Chapter 14
1. The air inside pushes outward with a force given by piA, where pi is the pressure inside the room and A is the area of the window. Similarly, the air on the outside pushes inward with a force given by poA, where po is the pressure outsi
Chapter 35
1. Comparing the light speeds in sapphire and diamond, we obtain
v v s vd 2 c.998
F 1I 1 G nJ n H K F1 10 m sh G H 177 .
c
s d 8
1 2.42
I 4.55 J K
107 m s .
2. (a) The frequency of yellow sodium light is
f
c
2.998 108 m s 589 10 9
Chapter 36
1. The condition for a minimum of a single-slit diffraction pattern is
a sin m
where a is the slit width, is the wavelength, and m is an integer. The angle is measured from the forward direction, so for the situation described in the p
Chapter 37
1. From the time dilation equation t = t0 (where t0 is the proper time interval,
2
1/ 1
, and
= v/c), we obtain
1
Ft I . Gt J HK
2 0
The proper time interval is measured by a clock at rest relative to the muon. Specifically, t0 = 2.2
Chapter 38
1. (a) Let E = 1240 eVnm/
min
= 0.6 eV to get
= 2.1
103 nm = 2.1 m.
(b) It is in the infrared region. 2. The energy of a photon is given by E = hf, where h is the Planck constant and f is the frequency. The wavelength is related to the
Chapter 39
1. According to Eq. 39-4 En L 2. As a consequence, the new energy level E'n satisfies
En En L L FI FI GJ GJ HK HK L L
2 2
1 , 2
which gives L
2 L. Thus, the ratio is L / L
2 1.41.
2. (a) The ground-state energy is
E1
h2 n2 2 8me L
Chapter 40
1. (a) For a given value of the principal quantum number n, the orbital quantum number ranges from 0 to n 1. For n = 3, there are three possible values: 0, 1, and 2. (b) For a given value of , the magnetic quantum number m ranges from
Chapter 41
1. The number of atoms per unit volume is given by n d / M , where d is the mass density of copper and M is the mass of a single copper atom. Since each atom contributes one conduction electron, n is also the number of conduction electrons
Chapter 42
1. Our calculation is similar to that shown in Sample Problem 42-1. We set K 5.30 MeV=U 1/ 4 0 q qCu / rmin and solve for the closest separation, rmin:
rmin
q qCu 4 0K 1.58 10
kq qCu 4 0K
14
2e 29 1.60 10
19
C 8.99 109 V m/C
5.30 10
Chapter 43
1. If R is the fission rate, then the power output is P = RQ, where Q is the energy released in each fission event. Hence, R = P/Q = (1.0 W)/(200 106 eV)(1.60 10 19 J/eV) = 3.1 1010 fissions/s.
2. We note that the sum of superscripts (mas
Chapter 44
1. Conservation of momentum requires that the gamma ray particles move in opposite directions with momenta of the same magnitude. Since the magnitude p of the momentum of a gamma ray particle is related to its energy by p = E/c, the partic
Formula Sheet for Homework and Exams Page 1 of 2
1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16.
vf v0 a tf t0
23.
1 2
U
Fcons dx
xf
xf
x0
x0
v 0 (t f
1 2
t0 )
v f )(t f
1 2
a (t f
t0 )
t 0 )2 t 0 )2
24. 25. 26. 27. 28. 29. 30. 31
Chapter 2
1. The speed (assumed constant) is (90 km/h)(1000 m/km) (3600 s/h) = 25 m/s. Thus, during 0.50 s, the car travels (0.50)(25) 13 m. 2. Huber's speed is v0=(200 m)/(6.509 s)=30.72 m/s = 110.6 km/h, where we have used the conversion factor 1 m
Chapter 3
1. The x and the y components of a vector a lying on the xy plane are given by
ax a cos ,
ay a sin
where a | a | is the magnitude and is the angle between a and the positive x axis. (a) The x component of a is given by ax = 7.3
Chapter 6
1. An excellent discussion and equation development related to this problem is given in Sample Problem 6-3. We merely quote (and apply) their main result (Eq. 6-13)
tan 1 s tan 1 0.04 2 .
