2 θ
1cos θ
√1 + tan2 θ
cot θ =√
1−sin2 θ
sin θcos θ√
1−cos2 θ1
tan θ
sin θ = 1csc θ
√sec2 θ−1sec θ
1√1+cot2 θ
cos θ =√
csc2 θ−1csc θ
1sec θ
cot θ√1+cot2 θ
tan θ = 1√csc2 θ−1
√sec2 θ − 1 1
cot θ
csc θ = csc θ sec θ√sec2 θ−1
√1 + cot2 θ
sec θ = csc θ√csc2 θ−1
sec θ√
1+cot2 θcot θ
cot θ =√
csc2 θ − 1 1√sec2 θ−1
cot θ
Law of sines, law of cosines, area of a triangle
HH
HHHH
A
B
C
a
b
c
a2 = b2 + c2 − 2bc cosAb2 = a2 + c2 − 2ac cosBc2 = a2 + b2 − 2ab cosC
a
sinA=
b
sinB=
c
sinC
let s = 12 (a+ b+ c)
Area=√s(s− a)(s− b)(s− c)
Area= 12bc sinA = 1
2ac sinB = 12ab sinC
VectorsA vector is a quantity with both magnitude and direction,such as displacement or velocity. Your textbook indicatesa vector in bold-face type as V and in class we have beenusing ~V . Both notations are equivalent.
A scalar is a quantity with only magnitude. This caneither be a quantity that is directionless such as time ormass, or it can be the magnitude of a vector quantity suchas speed or distance traveled. Your textbook indicates ascalar in italic type as V , in class we have not done any-thing to distinguish a scalar quantity. The magnitude of ~Vis written as V or |~V |.
A unit vector is a vector with magnitude 1 (a dimen-sionless constant) pointing in some significant direction. Aunit vector pointing in the direction of the vector ~V is in-dicated as V and would commonly be called V -hat. Anyvector can be normalized into a unit vector by dividing it byits magnitude, giving V = ~V
V . Three special unit vectors, ı,, and k are introduced with chapter 3. They point in thedirections of the positive x, y, and z axes, respectively (asshown below).
6
?
-
y+
−
x+−
z+
−6
-
ı
k
Vectors can be added to other vectors of the same di-mension (i.e. a velocity vector can be added to anothervelocity vector, but not to a force vector). The sum of allvectors to be added is called the resultant and is equivalentto all of the vectors combined.
Multiplying VectorsAny vector can be multiplied by any scalar, this has theeffect of changing the magnitude of the vector but not itsdirection (with the exception that multiplying a vector bya negative scalar will reverse the direction of the vector).As an example, multiplying a vector ~V by several scalarswould give:
- -
-
~V 2~V
.5~V −1~V
In addition to scalar multiplication there are also twoways to multiply vectors by other vectors. They will not bedirectly used in class but being familiar with them may helpto understand how some physics equations are derived. Thefirst, the dot product of vectors ~V1 and ~V2, represented as~V1 · ~V2 measures the tendency of the two vectors to point inthe same direction. If the angle between the two vectors isθ the dot product yields a scalar value as
~V1 · ~V2 = V1V2 cos θ
The second method of multiplying two vectors, thecross product, (represented as ~V1× ~V2) measures the ten-dency of vectors to be perpendicular to each other. It yieldsa third vector perpendicular to the two original vectors withmagnitude
|~V1 × ~V2| = V1V2 sin θ
The direction of the cross product is perpendicular to thetwo vectors being crossed and is found with the right-handrule — point the fingers of your right hand in the directionof the first vector, curl them toward the second vector, andthe cross product will be in the direction of your thumb.
Adding Vectors GraphicallyThe sum of any number of vectors can be found by drawingthem head-to-tail to scale and in proper orientation thendrawing the resultant vector from the tail of the first vector
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to the point of the last one. If the vectors were drawn ac-curately then the magnitude and direction of the resultantcan be measured with a ruler and protractor. In the ex-ample below the vectors ~V1, ~V2, and ~V3 are added to yield~Vr
-
AAAAAAU
~V1
~V2 ~V3
~V1, ~V2, ~V3
-AAAAAAU
~V1
~V2
~V3
~V1 + ~V2 + ~V3
-AAAAAAU
~V1
~V2
~V3
-~Vr
~Vr = ~V1 + ~V2 + ~V3
Adding Parallel VectorsAny number of parallel vectors can be directly added byadding their magnitudes if one direction is chosen as pos-itive and vectors in the opposite direction are assigned anegative magnitude for the purposes of adding them. Thesum of the magnitudes will be the magnitude of the resul-tant vector in the positive direction, if the sum is negativethen the resultant will point in the negative direction.
Adding Perpendicular VectorsPerpendicular vectors can be added by drawing them as aright triangle and then finding the magnitude and directionof the hypotenuse (the resultant) through trigonometry andthe Pythagorean theorem. If ~Vr = ~Vx + ~Vy and ~Vx ⊥ ~Vy
then it works as follows:
-~Vx
6
~Vy
-~Vx
6
~Vy
~Vr
θ
Since the two vectors to be added and the resultant forma right triangle with the resultant as the hypotenuse thePythagorean theorem applies giving
Vr = |~Vr| =√V 2
x + V 2y
The angle θ can be found by taking the inverse tangent ofthe ratio between the magnitudes of the vertical and hori-zontal vectors, thus
θ = tan−1 Vy
Vx
As was mentioned above, care must be taken to ensure thatthe angle given by the calculator is in the appropriate quad-rant for the problem, this can be checked by looking at the
diagram drawn to solve the problem and verifying that theanswer points in the direction expected, if not then makean appropriate correction.
Resolving a Vector Into ComponentsJust two perpendicular vectors can be added to find asingle resultant, any single vector ~V can be resolved intotwo perpendicular component vectors ~Vx and ~Vy so that~V = ~Vx + ~Vy.
~V
θ ~Vx
~Vy
~V
θ
-
6
As the vector and its components can be drawn as aright triangle the ratios of the sides can be found withtrigonometry. Since sin θ = Vy
V and cos θ = Vx
V it followsthat Vx = V cos θ and Vy = V sin θ or in a vector form,~Vx = V cos θı and ~Vy = V sin θ. (This is actually an appli-cation of the dot product, ~Vx = (~V · ı)ı and ~Vy = (~V · ),but it is not necessary to know that for this class)
Adding Any Two Vectors AlgebraicallyOnly vectors with the same direction can be directly added,so if vectors pointing in multiple directions must be addedthey must first be broken down into their components, thenthe components are added and resolved into a single resul-tant vector — if in two dimensions ~Vr = ~V1 + ~V2 then
*
~V1
~V2
~Vr
~Vr = ~V1 + ~V2
*
-
-6
6
~V1
~V2
~Vr
~V1x
~V2x
~V1y
~V2y
~V1 = ~V1x + ~V1y
~V2 = ~V2x + ~V2y
~Vr = (~V1x + ~V1y) + (~V2x + ~V2y)
*
- -
6
6
~V1
~V2
~Vr
~V1x~V2x
~V1y
~V2y
~Vr = (~V1x + ~V2x) + (~V1y + ~V2y)
Once the sums of the component vectors in each direc-tion have been found the resultant can be found from themjust as an other perpendicular vectors may be added. Since
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39
from the last figure ~Vr = (~V1x+~V2x)+(~V1y +~V2y) and it waspreviously established that ~Vx = V cos θı and ~Vy = V sin θit follows that
~Vr = (V1 cos θ1 + V2 cos θ2) ı+ (V1 sin θ1 + V2 sin θ2)
and
Vr =√
(V1x + V2x)2 + (V1y + V2y)2
=√
(V1 cos θ1 + V2 cos θ2)2 + (V1 sin θ1 + V2 sin θ2)
2
with the direction of the resultant vector ~Vr, θr, being foundwith
θr = tan−1 V1y + V2y
V1x + V2x= tan−1 V1 sin θ1 + V2 sin θ2
V1 cos θ1 + V2 cos θ2
Adding Any Number of Vectors Algebraically
For a total of n vectors ~Vi being added with magnitudes Vi
and directions θi the magnitude and direction are:
*
- - -
6
6
6
~V1
~V2
~V3
~Vr
~V1x~V2x
~V3x
~V1y
~V2y
~V3y
~Vr =
(n∑
i=1
Vi cos θi
)ı+
(n∑
i=1
Vi sin θi
)
Vr =
√√√√( n∑i=1
Vi cos θi
)2
+
(n∑
i=1
Vi sin θi
)2
θr = tan−1
n∑i=1
Vi sin θi
n∑i=1
Vi cos θi
(The figure shows only three vectors but this method willwork with any number of them so long as proper care istaken to ensure that all angles are measured the same wayand that the resultant direction is in the proper quadrant.)
CalculusAlthough this course is based on algebra and not calculus,it is sometimes useful to know some of the properties ofderivatives and integrals. If you have not yet learned cal-culus it is safe to skip this section.
DerivativesThe derivative of a function f(t) with respect to t is afunction equal to the slope of a graph of f(t) vs. t at everypoint, assuming that slope exists. There are several waysto indicate that derivative, including:
ddtf(t)
f ′(t)
f(t)
The first one from that list is unambiguous as to the in-dependent variable, the others assume that there is only onevariable, or in the case of the third one that the derivativeis taken with respect to time. Some common derivativesare:
ddttn = ntn−1
ddt
sin t = cos t
ddt
cos t = − sin t
ddtet = et
If the function is a compound function then there are afew useful rules to find its derivative:
ddtcf(t) = c
ddtf(t) c constant for all t
ddt
[f(t) + g(t)] =ddtf(t) +
ddtg(t)
ddtf(u) =
ddtu
dduf(u)
IntegralsThe integral, or antiderivative of a function f(t) withrespect to t is a function equal to the the area under agraph of f(t) vs. t at every point plus a constant, assumingthat f(t) is continuous. Integrals can be either indefinite ordefinite. An indefinite integral is indicated as:∫
f(t) dt
while a definite integral would be indicated as∫ b
a
f(t) dt
to show that it is evaluated from t = a to t = b, as in:∫ b
a
f(t) dt =∫f(t) dt|t=b
t=a =∫f(t) dt|t=b −
∫f(t) dt|t=a
Some common integrals are:∫tn dt =
tn+1
n+ 1+ C n 6= −1
Honors Physics 2008–2009 Mr. Strong
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∫t−1 dt = ln t+ C∫
sin tdt = − cos t+ C∫cos tdt = sin t+ C∫et dt = et + C
There are rules to reduce integrals of some compoundfunctions to simpler forms (there is no general rule to reducethe integral of the product of two functions):∫
cf(t) dt = c
∫f(t) dt c constant for all t∫
[f(t) + g(t)] dt =∫f(t) dt+
∫g(t) dt
Series Expansions
Some functions are difficult to work with in their normalforms, but once converted to their series expansion can bemanipulated easily. Some common series expansions are:
sinx = x− x3
3!+x5
5!− x7
7!+ · · · =
∞∑i=0
(−1)i x2i+1
(2i+ 1)!
cosx = 1− x2
2!+x4
4!− x6
6!+ · · · =
∞∑i=0
(−1)i x2i
(2i)!
ex = 1 + x+x2
2!+x3
3!+x4
4!+ · · · =
∞∑i=0
xi
(i)!
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Physics Using Calculus
In our treatment of mechanics the majority of the equa-tions that have been covered in the class have been eitherapproximations or special cases where the acceleration orforce applied have been held constant. This is requiredbecause the class is based on algebra and to do otherwisewould require the use of calculus.
None of what is presented here is a required part of theclass, it is here to show how to handle cases outside theusual approximations and simplifications. Feel free to skipthis section, none of it will appear as a required part of anyassignment or test in this class.
Notation
Because the letter ‘d’ is used as a part of the notation ofcalculus the variable ‘r’ is often used to represent the posi-tion vector of the object being studied. (Other authors use‘s’ for the spatial position or generalize ‘x’ to two or moredimensions.) In a vector form that would become ~r to giveboth the magnitude and direction of the position. It is as-sumed that the position, velocity, acceleration, etc. can beexpressed as a function of time as ~r(t), ~v(t), and ~a(t) butthe dependency on time is usually implied and not shownexplicitly.
Translational Motion
Many situations will require an acceleration that is not con-stant. Everything learned about translational motion sofar used the simplification that the acceleration remainedconstant, with calculus we can let the acceleration be anyfunction of time.
Since velocity is the slope of a graph of position vs. timeat any point, and acceleration is the slope of velocity vs.time these can be expressed mathematically as derivatives:
~v =ddt~r
~a =ddt~v =
d2
dt2~r
The reverse of this relationship is that the displacementof an object is equal to the area under a velocity vs. timegraph and velocity is equal to the area under an accelera-tion vs. time graph. Mathematically this can be expressedas integrals:
~v =∫~adt
~r =∫~v dt =
∫∫~adt2
Using these relationships we can derive the main equa-tions of motion that were introduced by starting with the
integral of a constant velocity ~a(t) = ~a as∫ tf
ti
~adt = ~a∆t+ C
The constant of integration, C, can be shown to be the ini-tial velocity, ~vi, so the entire expression for the final velocitybecomes
~vf = ~vi + ~a∆t
or~vf (t) = ~vi + ~at
Similarly, integrating that expression with respect to timegives an expression for position:∫ tf
ti
(~vi + ~at) dt = ~vi∆t+12~a∆t2 + C
Once again the constant of integration, C, can be shown tobe the initial position, ~ri, yielding an expression for positionvs. time
~rf (t) = ~ri + ~vit+12~at2
While calculus can be used to derive the algebraic formsof the equations the reverse is not true. Algebra can han-dle a subset of physics and only at the expense of needingto remember separate equations for various special cases.With calculus the few definitions shown above will sufficeto predict any linear motion.
Work and PowerThe work done on an object is the integral of dot product ofthe force applied with the path the object takes, integratedas a contour integral over the path.
W =∫
C
~F · d~r
The power developed is the time rate of change of thework done, or the derivative of the work with respect totime.
P =ddtW
MomentumNewton’s second law of motion changes from the familiar~F = m~a to the statement that the force applied to an objectis the time derivative of the object’s momentum.
~F =ddt~p
Similarly, the impulse delivered to an object is the in-tegral of the force applied in the direction parallel to themotion with respect to time.
∆~p =∫
~F dt
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Rotational MotionThe equations for rotational motion are very similar tothose for translational motion with appropriate variablesubstitutions. First, the angular velocity is the time deriva-tive of the angular position.
~ω =d~θdt
The angular acceleration is the time derivative of the an-gular velocity or the second time derivative of the angularposition.
~α =d~ωdt
=d2~θ
dt2
The angular velocity is also the integral of the angularacceleration with respect to time.
~ω =∫~α dt
The angular position is the integral of the angular veloc-ity with respect to time or the second integral of the angularacceleration with respect to time.
~θ =∫~ω dt =
∫∫~α dt2
The torque exerted on an object is the cross product ofthe radius vector from the axis of rotation to the point ofaction with the force applied.
~τ = ~r × ~F
The moment of inertia of any object is the integral ofthe square of the distance from the axis of rotation to eachelement of the mass over the entire mass of the object.
