HW2solution - Version PREVIEW HW 2 Savrasov (39824) 1 This...

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Unformatted text preview: Version PREVIEW HW 2 Savrasov (39824) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. Echo From a Distant Wall 001 10.0 points Karen claps her hand and hears the echo from a distant wall 0 . 646 s later. How far away is the wall? The speed of sound in air is 343 m / s. Correct answer: 110 . 789 m. Explanation: Let : v = 343 m / s , and t = 0 . 646 s . Karen hears the sound after it had traveled to the wall and back, so 2 d = v t d = v t 2 = (343 m / s) (0 . 646 s) 2 = 110 . 789 m . Pressure Amplitude 02 002 10.0 points Calculate the pressure amplitude of a 1 . 02 kHz sound wave in air if the displace- ment amplitude is 2 . 69 10- 8 m. Take the velocity of sound in air to be 343 m / s and the density of air to be 1 . 22 kg / m 3 . Correct answer: 0 . 0721417 Pa. Explanation: Let : f = 1 . 02 kHz , A = 2 . 69 10- 8 m , v s = 343 m / s , and = 1 . 22 kg / m 3 . The pressure amplitude is P = v s A = (2 f ) v s A = (1 . 22 kg / m 3 ) 2 (1 . 02 kHz) (343 m / s) (2 . 69 10- 8 m) = . 0721417 Pa ....
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This note was uploaded on 02/16/2009 for the course PYHS 191 taught by Professor Curtise during the Spring '08 term at Abilene Christian University.

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HW2solution - Version PREVIEW HW 2 Savrasov (39824) 1 This...

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