Physics111 Homework, some Answers and Notes

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Chapter 20:

6: a) 14.6 J/K, b) 30.4 J/K
23:
24: a) e=0.31, b) W = 16 kJ
27:
30: 500 W of useful power output. e=0.107, Heat in = 4670 J/s, b) exhaust = 4170 J/s

Chapter 19:

Chapter 18:

Chapter 17:

5:
6: 44.1 m
7: Hint: d = v t_u and d = 1/2 g t_d²
11:
19:
25:
31:
39:
44: a) λ=4 m, f=86 Hz, b) yes, low frequency, c) shorter passage results in higher freq
50: a) f=71.5 Hz, b) tension=64.7 N
55:
59:

Chapter 16:

Chapter 15:

1:
2: a) 10.1 N, b) 118 N/m
3:
14: a) x_m=0.50 m, b) at t=0 x=-0.251 m, c) v=-3.06 m/s
21:
28: a) No, it won't get there, not enough KE to get to that PE. b) stops and turns around at x=0.12 m
30: m=1.39 kg, k=200 N/m, f=1.91 Hz
33:
41:
47:
57:
70: a) r/R(k/m)^1/2, b) if r=R, ω=(k/m)^1/2, if r=0, ω=0
84:

Chapter 14:

2: If they really mean 77 m², then there is only a pressure difference of 6.2 Pa, if they mean 77 cm², then the pressure difference is 62 kPa. In either case, subtract the difference from the outside pressure.
3:
8: Pressure 20 m under water is 2.007 x 10⁵ Pa, just due to the water. Then the air pressure at the surface of the water decreases with height. What is the pressure due to being under 7600 m of air? That is 6.48 x 10⁴ Pa. So the total pressure difference is 2.007 x 10⁵ Pa + 6.48 x 10⁴ Pa = 2.66 x 10⁵ Pa.
19: Requires integration
26: The pressure in your mouth must be 392 Pa less than atmospheric pressure
27:
28: a) f=Fa/A, and b) f=103 N
31:
37:
49:
51:
61:
69:

Chapter 13:

1: take dF/dm=0...
4: F_sun/F_earth=2.1
5:
6: note that m₁ and m₄ forces cancel, so don't waste time calculating them,
the net force is at 45^o, so each component is the same, (1.2 x 10^-14i + 1.2 x 10^-14j) N
13:
18: 8.2 μm (the dangle angle is 9.4 x 10^-4degrees)
20: 0.4 N less weight for an 80 kg person
21:
31: a) Mars is 0.74 as dense as earth. Note also that taking ratios saves a lot of button pushing on calculators.
41:
61: a) the height is just R/2, where R=earth's radius. Solve it symbolically. It's easier.

Chapter 12:

3:
10: T1=32.8N, T2=18.8 N, T3=53.4 N, θ=20.6^o
11:
13:
26: You can write torque about ANY of the three places there are forces, and you will get a solvable equation for μ = 0.216
28: F_H = 433 N, F_V = 250 N
49:
52: 6.9 x 10⁹ N/m²
80: 44 N

Chapter 11:

Chapter 10:

1:
4: a) 2.0 rad, b) 0, c) 112 rad/s, d) 28 rad/s², e) α is not constant
9:
18: a) α=-0.0727 rad/sec/yr = -2.31 x 10^-9 rad/s², b) 2600 years, c) in year 1054, ω=260 rad/s, T = 0.024 s
20: a) a_t=0.064 m/s², b) a_r=0.032 m/s²
33:
35:
41:
45:
50: I=1.28 kg m²,
58: a) 1.96 m/s, b) no change
63:

Chapter 9:

1:
3: Hint: look closely at symmetry, and minimize calculations. Also, note the coordinate system they give.
11:
13:
18: 4.9 kg m/s
23:
28: 1800 N
39:
45:
49:
54: it rises 2.6 m after the collision
71: note that y-momentum is zero.
76: 108 m/s (it gains a bit over 3 m/s)

Chapter 8:

2: a) 0, b) 1.7 x 10⁵ J, c) 3.4 x 10⁵ J, d) 1.7 x 10⁵ J, e) 3.4 x 10⁵ J, f) ΔPE doubles
6: a) 0.15 J, b) 0.11 J, c) 0.19 J, d) 0.15 J, e) 0.11 J, f) no change in PE and work.
12: a) 21 m/s, b) no change, c) no change
20: a) 1.0 m/s, b) θ = 79°, c)64 J
24:0.10 m (hint: grav PE comes from dist spring compresses + 0.40 m)
39:
42: a) 556 J, b) 556 J
48: 75 J
57:

Chapter 7:

3:
4: a) 1.25 x 10⁵ MT, b) 9600 bombs
9:
15:
19:
27:
33: a) tricky... remember, the net work is zero, and it is the work done by the sum of the forces.
37:
44: the work done is against gravity, so the speed doesn't change it, 900 J
b) 113 W, c) 225 W
49:
62: a) 0.294 J, b) -18 J, c) 11.9 m/s, d) 0.239 m

Chapter 6:

1:
2: a=1.42 m/s², μ_k=0.58
10: a) a=0 because pull is not strong enough to overcome static friction, b) a=2.17 m/s²
14: a) it doesn't slide because static friction is greater than Wx (the downhill component of the weight) b) 2.98 x10⁷ N
36: area ratio = 3.75
39:
42: 13.4 m/s
47: Test your formula, what if v=0? normal force should be mg.
57:
59: Hint: There are only three forces on the mass, two tensions and the weight.

Chapter 5:

Chapter 4:

Chapter 3:

Chapter 2:

2: a) 1.74 m/s, b) 2.14 m/s c) the slope of the line connecting the origin to the final point is Δx/Δt.
5:
11: to arrive at 11:15 in spite of the traffic, must go 128 km/hr.
15:
18: v = 24t - 6t², a = 24 - 12t
19:
23:
28: 5.00 sec, b) 61.5 m
38: a) 32.9 m/s, b) 49.0 s, c) 11.7 m/s
45:
69:
73: Car catches truck at 82.0 m, at which time the car is going 19.0 m/s