I =∫r2 dm
The torque applied to an object is the time derivativeof the object’s angular momentum.
~τ =ddt~L
The integral of the torque with respect to time is theangular momentum of the object.
~L =∫~τ dt
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43
Data Analysis
Comparative MeasuresPercent ErrorWhen experiments are conducted where there is a calcu-lated result (or other known value) against which the ob-served values will be compared the percent error betweenthe observed and calculated values can be found with
percent error =∣∣∣∣observed value − accepted value
accepted value
∣∣∣∣× 100%
Percent DifferenceWhen experiments involve a comparison between two ex-perimentally determined values where neither is regardedas correct then rather than the percent error the percentdifference can be calculated. Instead of dividing by the ac-cepted value the difference is divided by the mean of thevalues being compared. For any two values, a and b, thepercent difference is
percent difference =∣∣∣∣ a− b
12 (a+ b)
∣∣∣∣× 100%
Dimensional AnalysisUnit ConversionsWhen converting units you need to be careful to ensure thatconversions are done properly and not accidentally reversed(or worse). The easiest way to do this is to be careful toalways keep the units associated with each number and ateach step always multiply by something equal to 1.
As an example, if you have a distance of 78.25 km andwant to convert it to meters the immediate reaction is tosay that you can multiply by 1000, or is that divide by1000. Knowing that 1 km = 1000 m it is obvious then that1 km
1000 m = 1 = 1000 m1 km . Since both of them are equal to 1
they can be multiplied by what you wish to convert with-out changing the quantity itself, it will only change the unitsin which it is shown. So, to convert 78.25 km to meters youwould multiply by 1000 m
1 km :
(78.25 km)(
1000 m1 km
)= 78 250 m
As you can see, the km in the original quantity and in thedenominator cancel leaving meters. If you were to reversethe conversion accidentally you would get this:
(78.25 km)(
1 km1000 m
)= 0.07825
km2
m
Instead of canceling the km and leaving the measurementin meters the result is in km2
m , a meaningless quantity. Thenonsensical result is an indication that the conversion wasreversed and should be redone. If the conversion was donewithout the units then it could be reversed without realizingthat it happened.
Units on ConstantsConsider a mass vibrating on the end of a spring. The equa-tion relating the mass to the frequency on graph might looksomething like
y =3.658x2− 1.62
This isn’t what we’re looking for, a first pass would be tochange the x and y variables into f and m for frequencyand mass, this would then give
m =3.658f2
− 1.62
This is progress but there’s still a long way to go. Massis measured in kg and frequency is measured in s−1, sincethese variables represent the entire measurement (includ-ing the units) and not just the numeric portion they do notneed to be labeled, but the constants in the equation doneed to have the correct units applied.
To find the units it helps to first re-write the equationusing just the units, letting some variable such as u (or u1,u2, u3, etc.) stand for the unknown units. The exampleequation would then become
kg =u1
(s−1)2+ u2
Since any quantities being added or subtracted must havethe same units the equation can be split into two equations
kg =u1
(s−1)2and kg = u2
The next step is to solve for u1 and u2, in this case u2 = kgis immediately obvious, u1 will take a bit more effort. Afirst simplification yields
kg =u1
s−2
then multiplying both sides by s−2 gives
(s−2)(kg) =( u1
s−2
)(s−2)
which finally simplifies and solves to
u1 = kg · s−2
Inserting the units into the original equation finally yields
m =3.658 kg · s−2
f2− 1.62 kg
which is the final equation with all of the units in place.Checking by choosing a frequency and carrying out the cal-culations in the equation will show that it is dimensionallyconsistent and yields a result in the units for mass (kg) asexpected.
Honors Physics 2008–2009 Mr. Strong
44
Final Exam Description
• 95 Multiple Choice Items (22 problems, 73 conceptual questions)
• Calculators will not be permitted
• Exam items are from the following categories:
– Chapter 1 — The Science of Physicsunits of measurementsignificant figures
– Chapter 2 — Motion in One Dimensiongraphsequations of motionacceleration
– Chapter 3 — Two-Dimensional Motion andVectorsvectors and scalarsvector additioncomponentsprojectilesrelative motion
– Chapter 4 — Forces and the Laws of Motioncauses of motioninertiabalanced forcesaccelerated motionfriction
– Chapter 5 — Work and Energyworktypes of energyconservation of energypower
– Chapter 6 — Momentum and Collisionsmomentum and impulseconservation of momentumtypes of collisions
– Chapter 7 — Rotational Motion and the Lawof Gravityrotational motion equationscentripetal force and accelerationcauses of centripetal forcegravitational force equation
– Chapter 8 — Rotational Equilibrium andDynamicstorquecenter of gravityequilibrium
– Chapter 12 — Vibrations and Wavessimle harmonic motionwave propertiesinterferencestanding waves
– Chapter 13 — Soundtypes of waveswave speedinterferenceDoppler shiftresonance
– Chapter 14 — Light and Reflectioncharacteristics of lightflat mirrorscurved mirrorspolarization
– Chapter 15 — Refractionindex of refractionSnell’s lawlenses
– Chapter 16 — Interference and Diffractioninterferencediffraction
– Chapter 17 — Electric Forces and Fieldselectric chargemethods of producing net chargeconductors and insulatorselectric forces
– Chapter 18 — Electrical Energy and Capac-itanceelectric potential energypotential difference
– Chapter 19 — Current and ResistancecurrentresistanceOhm’s lawpower
– Chapter 20 — Circuits and Circuit Elementsseries and parallel resistorsresistor combinations
– Chapter 21 — Magnetismcauses of magnetismmagnetic fieldscurrent and magnetismmagnetic forces
– Chapter 22 — Induction and AlternatingCurrentinduced currentLenz’s lawmotors and generatorstransformers
Honors Physics 2008–2009 Mr. Strong
45
Final Exam Equation Sheet
This is a slightly reduced copy of the front and back of the equation sheet that you will receive with the final exam.
Honors Physics – 2nd Semester Final Constants and Equations
KinematicsLinear
avd v t=
( )12 i fd v v t= + Δ
( ) 21
2id v t a t= Δ + Δ
f iv v a t= + Δ
2 2 2f iv v ad= +
x xd v t= Δ
( ) 21
2y y yd v t a t= Δ + Δ
sinyv v θ=
cosxv v θ=
KinematicsRotational
dr
θ =
vr
ω =
a
rα =
av tθ ω= Δ
( ) 21
2i t tθ ω α= Δ + Δ
( )12 i f tθ ω ω= + Δ
f i tω ω α= + Δ
2 2f i 2ω ω αθ= +
DynamicsLinear
F ma=∑gF mg=
cosF w θ =
sinF w θ=
Nf Fμ=
tanμ θ=
springF kx= −
1 22grav
m mF G
r=
DynamicsCircular
2 rv
Tπ
=
2 2
2
4c
v ra
r Tπ
= =
2 2
2
4c
v rF m m
r Tπ
= =
1T f=
DynamicsRotational
Iτ α=∑2
point massI mr=
counterclockwise clockwiseτ τ=∑ ∑
Work, Power, EnergyLinear
cosW Fd θ=
gPE mgh=
212elasticPE kx=
212KE mv=
0PE KE WΔ + Δ + =
WP Fv
t= =
Work, Power, EnergyRotational
W τθ=21
2KE Iω=
P τω=
MomentumLinear
P mv=
1 1 2 2 1 1 2 2i f f fm v m v m v m v+ = +
F t m vΔ = Δ
Rotational
L Iω=
Constants
210. msg =
2
2
11 N mkg
6.67 10G − = ×
8 ms3.00 10c = ×
2
2
9 N mC
9.00 10K = ×
Honors Physics 2008–2009 Mr. Strong
46
Simple Harmonic Motion
2m
Tk
π=
2pendulumTg
π=
( )cosx A tω=
( )sinv A tω ω= −
( )2 cosa A tω ω= −
ka x
m= −
( )2 2kv A x
m= ± −
22 f
Tπ
ω π= =
1T
f=
Waves and Sound
v f λ=
20.1 331 1273
cmsk
Tv T= = +
331 .6ms cv T= +
os
s
v vf f
v v
±=
1 2# Beats f f= −
Light
i r =
in vacuum
in medium
light
light
vn
v
sinsin
c in
v r
= =
sin sini rn i n r =
sinsin
i
r
v iv r
=
1sin ci n =
1 1 1
o if d d= +
i i
o o
s dM
s d= =
Electrostatics
1 22
q qF k
r=
2o
F qE k
q r= =
eB A
PEV V V
q
ΔΔ − =
qV k
r=
( )B AW q V q V V= − Δ = − −
ePE qEd=
Currents and Circuits
qI
t
ΔΔ
V IR=2
2 VP VI I R
R= = =
qC
V
Series Circuits
1 2 3 ...tI I I I= = = =
1 2 3 ...tV V V V= + + +
1 2 3 ...eqR R R R= + + +
1 2 3
1 1 1 1...
eqC C C C= + + +
Parallel Circuits
1 2 3 ...tI I I I= + + +
1 2 3 ...tV V V V= = = =
1 2 3
1 1 1 1...
eqR R R R= + + +
1 2 3 ...eqC C C C= + + +
Magnetism
sinF qvB θ=
sinF I B θ=
V El Blv= =
Transformers
p s
p s
V V
N N=
p p s sV I V I=
Note that some equations may be slightly different than the ones we have used in class, the sheet that you receivewill have these forms of the equations, it will be up to you to know how to use them or to know other forms that youare able to use.
Honors Physics 2008–2009 Mr. Strong
47
Mix
edRev
iew
Exe
rcises
HRW material copyrighted under notice appearing earlier in this book.
Hol
t Phy
sics
Sec
tion
Revi
ew W
orks
heet
s4
The
Scie
nce
ofPh
ysic
s
Mix
ed R
evie
wH
OLT
PH
YS
ICS
Ch
apte
r
1
1.C
onve
rt t
he
follo
win
g m
easu
rem
ents
to t
he
un
its
spec
ifie
d.
a.2.
5 da
ys to
sec
onds
b.35
km
to m
illim
eter
s
c.43
cm
to k
ilom
eter
s
d.22
mg
to k
ilogr
ams
e.67
1 kg
to m
icro
gram
s
f.8.
76×
107
mW
to g
igaw
atts
g.1.
753
×10
–13
s to
pic
osec
onds
2.A
ccor
din
g to
th
e ru
les
give
n in
Ch
apte
r 1
ofyo
ur
text
book
,how
man
y
sign
ific
ant
figu
res
are
ther
e in
th
e fo
llow
ing
mea
sure
men
ts?
a.0.
0845
kg
b.37
.00
h
c.8
630
000.
000
mi
d.0.
000
000
0217
g
e.75
0 in
.
f.0.
5003
s
Pow
erP
refi
xA
bbre
viat
ion
10–1
deci
-d
101
deka
-da
103
kilo
-k
106
meg
a-M
109
giga
-G
1012
tera
-T
1015
peta
-P
1018
exa-
E
Pow
erP
refi
xA
bbre
viat
ion
10–1
8at
to-
a
10–1
5fe
mto
-f
10–1
2pi
co-
p
10–9
nan
o-n
10–6
mic
ro-
m
10–3
mill
i-m
10–2
cen
ti-
c
HRW material copyrighted under notice appearing earlier in this book.
Chap
ter 1
5
3.W
ith
out c
alcu
lati
ng
the
resu
lt,f
ind
the
nu
mbe
r of
sign
ific
ant
figu
res
in
the
follo
win
g pr
odu
cts
and
quot
ien
ts.
a.0.
0050
32×
4.00
09
b.0.
0080
750
÷10
.037
c.(3
.52
×10
–11 )
×(7
.823
×10
11)
4.C
alcu
late
a+
b,a
−b,
a×
b,an
da
÷b
wit
h t
he
corr
ect
nu
mbe
r of
sign
ific
ant
figu
res
usi
ng
the
follo
win
g n
um
bers
.
a.a
=0.
005
078;
b=
1.00
03
a+
b=
a−
b=
a×
b=
a÷
b=
b.a
=4.
231
19 ×
107 ;b
= 3
.654
×10
6
a+
b=
a−
b=
a×
b=
a÷
b=
5.C
alcu
late
th
e ar
ea o
fa
carp
et 6
.35
m lo
ng
and
2.50
m w
ide.
Exp
ress
you
r
answ
er w
ith
th
e co
rrec
t n
um
ber
ofsi
gnif
ican
t fi
gure
s.
6.T
he
tabl
e be
low
con
tain
s m
easu
rem
ents
of
the
tem
pera
ture
an
d vo
lum
e of
an a
ir b
allo
on a
s it
hea
ts u
p.
In t
he
grid
at
righ
t,sk
etch
a g
raph
th
at b
est
desc
ribe
s th
ese
data
.
Tem
per
atu
re
Vol
um
e (°
C)
(m3 )
20.
0502
270.
0553
520.
0598
770.
0646
102
0.07
04
127
0.07
48
152
0.07
96
025
5075
100
125
150
175
0.08
00
0.07
50
0.07
00
0.06
50
0.06
00
0.05
50
0.05
00
Volume (m3)
Tem
pera
ture
(°C
)
HO
LT P
HY
SIC
S
Mix
ed R
evie
w co
ntin
ued
Ch
apte
r
1
Honors Physics 2008–2009 Mr. Strong
48HRW material copyrighted under notice appearing earlier in this book.
Mot
ion
in O
ne D
imen
sion
Mix
ed R
evie
wH
OLT
PH
YS
ICS
Ch
apte
r
2
Chap
ter 2
9
1.D
uri
ng
a re
lay
race
alo
ng
a st
raig
ht
road
,th
e fi
rst
run
ner
on
a t
hre
e-
pers
on te
am r
un
s d 1
wit
h a
con
stan
t ve
loci
ty v
1.T
he
run
ner
th
en h
ands
off
the
bato
n to
th
e se
con
d ru
nn
er,w
ho
run
s d 2
wit
h a
con
stan
t ve
loci
ty
v 2.T
he
bato
n is
th
en p
asse
d to
th
e th
ird
run
ner
,wh
o co
mpl
etes
th
e ra
ce
by t
rave
ling
d 3w
ith
a c
onst
ant
velo
city
v3.
a.In
term
s of
dan
dv,
fin
d th
e ti
me
it t
akes
for
each
ru
nn
er to
com
plet
e
a se
gmen
t of
the
race
.
3 ren
nu
R2 re
nn
uR
1 ren
nu
R
b.W
hat
is t
he
tota
l dis
tan
ce o
fth
e ra
ce c
ours
e?
c.W
hat
is t
he
tota
l tim
e it
tak
es t
he
team
to c
ompl
ete
the
race
?
2.T
he e
quat
ion
s be
low
incl
ude
the
equ
atio
ns
for
stra
ight
-lin
e m
otio
n.
For
each
of
the
follo
win
g pr
oble
ms,
indi
cate
whi
ch e
quat
ion
or
equ
atio
ns
you
wou
ld u
se to
sol
ve th
e pr
oble
m,b
ut d
o n
ot a
ctu
ally
per
form
the
calc
ula
tion
s.
a.D
uri
ng
take
off,
a pl
ane
acce
lera
tes
at 4
m/s
2an
d ta
kes
40 s
to r
each
take
off
spee
d.W
hat
is t
he
velo
city
of
the
plan
e at
tak
eoff
?
b.A
car
wit
h a
n in
itia
l spe
ed o
f31
.4 k
m/h
acc
eler
ates
at
a u
nif
orm
rat
e
of1.
