• 2: 3400 Hz
• 3: 332 m/s
• 6: 363 m/s
• 8: 924 m away
• 12: 3.16 x 10^-4 w/m²
• 17: yup, they are the same.
• 18: a) 93 dB is twice as loud, b) 83 dB is one fifth as loud
• 30: 878 Hz when approaching, and 735 Hz when receeding
• 44: 180 Hz, 270 Hz, 360 Hz
• 48: a) 33.4 cm long, b) 259 Hz
• 50: v=353 m/s so f = 4900 Hz to 3390 Hz
• 61: reading the 0 phon line, 60 Hz(48dB), 400 Hz (9 dB), 1000 Hz (0 dB), 4000 Hz (-8 dB), 15,000 Hz (20 dB)

• 1: a) 1230 N/m, b) 6.88 kg, c) 4.0 mm
• 3: a) 889 N/m, b) 133 N
• 6: 29.4 n/m, b) 10.0 cm
• 9: 400 Hz
• 11: 1500 Hz
• 13: 2.37 N/m
• 16: total M=0.88989 kg, so add 0.3889 kg, for T=2.01 sec add 8.9 g more, for T=1.99 s use 8.9 g less. This is +/- 1%
• 49: 40 Hz
• 53: 700 m
• 56: for 0.1 sec uncertainty in time get 400 m uncert for p-wave and 720 m uncert for s-wave
• 58: the other stricg might be either 261.5 Hz or 258.5 Hz
• 63: 0.225 W
• 64: 0.89 m away

• 20: 54.1%
• 24: a) 3.8 x 10⁹ J, b) 0667 barrells of fuel
• 25: a) Qc = 7.25 x 10¹³ J, b) Qc/W = 1.38, c) Work = 5.25 x 10¹³ J
• 28:T_H = 676K
• 31: a) 579 K, b) if T_c goes down to 150^oC, then e = 39.4%.
• 34: claim efficiency = 12/25 = 48%, best possible is e = 1-300/600 = 50%. We have not yet gotten that close to ideal, but the claim at least does not exceed the maximum possible efficiency. Still, it is so close to require skepticism.

• 1: 5.02 x 10⁸ J
• 4:a) 21^o C, b) 25^o C, c) 29.3^o C, d) 50.1^o C
• 6: c = 387 J/kg^oC, probably copper
• 9: must lose 467 W for 30 min, PLUS the 150 W they continue to produce, total is 617 W to be radiated.
• 12: a) 1.005 x 10⁵ J, b) 3.68 x 10⁵ J, c) Ice must melt before its temp can increase. The net is much more heat absorbed if you start with ice at 0^oC than with water at that temperature.
• 14: If you use Lv for boiling water (which this is not) you get 54 grams, but this is water condensing at a much lower temp, so, using the Lv for lower temp water (the book lists Lv for water at 37 C), you get 58 grams of ice melting.
• 16: 0.26% will evaporate
• 61: a) 1.3% increase of radiation for 34 vs. 33^oC, b) 21% increase for 34 vs. 20^oC
• 62: a) sun's surface is at 5740K, b) 6.17 x 10⁷ W/m², c) 1340 W/m²

• 5: 9890^oF
• 8: a) -40^oC = -40^oF, b) 575^oK = 575^oF
• 9: Washington monument shrinks by 1.9 cm
• 16: a) at 22^oC the steel one is 0.24 mm longer, b) the 30 m long tape is 7.3 mm longer
• 39: Argon atoms are 39.948 kg per 6.02 x 10²⁶ atoms = 6.64 x 10^-26 kg each, so v = 1250 m/s
• 42: a) 11.2 km/s = 11,200 m/s = 40,300 km/hr = 25,100 miles/hr, b) temperature for oxygen molecules to have this speed T = 161,000 K (I used 11.2 km/s)
• 44: 3.1 billion K.

• 1: 2.78 x 10^-3 liters/s = 2.78 cm³/s
• 2: 83.3 cm³/s = 8.33 x 10^-5 m³/s
• 3: 27 cm/s
• 5: a) 0.75 m/s, b) 13 cm/s
• 8: 1.7 mm/s
• 17: P=F/A = N/m² = (kg m/s²)/m² x m/m = (kg m²/s²)/m³ = J/m³
• 20: Remember v = Sqrt(2 g h), so, using the speeds from example 12.2 solve for h, and get a) h = 33.1 m, b) 0.196 m
• 22: 15.3 N on each square meter of sail (by finding the pressure difference. If you have a lot of square meters of sail, this can add up.
• 25: a) this is power = ρgh x Q = 960 MW. b) 960/680 = 1.4 times.

