Quiz 3, Phyx 103-0, Spring 2013, Solutions
_ 1) The majority of the radiation emitted to the general public during the normal operation of
a nuclear power plant comes from:
A. cooling water discharged to a lake or a river.
B. fission-fragment gases that e
Quiz #2, Physics 103-0, Spring 2013
_ 1) About how much current is drawn by a 1500 W appliance operating at 125 V?
A. 2 amps
B. 8 amps
C. 12 amps D. 15 amps E. 24 amps F. 30 amps
Since P = IV, the current will be I = P/V = 1500 W / 125 V = 12 amps.
Quiz #1, Physics 103-0, Spring 2013
Ans: D 1) A net force of 30 newtons is applied to a block of mass 10 kg. The force that must be
applied to a block of mass 5 kg to give it an equal acceleration is:
a. 2.5 N b. 5 N c. 10 N d. 15 N e. 20 N f. 30
Phyx 103-0 Midterm, Spr 2013
D 1) If a constant nonzero net force is applied to an object, its velocity will always:
A. go to zero
B. be nonzero
C. be constant
E. equal the acceleration
F. remain at constant speed, but change direction
51. (a) Setting h / p h /
b E / cg m c , we solve for K = E m c :
K me2 c 4 me c 2
10 10 nm
0.015 MeV 15 keV.
(b) Using the value hc 1240eV nm
hc 1240 eV nm
1.2 105 eV 120 keV.
10 103 nm
. 1 eV = 1.60 10 19 J . For a photon, E = hf = . h = 6.63 10 34 J s.
EXECUTE: (a) energy = Pavt = (0.600 W)(20.0 103 s) = 1.20 10 2 J = 7.5 1016 eV
IDENTIFY and SET UP:
(6.63 1034 J s)(3.00 108 m/s)
= 3.05 1019 J = 1.91 eV
652 109 m
36. (a) Eq. 37-36 leads to a speed of
c (0.004)(3.0 108 m/s) 1.2 106 m/s 1 106 m/s.
(b) The galaxy is receding.
37. We obtain
620 nm 540 nm
43. (a) The work-kinetic energy theorem applies as well to relativistic physics as to
9. (a) The rest length L0 = 130 m of the spaceship and its length L as measured by the
timing station are related by Eq. 37-13. Therefore,
L L0 1 (v / c) 2 130 m 1 0.740 87.4 m.
(b) The time interval for the passage of the spaceship is
4. Due to the time-dilation effect, the time between initial and final ages for the daughter
is longer than the four years experienced by her father:
tf daughter ti daughter = (4.000 y)
where is Lorentz factor (Eq. 37-8). Letting T denote the age of the f
19. (a) Using the notation of Sample Problem Pointillistic paintings use the diffraction
of your eye,
2(50 106 m)(1.5 103 m)
019 m .
122(650 109 m)
(b) The wavelength of the blue light is shorter so Lmax 1 will be larger.
20. Using the no
4. (a) Eq. 36-3 and Eq. 36-12 imply smaller angles for diffraction for smaller wavelengths.
This suggests that diffraction effects in general would decrease.
(b) Using Eq. 36-3 with m = 1 and solving for 2 (the angular width of the central
36. (a) On both sides of the soap is a medium with lower index (air) and we are
examining the reflected light, so the condition for strong reflection is Eq. 35-36. With
lengths in nm,
3360 for m = 0
1120 for m = 1
for m = 2
for m = 3
14. (a) For the maximum adjacent to the central one, we set m = 1 in Eq. 35-14 and obtain
1 sin 1
(b) Since y1 = D tan 1 (see Fig. 35-10(a), we obtain
y1 = (500 mm) tan (0.010 rad) = 5.0 mm.
