s17 - P17.3 Sound takes this time to reach the man: so the...

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P17.3 Sound takes this time to reach the man: ( ) 2 20.0 m1 . 75 m 5.32 10 s 343 m s so the warning should be shouted no later than 2 0.300 s5 .32 10 s0 .353 s = before the pot strikes. Since the whole time of fall is given by 2 1 2 y gt = : ( ) 22 1 18.25 m9 . 80 m s 2 t = 1.93 s t = the warning needs to come 1.93 .353 s1 .58 s = into the fall, when the pot has fallen ( ) () 2 2 1 9.80 m s 1.58 2 . 2 m 2 = to be above the ground by 2 . 2 m7 . 82 m −= P17.4 (a) At 9 000 m, 9000 1.00 C 60. 0C 150 T ⎛⎞ Δ= − ° = − ° ⎜⎟ ⎝⎠ so 30. T = −° . Using the chain rule: 1 0.607 247 dv dv dT dx dv dT v vv dt dT dx dt dT dx == = = , so 247 s v = 0 331. 50 .607 30.0 247 sln f i v t v f i v v t v = + +− ∫∫ 27.2 s t = for sound to reach ground. (b) 9 000 25.7 s h t v = ⎡+ ⎣⎦ It takes longer when the air cools off than if it were at a uniform temperature. P17.6 It is easiest to solve part (b) first: (b) The distance the sound travels to the plane is 2 2 5 2 2 s hh dh =+ = . The sound travels this distance in 2.00 s, so 5 343 m s 2.00 s6 8 6 m 2 s h d = giving the altitude of the plane as ( ) 2 686 m 614 m 5 h . (a) The distance the plane has traveled in 2.00 s is 2.00 s 307 m 2 h v . Thus, the speed of the plane is: 153 m s v . 497
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498 Sound Waves P17.12 (a) () 340 1.27 Pa sin ms x P t π ⎛⎞ Δ= ⎜⎟ ⎝⎠ (SI units) The pressure amplitude is: max 1.27 Pa P . (b) 23 4 0 f s ω ππ == , so 170 Hz f = (c) 2 m k λ , giving 2.00 m = (d) ( ) 170 H z 340 m s vf = P17.15 2 v Pv svs ρω ρ = ( ) 2 6 2 21 . 20 343 5.50 10 2 5.81 m 0.840 vs P πρ × = Δ P17.20 (a) 1 22 70.0 dB 10log 1.00 10 W m I = × Therefore, ( ) 70.010 12 2 5 2 10 1.00 10 W m I −− = × . (b) 2 2 P I v Δ = , so ( ) ( ) 3 52 1 . 20 kg m 343 m s 1.00 10 W m 90.7 mPa I P = × *P17.26 (a) We have v f = and f is the same for all three waves. Since the speed is smallest in air, is smallest in air. It is larger by 1493 m s 4.51 times
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s17 - P17.3 Sound takes this time to reach the man: so the...

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