Physics102 Homework Answers and Notes

DW's Homework Answers
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Chapter 32:

8: a) 100 mSv, b) 80 mSv, c) 15-30 mSv, depending on α energy
10: 2-4 Gy depending on the RBE of the α used.
16: Using RBE=20, we get 7.44 x 10⁸ nuclei
48: a) Z=90, N=142, b) product is ₉₀²³³Th, c) two β decays produce ₉₁²³³Pa, and ₉₂²³³U successively, d) the U has N=141, e) half life is 1590 years, plenty long to have a bomb sit around for a while before you use it. NOTE: There may be a typo in the parenthetical statement, which should say, (much as ²³⁸ U is bred into ²³⁹ Pu)
49: Thermal output is 2570 MW, b) 8.04 x 10¹⁹ fissions/sec, which is 2.53 x 10²⁷ fissions/year, a total mass of U of 990 kg has fissioned
52: 46 grams
54: 4.67 x 10¹² kWh, b) 4.67 x 10¹¹ dollars

Chapter 31:

9: a) 4.6 fm, b) 0.61
10: Using E=mc², 1 u gives 1.4924 x 10^-10 J, which is 931.5 MeV
18: produces ₂₀⁴⁰Ca and an anti-neutrino (but I don't care much about neutrinos)
20: produces ₂₅⁵²Mn
24: ²²⁶Ra → ₂⁴α + ₈₆²²²Rn
34: a) ₈₂²⁰⁸Pb, b) energy released = 33.05 MeV
45: a) 0.50 mCi, b) this is 3 half lives, = 15.81 years
56: it decayed to 0.7 of its initial activity, which takes 2.71 years

Chapter 29:

21: FM radio 90 MHz, a) 5.96 x 10^-26 J, = 3.72 x 10^-7 eV, b) at 50 kW of power this is 8.38 x 10²⁹ photons/sec
26: IR photon: a) E= 0.0827 eV, b) say 5 eV to break a molecule, this will take about 60 photons, c) γ-ray: 1.24 MeV, d) will break 250,000 molecules
35: at 650 kHz a 50kW radio station emits 1.16 x 10³² photons/sec. Spread out over a sphere with one photon per cm² per second, the sphere will have a radius of 3.04 x 10¹³ m, which is much less than a light year (which is about 10¹⁶ m), and the nearest star is about 4 light years away, and stars that may have alien civilizations are likely much farther away than that.
38: a) 1.67 x 10^-32 kg m/s
39: a) 6.63 x 10^-23 kg m/s, b) 0.124 MeV
50: not relativity, too slow, so λ = 8.08 x 10^-11 m
58: a) v = 728 m/s

Chapter 28:

1: γ = 1.033 and 1.15
3: 5.96 x 10^-8 s
6: v = 0.90 c
8: 0.24 c
16: Time as seen by us = 4.302 years. time as seen by astronaut = 0.143 years. The ratio is γ
36: 1.35 x 10^-21 kg m/s
40: 2.0 x 10⁸ m/s
44: 1.50 x 10^-10 J = 939 MeV
46: 1.4 x 10²⁰ stars can be formed if half of the total energy available is converted to mass.
47: a) mass lost = 1.11 x 10²⁷ kg, which is, b) 5.56 x 10^-5 of the original mass
67: for 1000 MW of electric output produced at 35% efficiency, a total power of 2.86 x 10⁹ W is needed to be produced from the reactor. For an entire year, this amounts to a total energy output of 9.0 x 10¹⁶ J, which has a mass equivalent of 1 kg. So only 1 kg of mass was converted to energy. Hard to notice 1 kg missing out of 10,000 kg.