2 m
/s2
for
1.3
s.W
hat
is t
he
fin
al s
peed
an
d di
spla
cem
ent
ofth
e
car
duri
ng
this
tim
e?
∆x=
1 2 (v i
+v f
)∆t
∆x=
1 2 (v f
)∆t
∆x=v i
(∆t)
+1 2 a
(∆t)
2∆x
=1 2 a
(∆t)
2
v f=v i
+a(
∆t)
v f=
a(∆t
)
v f2
=v i
2+
2a∆x
v f2
=2a
∆x
HRW material copyrighted under notice appearing earlier in this book.
Hol
t Phy
sics
Sec
tion
Revi
ew W
orks
heet
s103.
Bel
ow is
th
e ve
loci
ty-t
ime
grap
h o
fan
obj
ect
mov
ing
alon
g a
stra
igh
t
path
.Use
th
e in
form
atio
n in
th
e gr
aph
to fi
ll in
th
e ta
ble
belo
w.
HO
LT P
HY
SIC
S
Mix
ed R
evie
w co
ntin
ued
Ch
apte
r
2
Tim
eM
otio
nv
ain
terv
alA B C D E
4.A
bal
l is
thro
wn
upw
ard
wit
h a
n in
itia
l vel
ocit
y of
9.8
m/s
from
th
e to
p
ofa
build
ing.
a.Fi
ll in
th
e ta
ble
belo
w s
how
ing
the
ball’
s po
siti
on,v
eloc
ity,
and
acce
l-
erat
ion
at
the
end
ofea
ch o
fth
e fi
rst
4 s
ofm
otio
n.
Tim
eP
osit
ion
Vel
ocit
yA
ccel
erat
ion
(s)
(m)
(m/s
)(m
/s2 )
1 2 3 4
b.In
wh
ich
sec
ond
does
th
e ba
ll re
ach
th
e to
p of
its
flig
ht?
c.In
wh
ich
sec
ond
does
th
e ba
ll re
ach
th
e le
vel o
fth
e ro
of,o
n t
he
way
dow
n?
For
each
oft
he le
tter
ed in
terv
als
belo
w,i
ndi
cate
the
mot
ion
oft
he o
bjec
t
(whe
ther
it is
spe
edin
g up
,slo
win
g do
wn
,or
at r
est)
,the
dir
ecti
on o
fthe
velo
city
(+,
−,or
0),
and
the
dire
ctio
n o
fthe
acc
eler
atio
n (
+,−,
or 0
).
Velocity (m/s)
time
(s)
15 10 5 070
100
AB
CD
E30
5020
6040
Honors Physics 2008–2009 Mr. Strong
49
Two-
Dim
ensi
onal
Mot
ion
and
Vect
ors
Mix
ed R
evie
wH
OLT
PH
YS
ICS
Ch
apte
r
3
Chap
ter 3
15
1.T
he
diag
ram
bel
ow in
dica
tes
thre
e po
siti
ons
to
whi
ch a
wom
an tr
avel
s.Sh
e st
arts
at p
osit
ion
A,
trav
els
3.0
km to
the
wes
t to
poin
t B,t
hen
6.0
km
to
the
nor
th to
poi
nt C
.She
then
bac
ktra
cks,
and
trav
els
2.0
km to
the
sou
th to
poi
nt D
.
a.In
th
e sp
ace
prov
ided
,dia
gram
th
e di
spla
cem
ent
vect
ors
for
each
seg
men
t of
the
wom
an’s
tri
p.
b.W
hat i
s th
e to
tal d
ispl
acem
ent o
fth
e w
oman
from
her
init
ial p
osit
ion
,A,t
o he
r fi
nal
pos
itio
n,D
?
c.W
hat i
s th
e to
tal d
ista
nce
trav
eled
by
the
wom
an
from
her
init
ial p
osit
ion,
A,t
o he
r fin
al p
osit
ion,
D?
2.Tw
o pr
ojec
tile
s ar
e la
un
ched
from
th
e gr
oun
d,an
d bo
th r
each
th
e sa
me
vert
ical
hei
ght.
How
ever
,pro
ject
ile B
tra
vels
tw
ice
the
hor
izon
tal d
ista
nce
as p
roje
ctile
A b
efor
e h
itti
ng
the
grou
nd.
a.H
ow la
rge
is t
he
vert
ical
com
pon
ent
ofth
e in
itia
l vel
ocit
y of
proj
ec-
tile
B c
ompa
red
wit
h t
he
vert
ical
com
pon
ent
ofth
e in
itia
l vel
ocit
y
ofpr
ojec
tile
A?
b.H
ow la
rge
is t
he
hor
izon
tal c
ompo
nen
t of
the
init
ial v
eloc
ity
ofpr
o-
ject
ile B
com
pare
d w
ith
th
e h
oriz
onta
l com
pon
ent
ofth
e in
itia
l
velo
city
of
proj
ecti
le A
?
c.Su
ppos
e pr
ojec
tile
A is
lau
nch
ed a
t an
an
gle
of45
°to
th
e h
oriz
onta
l.
Wh
at is
th
e ra
tio,v B
/vA
,of
the
spee
d of
proj
ecti
le B
,vB
,com
pare
d
wit
h t
he
spee
d of
proj
ecti
le A
,vA
?
HRW material copyrighted under notice appearing earlier in this book.
C D BA
HRW material copyrighted under notice appearing earlier in this book.
Hol
t Phy
sics
Sec
tion
Revi
ew W
orks
heet
s163.
A p
asse
nge
r at
an
air
port
ste
ps o
nto
a m
ovin
g si
dew
alk
that
is 1
00.0
m
lon
g an
d is
mov
ing
at a
spe
ed o
f1.
5 m
/s.T
he
pass
enge
r th
en s
tart
s w
alk-
ing
at a
spe
ed o
f1.
0 m
/s in
th
e sa
me
dire
ctio
n a
s th
e si
dew
alk
is m
ovin
g.
Wh
at is
th
e pa
ssen
ger’
s ve
loci
ty r
elat
ive
to t
he
follo
win
g ob
serv
ers?
a.A
per
son
sta
ndi
ng
stat
ion
ary
alon
gsid
e to
th
e m
ovin
g si
dew
alk.
b.A
per
son
sta
ndi
ng
stat
ion
ary
onth
e m
ovin
g si
dew
alk.
c.A
per
son
wal
kin
g al
ongs
ide
the
side
wal
k w
ith
a s
peed
of
2.0
m/s
an
d
in a
dir
ecti
on o
ppos
ite
the
mot
ion
of
the
side
wal
k.
d.A
per
son
rid
ing
in a
car
t alo
ngs
ide
the
side
wal
k w
ith
a sp
eed
of5.
0 m
/s
and
in th
e sa
me
dire
ctio
n in
whi
ch th
e si
dew
alk
is m
ovin
g.
e.A
per
son
rid
ing
in a
car
t w
ith
a s
peed
of
4.0
m/s
an
d in
a d
irec
tion
perp
endi
cula
r to
th
e di
rect
ion
in w
hic
h t
he
side
wal
k is
mov
ing.
4.U
se t
he
info
rmat
ion
giv
en in
item
3 to
an
swer
th
e fo
llow
ing
ques
tion
s:
a.H
ow lo
ng
does
it t
ake
for
the
pass
enge
r w
alki
ng
on t
he
side
wal
k to
get
from
on
e en
d of
the
side
wal
k to
th
e ot
her
en
d?
b.H
ow m
uch
tim
e do
es th
e pa
ssen
ger
save
by
taki
ng
the
mov
ing
side
-
wal
k in
stea
d of
wal
kin
g al
ongs
ide
it?
HO
LT P
HY
SIC
S
Mix
ed R
evie
w co
ntin
ued
Ch
apte
r
3
Honors Physics 2008–2009 Mr. Strong
50
Forc
es a
nd th
e La
ws
ofM
otio
n
Mix
ed R
evie
wH
OLT
PH
YS
ICS
Ch
apte
r
4
Chap
ter 4
21
1.A
cra
te r
ests
on
the
hori
zont
al b
ed o
fa p
icku
p tr
uck.
For
each
sit
uati
on d
e-
scri
bed
belo
w,i
ndic
ate
the
mot
ion
ofth
e cr
ate
rela
tive
to th
e gr
ound
,the
mot
ion
ofth
e cr
ate
rela
tive
to th
e tr
uck,
and
whe
ther
the
crat
e w
ill h
it th
e
fron
t wal
l oft
he tr
uck
bed,
the
back
wal
l,or
nei
ther
.Dis
rega
rd fr
icti
on.
a.St
arti
ng
at r
est,
the
tru
ck a
ccel
erat
es to
th
e ri
ght.
b.T
he
crat
e is
at
rest
rel
ativ
e to
th
e tr
uck
wh
ile t
he
tru
ck m
oves
to t
he
righ
t w
ith
a c
onst
ant
velo
city
.
c.T
he
tru
ck in
item
b s
low
s do
wn
.
2.A
bal
l wit
h a
mas
s of
mis
th
row
n t
hro
ugh
th
e ai
r,as
sh
own
in t
he
figu
re.
HRW material copyrighted under notice appearing earlier in this book.
a.W
hat
is t
he
grav
itat
ion
al fo
rce
exer
ted
on t
he
ball
by E
arth
?
b.W
hat
is t
he
forc
e ex
erte
d on
Ear
th b
y th
e ba
ll?
c.If
the
surr
oun
din
g ai
r ex
erts
a fo
rce
on t
he
ball
that
res
ists
its
mot
ion
,
is t
he
tota
lfor
ce o
n t
he
ball
the
sam
e as
th
e fo
rce
calc
ula
ted
in p
art
a?
d.If
the
surr
oun
din
g ai
r ex
erts
a fo
rce
on t
he
ball
that
res
ists
its
mot
ion
,
is t
he
grav
itat
iona
l for
ce o
n t
he
ball
the
sam
e as
th
e fo
rce
calc
ula
ted
in
part
a?
HRW material copyrighted under notice appearing earlier in this book.
Hol
t Phy
sics
Sec
tion
Revi
ew W
orks
heet
s223.
Two
bloc
ks o
fm
asse
s m
1an
dm
2,r
espe
ctiv
ely,
are
plac
ed in
con
tact
wit
h
each
oth
er o
n a
sm
ooth
,hor
izon
tal s
urf
ace.
A c
onst
ant
hor
izon
tal f
orce
F
to t
he
righ
t is
app
lied
to m
1.A
nsw
er t
he
follo
win
g qu
esti
ons
in te
rms
of
F,m
1,an
dm
2.
a.W
hat
is t
he
acce
lera
tion
of
the
two
bloc
ks?
b.W
hat
are
th
e h
oriz
onta
l for
ces
acti
ng
on m
2?
c.W
hat
are
th
e h
oriz
onta
l for
ces
acti
ng
on m
1?
d.W
hat
is t
he
mag
nit
ude
of
the
con
tact
forc
e be
twee
n t
he
two
bloc
ks?
4.A
ssu
me
you
hav
e th
e sa
me
situ
atio
n a
s de
scri
bed
in it
em 3
,on
ly t
his
tim
e th
ere
is a
fric
tion
al fo
rce,
F k,b
etw
een
th
e bl
ocks
an
d th
e su
rfac
e.
An
swer
th
e fo
llow
ing
ques
tion
s in
term
s of
F,F k
,m1,
and
m2.
a.W
hat
is t
he
acce
lera
tion
of
the
two
bloc
ks?
b.W
hat
are
th
e h
oriz
onta
l for
ces
acti
ng
on m
2?
c.W
hat
are
th
e h
oriz
onta
l for
ces
acti
ng
on m
1?
d.W
hat
is t
he
mag
nit
ude
of
the
con
tact
forc
e be
twee
n t
he
two
bloc
ks?
HO
LT P
HY
SIC
S
Mix
ed R
evie
w co
ntin
ued
Ch
apte
r
4
Honors Physics 2008–2009 Mr. Strong
51
Wor
k an
d En
ergy
Mix
ed R
evie
wH
OLT
PH
YS
ICS
Ch
apte
r
5
Chap
ter 5
27
1.A
bal
l has
a m
ass
of3
kg.W
hat
is t
he
wor
k do
ne
on t
his
bal
l by
the
grav
i-
tati
onal
forc
e ex
erte
d by
Ear
th if
the
ball
mov
es 2
m a
lon
g ea
ch o
fth
e
follo
win
g di
rect
ion
s?
a.do
wnw
ard
(alo
ng
the
forc
e)
b.u
pwar
d (o
ppos
ite
the
forc
e)
2.A
sto
ne
wit
h a
mas
s of
mis
th
row
n o
ffa
build
ing.
As
the
ston
e pa
sses
poin
tA
,it
has
a s
peed
ofv A
at a
n a
ngl
e of
qto
th
e h
oriz
onta
l.T
he
ston
e
then
tra
vels
a v
erti
cal d
ista
nce
hto
poi
nt
B,w
her
e it
has
a s
peed
vB
.
HRW material copyrighted under notice appearing earlier in this book.
a.W
hat
is t
he
wor
k do
ne
on t
he
ston
e by
th
e gr
avit
atio
nal
forc
e du
e to
Ear
th w
hile
th
e st
one
mov
es fr
om A
to B
?
b.W
hat
is t
he
chan
ge in
th
e ki
net
ic e
ner
gy o
fth
e st
one
as it
mov
es fr
om
Ato
B?
c.W
hat
is t
he
spee
d v B
ofth
e st
one
in te
rms
ofv A
,g,a
nd
h?
d.D
oes
the
chan
ge in
th
e st
one’
s sp
eed
betw
een
Aan
dB
depe
nd
on t
he
mas
s of
the
ston
e?
e.D
oes
the
chan
ge in
th
e st
one’
s sp
eed
betw
een
Aan
dB
depe
nd
on t
he
angl
e q
?
h
A
Bθ
HRW material copyrighted under notice appearing earlier in this book.
Hol
t Phy
sics
Sec
tion
Revi
ew W
orks
heet
s283.
An
em
pty
coff
ee m
ug
wit
h a
mas
s of
0.40
kg
gets
kn
ocke
d of
fa
tabl
etop
0.75
m a
bove
th
e fl
oor
onto
th
e se
at o
fa
chai
r 0.
45 m
abo
ve t
he
floo
r.
Ass
um
e th
at t
he
grav
itat
ion
al p
oten
tial
en
ergy
,PE
g,is
mea
sure
d u
sin
g
the
floo
r as
th
e ze
ro e
ner
gy le
vel.
a.W
hat
is t
he
init
ial g
ravi
tati
onal
pot
enti
al e
ner
gy a
ssoc
iate
d w
ith
th
e
mu
g’s
posi
tion
on
th
e ta
ble?
b.W
hat
is t
he
fin
al g
ravi
tati
onal
pot
enti
al e
ner
gy a
ssoc
iate
d w
ith
th
e
mu
g’s
posi
tion
on
th
e ch
air
seat
?
c.W
hat
was
th
e w
ork
don
e by
th
e gr
avit
atio
nal
forc
e as
it fe
ll fr
om t
he
tabl
e to
th
e ch
air?
d.Su
ppos
e th
at z
ero
leve
l for
th
e en
ergy
was
tak
en to
be
the
ceili
ng
of
the
room
rat
her
th
an t
he
floo
r .W
ould
th
e an
swer
s to
item
s a
to c
be
the
sam
e or
dif
fere
nt?
