# PHYS102hw6 - HOMEWORK CH 14 4,8,12,16,20,24,29,33,35,51,53...

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HOMEWORK CH 14 4,8,12,16,20,24,29,33,35,51,53 14.4 The speed of sound in seawater at 25°C is 1 530 m s . Therefore, the time for the sound to reach the sea floor and return is 2 2(150 m) 0.196 s 1530 m/s d t v 14.8 At a temperature of T = 10.0°C = 283 K, the speed of sound in air is      283 K 331 m s 331 m s 337 m s 273 K 273 K T v The elapsed time between when the stone was released and when the sound is heard is the sum of the time t 1 required for the stone to fall distance h and the time t 2 required for sound to travel distance h in air on the return up the well. That is, t 1 + t 2 = 2.00 s. The distance the stone falls, starting from rest, in time t 1 is 2 1 2 gt h Also, the time for the sound to travel back up the well is 21 2.00 s h tt v Combining these two equations yields     2 11 2.00 s 2 g v With  2 337 m s and 9.80 m s vg , this becomes   2 -1 2 1.45 10 s 2.00 s 0. Applying the quadratic formula yields one positive solution of t 1 = 1.95 s, so the depth of the well is     2 2 2 1 9.80 m s 1.95 s 18.6 m 22 gt h 14.12 The decibel level due to the first siren is

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   2 1 -12 2 100.0 W m 10 log 140 dB 1.0 10 W m . Thus, the decibel level of the sound from the ambulance is  21 10 dB 140 dB 10 dB 150 dB 14.16 (a) From the defining equation of the decibel level,    0 10 log II . We solve for the intensity as 10 0 10 and find that   12 2 115 10 12 11.5 2 0.5 2 2 1.0 10 W m 10 1.0 10 W m 10 W m 0.316 W m I (b)
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## This note was uploaded on 03/07/2011 for the course PHYS 102 taught by Professor Armenn.kocharian during the Winter '11 term at California State University Los Angeles .

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PHYS102hw6 - HOMEWORK CH 14 4,8,12,16,20,24,29,33,35,51,53...

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