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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:

- 18: 9.5 x 10
^{6}m/s - 19:
- 21:
- 29:
- 35:

Chapter 18:

- 4: a) -96
^{o}F, b) 57^{o}C - 5:
- 9:
- 10: 1.1 cm taller
- 23:
- 25:
- 40: 412 J/kg
^{o}C - 51:

Chapter 17:

- 5:
- 6: 44.1 m
- 7: Hint: d = v t
_{u}and d = 1/2 g t_{d}^{2} - 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:

- 2: a) 21.9 seats/sec, b) 39 seats
- 4: 30 cm
- 15:
- 17:
- 41:
- 52: a) 4.0 m, b) 24 m/s, c) 0.347 kg/m, d) 0.67 sec

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 m
^{2}, then there is only a pressure difference of 6.2 Pa, if they mean 77 cm^{2}, 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 10
^{5}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 10^{4}Pa. So the total pressure difference is 2.007 x 10^{5}Pa + 6.48 x 10^{4}Pa = 2.66 x 10^{5}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 m
_{1}and m_{4}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^{-14}i + 1.2 x 10^{-14}j) N - 13:
- 18: 8.2 μm (the dangle angle is 9.4 x 10
^{-4}degrees) - 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 10
^{9}N/m^{2} - 80: 44 N

Chapter 11:

- 2: a) 59.3 rad/s, b) α = 9.31 rad/s
^{2}, c) 70.7 m - 3:
- 7: Oh nooooo! projectile motion.
- 10: a) 8 J, b) 3.0 m/s, c) 6.9 J, d) 1.8 m/s
- 21:
- 28: a) 600 kg m
^{2}/s in the +z-direction, b) 120 kg m^{2}/s in the -z-direction - 31:
- 36: 1024
- 41:
- 43:
- 57:
- 62: 20.2 rad/s
- 69:

Chapter 10:

- 1:
- 4: a) 2.0 rad, b) 0, c) 112 rad/s, d) 28 rad/s
^{2}, e) α is not constant - 9:
- 18: a) α=-0.0727 rad/sec/yr = -2.31 x 10
^{-9}rad/s^{2}, b) 2600 years, c) in year 1054, ω=260 rad/s, T = 0.024 s - 20: a) a
_{t}=0.064 m/s^{2}, b) a_{r}=0.032 m/s^{2} - 33:
- 35:
- 41:
- 45:
- 50: I=1.28 kg m
^{2}, - 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 10
^{5}J, c) 3.4 x 10^{5}J, d) 1.7 x 10^{5}J, e) 3.4 x 10^{5}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 10
^{5}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/s
^{2}, μ_{k}=0.58 - 10: a) a=0 because pull is not strong enough to overcome static friction, b) a=2.17 m/s
^{2} - 14: a) it doesn't slide because static friction is greater than Wx (the downhill component of the weight) b) 2.98 x10
^{7}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:

- 1:
- 4: F
_{2}= - F_{1}= -2i + 6j - 13:
- 14: 2N, down
- 17:
- 29:
- 37:
- 47:
- 63: What does the area under each segment tell you?
- 88: W = 3260 N, m = 2720 kg, "g" = 1.2 m/s
^{2}

Chapter 4:

- 3:
- 5:
- 8: a) 1080 km, b) 63.4
^{o}S of E, c) 480 km/h, d) same as b, e) 644 km/h - 12: 56.6 m 45
^{o}E of N c) 1.89 m/s, d) same as b, e) 0.471 m/s^{2}, 45^{o}N of E - 16: a) -18 m/s in x-dir, b) a=0 at t=0.75 sec, c) never, d) t = 2.19 s
- 21:
- 26: a) 15.32 m/s b) 2.07 m/s, d) -4.80 m/s, f) -36.2 m/s
- 28: a) 51.8 m,b) 27.4 m/s, c) 67.4 m
- 33:
- 46: 71% of the time in the top half
- 56: Be careful what value you use for R...
- 67:
- 81:
- 83:
- 85: draw a careful diagram for this. Then it should be easy.
- 96: a) 20.3 m/s, b) 21.7 m/s
- 108: a) 7.3 km, b) 22 m/s
- 121:

Chapter 3:

- 2: r
_{x}=13.0 m, r_{y}=7.5 m - 3:
- 8: 3.2 km at 41
^{o}south of west. - 10: r
_{x}=11.8m, r_{y}=-5.8 m r_{z}= -2.8 m, so r = 13.4 m - 15:
- 33:
- 34: 2k;
- 36: 0
- 63:

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 - 6t
^{2}, 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

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