4.A
car
ton
of
shoe
s w
ith
a m
ass
ofm
slid
es w
ith
an
init
ial s
peed
ofv i
m/s
dow
n a
ram
p in
clin
ed a
t an
an
gle
of23
°to
the
hor
izon
tal.
Th
e ca
rton
’s
init
ial h
eigh
t is
hi,
and
its
fin
al h
eigh
t is
hf,
and
it t
rave
ls a
dis
tan
ce o
f
ddo
wn
th
e ra
mp.
Th
ere
is a
fric
tion
al fo
rce,
F k,b
etw
een
th
e ra
mp
and
the
cart
on.
a.W
hat i
s th
e in
itia
l mec
han
ical
en
ergy
,ME
i,of
the
cart
on?
(Hin
t:
App
ly th
e la
w o
fco
nse
rvat
ion
of
ener
gy.)
b.Ifm
is t
he
coef
fici
ent
offr
icti
on b
etw
een
th
e ra
mp
and
the
cart
on,
wh
at is
Fk?
c.E
xpre
ss t
he
fin
al s
peed
,vf,
ofth
e ca
rton
in te
rms
ofv i
,g,d
,an
dm.
HO
LT P
HY
SIC
S
Mix
ed R
evie
w co
ntin
ued
Ch
apte
r
5
Honors Physics 2008–2009 Mr. Strong
52
Hol
t Phy
sics
Sec
tion
Revi
ew W
orks
heet
s32
Mom
entu
m a
nd C
ollis
ions
Mix
ed R
evie
wH
OLT
PH
YS
ICS
Ch
apte
r
6
1.A
pit
cher
th
row
s a
soft
ball
tow
ard
hom
e pl
ate.
Th
e ba
ll m
ay b
e h
it,s
end-
ing
it b
ack
tow
ard
the
pitc
her
,or
it m
ay b
e ca
ugh
t,br
ingi
ng
it to
a s
top
in
the
catc
her
’s m
itt.
a.C
ompa
re t
he
chan
ge in
mom
entu
m o
fth
e ba
ll in
th
ese
two
case
s.
b.D
iscu
ss t
he
mag
nit
ude
of
the
impu
lse
on t
he
ball
in t
hes
e tw
o ca
ses.
c.In
th
e sp
ace
belo
w,d
raw
a v
ecto
r di
agra
m fo
r ea
ch c
ase,
show
ing
the
init
ial m
omen
tum
of
the
ball,
the
impu
lse
exer
ted
on t
he
ball,
and
the
resu
ltin
g fi
nal
mom
entu
m o
fth
e ba
ll.
2.a.
Usi
ng
New
ton’
s th
ird
law
,exp
lain
why
th
e im
puls
e on
on
e ob
ject
in a
colli
sion
is e
qual
in m
agn
itu
de b
ut
oppo
site
in d
irec
tion
to t
he
im-
puls
e on
th
e se
con
d ob
ject
.
b.E
xten
d yo
ur
disc
uss
ion
of
impu
lse
and
New
ton’
s th
ird
law
to t
he
case
ofa
bow
ling
ball
stri
kin
g a
set
of10
bow
ling
pin
s.
HRW material copyrighted under notice appearing earlier in this book.
HRW material copyrighted under notice appearing earlier in this book.
Chap
ter 6
33
3.St
arti
ng
wit
h t
he
con
serv
atio
n o
fto
tal m
omen
tum
,pf
=p
i,sh
ow
that
th
e fi
nal
vel
ocit
y fo
r tw
o ob
ject
s in
an
inel
asti
c co
llisi
on is
v f=
m1m +1 m
2
v 1,i
+ m
1m +2 m2
v 2
,i.
4.Tw
o m
ovin
g bi
lliar
d ba
lls,e
ach
wit
h a
mas
s of
M,u
nde
rgo
an e
last
ic
colli
sion
.Im
med
iate
ly b
efor
e th
e co
llisi
on,b
all A
is m
ovin
g ea
st a
t
2 m
/s a
nd
ball
B is
mov
ing
east
at
4 m
/s.
a.In
term
s of
M,w
hat i
s th
e to
tal m
omen
tum
(m
agn
itu
de a
nd
dire
c-
tion
) im
med
iate
ly b
efor
e th
e co
llisi
on?
b.T
he
fin
al m
omen
tum
,M(v
A,f
+v B
,f),
mu
st e
qual
th
e in
itia
l mom
en-
tum
.If
the
fin
al v
eloc
ity
ofba
ll A
incr
ease
s to
4 m
/s e
ast
beca
use
of
the
colli
sion
,wh
at is
th
e fi
nal
mom
entu
m o
fba
ll B
?
c.Fo
r ea
ch b
all,
com
pare
th
e fi
nal
mom
entu
m o
fth
e ba
ll to
th
e in
itia
l
mom
entu
m o
fth
e ot
her
ball.
Th
ese
resu
lts
are
typi
cal o
fh
ead-
on
elas
tic
colli
sion
s.W
hat
gen
eral
izat
ion
abo
ut
hea
d-on
ela
stic
col
lisio
ns
can
you
mak
e?
HO
LT P
HY
SIC
S
Mix
ed R
evie
w co
ntin
ued
Ch
apte
r
6
Honors Physics 2008–2009 Mr. Strong
53HRW material copyrighted under notice appearing earlier in this book.
Rota
tiona
l Mot
ion
and
the
Law
ofG
ravi
ty
Mix
ed R
evie
wH
OLT
PH
YS
ICS
Ch
apte
r
7
Chap
ter 7
37
1.C
ompl
ete
the
follo
win
g ta
ble.
s(m
)r(
m)
q(r
ad)
t(s)
w(r
ad/s
)v t
(m/s
)a c
(m/s
2 )
a.4.
51.
50.
50
b.5.8
5.805.0
c.85
02.02.3
d.
1250
2.0
17
e.68
0570573
2.D
escr
ibe
the
forc
e th
at m
ain
tain
s ci
rcu
lar
mot
ion
in t
he
follo
win
g ca
ses.
a.A
car
exi
ts a
free
way
an
d m
oves
aro
un
d a
circ
ula
r ra
mp
to r
each
th
e
stre
et b
elow
.
b.T
he
moo
n o
rbit
s E
arth
.
c.D
uri
ng
gym
cla
ss,a
stu
den
t h
its
a te
ther
bal
l on
a s
trin
g.
3.D
eter
min
e th
e ch
ange
in g
ravi
tati
onal
forc
e u
nde
r th
e fo
llow
ing
chan
ges.
a.on
e of
the
mas
ses
is d
oubl
ed
b.bo
th m
asse
s ar
e do
ubl
ed
c.th
e di
stan
ce b
etw
een
mas
ses
is d
oubl
ed
d.th
e di
stan
ce b
etw
een
mas
ses
is h
alve
d
e.th
e di
stan
ce b
etw
een
mas
ses
is t
ripl
ed
HRW material copyrighted under notice appearing earlier in this book.
Hol
t Phy
sics
Sec
tion
Revi
ew W
orks
heet
s384.
Som
e pl
ans
for
a fu
ture
spa
ce s
tati
on m
ake
use
ofro
tati
onal
forc
e to
sim
u-
late
gra
vity
.In
ord
er to
be
effe
ctiv
e,th
e ce
ntr
ipet
al a
ccel
erat
ion
at t
he o
uter
rim
oft
he s
tati
on s
houl
d eq
ual a
bout
1 g
,or
9.81
m/s
2 .How
ever
,hum
ans
can
wit
hsta
nd
a di
ffer
ence
ofo
nly
1/1
00 g
bet
wee
n th
eir
head
an
d fe
et
befo
re th
ey b
ecom
e di
sori
ente
d.A
ssum
e th
e av
erag
e hu
man
hei
ght i
s
2.0
m,a
nd
calc
ulat
e th
e m
inim
um r
adiu
s fo
r a
safe
,eff
ecti
ve s
tati
on.
(Hin
t:T
he r
atio
oft
he c
entr
ipet
al a
ccel
erat
ion
ofa
stro
nau
t’s fe
et to
the
cen
trip
etal
acc
eler
atio
n o
fthe
ast
ron
aut’s
hea
d m
ust b
e at
leas
t 99/
100.
)
5.A
s an
ele
vato
r be
gin
s to
des
cen
d,yo
u fe
el m
omen
tari
ly li
ghte
r.A
s th
e
elev
ator
sto
ps,y
ou fe
el m
omen
tari
ly h
eavi
er.S
ketc
h t
he
situ
atio
n,a
nd
expl
ain
th
e se
nsa
tion
s u
sin
g th
e fo
rces
in y
our
sket
ch.
6.Tw
o ca
rs s
tart
on
opp
osit
e si
des
ofa
circ
ula
r tr
ack.
On
e ca
r h
as a
spe
ed
of0.
015
rad/
s;th
e ot
her
car
has
a s
peed
of
0.01
2 ra
d/s.
Ifth
e ca
rs s
tart
pra
dian
s ap
art,
calc
ula
te t
he
tim
e it
tak
es fo
r th
e fa
ster
car
to c
atch
up
wit
h t
he
slow
er c
ar.
HO
LT P
HY
SIC
S
Mix
ed R
evie
w co
ntin
ued
Ch
apte
r
7
Honors Physics 2008–2009 Mr. Strong
54
Rota
tiona
l Equ
ilibr
ium
and
Dyn
amic
s
Mix
ed R
evie
wH
OLT
PH
YS
ICS
Ch
apte
r
8
Chap
ter 8
43
1.a.
On
som
e do
ors,
the
door
knob
is in
th
e ce
nte
r of
the
door
.Wh
at
wou
ld a
phy
sici
st s
ay a
bou
t th
e pr
acti
calit
y of
this
arr
ange
men
t? W
hy
wou
ld p
hysi
cist
s de
sign
doo
rs w
ith
kn
obs
fart
her
from
th
e h
inge
?
b.H
ow m
uch
mor
e fo
rce
wou
ld b
e re
quir
ed to
ope
n t
he
door
from
th
e
cen
ter
rath
er t
han
from
th
e ed
ge?
2.Fi
gure
ska
ters
com
mon
ly c
han
ge t
he
shap
e of
thei
r bo
dy in
ord
er to
ach
ieve
spi
ns
on t
he
ice.
Exp
lain
th
e ef
fect
s on
eac
h o
fth
e fo
llow
ing
quan
titi
es w
hen
a fi
gure
ska
ter
pulls
in h
is o
r h
er a
rms.
a.m
omen
t of
iner
tia
b.an
gula
r m
omen
tum
c.an
gula
r sp
eed
3.Fo
r th
e fo
llow
ing
item
s,as
sum
e th
e ob
ject
s sh
own
are
in r
otat
ion
al e
quili
briu
m.
a.W
hat
is t
he
mas
s of
the
sph
ere
to t
he
righ
t?
b.W
hat
is t
he
mas
s of
the
port
ion
of
the
met
er-
stic
k to
th
e le
ft o
fth
e pi
vot?
(H
int:
20%
of
the
mas
s of
the
met
erst
ick
is o
n t
he
left
.How
mu
ch m
ust
be
on t
he
righ
t?)
HRW material copyrighted under notice appearing earlier in this book.
1.0
kg
0 cm
50 c
m25
cm
75 c
m10
0 cm
1.0
kg0 cm
50 c
m25
cm
75 c
m10
0 cm
HRW material copyrighted under notice appearing earlier in this book.
Hol
t Phy
sics
Sec
tion
Revi
ew W
orks
heet
s444.
A fo
rce
of25
N is
app
lied
to t
he
end
ofa
un
ifor
m r
od t
hat
is 0
.50
m lo
ng
and
has
a m
ass
of0.
75 k
g.
a.Fi
nd
the
torq
ue,
mom
ent o
fin
erti
a,an
d an
gula
r ac
cele
rati
on if
the
rod
is a
llow
ed to
piv
ot a
rou
nd
its
cen
ter
ofm
ass.
b.Fi
nd
the
torq
ue,
mom
ent o
fin
erti
a,an
d an
gula
r ac
cele
rati
on if
the
rod
is a
llow
ed to
piv
ot a
rou
nd
the
end,
away
from
the
appl
ied
forc
e.
5.A
sat
ellit
e in
orb
it a
rou
nd
Ear
th is
init
ially
at
a co
nst
ant
angu
lar
spee
d of
7.27
×10
–5ra
d/s.
Th
e m
ass
ofth
e sa
telli
te is
45
kg,a
nd
it h
as a
n o
rbit
al
radi
us
of4.
23 ×
107
m.
a.Fi
nd
the
mom
ent
ofin
erti
a of
the
sate
llite
in o
rbit
aro
un
d E
arth
.
b.Fi
nd
the
angu
lar
mom
entu
m o
fth
e sa
telli
te.
c.Fi
nd
the
rota
tion
al k
inet
ic e
ner
gy o
fth
e sa
telli
te a
rou
nd
Ear
th.
d.Fi
nd
the
tan
gen
tial
spe
ed o
fth
e sa
telli
te.
e.Fi
nd
the
tran
slat
ion
al k
inet
ic e
ner
gy o
fth
e sa
telli
te.
6.A
ser
ies
oftw
o si
mpl
e m
ach
ines
is u
sed
to li
ft a
133
00 N
car
to a
hei
ght
of3.
0 m
.Bot
h m
ach
ines
hav
e an
eff
icie
ncy
of
0.90
(90
per
cen
t).M
ach
ine
A m
oves
th
e ca
r,an
d th
e ou
tpu
t of
mac
hin
e B
is t
he
inpu
t to
mac
hin
e A
.
a.H
ow m
uch
wor
k is
don
e on
th
e ca
r?
b.H
ow m
uch
wor
k m
ust
be
don
e on
mac
hin
e A
in o
rder
to a
chie
ve t
he
amou
nt
ofw
ork
don
e on
th
e ca
r?
c.H
ow m
uch
wor
k m
ust
be
don
e on
mac
hin
e B
in o
rder
to a
chie
ve t
he
amou
nt
ofw
ork
from
mac
hin
e A
?
d.W
hat
is t
he
over
all e
ffic
ien
cy o
fth
is p
roce
ss?
HO
LT P
HY
SIC
S
Mix
ed R
evie
w co
ntin
ued
Ch
apte
r
8
Honors Physics 2008–2009 Mr. Strong
55
Vibr
atio
ns a
nd W
aves
Mix
ed R
evie
wH
OLT
PH
YS
ICS
Chap
ter 1
267
1.A
pen
dulu
m w
ith
a m
ass
of0.
100
kg w
as r
elea
sed.
