Dept. of Physics & Astronomy

Office: Watanabe Hall, Rm. 426, (808) 956-2922

E-mail: nassir @ hawaii.edu

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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