The separation is y = y
40. We use Eq. 34-8 (and Fig. 34-11(d) is useful), with n1 = 1.6 and n2 = 1 (using the
rounded-off value for air):
1.6 1 1 16
Using the sign convention for r stated in the paragraph following Eq. 34-8 (so that
r 5.0 cm ), we obtain i = 2.4 cm for o
47. The law of refraction states
n1 sin 1 n2 sin 2 .
We take medium 1 to be the vacuum, with n1 = 1 and 1 = 32.0. Medium 2 is the glass,
with 2 = 21.0. We solve for n2:
51. (a) Approximating
33. Let I0 be the intensity of the unpolarized light that is incident on the first polarizing
sheet. The transmitted intensity is I1 2 I 0 , and the direction of polarization of the
transmitted light is 1 = 40 counterclockwise from the y axis in the dia
55. We use vS = r (with r = 0.600 m and = 15.0 rad/s) for the linear speed during
circular motion, and Eq. 17-47 for the Doppler effect (where f = 540 Hz, and v = 343 m/s
for the speed of sound).
(a) The lowest frequency is
526 Hz .
38. The frequency is f = 686 Hz and the speed of sound is vsound = 343 m/s. If L is the
length of the air-column, then using Eq. 1741, the water height is (in unit of meters)
h 1.00 L 1.00
(1.00 0.125n) m
where n = 1, 3, 5, wi
6. Let be the length of the rod. Then the time of travel for sound in air (speed vs) will
be ts / vs . And the time of travel for compressional waves in the rod (speed vr) will be
tr / vr . In these terms, the problem tells us that
ts tr 0.12s .
18. The volume of a cylinder of height is V = r2 = d2 /4. The strings are long,
narrow cylinders, one of diameter d1 and the other of diameter d2 (and corresponding
linear densities 1 and 2). The mass is the (regular) density multiplied by the volume: m
4. The distance d between the beetle and the scorpion is related to the transverse speed vt
and longitudinal speed v as
d vt tt v t
where tt and t are the arrival times of the wave in the transverse and longitudinal
directions, respectively. With vt 50 m/
IDENTIFY: Use = Fl = rF sin to calculate the magnitude of each torque and use the right-hand rule to
determine the direction of each torque. Add the torques to find the net torque.
SET UP: Let counterclockwise torques be positive. For the 11.9 N forc
IDENTIFY: Use the equations in Table 9.2. I for the rod is the sum of I for each segment. The parallel-axis
theorem says I p = I cm + Md 2 .
SET UP: The bent rod and axes a and b are shown in Figure 9.59. Each segment has length L / 2 and mass M / 2
z1 + z 2
. 0 = av-z t . When z is linear in t, av-z for the time interval t1 to t2 is
SET UP: From the information given, z (t ) = 6.00 rad/s + (2.00 rad/s 2 )t
EXECUTE: (a) The angular acceleration is positive, sin
IDENTIFY: Eqs. 8.24 and 8.25 apply, with object A being the neutron.
SET UP: Let +x be the direction of the initial momentum of the neutron. The mass of a neutron is mn = 1.0 u .
1.0 u 2.0 u
EXECUTE: (a) v A 2 x = A
v A1x = v A1x / 3.0 . The s
IDENTIFY: Apply Eq. 8.9 to relate the change in momentum of the momentum to the components of the average
force on it.
SET UP: Let +x be to the right and +y be upward.
EXECUTE: (a) J x = px = mv2 x mv1x = (0.145 kg)( [65.0 m/s]cos30 50.0 m/s) = 15.4
IDENTIFY: For the system of two blocks, only gravity does work. Apply Eq.(7.5).
SET UP: Call the blocks A and B, where A is the more massive one. v A1 = vB1 = 0 . Let y = 0 for each block to be
at the initial height of that block, so y A1 = yB1 = 0 .
IDENTIFY: All forces are constant and each block moves in a straight line. so W = Fs cos . The only direction the
system can move at constant speed is for the 12.0 N block to descend and the 20.0 N block to move to the right.
SET UP: Since the 12.0 N