• 2: 2.54 x 10^-3 m³ = 2.54 liters
• 4: 2700 kg/m³
• 10: a) 10¹⁸ kg/m³, b) about 17 km
• 11: 3.59 x 10⁶ Pa
• 15: 1.1 x 10⁸ Pa, about 1100 atmospheres
• 16: both have units that reduce to kg/(m s²)
• 25: 136 N
• 33: blood pressure is 159/119 when measured 0.5 m below heart level. (1 Pa = 0.0075006 mm Hg)
• 34: 40,000 newtons (about 10,000 lbs!)
• 41: a) 198 grams, b) 198 cm³, c) 540 grams/198 cm³ = 2.73 g/cm³ = 2730 kg/m³
• 77: a) 4 x 10⁶ Pa, b) it compresses 2.1 x 10^-3 cm (remember Young's modulus?)
• 81: a) 2.01 x 10⁴ N, b) 0.12 cm, c) 25 GPa

• 1: ω = 0.737 rev/s
• 2: a) α = 87.3 rad/s², b) a_T = 8.29 m/s², c) a_c = 1.04 x 10⁷ m/s² = 1.06 x 10⁶ g's
• 5: a) α = 80 rad/s², b) number of revs = 1.0 rev
• 6: recall dist = vt, but v = rω, so dist = 405 m
• 8: a) α = -25.0 rad/s², b) 28.7 rev, c) stop time = 3.80 s, d) dist = 50.5 m, e) looking up braking distances that looks ok, although a little short.
• 22: hoop's final speed = 7.0 m/s
• 26: a) ω_leg 9.66 rad/s, b) v = 10.1 m/s, c) conservation of momentum, and don't forget the ball is moving down too.
• 28: a) if the ball is hollow, h = 5.44 m up the hill, if the "ball" slides, not rolls, h = 3.27 m. The rolling ball has more KE at the start, so there is more PE at the end.

• 1: a) 47 N, b) height is of no consequence
• 3: 23.3 N
• 5: a) 1.36 m, b) 686 N
• 9: 392 N up and 196 N down
• 13: a) for two cables, in each T = 6600 N, b) hinge force is 26,800 N at 68^o ccw up from the direction of the bridge roadway.
• 29: muscle must exert 1715 N of force
• 31: bicep muscle must exert 470 N of force.
• 33: Achilles tendon tension is 2205 N, leg bones push down with 2940 N.
• 37: a ) 147 N, b) F_tri (1.75) = F_r (20), so triceps tension is 1680 N, c) work to push up is 118 J, so , d) for 25 in one minute, 25(118)/60 = 49 watts.

• 1: a) 15,000 kg m/s, b) 24 kg m/s, c) 666 kg m/s
• 2: a) ship m = 1.2 x 10⁸ kg, b) artillery shell p = 1.32 x 10⁶ kg m/s
• 7: 9000 N
• 12: 6670 N
• 15: 4.69 x 10⁵ N
• 32: a) together their speed is 1.78 m/s, b) they lost 267 J of KE.
• 36: a) final v = 0.182 m/s, b) lost KE = 8520 J
• 38: a) rifle recoils at 4.58 m/s, b) rifle KE = 31.5 J, c) tightly held rifle recoil v = 0.49 m/s, d) KE = 3.38 J, e) player p = 880 kg m/s, ball p = 10.3 kg m/s, which isn't much compared to the player. Probably won't knock him down!

• 2: 1840 J
• 5: 3140 J
• 6: 1300 J
• 13: 2720 N
• 14: a) 4670 N, b) 17,500 N, c) duh... yes, a) is bad and b) is much worse.
• 20: KE initial = KE final + mgh, v_f = 0.687 m/s
• 21: a) vinal v = 26.2 m/s in a time of 5.35 sec, b) final v = 26.3 m/s in a time of 4.86 sec.
    Not much speed difference, but plenty of time difference in races that are decided by hundredtsh of a second.
• 28: 10,500 bombs
• 32: a) 40 people will power your dryer, b) 8 million people will have the output of a medium sized power plant.
• 43: a) sun intensity at the top of the atmosphere, the sun's output is spread out over a sphere the radius of the earth-sun distance. b) 2% efficiency will require 28.8 km² of solar collectors to replace one medium sized power plant. (That is a square less than 3.5 miles on a side.)
  b) 128,000 km² to provide all the needs of the USA, Aussies can do it with only 6600 km², and China can do it with 77,000 km²
• 44: about 9 minutes, which is about 66 floors (I roughed it out with 4 J = 1 cal)
• 52: a) 31 kN, b) 1570 N, so an extra 40 times your weight stiff-legged, versus an extra 2 times your weight if your knees bend.
• 54: a) total work done is 108 kJ, b) power is 599 W

• 1: 723 km
• 2: 0.10 rev/s = 0.628 rad/s
• 8: a) 33.3 rad/s, b) 500 N, c) 40.8 m
• 15: a) a_c = 3950 m/s², b) 37 percent of sound speed
• 19: 0.313 rad/s = 2.99 rev/min
• 23: a) 483 N, b) 17.4 N, c) 2.2 times her weight, 8.1 percent of her weight
• 28: b) θ = 26.1^o
• 33: M = 5.978 x 10²⁴ kg
• 34: a) at earth: g_moon = 3.32 x 10^-5 m/s², b) at earth: g_sun = 5.93 x 10^-3 m/s², c) g_moon/g_sun = 0.56 percent
• 39:a) F_dad = 7 x 10^-7 N, b) F_Jupiter = 1.3 x 10^-6 N, ratio ≈0.5 so they are similar "effects".

