Chapter 35
1. The fact that wave W2 reflects two additional times has no substantive effect on the
calculations, since two reflections amount to a 2(/2) = phase difference, which is
effectively not a
HW 8 Solutions
Chapter 15 Problem 7
In exercising, a weight lifter loses 0.150 kg of water through evaporation, the heat required to
evaporate the water coming from the weight lier's body. The work do
Chapter 1:
MEASUREMENT
1. The SI standard of time is based on:
A. the daily rotation of the earth
B. the frequency of light emitted by Kr86
C. the yearly revolution of the earth about the sun
D. a pre
Chapter 32
1. We use
6
n1
Bn 0 to obtain
5
B 6 Bn 1Wb 2 Wb 3Wb 4 Wb 5 Wb 3Wb .
n1
2. (a) The flux through the top is +(0.30 T)r2 where r = 0.020 m. The flux through the
bottom is +0.70 mWb as given
Chapter 33
1. Since , we find f is equal to
8
9
c c (3.0 10 m/s)(0.0100 10 m)
2
7.49 109 Hz.
9
2
(632.8 10 m)
2. (a) The frequency of the radiation is
f
c
3.0 108 m / s
4.7 103 Hz.
5
6
(10 10
Chapter 34
1. The bird is a distance d2 in front of the mirror; the plane of its image is that same
distance d2 behind the mirror. The lateral distance between you and the bird is d3 = 5.00
m. We deno
Chapter 28
1. THINK The magnetic force on a charged parti014cle is given by FB qv B, where
v is the velocity of the charged particle and B is the magnetic field.
EXPRESS The magnitude of the magnetic
Chapter 38
1. (a) With E = hc/min = 1240 eVnm/min = 0.6 eV, we obtain = 2.1 103 nm = 2.1
m.
(b) It is in the infrared region.
2. Let
1
hc
me v 2 E photon
2
and solve for v:
v
2hc
2hc 2
2hc
c c
2
me
Chapter 16
1. Let y1 = 2.0 mm (corresponding to time t1) and y2 = 2.0 mm (corresponding to time t2).
Then we find
kx + 600t1 + = sin1(2.0/6.0)
and
kx + 600t2 + = sin1(2.0/6.0) .
Subtracting equations
Chapter 37
1. From the time dilation equation t = t0 (where t0 is the proper time interval,
1 / 1 2 , and = v/c), we obtain
FG t IJ .
H t K
2
1
0
The proper time interval is measured by a clock at r
Chapter 36
1. (a) We use Eq. 36-3 to calculate the separation between the first (m1 = 1) and fifth
(m2 5) minima:
D
m D
y D sin D
m
m2 m1 .
a
a a
Solving for the slit width, we obtain
b
g b
gc
hb
Chapter 31
1. (a) All the energy in the circuit resides in the capacitor when it has its maximum
charge. The current is then zero. If Q is the maximum charge on the capacitor, then the
total energy is
Chapter 29
1. (a) The magnitude of the magnetic field due to the current in the wire, at a point a
distance r from the wire, is given by
i
B 0 .
2r
With r = 20 ft = 6.10 m, we have
c4
B
hb
2 b mg
T m
Chapter 30
1. The flux B BA cos does not change as the loop is rotated. Faradays law only
leads to a nonzero induced emf when the flux is changing, so the result in this instance is
zero.
2. Using Far
Chapter 19
1. Each atom has a mass of m = M/NA, where M is the molar mass and NA is the
Avogadro constant. The molar mass of arsenic is 74.9 g/mol or 74.9103 kg/mol.
Therefore, 7.50 1024 arsenic atoms
Chapter 20
1. THINK If the expansion of the gas is reversible and isothermal, then theres no change
in internal energy. However, if the process is reversible and adiabatic, then there would
be no chan
Chapter 21
1. THINK After the transfer, the charges on the two spheres are Q q and q.
EXPRESS The magnitude of the electrostatic force between two charges q1 and q2
separated by a distance r is given
Chapter 25
1. (a) The capacitance of the system is
C
q
70 pC
35 pF.
.
V
20 V
(b) The capacitance is independent of q; it is still 3.5 pF.
(c) The potential difference becomes
V
q 200 pC
57 V.
C 35
Chapter 22
1. We note that the symbol q2 is used in the problem statement to mean the absolute value
of the negative charge that resides on the larger shell. The following sketch is for q1 q2 .
The fo
Chapter 23
1. THINK This exercise deals with electric flux through a square surface.
EXPRESS The vector area A and the electric field E are shown on the diagram below.
The electric flux through the su
Chapter 24
1. THINK Ampere is the SI unit for current. An ampere is one coulomb per second.
EXPRESS To calculate the total charge through the circuit, we note that 1 A 1 C/s and
1 h 3600 s.
ANALYZE (a
Chapter 26
1. (a) The charge that passes through any cross section is the product of the current and
time. Since t = 4.0 min = (4.0 min)(60 s/min) = 240 s,
q = it = (5.0 A)(240 s) = 1.2 103 C.
(b) The
Chapter 27
1. THINK The circuit consists of two batteries and two resistors. We apply Kirchhoffs
loop rule to solve for the current.
EXPRESS Let i be the current in the circuit and take it to be posit
Chapter 39
1. According to Eq. 39-4, En L 2. As a consequence, the new energy level E'n satisfies
FG IJ FG L IJ
H K H L K
En
L
En
L
2
2
1
,
2
which gives L 2 L. Thus, the ratio is L / L 2 1.41.
2. (a)
Chapter 40
1. The magnitude L of the orbital angular momentum L is given by Eq. 40-2:
L ( 1) . On the other hand, the components Lz are Lz m , where m ,. .
Thus, the semi-classical angle is cos Lz / L
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Exam 3: Form A
Introductory Physics II: Spring 2015
NOTE: you may need the mass of the electron and proton me = 9.11
1031 kg and mp = 1.67 1027 kg. If any other constant or equation is
needed, please
PHYS:1611 Discussion:
Week 1
January 21, 2018
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January 21, 2018
1/9
Things to know:
My name: Kenny Heitritter
PHYS:1611 Discussion: Week 1
January 21, 2018
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Things to kn
PHYS:1611 Discussion:
Week 2
February 1, 2018
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Before we begin
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not what you generally
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