Chapter 27:

2: for visible range of 400nm - 700nm, λ in crown glass is 263nm - 460nm, for 380-760 nm it is 250-500 nm in crown glass
8: d = 1.22 μm
11: 577 nm
13: m=62 (set Sinθ=1, solve for m=62.5, but the biggest integer value of m=62)
23: 8990 lines/cm (solve for d, which is meters/line, convert to cm/line, and invert for lines/cm)
25: 5000 lines/cm gives d = 2.00 μm, λ = 707 nm
45: a) D=1.35 μm, b) θ=69.7^o
51: a) You can't use the small angle approximation, because the angles are 17.1306^o and 17.1455^o, 0.0150^o different, b) And you cannot use the small angle approx to determine the separation, because it depends on the angles themselves. Find each y-value, then take the difference to get 0.273 mm
58: 107 m
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/m²
89: Final intensity is 1/4 the original intensity. If middle polarizer is removed, final intensity is zero.

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. (-5.6 D for glasses at 2 cm from eye)

Chapter 25:

2: Show that the change of direction equals 180 degrees
5: 2.25 x 10⁸ m/s and 2.04 x 10⁸ m/s
8: ice, n=1.31
13: n=1.457, so probably fused quartz (which is a fairly commonly used material)
14: the time delay for two trips through the atmosphere is 5.86 x 10^-8 s, which is 2.29 x 10^-6% of the round trip time.
22: 66.3^o
29: θ_red = 46.5^o, θ_violet = 46.0^o
36: 20.0 diopters
41: Screen is 3.43 m away. Mag = 33.3, so image is 80 cm x 120 cm.
42: a) -135 cm, that is, 135 cm from lens on the same side as the mole, b) m = 10, c) 50.0 mm image size
57: f = 36.0 cm
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:

1: yup, you get the speed of light
3: 1.50 x 10⁵ V/m
7: a) wavelengths from 188 to 556 m for AM, and 2.78 to 3.41 m for FM
12: λ = 0.25 μm, which is about the size of the smallest detail you could see with the UV microscope.
18: a) 9.46 x 10¹⁵ m, b) 1.9 x 10²² m, c) 1.1 x 10²⁶ m
23: a) 193 nm has f = 1.55 x 10¹⁵ Hz, the resolution ratio will be the ratio of the wavelengths, so for 400 nm light, 400/193 ≈2 two times better resolution.
26: in 0.1 ns light goes 3 cm, which is the round trip uncertainty. Half of that is 1.5 cm, the uncertainty of the distance to the moon. b) 3.9 x 10^-9 percent
32: Intensity, I = Power/area, 318 W/m², b) B = 1.6 μT, c) E = 490 V/m
42: a) 1.50 μm IR light, 200 W, in a 25.0 cm dia circle on your shoulder. Intensity, I=P/A = 4074 W/m², b) E = 1750 V/m, c) B = 5.84 x 10^-6 T, d) to heat 4.00 kg of shoulder requires 2.78 x 10⁴ J, and power = energy/time, so t = E/P = 139 seconds.

Chapter 23:

4: a) ccw, b) cw c) cw
9: a) 3.04 mV, b) pretty low power, not much temperature increase. Even if R = 0.001 Ω the power is only 9 mW for a short time.
12: B = 0.425 T
17: 0.63 V, not much
27: a) I=0, b) I = cw, c) I = 0, d) I = ccw, e) I=0
44: a) 30 turns, (b) 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)
67: L = 200 H, b) temperature increases by 5^oC

Chapter 22:

2: a) right, b) Perp. out of page, c) down, to bottom of page, d) 0, e) left, f) up, to top of page
3: a) right, b) perp. into page, c) down, to bottom of page
10: 10.1^o or 169.9^o
14: 0.261 T, easily achieved even by permanent magnets.
32: a) left, b) perp. out of page, c) up to top of page
37: B = 1.80 T
48: τ = 0.471 N m, viewed from above, rotation will be cw.
57: F = 7.5 x 10^-5 N toward the bottom of the page, in line with the right side of the square.
68: Need I₂ going down. I₁/I₂ = 1/π
84: need F_B = mg, so you need to figure out m. I looked up the density of copper, which I won't tell you so you also have the pride of that accomplishment. Ultimately, I got I = 1730 A. If earth contributes a field of 3 x 10^-5 T, this is 11% of the field contributed by the lower wire, so decrease its current by 11% to 1540 A.

