Physics102 Homework Answers and Notes

DW's Homework Answers
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Chapter 30:

12: a) ¹³₇N, b) ¹³₆C, c) ¹⁴₇N, d) ¹⁵₈O, e) ¹⁵₇N, f) ¹²₆C
39: 3.786 kg of H2O has 421 g of H atoms, which is 2.5 x 10²⁶ atoms, 1/6500 of them are ²₁D, which is 3.9 x 10²² D atoms, each pair of them fuse releasing 3.27 MeV of energy, which is 1 x 10¹⁰J out of 1 gallon of water. This will power each of us for 11.6 days.

Chapter 29:

16: R = 8750 Bq
20: three half lives, = 17,190 years
24: ¹²₆C, ⁴₂α,¹⁴₆C,
44: 2000 x-rays per year = 2.5 mrem per x-ray. 5 rem/year is about 40 times background dose.
54: 0.234 mg of ⁴⁰K per liter of milk = 3.53 x 10¹⁸ atoms of ⁴⁰K per liter of milk. half-life is 1.28 Gy so λ_K = 1.72 x 10^-17/s thus the decay rate = 61 Bq. Need 2000 Bq of ¹³¹I to decay down to 61 Bq, with a half-life of 8.04 days, gives a time of 40.5 days.

Chapter 27:

12: λ=400nm has E=3.1eV. so YES to PE for Li, which would have KE_max = 3.1 - 2.3 = 0.8eV, NO PE for Be, Hg, because the photon has less energy than the work function for them.
28: a) λ = 1.98 x 10^-11m, b) γ = 1.00223, so either classical or relativistic (correct) answer are 1.98 x 10^-14m
32: found λ = 5.25 x 10^-11m, which gave E = 3.79 x 10^-15J = 23.7 keV.

Chapter 26:

2: γ = 3.20 so a) 4.42 years/γ = 1.38 years b) distance = 4.2 light years/γ = 1.31 light years
22: m1 = 2.512 x 10^-28kg, m2 = 8.825 x 10^-28kg
30: γ = 1.065 x 10¹⁰ a) time = our time/γ = (dist/speed)/γ = 10⁵y/γ = 9.39 x 10^-6 y = 296 seconds b) dist = 10⁵c y)/γ = 9.39 x 10^-6 light years = 8.88 x 10¹⁰ m
39: The answer in the book is wrong. When the Astronauts get there and compare clocks and calendars the trip for Speedo has taken 6.58 years and for Goslo has taken 17.66 years, so the difference is 11.1 years.

Chapter 25:

Chapter 24:

6: 1.5 mm
8: y₁ = 3 cm, y₂ = 6 cm, the diff between them is 3 cm
28: w/o using small angle approximation -- y = 0.903m
using small angle approx. -- y = 0.912 m
34: m=1 gives θ = 4.3^o. max m = 13
48: λ₂ = 4/5 λ₁ = 432 nm
Loud sound at 29.1^o and 76.3^o
weak sound at 14.1^o and 46.8^o

Chapter 23:

6: R = -0.79 cm
30: p = 5.36 cm, q = -18.75 cm, magnifying glass
36: for lens #1: q = -100 cm, M1 = +5
for lens #2: p2 = 125 cm, q2 = -9.26 cm, M2 = 0.0741,
so net M = 0.37

Chapter 22:

6: distance = 1.945 m, the return direction is parallel to incoming direction
14: θ = 36.6^o
16: θ = 19.47^o just inside first interface, and before it hits second interface
then θ = 30.0^o just after leaving second interface
28: First interface: θ_2red = 28.22^o, θ_2violet = 27.48^o
Second interface: θ_3red = 31.78^o, θ_3violet = 32.52^o
Second interface: θ_4red = 58.56^o, θ_4violet = 63.18^o
Sooooo... the angular dispersion is the difference, 4.62^o
34: θ = 67.18^o
50:

Chapter 21:

4: for the 150W lamps, I = 1.25A, R = 96 Ω, the 100W lamp, I = 0.83A, R = 145&Omega
34: N₂ = 18 turns, b) P = 3.6 W
44: λ = 11 m

Chapter 20:

2 a) flux = 1 x 10^-7 Wb, b) 8.66 x 10^-8 Wb, c) 0 Wb
8: 0.1 mV
10: 5.2 x 10^-5 T
14: ΔB = 2.2T, and area = 0.01m², so I = 8.8A
44: a) I goes from left to right in the resistor , b) no I , c)I goes from right to left in the resistor

Chapter 19:

16: a) 9 x 10^-3N, 15⁰ above the horizontal b) 2.33 x 10^-3N, due west
22: 9.98 Nm, rotate ccw when viewed from above.
30: a) I is to the left, b) I is out of page, c) I is up and to the right
40: F = 2.7 x 10^-5N
42: the wire will be 57.75 m long, the solenoid will have 919 turns, and be 0.116 m long

Chapter 18:

2: a) R_eq=24Ω b) I=11.0A,
c) R_eq=2.18Ω , I₄=6 A, I₈=3 A, I₁₂=2 A
8: a) R_eq=5.13Ω b) E = 4.53 V
12: P₂=14.2 W, P₃=1.33 W
P₁= 4 W, P₄=28.5 W
14: V₅=10.0 V, so V₇=5 V, thus I₇=0.714 A and I₂=1.286 A, so V₂= 2.57 V, and 10 = E - 2.57 thus E= 12.57 V
26: Q = 1.8 x 10^-4 C, q = 1.14 x 10^-4 C
32: a) 8.00A, b) 120V, c) 0.8 A, d) 576 W

Chapter 17:

⁷

Chapter 16:

Chapter 15: