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Unformatted text preview: Zzztayloe, Rex Homework 13 Heinz 81204 Nov 04, 2004 1 This printout should have 15 questions. Multiplechoice 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.
 Fall '09
 WOODAHL
 Physics, Work

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