• 4: Max friction = 588 N, then accel = 1.96 m/s²
• 10: 1.8 m/s² directed downhill
• 18: a) 51 N, b) a = 0.718 m/s²
• 19: a) 46.5 N, b) a = 0.656 m/s²
• 24: car drag increases by 2.9 times (take the ratio: drag at 110 over drag at 65)
• 25: a) with no drag, the drop would hit the ground at 313 m/s,
      b) use regular drag (not Stokes), v = 9.8 m/s is the speed at which the drop's weight equals the drag force.
       also recall that mass is density times volume
• 28: η = 0.76 kg s/m
• 30: the bone shortens by 0.045 mm

• 10: 1260 N
• 13: g_moon=1.67 m/s², so m = 150 kg, W_earth=1500 N
• 15: 26 million newtons, equal and oppoosite forces on the shell and the ship
• 19: a) 7.84 x 10^-4 N, b) tension = 1.88 x 10^-3 N, 2.4 times as much.
• 31: Net force is 14.3 - 7.5=6.8 N, 12.4^o N of E. Accel = 0.139 m/s²
• 41: The upward component for each muscle is 188 N each, times 2 = 376 N up. The horizontal components cancel. The effect is to raise your heel.
• 42: T1 = 736 N, T2 = 194 N.

• 5: 19.5 m 4.7^o S of W
• 10: v_A=3.45 m/s, v_B=3.94 m/s (Note: A tedious two equations in two unknowns algebra problem)
• 15: 87 km E and 87 km N
• 20: 92.3 m, 53.7^o S of W
• 25: 130 m horizontally, and 30.9 m high
• 40: Fish is going a total velocity of 10.3 m/s at an angle of 73.2^o below the horizontal
• 45: 15.05 m/s
• 54: a) The behind runner is going 0.7 m/s relative to the ahead runner. b) The behind runner wins, that is, the initial runner takes 71.43 seconds to go 250 m, while the other runner takes 70.24 seconds to go 295 m. c) the gap at the finish is 4.17 m
• 65: It takes 50 seconds to swim across, so in 50 sec the swimmedrifts 40 m downstream. So the water is going 0.8 m/s. the net v of the swimmer is 0.94 m/s with respect to the ground.

• 3: (note: go to the tip of the arrow) a) distance = 8+2+3=13 m, b) magnitude of displacement = 9 m, c) Δx=+9 m
• 4: a) 8 m, b) 4 m, c) - 4 m
• 9: 124.88 km/hr, 34.689 m/s
• 14: a) v=+6 m/s, -1.7 m/s, +4.04 m/s, b) +3.49 m/s
• 17: accel = 56.4 m/s² = +5.76 g, b) accel = -201 m/s² = -20.6 g
• 20: a) v=10.8 m/s. Maybe draw the v(t) graph first to help you think about the x(t) graph. v(t) is a straight line. x(t) is parabolic.
• 22: v=502 m/s, An additional question you can ask and answer is how long is the barrel? It is 20 cm long.
• 47: a) The cliff is 8.26 m high. b) thrown straight down the ball will take 0.717 seconds to hit. (I used the quadratic formula...)
• 52: distance in 1 second = 4.9 m, b) v_f=38.3 m/s, c) distance during last second = 33.4 m. (There is more than one way to do this.)
• 66: Drawing the v(t) graph should be easy. Drawing a(t) is icky and in this case I don't care much. Strictly speaking the "kink" in the x(t) graph makes it mathematically ill-defined to find a(t).

Math Review Sheet

• 1a: x⁸, 10⁴, 5 x 10^-5, 10, 3 x 10^-6
• 1b: 3.26 x 10⁴, 8.31 x 10^-1, 2 x 10^-4, 1.006 x 10³, 1.9 x 10^-10
• 2a: angle=65 degrees, sides are 9.06 cm and 4.23 cm long
• 2b: AC = 31.8 m long, side AD = 24.7 m long
• 3: 120, -2 and 3/2, -50, 99.99, 4.1 x 10^-18, 0.0623
• 4: 0.00358 g, 0.0000023 g, 15.7 MN