Chapter 21:

3: Series gives the max R_s=786 Ω, parallel gives the min,R_p=20.3 Ω
6: a) I=0.400 A, P₂₄=3.84 W, P₉₆=15.4 W, b) R_p=19.2 Ω, I=2.50 A, 2.0 A through R₂₄ and 0.5 A through R₉₆, P₂₄=96 W, P₉₆=24 W
16: 2.9994 V
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

The additional problems handed out, calling the circuits (A), (B) and (C) clockwise from the top
For Resistors
(A) R_eq=8.75 Ω total power P=9.26 W, I=1.03 A
Note: R₂ and R₃ are parallel, then R₂₃ is series with R₁
(B) R_eq=3.58 Ω total power P=22.6 W, I=2.52 A
Note: R₃, R₂, R₄ are parallel , then R₃₂₄is series with R₂ ...
(C) R_eq=3.27 Ω total power P=24.8 W, I=2.75 A

If instead they are capacitors

(A) C_eq=3.81 μF, Q=34.3 μC, E=1.54 x 10^-4 J
(B) C_eq=12.06 μF, Q=108 μC, E=4.88 x 10^-4 J
(C) C_eq=12.8 μF, Q=115 μC, E=5.2 x 10^-4 J

Chapter 20:

4: 4.00 mA
9: 0.12 C, 7.5 x 10¹⁷ electrons
23: 0.20 mA
25: 0.322 Ω
27: the Al will be 1.24 times the diameter of the Cu wire
40: P = 2 x 10¹² W. But not for very long.
49: beam power is 1500 W
56: $3.94 billion
69: a) = 2.08 x 10⁵ A, for that current, b) the waste heat will be I²R= 4.3 x 10¹⁰ W. c) but that is greater than the power output of the generator! So it cannot happen. d) Don't confuse the idea of how much power you want to get to the destination with how much power you need to produce at the source.
73: 480 V
76: R = 576 Ω, power during melt down is 100 W.

Chapter 19:

2: eV=1/2 mv², so v=1.2 x 10⁸ m/s
5: 1940 V
7: a) 2 GJ, b) 765 kg of water can evaporate (mcΔT + mL_f... remember?)
13: yup, they are equivalent
17: a) E=2.5 MV/m (mega volts/meter), so no breakdown (no spark), b) at d=1.7 mm you get a big enough E-field.
29: Q = +0.833 μC
38: sketch so equipotentials are perp to E-field. Higher potential is in direction opposite the arrows of E-field
50: 20,000 V
61: C_net = 11.4 μF. (a couple of intermediate combinations give 2.06 and 1.37 μF)
66: series: 14.2 μC and 63.8 μJ, parallel: 84.6 μC and 381 μJ

Chapter 18:

1: a) 1.25 x 10¹⁰ electrons b) 3.13 x 10¹² electrons
6: a) 1.88 x 10⁶ electrons short of a full load, b) 1.88 x 10^-10
16: 14.4 N away from Qa, (or in 1 sig fig, 10 N)
17: 3.45 x 10¹⁶ m/s²
20: 1.04 nC
25: 85.9 cm right of negative charge, or, if both positive 10.9 cm left of +3 μC
30: 69.4 nC and 6.25 N/C
43: a) 4 x 10⁷N/C, b) 7 cm, in the middle, c)No, there is no suitable place 0-8 cm, d) The net charge in fig (a) is zero, but the net charge in fig (b) is +1, so E goes to zero faster in the case of fig. (a), and e) Add the electric fields due to the three charges for some point right of 11 cm. This is ugly algebra, which is why they suggest looking at the graph of the sum of the E-fields (if you have some way to make a graph of a function), then the graph of the sum of the E-fields crosses the zero at x=30.6 cm
48: Vector addition! All the x-components cancel, and the net vector is 0.102 N in the negative y-direction
57: remember circular motion? F=ma=mv²/r, and v=rω, but F=kqQ/r², put it all together and get ω = 4.12 x 10¹⁶ rad/sec (and from v=rω get v=2.2 x 10⁶ m/s, if you are interested)

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