2. The free-body diagram for the player is
Chapter 7
1. With speed v = 11200 m/s, we find
K
1 2 mv 2
1 (2.9 105 ) (11200) 2 2
18 1013 J. .
2. (a) The change in kinetic energy for the meteorite would be
K
Kf
Ki
Ki
1 mi vi2 2
1 4 106 kg 15 103 m/s 2
2
5 1014 J ,
or | K | 5 1014 J
Chapter 8
1 1. The potential energy stored by the spring is given by U 2 kx 2 , where k is the spring constant and x is the displacement of the end of the spring from its position when the spring is in equilibrium. Thus
k
2U x2
bg 0 b.075 mg
2 25
Spreadsheet for Magnetic Field of a Bar Magnet Rev. 17-Nov-04 GB Fill in boxes for L and W in cm: L= 7.5 W= 1.5 D= 3 This is calcuated from Point #1:
Qm 4
191.3474
Fill in measured values of magnetic field (Bx) in Gauss: Point # 1 2 3 4 5 6 7 8 X
Chapter 34
1. The image is 10 cm behind the mirror and you are 30 cm in front of the mirror. You must focus your eyes for a distance of 10 cm + 30 cm = 40 cm. 2. The bird is a distance d2 in front of the mirror; the plane of its image is that same di
Chapter 33
1. In air, light travels at roughly c = 3.0 distance of
d ct
108 m/s. Therefore, for t = 1.0 ns, we have a
(3.0 108 m / s) (1.0 10
9
s)
0.30 m.
2. (a) From Fig. 33-2 we find the smaller wavelength in question to be about 515 nm, (b)
Chapter 15
1. (a) The amplitude is half the range of the displacement, or xm = 1.0 mm. (b) The maximum speed vm is related to the amplitude xm by vm = xm, where is the angular frequency. Since = 2f, where f is the frequency,
vm = 2 fxm = 2 120 Hz
Chapter 16
1. (a) The motion from maximum displacement to zero is one-fourth of a cycle so 0.170 s is one-fourth of a period. The period is T = 4(0.170 s) = 0.680 s. (b) The frequency is the reciprocal of the period:
f
1 T
1 1.47 Hz. 0.680s
(c) A
Chapter 17
1. The time it takes for a soldier in the rear end of the column to switch from the left to the right foot to stride forward is t = 1 min/120 = 1/120 min = 0.50 s. This is also the time for the sound of the music to reach from the musician
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 74.9 103 kg/mol. 7.50 1024 arsenic atoms have a total mass of (7.50 1024) (74.9 103 kg/mol)/(6.02 10
Chapter 20
1. (a) Since the gas is ideal, its pressure p is given in terms of the number of moles n, the volume V, and the temperature T by p = nRT/V. The work done by the gas during the isothermal expansion is
W
V2 V1
p dV
n RT
V2 V1
dV V
n R
Chapter 21
1. (a) With a understood to mean the magnitude of acceleration, Newtons second and third laws lead to
m2 a2
m1a1
m2
6 c.3
10 7 kg 7.0 m s2 9.0 m s
2
c h
h 4.9
2
10 7 kg.
(b) The magnitude of the (only) force on particle 1 is
F m1a
Chapter 22
1. We note that the symbol q2 is used in the problem statement to mean the absolute value of the negative charge which resides on the larger shell. The following sketch is for q1 q2 .
The following two sketches are for the cases q1 > q2 (
Chapter 23
1. The vector area A and the electric field E are shown on the diagram below. The angle between them is 180 35 = 145, so the electric flux through the area is
2 E A EA cos 1800 N C 3.2 103 m cos145 1.5 102 N m 2 C.
2
Chapter 24
1. (a) An Ampere is a Coulomb per second, so
84 A h
F C h IF s I 3.0 84 G s J3600 h J G H K K H
109 eV = 1.2 GeV.
105 C.
106 J.
(b) The change in potential energy is U = q V = (3.0 2. The magnitude is U = e V = 1.2
105 C)(12 V) = 3.6
Chapter 25
1. Charge flows until the potential difference across the capacitor is the same as the potential difference across the battery. The charge on the capacitor is then q = CV, and this is the same as the total charge that has passed through th