81204hw13Sol

81204hw13Sol - Zzztayloe, Rex Homework 13 Heinz 81204 Nov...

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Unformatted text preview: Zzztayloe, Rex Homework 13 Heinz 81204 Nov 04, 2004 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Find the speed of sound in tungsten, which has a bulk modulus of approximately 2 10 11 N / m 2 and a density of 19250 kg / m 3 . Correct answer: 3 . 22329 km / s. Explanation: Given : B = 2 10 11 N / m 2 and = 19250 kg / m 3 . The speed of sound caused by tungstens com- pressibility is v = s B = s 2 10 11 N / m 2 19250 kg / m 3 1 km 1000 m = 3 . 22329 km / s . 002 (part 1 of 1) 10 points Some studies suggest that the upper fre- quency limit of hearing is determined by the diameter of the eardrum. The wavelength of the sound wave and the diameter of the eardrum are approximately equal at this up- per limit. If this is precisely true, what is the diameter of the eardrum of a person capable of hearing 21393 Hz? (Assume a body temperature of 36 . 2 C.) The speed of sound in air at 0 C is 331 m / s. Correct answer: 1 . 64663 cm. Explanation: Given : f = 21393 Hz . v = 331 m / s , and T = 36 . 2 C = 309 . 2 K . The speed of sound in air is v = f = v r T 273 K = v f r T 273 K = 331 m / s 21393 Hz r 309 . 2 K 273 K 100 cm m = 1 . 64663 cm . 003 (part 1 of 3) 4 points An outside loudspeaker (considered a small source) emits sound waves with a power out- put of 79 W. Find the intensity 11 . 9 m from the source. Correct answer: 0 . 0443939 W / m 2 . Explanation: Given : P = 79 W and r = 11 . 9 m . The wave intensity is I = P 4 r 2 = 79 W 4 (11 . 9 m) 2 = . 0443939 W / m 2 ....
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This note was uploaded on 07/02/2011 for the course PHYS 201 taught by Professor Woodahl during the Fall '09 term at Indiana.

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81204hw13Sol - Zzztayloe, Rex Homework 13 Heinz 81204 Nov...

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