Th
e st
rin
g m
ade
a
7.0°
angl
e w
ith
th
e ve
rtic
al.T
he
bob
ofth
e pe
ndu
lum
ret
urn
s to
its
low
est
poin
t ev
ery
0.10
s.
a.W
hat
is it
s pe
riod
? W
hat
is it
s fr
equ
ency
?
b.T
he
pen
dulu
m is
rep
lace
d by
on
e w
ith
a m
ass
of0.
300
kg a
nd
set
to
swin
g w
ith
a 1
5°an
gle.
Do
the
follo
win
g qu
anti
ties
incr
ease
,dec
reas
e,
or r
emai
n t
he
sam
e?
per
iod
freq
uen
cy
tota
l en
ergy
spee
d at
th
e lo
wes
t po
int
2.A
nar
row
,fla
t st
eel r
od is
an
chor
ed a
t it
s lo
wer
en
d,w
ith
a 0
.500
kg
ball
wel
ded
to t
he
top
end.
A fo
rce
of6.
00 N
is r
equ
ired
to h
old
the
ball
10.0
cm
aw
ay fr
om it
s ce
ntr
al p
osit
ion
.
Ifth
is a
rran
gem
ent
is m
odel
ed a
s an
osc
illat
ing
hor
izon
tal m
ass-
spri
ng
syst
em,v
ibra
tin
g w
ith
a s
impl
e h
arm
onic
mot
ion
,fin
d
a.th
e fo
rce
con
stan
t,k,
ofth
e sp
rin
g.
b.th
e pe
riod
an
d fr
equ
ency
of
the
osci
llati
ons.
3.Fi
nd
the
acce
lera
tion
du
e to
gra
vity
at
a pl
ace
wh
ere
a si
mpl
e pe
ndu
lum
0.15
0 m
lon
g co
mpl
etes
1.0
0 ×
102
osci
llati
ons
in 3
.00
×10
2se
con
ds.
Cou
ld t
his
pla
ce b
e on
Ear
th?
Ch
apte
r
12HRW material copyrighted under notice appearing earlier in this book.
AC
B
Hol
t Phy
sics
Sec
tion
Revi
ew W
orks
heet
s684.
Con
side
r th
e fi
rst
two
cycl
es o
fa
pen
dulu
m s
win
gin
g fr
om p
osit
ion
A
wit
h a
per
iod
of2.
00 s
.
a.A
t w
hic
h t
imes
is t
he
bob
fou
nd
at p
osit
ion
s A
,B,a
nd
Cdu
rin
g th
e
firs
t tw
o cy
cles
?
b.A
t w
hic
h t
imes
an
d lo
cati
ons
is g
ravi
tati
onal
pot
enti
al e
ner
gy a
t a
max
imu
m?
At
wh
ich
tim
es is
kin
etic
en
ergy
at
a m
axim
um
?
c.A
t w
hic
h t
imes
an
d lo
cati
ons
is t
he
velo
city
at
a m
axim
um
? th
e
rest
orin
g fo
rce?
th
e ac
cele
rati
on?
5.T
he fr
eque
ncy
ofa
pre
ssur
e w
ave
is 1
.00
×10
2H
z.It
s w
avel
engt
h is
3.0
0 m
.
Fin
d th
e sp
eed
ofw
ave
prop
agat
ion
.
6.A
pre
ssu
re w
ave
of0.
50 m
wav
elen
gth
pro
paga
tes
thro
ugh
a 3
.00
m lo
ng
coil
spri
ng
at a
spe
ed o
f2.
00 m
/s.H
ow lo
ng
does
it t
ake
for
the
wav
e to
trav
el fr
om o
ne
end
ofth
e co
il to
th
e ot
her
? H
ow m
any
wav
elen
gth
s fi
t
in t
he
coil?
HO
LT P
HY
SIC
S
Mix
ed R
evie
w co
ntin
ued
Ch
apte
r
12
HRW material copyrighted under notice appearing earlier in this book.
AC
B
Honors Physics 2008–2009 Mr. Strong
56
Hol
t Phy
sics
Sec
tion
Revi
ew W
orks
heet
s72
Soun
d
Mix
ed R
evie
wH
OLT
PH
YS
ICS
Ch
apte
r
13
1.T
he s
peed
ofs
ound
incr
ease
s w
ith
tem
pera
ture
.It i
s 33
1 m
/s in
air
at 0
°Can
d 34
3 m
/s in
air
at 2
0°C
.A g
lass
pip
e vi
brat
es w
ith
a fr
eque
ncy
of15
1 H
z.
a.W
hat
is t
he
wav
elen
gth
of
the
sou
nd
prod
uce
d by
th
e co
lum
n o
fai
r
in t
he
pipe
on
a c
old
day
(0°C
) an
d on
a w
arm
er d
ay (
20°C
)?
b.H
ow d
oes
air
tem
pera
ture
aff
ect t
he w
avel
engt
h of
the
soun
d pr
oduc
ed
by th
e pi
pe?
2.T
he
driv
er o
fan
am
bula
nce
tu
rns
on it
s si
ren
as
the
ambu
lan
ce h
eads
east
at
30 m
ph.A
pol
ice
car
is fo
llow
ing
the
ambu
lan
ce a
t 30
mph
.A
tru
ck b
ehin
d th
e po
lice
car
is m
ovin
g at
20
mph
.A v
an is
tra
velin
g w
est
in t
he
oppo
site
lan
e at
20
mph
.A s
mal
l car
is s
topp
ed a
t th
e si
de o
fth
e
road
.Th
e ve
hic
les
are
posi
tion
ed a
s sh
own
.
a.O
n t
he
diag
ram
,ske
tch
an
d la
bel a
rrow
s to
indi
cate
th
e ve
loci
ty o
f
each
veh
icle
.
HRW material copyrighted under notice appearing earlier in this book.
b.R
ank
the
sou
nds
per
ceiv
ed b
y th
e pa
ssen
gers
in e
ach
of
the
veh
icle
s
in o
rder
of
decr
easi
ng
freq
uen
cy.
Tru
ckPo
lice
car
Am
bula
nce
Van
Smal
lca
r
Chap
ter 1
373
3.A
330
Hz
tun
ing
fork
is v
ibra
tin
g af
ter
bein
g st
ruck
.It i
s pl
aced
on
a ta
ble
nea
r bu
t not
dir
ectl
y to
uchi
ng
othe
r ob
ject
s,in
clud
ing
othe
r tu
nin
g fo
rks.
Even
tual
ly o
ne
glas
s an
d on
e ot
her
tun
ing
fork
sta
rt v
ibra
tin
g.E
xpla
in w
hy
this
hap
pen
s.
4.T
he
firs
t h
arm
onic
in a
pip
e cl
osed
at
one
end
is 4
87 H
z.
a.Fi
nd
the
nex
t tw
o h
arm
onic
freq
uen
cies
th
at w
ill o
ccu
r in
th
is p
ipe.
b.W
hat
are
th
e co
rres
pon
din
g w
avel
engt
hs
ofth
e fi
rst
thre
e h
arm
onic
s?
(Hin
t:as
sum
e th
e sp
eed
ofso
un
d is
345
m/s
.)
c.W
hat
is t
he
len
gth
of
this
pip
e?
d.R
epea
t th
is e
xerc
ise
for
a pi
pe o
pen
at
both
en
ds.
5.A
pia
no
tun
er u
ses
a 44
0 H
z tu
nin
g fo
rk to
tu
ne
a st
rin
g th
at is
cu
rren
tly
vibr
atin
g at
445
Hz.
a.H
ow m
any
beat
s pe
r se
con
d do
es h
e h
ear?
b.W
hat
oth
er fr
equ
ency
cou
ld p
rodu
ce t
he
sam
e so
un
d ef
fect
? E
xpla
in w
hy.
HO
LT P
HY
SIC
S
Mix
ed R
evie
w co
ntin
ued
Ch
apte
r
13
HRW material copyrighted under notice appearing earlier in this book.
Honors Physics 2008–2009 Mr. Strong
57
Hol
t Phy
sics
Sec
tion
Revi
ew W
orks
heet
s78
Ligh
t and
Re
ectio
n
Mix
ed R
evie
wH
OLT
PH
YS
ICS
Ch
apte
r
14
1.P
roxi
ma
Cen
tau
ri,t
he
nea
rest
sta
r in
ou
r ga
laxy
,is
4.30
ligh
t-ye
ars
away
.
Wh
at is
its
dist
ance
in m
eter
s?
2.R
adio
sig
nal
s em
itte
d fr
om a
nd
rece
ived
by
an a
irpl
ane
hav
e a
freq
uen
cy
of3.
00 ×
1012
Hz
and
trav
el a
t th
e sp
eed
oflig
ht.
a.H
ow lo
ng
is t
he
dela
y in
eac
h m
essa
ge g
oin
g fr
om t
he
con
trol
tow
er
to a
jet
flyi
ng
at 1
.00
×10
4m
of
alti
tude
?
b.W
hat
is t
he
wav
elen
gth
of
thes
e si
gnal
s?
3.A
lase
r be
am is
sen
t to
the
moo
n fr
om E
arth
.The
ref
lect
ed b
eam
is r
ecei
ved
on E
arth
aft
er 2
.56
seco
nds
.Wha
t is
the
dist
ance
from
Ear
th to
the
moo
n?
4.T
he
back
grou
nd
radi
atio
n in
th
e u
niv
erse
(be
lieve
d to
com
e fr
om t
he
Big
Ban
g) in
clu
des
mic
row
aves
wit
h w
avel
engt
hs
of0.
100
cm.W
hat
is t
he
freq
uen
cy o
fth
is r
adia
tion
?
5.Li
st fi
ve o
bjec
ts t
hat
ref
lect
ligh
t di
ffu
sely
.Lis
t th
ree
obje
cts
that
ref
lect
ligh
t sp
ecu
larl
y fo
r th
e m
ost
part
.
Dif
fuse
ref
lect
ion
Spec
ula
r re
flec
tion
HRW material copyrighted under notice appearing earlier in this book.
Chap
ter 1
479
6.A
mir
ror
door
is lo
cate
d n
ext
to a
larg
e w
all m
irro
r.
Th
e do
or is
clo
sed
to c
reat
e a
90°a
ngl
e w
ith
th
e
wal
l.Yo
u s
tan
d 2.
00 m
from
th
e do
or a
nd
1.00
m
from
th
e w
all.
a.O
n t
he
diag
ram
at
righ
t,sk
etch
a to
p-vi
ew d
ia-
gram
of
the
situ
atio
n a
t sc
ale.
Labe
l th
e ob
ject
(you
rsel
f) A
.
b.Lo
cate
you
r fi
rst
imag
e in
th
e m
irro
r on
th
e
door
.Lab
el it
B.L
ocat
e B
’s im
age
in t
he
mir
ror
on t
he
wal
l.La
bel i
t C
.
c.Lo
cate
you
r fi
rst
imag
e in
th
e m
irro
r on
th
e w
all a
nd
its
imag
e in
th
e
mir
ror
on t
he
door
.Lab
el t
hem
Dan
dE
.
d.W
her
e w
ill t
he
nex
t im
ages
of
the
imag
es b
e lo
cate
d?
7.A
n o
bjec
t lo
cate
d 36
.0 c
m fr
om a
con
cave
mir
ror
prod
uce
s a
real
imag
e
loca
ted
12.0
cm
from
th
e m
irro
r.
a.Fi
nd
the
foca
l len
gth
of
this
mir
ror
b.Fi
nd
the
loca
tion
,typ
e,an
d si
ze o
fth
e im
age
form
ed b
y a
6.00
cm
tal
l
obje
ct lo
cate
d 30
.0 c
m,2
4.0
cm,1
8.0
cm,1
2.0
cm,a
nd
6.00
cm
in
fron
t of
the
mir
ror.
8.T
he
con
cave
mir
ror
in t
he
prob
lem
abo
ve is
rep
lace
d by
a c
onve
x on
e
wit
h t
he
sam
e cu
rvat
ure
.Fin
d th
e lo
cati
on o
fth
e im
ages
pro
duce
d w
hen
the
obje
ct is
loca
ted
30.0
cm
,24.
0 cm
,18.
0 cm
,12.
0 cm
,an
d 6.
00 c
m in
fron
t of
the
mir
ror.
HO
LT P
HY
SIC
S
Mix
ed R
evie
w co
ntin
ued
Ch
apte
r
14
HRW material copyrighted under notice appearing earlier in this book.
wal
l
door
Honors Physics 2008–2009 Mr. Strong
58
Refractio
n
Mix
ed R
evie
wH
OLT
PH
YS
ICS
Chap
ter 1
583
1.Tw
o pa
ralle
l ray
s en
ter
an a
quar
ium
as
show
n.
Ray
1 fo
rms
a 70
.0°a
ngl
e w
ith
th
e n
orm
al to
the
surf
ace.
Ray
2 fo
rms
a 20
.0°a
ngl
e w
ith
the
nor
mal
to t
he
wal
l.(H
int:
the
inde
x of
refr
acti
on fo
r w
ater
is 1
.33.
)
a.C
alcu
late
th
e an
gle
ofre
frac
tion
of
each
ray
.
b.Tr
ace
the
path
of
each
ligh
t ra
y in
side
th
e w
ater
.
c.A
re t
he
refr
acte
d ra
ys in
side
th
e w
ater
sti
ll pa
ralle
l? W
ill t
hey
inte
rsec
t
in t
he
wat
er?
2.A
larg
e be
aker
con
tain
s la
yers
of
wat
er o
fin
crea
sin
g sa
linit
y,se
para
ted
by
a th
in p
last
ic p
late
.Th
e lo
wes
t la
yer
has
th
e h
igh
est
salin
ity
and
refr
acti
ve
inde
x,as
sh
own
in t
he
diag
ram
.A r
ay o
flig
ht
stri
kes
the
surf
ace
offr
esh
wat
er a
t th
e to
p,at
a 7
0.0°
angl
e fr
om t
he
nor
mal
.
Ch
apte
r
15HRW material copyrighted under notice appearing earlier in this book.
1
2
70.0
°
20.0
°
air
fres
h w
ater
salt
wat
er
hig
h s
alin
ity
n =
1.0
0
n =
1.3
3
n =
1.4
5
n =
1.5
7
70.0
°
a.Fi
nd
the
angl
es o
fre
frac
tion
an
d th
e an
gles
of
inci
den
ce a
t ea
ch
bou
nda
ry.
b.T
here
is a
flat
mir
ror
at th
e bo
ttom
oft
he c
onta
iner
.Tra
ce th
e pa
th o
f
one
light
ray
com
ing
from
the
air
to th
e bo
ttom
oft
he b
eake
r an
d ba
ck.
Hol
t Phy
sics
Sec
tion
Revi
ew W
orks
heet
s843.
An
obj
ect
loca
ted
36.0
cm
from
a t
hin
con
verg
ing
len
s h
as a
rea
l im
age
loca
ted
12.0
cm
from
th
e le
ns.
a.Fi
nd
the
foca
l poi
nt
ofth
is le
ns.
b.Fi
nd
the
loca
tion
,typ
e,an
d si
ze o
fth
e im
age
form
ed b
y a
6.00
cm
tall
obje
ct lo
cate
d 30
.0 c
m,2
4.0
cm,1
8.0
cm,1
2.0
cm,a
nd
6.00
cm
in fr
ont
ofth
e le
ns.
