Chapter 18 MC Explanations by TA Ajay Kailas
1.) Current, by default, goes in the opposite direction of electron movement. Current flows in the direction of positive charge. Therefore, if Chlorine is moving left (it's negative) current is moving right. Si
10.88: Assuming the blow to be concentrated at a point (or using a suitably chosen rFave, and L rFave t rJ . "average" point) at a distance r from the hinge, ave The angular velocity is then L rFave t l 2 Fave t 3 Fave t , 1 I I ml 2 2 ml 3 Where l is the
10.100:
a)
The distance from the center of the ball to the midpoint of the line joining the points where the ball is in contact with the rails is R 2 d 2 , so vcm R 2 d 2 4. R, the same as rolling on a flat surface. When when d 0, this reduces to vcm 0 fo
10.101:
a)
The friction force is f acceleration is gives
t
fR I k MgR 1 2 MR 2
k n
2 k g R
k Mg , so a
at
k g. The magnitude of the angular
R
0
. b) Setting v
t R and solving for t
R0 a R
R0 k g 2 k g
R0 , 3 k g
and
d 1 2 at 2 1 R0 k g 2 3 k g
2 2 R 20 .
10.102: Denoting the upward forces that the hands exert as FL and FR , the conditions that FL and FR must satisfy are FL FR w
I , r where the second equation is L, divided by r. These two equations can be solved for the forces by first adding and then sub
3/10/2011
Chapter 22: Electromagnetic
Waves
A magnetic dipole antenna
works by Faradays law:
changing magnetic flux
induces a current in the
loop.
Nature of EM waves
Antennae
The EM Spectrum
EM
Speed of EM Waves
Energy Transport
Polarization
Doppler Effec
2/24/2011
Unit 13, Chapter 20 Overview:
Electromagnetic Induction
Motional EMF
The motional EMF is
= vBL
where L is the separation between the rails.
The current in the rod is
Electric Generators
Faradays Law
I=
V vBL
=
R
R
R
where R is the resistance in
Chapter 21: Alternating Currents
Sinusoidal Voltages and Currents
Capacitors, Resistors, and Inductors in AC Circuits
Series RLC Circuits
Resonance
1
21.1 Sinusoidal Currents and
Voltage
2
A power supply can be set to give an EMF of form:
(t ) = 0 sin t
Chapter 28: Quantum Physics
The Hydrogen Atom
The Pauli Exclusion Principle
Electron Energy Levels in a Solid
The Laser
Quantum Mechanical Tunneling
28.6 The Hydrogen Atom: Wave
Functions and Quantum Numbers
In the quantum picture of the atom the electron
10.99: In Fig. (10.34(a), if the vector r , and hence the vector L are not horizontal but make an angle with the horizontal, the torque will still be horizontal (the torque must be perpendicular to the vertical weight). The magnitude of the torque will be
10.98: The velocity of the center of mass will change by velocity will change by
vcm
J m
, and the angular
The change is velocity of the end of the bat will J J ( x xcm ) xcm then be vend Setting vcm _ xcm m vend 0 allows cancellation of J , and gives ( x
10.97: From Eq. (10.36),
r
(50.0 kg)(9.80 m s2 )(0.040 m) 12.7 rad s, (0.085 kg m2 )(6.0 m s) (0.33 m)
or 13 rad s to two figures, which is quite large.
10.89: is I
a) The initial angular momentum is L
mv l 2 and the final moment of inertia
I0
m l 2 , so
mv l 2 M 3 l2 m l 2 5.46 rad s.
