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# ch17 - Chapter 17 1(a The time for the sound to travel from...

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727 Chapter 17 1. (a) The time for the sound to travel from the kicker to a spectator is given by d / v , where d is the distance and v is the speed of sound. The time for light to travel the same distance is given by d / c , where c is the speed of light. The delay between seeing and hearing the kick is t = ( d / v ) – ( d / c ). The speed of light is so much greater than the speed of sound that the delay can be approximated by t = d / v . This means d = v t . The distance from the kicker to spectator A is d A = v t A = (343 m/s)(0.27 s) = 93 m. (b) The distance from the kicker to spectator B is d B = v t B = (343 m/s)(0.12 s) = 41 m. (c) Lines from the kicker to each spectator and from one spectator to the other form a right triangle with the line joining the spectators as the hypotenuse, so the distance between the spectators is 2 2 2 2 93m 41m 100m A B D d d . 2. The density of oxygen gas is 3 3 0.0320kg 1.43kg/m . 0.0224m From / v B we find 2 2 3 5 317m/s 1.43kg/m 1.44 10 Pa. B v 3. (a) When the speed is constant, we have v = d / t where v = 343 m/s is assumed. Therefore, with t = 15/2 s being the time for sound to travel to the far wall we obtain d = (343 m/s) (15/2 s), which yields a distance of 2.6 km. (b) Just as the 1 2 factor in part (a) was 1/( n + 1) for n = 1 reflection, so also can we write  343 15 15s 343m/s 1 1 d n n d for multiple reflections (with d in meters). For d = 25.7 m, we find n = 199 2 2.0 10 .

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CHAPTER 17 728 4. The time it takes for a soldier in the rear end of the column to switch from the left to the right foot to stride forward is t = 1 min/120 = 1/120 min = 0.50 s. This is also the time for the sound of the music to reach from the musicians (who are in the front) to the rear end of the column. Thus the length of the column is 2 (343m/s)(0.50s) =1.7 10 m. l vt 5. If d is the distance from the location of the earthquake to the seismograph and v s is the speed of the S waves, then the time for these waves to reach the seismograph is t s . = d / v s . Similarly, the time for P waves to reach the seismograph is t p = d / v p . The time delay is t = ( d / v s ) – ( d / v p ) = d ( v p v s )/ v s v p , so 3 (4.5 km/s)(8.0km/s)(3.5min)(60s /min) 2.2 10 km. ( ) 8.0km/s 4.5km/s s p p s v v t d v v We note that values for the speeds were substituted as given, in km/s, but that the value for the time delay was converted from minutes to seconds. 6. Let be the length of the rod. Then the time of travel for sound in air (speed v s ) will be / s s t v . And the time of travel for compressional waves in the rod (speed v r ) will be / r r t v . In these terms, the problem tells us that 1 1 0.12s . s r s r t t v v Thus, with v s = 343 m/s and v r = 15 v s = 5145 m/s, we find 44m = .
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