4.T
he
conv
ergi
ng
len
s in
item
3 is
rep
lace
d by
a d
iver
gin
g le
ns.
Now
th
e
imag
e of
the
firs
t ob
ject
is lo
cate
d 12
.0 c
m in
fron
t of
the
len
s.Fi
nd
the
foca
l dis
tan
ce o
fth
e di
verg
ing
len
s an
d th
e lo
cati
on o
fth
e im
ages
pro
-
duce
d w
hen
th
e ob
ject
is p
lace
d at
th
e di
stan
ces
desc
ribe
d in
item
3b.
5.A
bu
g pl
aced
1.0
0 cm
un
der
a m
agn
ifyi
ng
glas
s ap
pear
s ex
actl
y si
x
tim
es la
rger
.
a.W
her
e is
th
e bu
g’s
imag
e lo
cate
d?
b.W
hat
is t
he
foca
l poi
nt
ofth
e le
ns
in t
he
mag
nif
yin
g gl
ass?
HO
LT P
HY
SIC
S
Mix
ed R
evie
w co
ntin
ued
Ch
apte
r
15
HRW material copyrighted under notice appearing earlier in this book.
Honors Physics 2008–2009 Mr. Strong
59
Hol
t Phy
sics
Sec
tion
Revi
ew W
orks
heet
s88
Inte
rfer
ence
and
Di
ract
ion
Mix
ed R
evie
wH
OLT
PH
YS
ICS
Ch
apte
r
16
1.T
he
seco
nd-
orde
r br
igh
t fr
inge
s of
inte
rfer
ence
are
obs
erve
d at
an
8.5
3°an
gle
in a
dou
ble-
slit
exp
erim
ent
wit
h li
ght
of5.
00 ×
102
nm
wav
elen
gth
.
a.D
eter
min
e th
e sl
its’
sepa
rati
on.
b.Fi
nd
the
angl
e of
the
ten
th-o
rder
bri
ght
frin
ge.
c.In
th
is e
xper
imen
t,th
e sc
reen
is 2
.00
m w
ide.
Its
dist
ance
from
th
e
sou
rce
is 1
.00
m.C
ould
th
e te
nth
-ord
er fr
inge
be
obse
rved
? W
hy o
r
why
not
?
2.D
iffr
acti
on o
fw
hit
e lig
ht
wit
h a
sin
gle
slit
pro
duce
s br
igh
t lin
es o
f
diff
eren
t co
lors
.
a.W
hic
h w
avel
engt
hs
are
mor
e di
ffra
cted
by
the
sam
e sl
it s
ize?
b.In
the
spac
e be
low
,ske
tch
a di
agra
m s
how
ing
the
loca
tion
ofr
ed,g
reen
and
blue
lin
es o
fthe
firs
t an
d se
con
d or
der.
Des
crib
e th
e se
quen
ce in
whi
ch th
e co
lors
app
ear,
begi
nn
ing
wit
h th
e co
lor
clos
est t
o th
e ce
nte
r.
c.W
hat
is t
he
colo
r of
the
cen
tral
imag
e?
HRW material copyrighted under notice appearing earlier in this book.
Chap
ter 1
689
3.Yo
u h
ave
thre
e di
ffra
ctio
n g
rati
ngs
.Gra
tin
g A
has
2.0
×10
5lin
es p
er m
eter
.
Gra
tin
g B
has
9.0
×10
6lin
es p
er m
eter
.Gra
tin
g C
has
3.0
×10
7lin
es
per
met
er.
a.W
hat
is t
he
slit
dis
tan
ce o
fea
ch g
rati
ng?
b.W
hic
h g
rati
ngs
can
dif
frac
t th
e fo
llow
ing:
•vi
sibl
e lig
ht
of50
0 n
m w
avel
engt
h
•X
ray
s of
5.00
nm
wav
elen
gth
•in
frar
ed li
ght
of50
00 n
m w
avel
engt
h
4.Yo
u d
rop
pebb
les
into
th
e w
ater
on
a r
ocky
bea
ch.W
hen
th
e w
aves
you
mad
e re
ach
th
e ro
cks,
new
wav
es a
ppea
r to
sta
rt in
th
e sp
aces
bet
wee
n
the
rock
s.
a.A
re t
hes
e w
aves
coh
eren
t?
b.H
ow is
th
is li
ke a
dou
ble
slit
illu
min
ated
by
a si
ngl
e lig
ht
sou
rce?
HO
LT P
HY
SIC
S
Mix
ed R
evie
w co
ntin
ued
Ch
apte
r
16
HRW material copyrighted under notice appearing earlier in this book.
Honors Physics 2008–2009 Mr. Strong
60
Elec
tric
For
ces
and
Fiel
ds
Mix
ed R
evie
wH
OLT
PH
YS
ICS
Chap
ter 1
793
Use
kC
=8.
99×
109
N•m
2 /C2 .
1.Tw
o sp
her
es,A
and
B,a
re p
lace
d 0.
60 m
apa
rt,a
s sh
own
.Sph
ere
Ah
as
+3.0
0mC
exc
ess
char
ge.S
pher
e B
has
+5.
00 m
C e
xces
s ch
arge
.
Ch
apte
r
17HRW material copyrighted under notice appearing earlier in this book.
AB
a.H
ow m
any
elec
tron
s ar
e m
issi
ng
on s
pher
e A
? on
sph
ere
B?
b.H
ow d
o th
e fo
rces
of
Bon
Aan
dA
onB
com
pare
? D
oes
the
grea
ter
char
ge e
xert
a g
reat
er fo
rce?
2.A
th
ird
sph
eric
al c
har
ge,C
,of
+2.
00 m
C,i
s pl
aced
on
th
e lin
e co
nn
ecti
ng
sph
eres
Aan
dB
.Fin
d th
e re
sult
ant
forc
e ex
erte
d by
Aan
dB
onC
wh
en
Cis
pla
ced
in t
he
follo
win
g lo
cati
ons.
a.0.
20 m
to t
he
left
of
A
b.0.
20 m
to t
he
righ
t of
Abe
twee
nA
and
B
c.ex
actl
y in
th
e m
iddl
e be
twee
n A
and
B
Hol
t Phy
sics
Sec
tion
Revi
ew W
orks
heet
s943.
Alp
ha
part
icle
s ar
e m
ade
oftw
o pr
oton
s an
d tw
o n
eutr
ons.
mp
=1.
673
×10
–27
kg;m
n=
1.67
5×
10–2
7kg
;qe
=1.
60×
10–1
9C
a.Fi
nd
the
elec
tric
forc
e ac
tin
g on
an
alp
ha
part
icle
in a
hor
izon
tal
elec
tric
fiel
d of
6.00
×10
2N
/C.
b.W
hat
is t
he
acce
lera
tion
of
this
alp
ha
part
icle
?
c.H
ow d
oes
this
acc
eler
atio
n c
ompa
re w
ith
gra
vity
? D
escr
ibe
the
part
i-
cle’
s tr
ajec
tory
.Will
it b
e cl
ose
to h
oriz
onta
l? to
ver
tica
l fre
e fa
ll?
4.A
2.0
0 mC
poi
nt
char
ge o
fm
ass
5.00
g is
su
spen
ded
on a
str
ing
and
plac
ed in
a h
oriz
onta
l ele
ctri
c fi
eld.
Th
e m
ass
is in
equ
ilibr
ium
wh
en t
he
stri
ng
form
s a
17.3
°an
gle
wit
h t
he
vert
ical
.
a.In
the
spac
e be
low
,ske
tch
a fr
ee-b
ody
diag
ram
oft
he p
robl
em.S
how
the
vert
ical
an
d ho
rizo
nta
l com
pon
ents
oft
he te
nsi
on fo
rce
in th
e st
rin
g.
b.Fi
nd
the
elec
tric
forc
e on
th
e ch
arge
in t
his
fiel
d.
c.Fi
nd
the
stre
ngt
h o
fth
e el
ectr
ic fi
eld.
5.H
ow m
any
elec
tron
s ar
e th
ere
in 1
.00
C?
How
man
y el
ectr
ons
are
ther
e
in 1
.00 mC
?
HO
LT P
HY
SIC
S
Mix
ed R
evie
w co
ntin
ued
Ch
apte
r
17
HRW material copyrighted under notice appearing earlier in this book.
Honors Physics 2008–2009 Mr. Strong
61
Curr
ent a
nd R
esis
tanc
e
Mix
ed R
evie
wH
OLT
PH
YS
ICS
Chap
ter 1
910
3
1.A
60.
0 cm
met
al w
ire
draw
s 0.
185
A fr
om a
36.
0 V
bat
tery
.Will
th
e
curr
ent
incr
ease
or
decr
ease
wh
en t
he
follo
win
g ch
ange
s ar
e pe
rfor
med
?
Exp
lain
wh
eth
er t
he
chan
ge is
du
e to
a c
han
ge in
res
ista
nce
,a c
han
ge in
pote
nti
al d
iffe
ren
ce,o
r ot
her
rea
son
s.
a.T
he
wir
e is
cu
t in
to fo
ur
piec
es,a
nd
only
on
e se
gmen
t is
use
d.
b.T
he
wir
e is
ben
t to
form
an
Msh
ape.
c.T
he
wir
e is
hea
ted
to 5
00°C
.
d.T
he
36.0
V b
atte
ry is
rep
lace
d by
a 2
4.0
V b
atte
ry.
2.A
25
Ωre
sist
ance
hea
ter
is c
onn
ecte
d to
a p
oten
tial
dif
fere
nce
of
120
V
for
5.00
h.
a.H
ow m
uch
cu
rren
t do
es t
he
hea
ter
draw
?
b.H
ow m
uch
ele
ctri
c ch
arge
tra
vels
th
rou
gh t
he
hea
tin
g el
emen
t du
rin
g
this
tim
e?
c.W
hat
is t
he
pow
er c
onsu
mpt
ion
of
the
hea
ter?
d.U
se th
e po
wer
an
d ti
me
to c
alcu
late
how
mu
ch e
ner
gy w
as c
onsu
med
.
Ch
apte
r
19HRW material copyrighted under notice appearing earlier in this book.
HRW material copyrighted under notice appearing earlier in this book.
Hol
t Phy
sics
Sec
tion
Revi
ew W
orks
heet
s10
4
3.T
he
labe
l on
a t
hre
e-w
ay li
ght
bulb
pac
kage
spe
cifi
es 1
00 W
,150
W,
250
W,1
20 V
.
a.H
ow m
uch
cu
rren
t doe
s th
e lig
ht b
ulb
dra
w in
eac
h of
the
thre
e w
ays?
(Ass
um
e th
ree
sign
ific
ant f
igu
res
in e
ach
ofth
ese
mea
sure
men
ts.)
b.W
hat
is t
he
bulb
’s r
esis
tan
ce in
eac
h w
ay?
c.C
ompa
re t
he
cost
of
usi
ng
the
ligh
t bu
lb fo
r 10
0.0
h in
eac
h w
ay.
(Ass
um
e th
at t
he
pric
e is
7.0
0 ¢/
kWh
.)
4.A
n e
lect
ric
hot p
late
dra
ws
6.00
A o
fcu
rren
t whe
n it
s re
sist
ance
is 2
4.0
Ω.
a.W
hat
is t
he
volt
age
acro
ss t
he
hot
pla
te’s
hea
tin
g el
emen
t?
b.H
ow m
uch
pow
er d
oes
it c
onsu
me?
c.Fo
r w
hat
len
gth
of
tim
e sh
ould
it b
e ke
pt o
n in
ord
er to
su
pply
9×
104
J to
a c
offe
epot
? (A
ssu
me
that
all
elec
tric
al e
ner
gy is
tra
ns-
ferr
ed to
th
e co
ffee
pot
by h
eat.
)
HO
LT P
HY
SIC
S
Mix
ed R
evie
w co
ntin
ued
Ch
apte
r
19
Honors Physics 2008–2009 Mr. Strong
62
Hol
t Phy
sics
Sec
tion
Revi
ew W
orks
heet
s10
8
Circ
uits
and
Circ
uit E
lem
ents
Mix
ed R
evie
wH
OLT
PH
YS
ICS
Ch
apte
r
20
1.C
onsi
der
the
circ
uit
sh
own
bel
ow.
HRW material copyrighted under notice appearing earlier in this book.
A
1
2
34
5
BC
D
a.D
o an
y of
the
bulb
s h
ave
a co
mpl
ete
circ
uit
wh
en a
ll th
e sw
itch
es a
re
open
? W
hic
h o
ne(
s)?
b.D
o an
y of
the
swit
ches
cau
se a
sho
rt c
ircu
it w
hen
clos
ed? W
hich
one
(s)?
c.W
hic
h s
wit
ches
sh
ould
be
kept
ope
n,a
nd
wh
ich
sh
ould
be
clos
ed fo
r
the
follo
win
g to
occ
ur?
•on
ly b
ulb
s A
and
Bar
e of
f
•on
ly b
ulb
s A
and
Car
e of
f
•on
ly b
ulb
s B
and
Car
e of
f
2.A
ligh
t bu
lb o
fu
nkn
own
res
ista
nce
is c
onn
ecte
d in
ser
ies
wit
h a
9.0
Ωre
sist
or to
a 1
2.0
V b
atte
ry.T
he
curr
ent
in t
he
bulb
is 0
.80
A.
a.In
th
e sp
ace
belo
w,s
ketc
h a
sch
emat
ic d
iagr
am o
fth
e ci
rcu
it.
b.Fi
nd
the
equ
ival
ent
resi
stan
ce o
fth
e ci
rcu
it.
c.Fi
nd
the
resi
stan
ce o
fth
e lig
ht
bulb
.
HRW material copyrighted under notice appearing earlier in this book.
Chap
ter 2
010
9
3.A
ligh
t bu
lb o
fu
nkn
own
res
ista
nce
is c
onn
ecte
d in
par
alle
l to
a 48
.0 Ω
resi
stor
an
d to
a 1
2.0
V b
atte
ry.T
he
curr
ent
thro
ugh
th
e ba
tter
y is
2.5
0 A
.
a.In
th
e sp
ace
belo
w,s
ketc
h a
sch
emat
ic d
iagr
am o
fth
e ci
rcu
it.
b.Fi
nd
the
pote
nti
al d
iffe
ren
ce a
cros
s th
e re
sist
or a
nd
acro
ss t
he
bulb
.
c.Fi
nd
the
curr
ent
in t
he
resi
stor
an
d in
th
e bu
lb.
d.Fi
nd
the
resi
stan
ce o
fth
e lig
ht
bulb
.
4.In
th
e ci
rcu
it b
elow
,fin
d th
e eq
uiv
alen
t re
sist
ance
for
the
follo
win
g
situ
atio
ns.
HO
LT P
HY
SIC
S
Mix
ed R
evie
w co
ntin
ued
Ch
apte
r
20
a.R
a=
Rb
=R
c=
Rd
=R
e=
Rf
=10
.0Ω
b.R
a=
10.0
Ω;R
b=
20.