2
2
b) M m gh 1 2 2 I , and after solving for h and substitution of numerical values, will be h 3.16 10 2 m. c) Rather t
10.90: Angular momentum is conserved, so I 0 0 I 2 2 , or, using the fact that for a common mass the moment of inertia is proportional to the square of the radius, 2 2 2 2 R020 R2 2 , or R0 0 R0 R2 0 ~ R0 0 2R0 R 0 R0 , where the terms in R and 2 have bee
10.91: The initial angular momentum is L1
K1
0 I A and the initial kinetic energy is
2 I A0 2. The final total moment of inertia is 4 I A , so the final angular velocity
is 1 4 0 and the final kinetic energy is 1 2 4I A 0 4 2 obtained more directly from K
10.92: The tension is related to the block's mass and speed, and the radius of the circle, v2 by T m . The block's angular momentum with respect to the hole is L mvr , so in r terms of the angular momentum,
T 1 mv r
2
m 2v 2 r 2 m r3
mvr mr 3
2
L2 . mr 3
10.93: The train's speed relative to the earth is 0.600 m s 0.475 m , so the total angular momentum is
0.600 m s
0.475 m 1.20 kg 0.475 m
1.00 m 1 2 7.00 kg 2
2
0,
from which 0.298 rad s ,with the minus sign indicating that the turntable moves clockwise,
10.94:
a), g)
b) Using the vector product form for the angular momentum, v1
v2 and r1
r2 , so
mr1 v1 , ^. Then, j so the angular momenta are the same. c) Let ^ ^ v1 r1 zi xk , and ^ L1 mr1 v1 m xR i x 2 y 2 ^ j
mr2 v2
^ xR k .
With x 2
y2
m 2R2 ( 2 m
a) The initial angular momentum with respect to the pivot is mvr, and the final total moment of inertia is I mr 2 , so the final angular velocity is mvr mr 2 I . b) The kinetic energy after the collision is 1 2 K mr 2 I M m gh, or 2 2 M m gh . mr 2 I c) S
10.96: The initial angular momentum is 1 mRv1 , with the minus sign indicating that runner's motion is opposite the motion of the part of the turntable under his feet. The final angular momentum is 2 ( mR 2 ), so 1 mRv1 2 mR 2
(80 kg m 2 )( 0.200 rad s) (
Einstein proposed a particle theory of light.
Light is absorbed in discrete quanta called photons.
Physics 1102 Overview Sections 27.1-28.5
Quantum Physics
The energy of a photon depends on frequency:
E = hf
where h is Plancks constant.
Quantization of li
Chapter 26: Relativity
The Postulates of Special Relativity
Simultaneous Events
Time Dilation
Length Contraction
Relativistic Velocity Addition
Relativistic Momentum
Relativistic Rest Mass Energy, Kinetic Energy, and Total Energy
Albert Einstein
26.1 Post
REVIEW AND SYNTHESIS: CHAPTERS 1618
Review Exercises
1. Strategy and Solution Since the spheres are identical, the charge will be shared evenly by the two spheres, so the spheres will have (18.0 C 6.0 C) 2 12.0 C of charge each. 5. Strategy The force on c
Chapter 16 multiple choice answers
1 j
2 c
3 e
4 a
5 c
6 b
7 d
8 c
9 b
10 b
Chapter 17 multiple choice answers
1 f
2 a
3 e
4 c
5 d
6 b
7 f
8 e
9 b
10 b
11 b
12 d
UNIT 15 CONNECT PROBLEMS
1.
(b)
(c)
2.
(b)
(c)
3.
4.
PROBLEM #4: enter answers for X, Y, and Z before
checking your work. They aren't treated as separate
parts. If you get it wrong, click on View Details of Last
Answer Check to see which of
Unit 13, Chapter 20 Overview:
Electromagnetic Induction
Motional EMF
Electric Generators
Faradays Law
Lenzs Law
Transformers
Eddy Currents
Mutual- and Self-Inductance
LR Circuits
1
2016 Alan Giambattista
20.1 Motional EMF
A conductor in a B-field
Chapter 21: Alternating Currents
Sinusoidal Voltages and Currents
Capacitors, Resistors, and Inductors in AC Circuits
Series RLC Circuits
Resonance
1
2016 Glenn Fletcher
21.1 Sinusoidal Currents and
Voltage
2
2016 Glenn Fletcher
A power supply can b
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