Return to PHYS 152 Past Exams & Answers
Part A
1a. F
1b. B
1c. E
2a. E
2b. B
2c. E
3a. B
3b. A
3c. A
4. E
5. C
6a. B
6b. D
7. D
Part B
8. C
9a. A
9b. D
10. E
11. D
12. D (correction)
13. D (correction)
14. B
15. E (correction)
16a. B
16b. A
17. D
18. C
19. A
Part C
1a. To minimize currrent drawn from the voltage source,
we want to maximize the total resistance of the 3 resistors (light
bulbs). This is accomplished when the three are all connected
in series with each other. (The equivalent resistance is
then just the sum of the 3 resistances: 792 ohms.) I =
0.152 A is drawn from the voltage source.
1b. To maximize current drawn from the voltage source,
we want to minimize the total resistance of the 3 resistors (light
bulbs). This is accomplished when the three are all connected
in parallel with each other. (The equivalent resistance
then becomes 82.3 ohms.) The current drawn from the voltage source
becomes 1.46 A.
1c. B = all light bulbs connected to the voltage source
in parallel is the situation in household circuits
1d. C, B, A
2a. upward-curving path: alpha & beta-plus
straight path: gamma
downward-curving path: beta-minus
2b. (q/m) = 1.8x10^11 C/kg. The upward-curving path could
be an alpha particle (q/m = 4.8x10^7 C/kg) or a beta-plus particle
(positron) (q/m = 1.76x10^11 C/kg). Thus, our value of q/m is
consistent with a positron.
2c. 2.27 hours or 136 min.
Part D
3a. +47 cm
3b. real, inverted, larger than object, same side of mirror
as object
3c. ray diagram from object (given) to inverted image,
-10 cm tall, located 47 cm to the left of the mirror
3d. virtual, upright, larger than object, opposite side
of mirror from object
4a. 1.960 eV
4b. each photon carries 1.960 eV = 3.139x10^-19 J, so to
emit 20 mW = 0.020 J/s, the laser must emit 6.4x10^16 photons/second
4c. 1.047x10^-27 (kg m)/s
4d. ("4e.") 66.3 nm, which lies in the ultraviolet