0 Ω
;Rc
=30
.0Ω
;Rd
=40
.0Ω
;Re
=50
.0Ω
;Rf
=60
.0Ω
Ra
Rb
Rd
Rc
Re
Rf
Honors Physics 2008–2009 Mr. Strong
63
Mag
netism
Mix
ed R
evie
wH
OLT
PH
YS
ICS
Chap
ter 2
111
3
1.A
wir
e fr
ame
carr
ies
an e
lect
ric
curr
ent i
n th
e di
rect
ion
sho
wn
.
Con
side
r th
e m
agn
etic
fiel
d co
ntr
ibut
ed b
y ea
ch s
egm
ent o
fthe
fram
e at
poi
nts
A,B
,C,D
,an
dE
.
a.U
se th
e co
nven
tion
sym
bols
(×,
•,an
d →
) to
rep
rese
nt th
e
dire
ctio
n of
mag
net
ic fi
elds
cre
ated
at p
oin
t Aby
the
vert
ical
segm
ents
oft
he fr
ame.
Do
they
hav
e th
e sa
me
dire
ctio
n?
the
sam
e st
ren
gth?
b.R
epea
t fo
r th
e h
oriz
onta
l seg
men
ts.
c.A
nsw
er it
ems
a an
d b
for
poin
ts B
,C,D
,an
dE
,an
d fi
ll in
th
e ta
ble
belo
w.
Ch
apte
r
21HRW material copyrighted under notice appearing earlier in this book.
EA
CB
D
d.D
o th
e co
ntr
ibu
tion
s of
each
seg
men
t to
th
e m
agn
etic
fiel
d ca
nce
l
out
at t
he
cen
ter?
Exp
lain
.
e.Is
th
e m
agn
etic
fiel
d re
sult
ing
from
th
e co
mbi
ned
eff
ects
of
the
fou
r
side
s of
the
fram
e st
ron
ger
insi
de o
r ou
tsid
e th
e fr
ame?
left
mos
tri
ghtm
ost
upp
erlo
wer
B C D E
HRW material copyrighted under notice appearing earlier in this book.
114
2.A
2.0
m lo
ng
con
duct
ing
wir
e h
as a
cu
rren
t of
5.0
in a
un
ifor
m m
agn
etic
fiel
d of
0.43
T.T
he
fiel
d is
par
alle
l to
the
x-ax
is.
HO
LT P
HY
SIC
S
Mix
ed R
evie
w co
ntin
ued
Ch
apte
r
21 IB
(a)
I
B
(b)
a.W
hat
is t
he
forc
e on
th
e w
ire
wh
en it
is v
erti
cal,
para
llel t
o th
e y-
axis
as s
how
n a
?
b.W
hat i
s th
e fo
rce
on th
e w
ire
whe
n it
is h
oriz
onta
l,pa
ralle
l to
the
x-ax
is
as s
how
n in
b?
3.T
he
wir
e in
item
2 is
ben
t to
form
a 0
.50
m ×
0.50
m s
quar
e ca
rryi
ng
the
sam
e 5.
0 A
cu
rren
t,w
ith
th
e po
siti
ve c
har
ges
mov
ing
cloc
kwis
e in
th
e
fram
e.T
he
fram
e is
in t
he
sam
e m
agn
etic
fiel
d (B
=0.
43 T
).
a.Sk
etch
a d
iagr
am o
fth
e si
tuat
ion
.Use
arr
ows
to in
dica
te t
he
dire
c-
tion
of
the
curr
ent
in e
ach
seg
men
t of
the
fram
e.
b.Fi
nd
the
forc
es a
ctin
g on
eac
h s
ide
ofth
e fr
ame.
Spec
ify
thei
r m
agn
i-
tude
an
d di
rect
ion
.
c.D
o th
e fo
rces
on
th
e fr
ame
can
cel e
ach
oth
er?
Will
th
e fr
ame
be a
ble
to m
ove?
Will
it b
e ab
le to
rot
ate?
Exp
lain
.
Hol
t Phy
sics
Sec
tion
Revi
ew W
orks
heet
s
Honors Physics 2008–2009 Mr. Strong
64
Hol
t Phy
sics
Sec
tion
Revi
ew W
orks
heet
s11
8
Indu
ctio
n an
d Al
tern
atin
g Cu
rren
t
Mix
ed R
evie
wH
OLT
PH
YS
ICS
Ch
apte
r
22
1.W
hic
h o
fth
e fo
llow
ing
acti
ons
will
indu
ce a
n e
mf
in a
con
duct
or?
a.M
ove
a m
agn
et n
ear
the
con
duct
or.
b.M
ove
the
con
duct
or n
ear
a m
agn
et.
c.R
otat
e th
e co
ndu
ctor
in a
mag
net
ic fi
eld.
d.C
han
ge t
he
mag
net
ic fi
eld
stre
ngt
h.
e.al
l of
the
abov
e
2.A
cir
cula
r lo
op (
10 t
urn
s) w
ith
a r
adiu
s of
29 c
m is
in a
mag
net
ic fi
eld
that
osc
illat
es u
nif
orm
ly b
etw
een
0.9
5 T
an
d 0.
45 T
wit
h a
per
iod
of1.
00 s
.
a.H
ow m
uch
tim
e is
req
uir
ed fo
r th
e fi
eld
to c
han
ge fr
om 0
.95
T
to 0
.45
T?
b.W
hat
is t
he
cros
s-se
ctio
nal
are
a of
one
turn
of
the
loop
?
c.A
ssu
min
g th
at t
he
loop
is p
erp
endi
cula
r to
th
e m
agn
etic
fiel
d,w
hat
is
the
indu
ced
emf
in t
he
loop
?
3.E
lect
ric
gen
erat
ors
conv
ert
mec
han
ical
en
ergy
into
ele
ctri
cal e
ner
gy.
a.W
hat
are
th
e re
quir
emen
ts fo
r ge
ner
atin
g em
f?
b.T
he
mec
han
ical
en
ergy
inpu
t is
usu
ally
rot
atio
nal
mot
ion
.Wh
at a
re
two
poss
ible
sou
rces
of
rota
tion
al m
otio
n?
HRW material copyrighted under notice appearing earlier in this book.
Chap
ter 2
211
9
4.A
250
-tu
rn g
ener
ator
wit
h c
ircu
lar
loop
s of
radi
us
15 c
m r
otat
es a
t
60.0
rpm
in a
mag
net
ic fi
eld
wit
h a
str
engt
h o
f1.
00 T
.
a.W
hat
is t
he
angu
lar
spee
d of
the
loop
s?
b.W
hat
is t
he
area
of
one
loop
?
c.W
hat
is t
he
max
imu
m e
mf?
d.W
hat
is t
he
rms
emf?
5.A
n e
lect
ric
mot
or is
som
etim
es c
alle
d a
gen
erat
or in
rev
erse
.Exp
lain
you
r u
nde
rsta
ndi
ng
ofth
is s
tate
men
t.
6.C
onsi
der
a tw
o-co
il tr
ansf
orm
er jo
ined
by
a co
mm
on ir
on c
ore.
a.If
the
curr
ent
in t
he
prim
ary
side
is in
crea
sed,
wh
at h
appe
ns
to t
he
mag
net
ic fi
eld
in t
he
core
?
b.W
hat
eff
ect
does
th
e an
swer
to it
em 6
a h
ave
on t
he
seco
nda
ry c
oil?
c.Fu
lly e
xpla
in t
he
effe
ct o
fre
duci
ng
the
curr
ent
to t
he
prim
ary
side
of
a tr
ansf
orm
er.
HO
LT P
HY
SIC
S
Mix
ed R
evie
w co
ntin
ued
Ch
apte
r
22
HRW material copyrighted under notice appearing earlier in this book.
Honors Physics 2008–2009 Mr. Strong
65
Mix
edRev
iew
Answ
ers
Sect
ion
Thre
e—Se
ctio
n Re
view
Wor
kshe
et A
nsw
ers
III–1
1.a.
2.2
×10
5s
b.3.
5×
107
mm
c.4.
3×
10−4
km
d.2.
2×
10−5
kg
e.6.
71×
1011m
g
f.8.
76×
10−5
GW
g.1.
753
×10
−1ps
2.a.
3
b.4
c.10
d.3
e.2
f.4
3.a.
4
b.5
c.3
4.a.
1.00
54;−
0.99
52;5
.080
×10
−3;
5.07
6×
10−3
b.4.
597
×10
7 ;3.8
66 ×
107 ;
1.54
6×
1014
;11.
58
5.15
.9 m
2
6.T
he g
raph
sho
uld
be a
str
aigh
t lin
e.
Chap
ter 1
Mix
ed R
evie
w
Hol
t Phy
sics
Sol
utio
n M
anua
lIII
–2
III
2.a.
v f=
a(∆t
)
b.v f
=v i
+a(
∆t);
∆x=
1 2 (v i
+v f
)∆t
or∆x
=v i
(∆t)
+1 2 a
(∆t)
2
1.a.
t 1=
d 1/v
1;t 2
=d 2
/v2;t
3=
d 3/v
3
b.to
tal d
ista
nce
=d 1
+d 2
+d 3
c.to
tal t
ime
=t 1
+t 2
+t 3
3.
Tim
e in
terv
alTy
pe o
fmot
ion
v(m
/s)
a(m
/s2 )
Asp
eedi
ng
up
++
Bsp
eedi
ng
up
++
Cco
nst
ant
velo
city
+0
Dsl
owin
g do
wn
+−
Esl
owin
g do
wn
+−
.b
.a
.4
1 s
c.2
sTi
me
(s)
Posi
tion
(m)
v(m
/s)
a(m
/s2 )
14.
90
−9.8
1
20
−9.8
−9.8
1
3−1
4.7
−19.
6−9
.81
4−3
9.2
−29.
4−9
.81
Chap
ter 2
Mix
ed R
evie
w
Hol
t Phy
sics
Sol
utio
n M
anua
lIII
–4
1.a.
Th
e di
agra
m s
hou
ld in
dica
te t
he
rela
tive
dis
tan
ces
and
dire
ctio
ns
for
each
seg
men
t of
the
path
.
b.5.
0 km
,slig
htl
y n
orth
of
nor
thw
est
c.11
.0 k
m
2.a.
Th
e sa
me
b.Tw
ice
as la
rge
c.1.
58
3.a.
2.5
m/s
,in
th
e di
rect
ion
of
the
side
wal
k’s
mot
ion
b.1.
0 m
/s,i
n t
he
dire
ctio
n o
fth
e si
dew
alk’
s m
otio
n
c.4.
5 m
/s,i
n t
he
dire
ctio
n o
fth
e si
dew
alk’
s m
otio
n
d.2.
5 m
/s,i
n t
he
dire
ctio
n o
ppos
ite
to t
he
side
wal
k’s
mot
ion
e.4.
7 m
/s,q
=32
°
4.a.
4.0
×10
1se
con
ds
b.6.
0×
101
seco
nds
Chap
ter 3
Mix
ed R
evie
w
Sect
ion
Thre
e—Se
ctio
n Re
view
Wor
kshe
et A
nsw
ers
III–5
1.a.
at r
est,
mov
es to
th
e le
ft,h
its
back
wal
l
b.m
oves
to t
he
righ
t (w
ith
vel
ocit
y v)
,at
rest
,nei
ther
c.m
oves
to t
he
righ
t,m
oves
to t
he
righ
t,h
its
fron
t w
all
2.a.
mg,
dow
n
b.m
g,u
p
c.n
o
d.ye
s
3.a.
a=
m1
+F
m2
b.m
2a
c.F
−m
2a
=m
1a
d. m
1m +1 m2
F
4.a.
a= mF 1− +F mk
2
b.m
2a
−F k
c.F
−m
2a
−F k
=m
1a−
F k
d. m
1m +1 m2
(F
−F k
)
Chap
ter 4
Mix
ed R
evie
w
Hol
t Phy
sics
Sol
utio
n M
anua
lIII
–6
1.a.
60 J
b.−6
0 J
2.a.
mgh
b.m
gh
c.v B
=v A
2+
2gh
d.n
o
e.n
o
3.a.
2.9
J
b.1.
8 J
c.1.
2 J
d.a,
b:di
ffer
ent;
c:sa
me
4.a.
1 2 mv i
2+
mgh
i=
1 2 mv f
2+
mgh
f+
F kd
b.F k
=mm
g(co
s 23
°)
c.v f
= mv i
2+
2g(d
sin
23°−
mco
s23
°)
Chap
ter 5
Mix
ed R
evie
w
Sect
ion
Thre
e—Se
ctio
n Re
view
Wor
kshe
et A
nsw
ers
III–7
III
Copyright ©by Holt, Rinehart and Winston.All rights reserved.
1.a.
Th
e ch
ange
du
e to
th
e ba
t is
gre
ater
th
an t
he
chan
gedu
e to
th
e m
itt.
b.T
he
impu
lse
due
to t
he
bat
is g
reat
er t
han
th
e im
-pu
lse
due
to t
he
mit
t.
c.C
hec
k st
ude
nt
diag
ram
s.B
at:v
ecto
r sh
owin
g in
itia
lm
omen
tum
an
d a
larg
er v
ecto
r in
th
e op
posi
te d
i-re
ctio
n s
how
ing
impu
lse
ofba
t,re
sult
is t
he
sum
of
the
vect
ors.
Mit
t:ve
ctor
sh
owin
g in
itia
l mom
entu
man
d an
equ
al le
ngt
h v
ecto
r sh
owin
g im
puls
e of
mit
t,re
sult
is t
he
sum
,wh
ich
is e
qual
to z
ero.
2.a.
Th
e im
puls
es a
re e
qual
,bu
t op
posi
te fo
rces
,occ
ur-
rin
g du
rin
g th
e sa
me
tim
e in
terv
al.
b.T
he
tota
l for
ce o
n t
he
bow
ling
ball
is t
he
sum
of
forc
es o
n p
ins.
Th
e fo
rce
on t
he
pin
s is
equ
al b
ut
op-
posi
te o
fto
tal f
orce
on
bal
l.
3.m
1v1i
+m
2v2i
=(m
1+
m2)v f
;
m1v
1i/(
m1
+m
2)+
m2v
2i/(
m1
+m
2)=v f
4.a.
M(6
m/s
)
b.2
m/s
c.ob
ject
s tr
ade
mom
entu
m;i
fm
asse
s ar
e eq
ual
,ob-
ject
s tr
ade
velo
citi
es
Chap
ter 6
Mix
ed R
evie
w
Hol
t Phy
sics
Sol
utio
n M
anua
lIII
–8
1.a.
3.0,
3.0,
9.0,
27
b.4.
3,1.
0,4.
3,37
c.16
,0.2
8,11
,6.0
×10
2
d.63
0,0.
11,7
4,8.
7
e.5.
0,44
,0.1
1,9.
9
2.a.
fric
tion
b.gr
avit
atio
nal
forc
e
c.te
nsi
on in
str
ing
3.a.
dou
bled
b.qu
adru
pled
c.re
duce
d to
1 4
d.qu
adru
pled
e.re
duce
d to
1 9
4.19
0 m
5.St
uden
t dia
gram
s sh
ould
sho
w v
ecto
rs fo
r w
eigh
t an
dn
orm
al fo
rce
from
ele
vato
r;de
scen
t sho
uld
show
nor
mal
forc
e le
ss th
an w
eigh
t;st
oppi
ng
shou
ld s
how
nor
mal
forc
e gr
eate
r th
an w
eigh
t;“w
eigh
tles
snes
s”fe
elin
g is
due
to a
ccel
erat
ion
.
6.10
50 s
(17
.5 m
in)
Chap
ter 7
Mix
ed R
evie
w
Honors Physics 2008–2009 Mr. Strong
66
Sect
ion
Thre
e—Se
ctio
n Re
view
Wor
kshe
et A
nsw
ers
III–9
1.a.
Ifth
e kn
ob is
fart
her
from
th
eh
inge
,tor
que
is in
crea
sed
torq
ue
for
a gi
ven
forc
e.
b.tw
ice
as m
uch
2.a.
Rot
atio
nal
iner
tia
is r
edu
ced.
b.A
ngu
lar
mom
entu
m r
emai
ns
the
sam
e.
c.A
ngu
lar
spee
d in
crea
ses.
3.a.
2.0
kg
b.0.
67 k
g
4.a.
6.2
N• m
,0.0
16 k
g• m
2 ,39
0 ra
d/s2
b.12
N• m
,0.0
62 k
g•m
2 ,190
rad
/s2
5.a.
8.1
×10
16kg
• m2
b.5.
9×
1012
kg• m
2 /s
c.2.
1×
108
J
d.3.
1×
103
m/s
e.2.
2×
108
J
6.a.
4.0
×10
4J
b.4.
4×
104
J
c.4.
9×
104
J
d.0.
81
Chap
ter 8
Mix
ed R
evie
w
Sect
ion
Thre
e—Se
ctio
n Re
view
Wor
kshe
et A
nsw
ers
III–1
3
1.a.
0.20
s;5
.0 H
z
b.sa
me,
sam
e,in
crea
se,i
ncr
ease
2.a.
60.0
N/m
b.0.
574
seco
nds
;1.7
4 H
z
3.6.
58 m
/s2 ;n
o
4.a.
A:0
s,2
s,4
s;B
:0.5
s,1
.5 s
,2.5
s,3
.5 s
;C:1
,3 s
b.P
E:0
s a
t A,1
s a
t C
,2 s
at A
,3 s
at
C,4
s a
t A;K
E:
0.5
s,1.
5 s,
2.5
s,3.
5 s
at B
c.0.
5 s,
2.5
s at
B to
th
e ri
ght
1.5
s,3.
5 s
at B
to t
he
left
;0
s,2
s,4
s at
A to
th
e ri
ght,
1 s,
3 s
at C
to t
he
left
5.3.
00×
102
m/s
6.3.
0 s;
6.0
Chap
ter 1
2 M
ixed
Rev
iew
Hol
t Phy
sics
Sol
utio
n M
anua
lIII
–14
III
Copyright ©by Holt, Rinehart and Winston.All rights reserved.
1.a.
2.19
m;2
.27
m
b.w
avel
engt
h in
crea
ses
wh
en te
mpe
ratu
re in
crea
ses
2.a.
arro
ws
poin
tin
g E
ast
on a
mbu
lan
ce,p
olic
e,an
dtr
uck
,Wes
t on
van
.
b.po
lice
and
ambu
lan
ce (
equ
al),
tru
ck,s
mal
l car
,van
3.T
hes
e ob
ject
s h
ad t
he
sam
e n
atu
ral f
requ
ency
of
330
Hz,
so r
eson
ance
occ
urr
ed.
4.a.
1460
Hz,
2440
Hz
b.70
.8 c
m,2
3.6
cm,1
4.1
cm
c.0.
177
m
d.97
4 H
z,14
60 H
z;70
.8 c
m,3
5.4
cm,2
3.6
cm;0
.354
m
5.a.
5
b.43
5 H
z,be
cau
se it
will
als
o pr
ovid
e a
diff
eren
ce
of5
Hz.
Chap
ter 1
3 M
ixed
Rev
iew
7.a.
9.00
cm
b.p
=30
.0 c
m;q
=12
.9 c
m;r
eal;
inve
rted
;2.5
8 cm
tal
l
p=
24.0
cm
;q=
14.4
cm
;rea
l,in
vert
ed;3
.60
cm t
all
p=
18.0
cm
;q=
18.0
cm
;rea
l;in
vert
ed;6
.00
cm t
all
p=
12.0
cm
;q=
36.0
cm
;rea
l;in
vert
ed;2
.00
cm t
all
p=
6.0
cm;q
=−1
8 cm
;vir
tual
;upr
igh
t;18
cm
tal
l
8.p
=30
.0 c
m;q
=−6
.92
cm;v
irtu
al;u
prig
ht;
1.38
cm
tal
l
p=
24.0
cm
;q=
−6.5
5 cm
;vir
tual
,upr
igh
t;1.
64 c
m t
all
p=
18.0
cm
;q=
−6.0
0 cm
;vir
tual
;upr
igh
t;2.
00 c
m t
all
p=
12.0
cm
;q=
−5.1
4 cm
;vir
tual
;upr
igh
t;2.
57 c
m t
all
p=
6.0
cm;q
=−3
.6 c
m;v
irtu
al;u
prig
ht;
3.6
cm t
all
1.4.
07×
1016
m
2.a.
3.33
×10
−5s
b.1.
00×
10−4
m
3.3.
84×
108
m
4.3.
00×
1011
Hz
5.D
iffu
se r
efle
ctio
n:(
non
shin
y su
rfac
es)
tabl
e to
p,fl
oor,
wal
ls,c
ar p
ain
t,po
ster
s (a
nsw
ers
will
var
y)
Spec
ula
r re
flec
tion
:met
allic
su
rfac
es,w
ater
,mir
rors
(an
swer
s w
ill v
ary)
6.a.
Ch
eck
stu
den
t dr
awin
gs fo
r ac
cura
cy.
b.B
is 4
m fr
om A
hor
izon
tally
,Cis
2 m
bel
ow B
vert
ical
ly
c.D
is 2
m b
elow
Ave
rtic
ally
,Eco
inci
des
wit
h C
d.th
ey w
ill o
verl
ap t
he
exis
tin
g im
ages
or
obje
cts
Chap
ter 1
4 M
ixed
Rev
iew
Sect
ion
Thre
e—Se
ctio
n Re
view
Wor
kshe
et A
nsw
ers
III–1
7
1.a.
Ray
1 a
t 45
°;R
ay 2
at
14.9
°
b.R
ays
shou
ld in
ters
ect
insi
de t
he
aqu
ariu
m.
c.B
ecau
se t
he
rays
are
no
lon
ger
para
llel,
they
will
in-
ters
ect
in t
he
wat
er.
2.a.
Firs
t bo
un
dary
:70.
0°,4
5.0°
Seco
nd
bou
nda
ry:4
5.0°
,40.
4°
Th
ird
bou
nda
ry:4
0.3°
,36.
8°
b.In
com
ing
rays
get
clo
ser
and
clos
er to
th
e n
orm
al.
Ref
lect
ed r
ays
get
fart
her
aw
ay fr
om t
he
nor
mal
wit
hth
e sa
me
angl
es.
3.a.
9.00
cm
b.12
.9 c
m,1
4.4
cm,1
8.0
cm,3
6.0
cm,−
18.0
cm
2.58
cm
,3.6
cm
,6.0
0 cm
,18.
0 cm
,−18
.0 c
m
real
,rea
l,re
al,r
eal,
virt
ual
4.18
.0 c
m,w
ith
all i
mag
es v
irtu
al a
nd o
n th
e le
ft o
fthe
len
s
−11.
2,−1
0.3,
−9.0
0,−7
.20,
−4.5
0
5.a.
6.00
cm
in fr
ont
ofth
e le
ns
b.0.
857
cm
Chap
ter 1
5 M
ixed
Rev
iew
Hol
t Phy
sics
Sol
utio
n M
anua
lIII
–18
III
1.a.
6.74
×10
−6m
b.47
.9°
c.T
he
max
imu
m a
ngl
e fo
r lig
ht
to r
each
th
e sc
reen
inth
is a
rran
gem
ent
is 4
5°.
2.a.
Lon
ger
wav
elen
gths
are
dif
frac
ted
wit
h a
grea
ter
angl
e.
b.Fi
rst
orde
r gr
oup
oflin
es:b
lue,
gree
n,r
ed;s
econ
dor
der:
the
sam
e
c.W
hit
e
3.a.
A=
5.0
×10
−6m
,B =
1.1
×10
−7m
,C =
3.3
×10
−8m
b.vi
sibl
e:A
;x-r
ay:A
,B,o
r C
;IR
:non
e
4.a.
Nei
ther
wou
ld w
ork
beca
use
th
ey w
ould
act
as
dif-
fere
nt
sou
rces
,so
even
wit
h t
he
sam
e fr
equ
ency
,th
eysh
ould
not
be
in p
has
e.
b.In
terf
eren
ce is
occ
urr
ing.
Chap
ter 1
6 M
ixed
Rev
iew
Sect
ion
Thre
e—Se
ctio
n Re
view
Wor
kshe
et A
nsw
ers
III–1
9
1.a.
A;1
.87
×10
13el
ectr
ons;
B:3
.12
×10
13el
ectr
ons
b.th
e fo
rces
are
equ
al a
nd
oppo
site
,no
2.a.
Res
ult
ant =
1.49
N,l
eft;
F(A
-C)
=1.
35 N
,lef
t;F
(B-C
)=
0.14
0 N
,lef
t
b.R
esu
ltan
t =0.
788
N,r
igh
t;F(
A-C
) =
1.35
N,r
igh
t;F
(B-C
)=
0.56
2 N
,lef
t
c.R
esu
ltan
t =0.
400
N,l
eft;
F(A
-C)
=0.
599
N,r
igh
t;F
(B-C
)=
0.99
9 N
,lef
t
3.a.
1.92
×10
16N
b.2.
87×
1010
m/s
2
c.9.
81 m
/s2 ;t
his
is n
eglig
ible
in c
ompa
riso
n w
ith
th
eac
cele
rati
on a
;alp
ha
part
icle
s w
ill m
ove
hor
izon
tally
4.a.
Ch
eck
stu
den
ts d
iagr
ams
for
accu
racy
.
b.1.
53×
10−2
N
c.7.
65×
103
N/C
5.1
C =
6.25
×10
18;1
mC
=6.
25×
1012
Chap
ter 1
7 M
ixed
Rev
iew
Sect
ion
Thre
e—Se
ctio
n Re
view
Wor
kshe
et A
nsw
ers
III–2
1
1.a.
Iin
crea
ses
beca
use
Rde
crea
ses
(sh
orte
r)
b.n
o ch
ange
c.I
decr
ease
s be
cau
se R
incr
ease
sw
ith
tem
pera
ture
d.I
decr
ease
s
2.a.
4.8
A
b.8.
64×
104
J
c.58
0 W
d.1.
0×
107
J
3.a.
0.83
3 A
;1.2
5 A
;2.0
8 A
b.14
4Ω
;96.
0 Ω
;57.
6 Ω
c.70
.0 ¢
;$1.
05;$
1.75
4.a.
144
V
b.86
4 W
c.10
4 se
con
ds
Chap
ter 1
9 M
ixed
Rev
iew
Hol
t Phy
sics
Sol
utio
n M
anua
lIII
–22
1.a.
D
b.sw
itch
5
c.•
swit
ches
1 a
nd
3 op
en,s
wit
ches
2,4
,an
d 5
clos
ed
• sw
itch
es 1
an
d 4
open
,sw
itch
es 2
,3,a
nd
5 cl
osed
• sw
itch
2 o
pen
,sw
itch
es 1
,3,4
,an
d 5
clos
ed;o
rsw
itch
es 3
an
d 4
open
,sw
itch
es 1
,2,a
nd
5 cl
osed
;or
sw
itch
es 2
,3,a
nd
4 op
en,s
wit
ches
1 a
nd
5 cl
osed
2.a.
Ch
eck
stu
den
ts’d
iagr
ams,
wh
ich
sh
ould
sh
ow a
bulb
an
d a
resi
stor
in s
erie
s w
ith
a b
atte
ry.
b.15
Ω
c.6
Ω
3.a.
Ch
eck
stu
den
ts d
iagr
ams.
b.12
.0 V
,12.
0 V
c.0.
25 A
,2.2
5 A
d.5.
33Ω
4.a.
R=
6.15
Ω
b.R
=30
.4Ω
Chap
ter 2
0 M
ixed
Rev
iew
Honors Physics 2008–2009 Mr. Strong
67
Sect
ion
Thre
e—Se
ctio
n Re
view
Wor
kshe
et A
nsw
ers
III–2
3
1.a.
Th
e m
agn
etic
fiel
d fr
om t
he
left
mos
t se
gmen
t is
•an
d st
ron
ger.
Th
e m
agn
etic
fiel
d fr
om t
he
righ
tmos
tse
gmen
t is
×an
d w
eake
r.
b.A
t A,b
oth
hor
izon
tal s
egm
ents
con
trib
ute
a ×
mag
net
ic fi
eld
ofeq
ual
str
engt
h
c.B
;×;×
wea
ker;
×;×
sam
e
C;×
;×sa
me;
×;×
sam
e
D;×
;×st
ron
ger;
×;×
sam
e
E;×
;•st
ron
ger;
×;×
sam
e
d.N
o.T
hey
rei
nfo
rce
each
oth
er in
th
e sa
me
dire
ctio
n.
e.in
side
2.a.
F=
4.3
N in
to t
he
page
b.F
=0
3.a.
Dia
gram
s sh
ould
sh
ow c
lock
wis
e cu
rren
t.
b.St
arti
ng
from
th
e le
ft s
ide:F
=1.
1 N
into
th
e pa
ge;
F=
0;F
=1.
1 N
ou
t of
the
page
;F=
0
c.Fo
rces
are
equ
al a
nd
oppo
site
,so
no
tran
slat
ion
alm
otio
n w
ill o
ccu
r,bu
t it
cou
ld r
otat
e ar
oun
d a
vert
i-ca
l axi
s.
Chap
ter 2
1 M
ixed
Rev
iew
Hol
t Phy
sics
Sol
utio
n M
anua
lIII
–24
1.e
2.a.
0.50
s
b.0.
26 m
2
c.2.
6 V
3.a.
mag
net
ic fi
eld,
con
duct
or,r
elat
ive
mot
ion
b.an
swer
s m
ay v
ary,
but
cou
ld in
clu
de t
he
follo
win
g:w
ater
wh
eel,
win
dmill
,ele
ctri
c m
otor
,com
bust
ion
engi
ne
4.a.
6.28
rad
/s
b.7.
1×
10−2
m2
c.11
0 V
d.78
V
5.A
mot
or c
onve
rts
elec
tric
en
ergy
to r
otat
ion
al e
ner
gy;
gen
erat
ors
conv
erts
rot
atio
nal
en
ergy
to e
lect
ric
ener
gy.
6.a.
incr
ease
s
b.in
duce
s cu
rren
t w
hile
ch
ange
occ
urs
c.It
dec
reas
es m
agn
etic
fiel
d w
hic
h w
ill in
duce
a c
ur-
ren
t w
hile
th
e ch
ange
occ
urs
.
Chap
ter 2
2 M
ixed
Rev
iew
Honors Physics 2008–2009 Mr. Strong
Recommended