# PHYSICS 152

##### University of Hawaii, Manoa Dept. of Physics & Astronomy

Instructor: Michael Nassir
Office: Watanabe Hall, Rm. 426, (808) 956-2922
E-mail: nassir @ hawaii.edu

## ANSWERS to FINAL EXAM from Fall 2004

Part A

1. D  (E = kQ/r^2, and V = kQ/r)
2. E  (F = kQQ/r^2, solve for Q)
3. B
4a. A
4b. B
5. E  (I_out = I_in * V_in / V_out)
6. C  (first: d_i = -m * d_o; then: 1/f = 1/d_o + 1/d_i)
7. F
8. A  (first: m=1 minimum occurs at sin(theta) = lambda/D; then: x_1 = L * tan(theta); finally, TOTAL width of central max = 2 * x_1)
9. A  (Malus's Law: I_2 = I_1 * (cos(theta))^2 )
10. B  (relativistic mass increase: m = m_0 / sqrt(1 - (v/c)^2), let m = 1.10*m_0 and solve for v)
11. B

Part B

12a. C  (Wien's Law: peak wavelength = 2.898^10_3m*K / T)
12b. B
13a. B
13b. B  (f = c / lambda)
13c. A  (E = h*c / lambda)
14. B
15. F  (E = m * c^2)
16. A  (beta-plus-decay and alpha-decay both increase the neutron/proton ratio; gamma-decay leaves it unchanged)
17. D  (1/64 of original sample remains undecayed, so 63/64 has decayed)
18. B
19a. E
19b. C  (N = N_0 * exp(lambda*t), where N = 0.081*N_0, lambda = (ln 2)/(half-life), and solve for t)
20. E
21. B
22. D
23. C

Part C

1a. I_A = 1.33A
I_B = 1.33 A
I_C = 2.67 A
V_A = 4.00 V
V_B = 4.00 V
V_C = 8.00 V
P_A = 5.33 W
P_B = 5.33 W
P_C = 21.33 W
We see power (wattage) as brightness, so bulb C looks brightest

1b. After re-solving the circuit without the bulb A branch, we find that now: P_B = 12.00 W and P_C = 12.00 W (and P_A is now zero!)
i. B
ii. A
iii. B

2a. 3.4x10^-15 N  (F = q*E, where E = V/d for uniform E-field, and q = +e for the He+ ion)
2b. B  (to the left)
2c.i. C  (q = +2e for the alpha-particle)
2c.ii. B  (q = +e for the proton)
2d. D  (in order to create a magnetic force on the ion to the right)
2e. 2.8 T  (F_electric = F_magnetic = q * v * B * sin(theta) )

Part D

3a. 5700 K  (Wien's Law)
3b. 3.6x10^17 photons/second  (each photon carries E = h*c/lambda = 3.9x10^-19 J of energy;  power = energy/second)
3c. 1.3x10^-27 kg*m/s  (p=h/lambda; or p=E/c)
3d. A  (F = delta(p)/delta(t) = momentum per photon * (number of photons/second that hit sail) )

4a. E
4b. B
4c. E  (this is true in general; "3p" is irrelevant)
4d. A
4e. wavelength A = 820.1 nm (infrared)  (lambda = h * c / [delta(energy)] )
wavelength B = 1141 nm (infrared)
wavelength C = 589.1 nm (visible - yellow) -- only this one is visible