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Chapter 32:

- 8: a) 100 mSv, b) 80 mSv, c) 30-45 mSv, depending on α energy
- 10: 2-4 Gy depending on the RBE of the α used.
- 13: a) cancer risk 30 in a million per year, b) genetic defect 10 in a million
- 19: a) 158 rem (1.58 Sv) to tumor, b) this is low, table 32.7 says 80+ Sv
- 35: 3.4 x 10
^{38}protons/sec, b) just for info, about 6 x 10^{14}neutrinos per second per square meter here. - 45:a) net mass loss is 0.193831 u, = 180.6 MeV from E=mc
^{2}, b) the reaction balances. - 52: 47 grams
- 54: 4.67 x 10
^{12}kWh, b) 4.67 x 10^{11}dollars - 61: a) 4.86 G watts, this is about the output of a big electric power station, b) 1.1 years

Chapter 31:

- 8: 7.4 fm
- 9: a) 4.6 fm, b) 0.61
- 18: produces
_{20}^{40}Ca and an anti-neutrino (but I don't care much about neutrinos) - 20: produces
_{25}^{52}Mn - 22:
_{48}^{106}Cd - 23:
_{82}^{206}Pb - 34: a)
_{82}^{208}Pb, b) energy released = 33.05 MeV - 45: a) 0.50 mCi, b) this is 3 half lives, = 15.81 years
- 57: Originally it was 23.2%
^{235}U, and 76.8%^{238}U - 64: I'll give a little help here, from chemistry...

the original number of atoms of^{238}Pu is N_{o}= 2.78 x 10^{25}atoms.

a) 6.966 x 10^{15}Bq.

b) Power = 6.23 kW originally, and after 12 years down to 5.67 kW

Chapter 29:

- 38: a) 1.666 x 10
^{-32}kg m/s, b) One reason is that you don't notice "impact" forces from them. - 39: a) 6.63 x 10
^{-23}kg m/s, b) 0.124 MeV - 41: a) λ = 8.28 x 10
^{-14}m, b) 15.0 MeV - 49: 7.28 x 10
^{-4}m/s - 52: 2.21 x 10
^{-32}m/s - 58: a) v = 728 m/s, b) use KE = q V, get V = 1.51 μV

Chapter 28:

- 2: γ=1.005 and 2.29
- 5: v = 0.8 c
- 7: v = 0.14 c
- 16: a) 4.3024 years, b) 0.144 years
- 36: 1.35 x 10
^{-21}kg m/s - 44: 1.50 x 10
^{-10}J = 939 MeV - 51: γ = 211, v = 0.999988 c
- 66: converting 1 gram of mass into energy will power your house for about 160 years.
- 69: Note: you need info from table 7.1, in chapter 7, about how much eneregy 1 kg of hydrogen will give you.

a) 6.25 x 10^{11}kg/s of hydrogen is fusing. b) 1.4 x 10^{18}s, = 45 billion years, c) 4.4 x 10^{9}kg/s, d) 0.3 percent of its total mass

Chapter 27:

- 2: for visible range of 400nm - 700nm, λ in crown glass is 263nm - 460nm
- 6: θ = 0.516
^{o} - 9: 1st min is m=0, so d = 290 nm
- 10: m=2, so λ = 600 nm
- 21: 2000 lines/cm gives d = 5 μm, θ=5.97
^{o} - 25: 5000 lines/cm gives d = 2.00 μm, λ = 707 nm
- 38: θ=21.8
^{o}, which is NOT a small angle, d = 1.70 μm - 45: a) 1.35 μm, b) θ = 69.9sup>o
- 47: 750 nm
- 57: a) 1.627 x 10
^{-4}radians, which is 0.00932^{o}, b) 325 light years apart - 62: a) 2.237 x 10
^{-4}radians, which is 0.0128^{o}, b)5.8 km, c) 179 μm d) that is narrower than a pencil line, about the width of two hairs. - 86: 0.046 W/m
^{2} - 89: Final intensity is 1/4 the original intensity.
- 91: θ
_{B}= 67.5^{o}

Chapter 26:

- 2: 50.3 diopters
- 4: calculate the magnification, get 5.3 x 10
^{-6}, then calc the distance = 3.75 km - 6: 2.0 m
- 9: Original power = 51 diopters. so 51 = 1/x + 1/0.02, giving x = 1.0 m
- 16: -5 diopters is the prescription for contacts
- 17: -5.4 diopters is the prescription for glasses.

Chapter 25:

- 1: the shortest mirror would be 89 cm tall
- 2: Show that the change of direction equals 180 degrees
- 6: 1.97 x 10
^{8}m/s - 7: n = 1.49, probably polystyrene
- 11: 1.03 ns
- 16: daisy chain Snell's Law...
- 22: 66.3
^{o} - 25: a) n = 1.43, fluorite, b) 44.2
^{o} - 29: θ
_{red}= 46.5^{o}, θ_{violet}= 46.0^{o} - 35: θ
_{red}= 53.5^{o}, θ_{violet}= 55.2^{o} - 39: -4.5 diopters gives f = -22.2 cm
- 43: f=1/P so 44.4 cm
- 47: power = 13.3 diopters, much stronger than almost any glasses
- 59: a) m = 0.111, b)image distance = -0.334 cm (behind the cornea), c) f = -0.376 cm, so R =2f = -0.752 cm

Chapter 24:

- 3: 150 kV/m
- 6: a) wavelengths of 33 cm and 12 cm. b) smaller wavelength gives smaller interference spots.
- 9: 3.9 x 10
^{14}Hz to 7.9 x 10^{14}Hz - 13: 9 km
- 18: a) 9.46 3.9 x 10
^{15}m, b) 1.9 x 10^{22}m, c) 1.1 x 10^{26}m - 26: in 0.1 ns light goes 3 cm, which is the round trip uncertainty. Half of that is 1.5 cm, the uincertainty of the distance to the moon. b) 3.9 x 10
^{-9}percent - 32: Intensity, I = Power/area, 318 W/m
^{2}, b) B = 1.6 μT, c) E = 490 V/m - 45: a) 8.44 x 10
^{-12}watts, b) 5.65 mV

Chapter 23:

- 3: a) ccw, b) cw c) zero
- 9: a) 3.04 mV, b) pretty low power, not much temperature increase. Even if R = 0.001 mΩ the power is only 9 mW for a short time.
- 14: B = 8.00 mT, EMF = 250 V
- 20: 198 T (a very strong field)
- 27: a) I=0, b) I = cw, c) I = 0, d) I = ccw, e) I=0
- 44: a) N
_{s}= 30, b) I_{p}= 97.5 mA - 50: a) 2.24, b) 0.447 (which is 1/2.24), c) loss is 20% of what it was before (0.447 squared)

Chapter 22:

- 1: a) left, b) Perp. into page, c) up, to top of page, d) 0, e) right, f) down, to bottom of page
- 3: a) right, b) perp. into page, c) down, to bottom of page
- 9: a) B = 3.01 x 10
^{-5}T, b) Yes, the earth's field is roughly 1 G. - 14: 0.261 T, easily achieved even by permanent magnets.
- 31: a) left, b) perp. into page, c) up to top of page, d) 0, e) right, f) down, to bottom of page
- 35: a) 2.5 N, small value, the weight of a couple of birds.
- 42: a) τ = 389 Nm, b) τ = 73.5 Nm
- 55: forces on the short sides cancel, Force on near long side is greater than that on far long side. F
_{net}= 2.06 x 10^{-4}N, repelled from the wire. - 68: Need I
_{2}going down. I_{1}/I_{2}= 1/π - 82: Electric force = 2 x 10
^{-10}N, use F = ma to find a, then deflection distance is from x=1/2 at^{2}= 4 x 10^{-10}m

Chapter 21:

- 2: a) 6,600 Ω b) 93.9 Ω
- 4: 15 A + 11.7 A + 0.63 A = 27.3 A, so the circuit is overloaded
- 23: a) 200 A, b) 10.0 V, c) 2000 W, d) 80 A, V
_{mot}= 4.0 V, P_{mot}= 320 W - 26: b) 1.67 A, c) 27.9 kW, d) 20 MΩ

A: R

B: R

C: R

Now considering the same arrangements, except the resistors are now capacitors.

A: C

B: C

C: C

Chapter 20:

- 3: 0.25 A
- 10: a) 2 x 10
^{7}s = 231 days b) 3.13 x 10^{15}elec/s - 22: a) 0.3 V, b) 1.5 V. Lower V to appliance means less power to it.
- 24: 0.104 Ω
- 39: a) Temp to double R = 5 x 10
^{5}^{o}C, hot enough to vaporize anything. b) to halve it -2.5 x 10^{5}^{o}C. Waaaay below absolute zero. - 40: P = 2 x 10
^{12}W. But not for very long. - 50: $438/year for hot water.
- 56: $3.94 billion
- 66: a) 1230 kg of Al per km, b) 2640 kg of Cu per km. More expensive wire and requires stronger supports.
- 73: 480 V
- 77: a) 4000 A, b) 16 MW just becomes heat in the wires, so 16% loss.

Chapter 19:

- 2: eV=1/2 mv
^{2}, so v=1.2 x 10^{8}m/s - 7: a) 2 x 10
^{9}J, b) to heat and boil water. Remember Q=mcΔT and Q=mL_{v}. These must add up to the energy available. get m = 765 kg of vaporized water. c) the hot expanding steam explodes the tree. (Be careful, the context of "Q" changes its meaning) - 15: a) 3.0 kV, b) 750 V
- 18: E = 8.9 MV/m
- 29: Q = +0.833 μC
- 46: Q = 21.6 mC
- 52: C = 0.667 nF
- 53: C = 0.93 μF
- 61: C
_{net}= 11.4 μF. (a couple of intermediate combinations give 2.06 and 1.37 μF) - 64: a) V = 3160 V, b) Q = 25.3 mC

Chapter 18:

- 1: a) 1.25 x 10
^{10}electrons b) 3.13 x 10^{12}electrons - 5: 1 x 10
^{12}+ 3.13 x 10^{10}= 1.0313 x 10^{12}elec - 14: r = 7.1 mm
- 16: 14.4 N away from Qa, (or in 1 sig fig, 10 N)
- 21: a) 4.16 x 10
^{42}, b) 1.24 x 10^{36} - 27:11.4 N/C down. Warning, it is dangerous to put in a negative sign when you specify a direction. Feel free to ask why.
- 31: F=ma (our old friend) results in a=F/m = qE/m = 4.8 x 10
^{14}m/s^{2} - 41: 12.8 N to the right
- 46: 2.04 x 10
^{7}N/C, up on the page - 49: 1.25 x 10
^{6}N/C 60^{o}cw from the vertical axis - 60: 6890 N/C

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