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 λK = 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 KEmax = 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:
- 30: D = 0.773 m
- 36: Minimum separation distance = 0.38 m
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 = 4.3o. max m = 13
- 48: λ2 = 4/5 λ1 = 432 nm
- Loud sound at 29.1o and 76.3o
weak sound at 14.1o and 46.8o
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 = 36.6o
- 16: θ = 19.47o just inside first interface, and before it hits second interface
then θ = 30.0o just after leaving second interface
- 28: First interface: θ2red = 28.22o, θ2violet = 27.48o
Second interface: θ3red = 31.78o, θ3violet = 32.52o
Second interface: θ4red = 58.56o, θ4violet = 63.18o
Sooooo... the angular dispersion is the difference, 4.62o
- 34: θ = 67.18o
- 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) Req=24Ω b) I=11.0A,
c) Req=2.18Ω , I4=6 A, I8=3 A, I12=2 A
- 8: a) Req=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:
- 16: Say an 800 W dryer, used 5 minutes a day...
energy/year = 8.76 x 107 J/year = 24 kwh/year at 8 cents/kwh = about $2 per year.
- 20: Thot = 1435oC
- 28: a) 29 cents, b) $2.64
Chapter 16:
- 2: 50 volts, 6.0 x 10-4 J
- 10: 45 V + 180 V = 225 V
- 12: potential at that point due to other two charges = 1.14 x 106V, W = 9.1 J
- 40: 9.8 kg of water can be boiled.
Chapter 15:
- 8: r = 5.1 m
- 10: FA = 46.8 N left
FB = 158 N right
FC = 111 N left
- 20: E = 16250 N/C
- 26: Use the "and another" link for Chap. 15 on class home page
- 38: q = 5.25 μC
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