hw13 - strong (ams4684) HW13 ditmire (58216) This print-out...

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strong (ams4684) – HW13 – ditmire – (58216) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points IF the Frequency oF sound is doubled, how will its speed change? How will its wavelength change? 1. Both speed and wavelength will double. 2. Its speed will halve, and its wavelength will double. 3. Its speed will double, and its wavelength will not change. 4. Its speed will not change, and its wave- length will halve. correct 5. There will be no change in its speed and wavelength. Explanation: v = = (2 f ) ± 1 2 λ ² The speed oF sound is constant in a given medium. IF the Frequency oF sound is doubled, its speed will not change, but its wavelength must be “compressed” to halF. The speed oF sound depends only on the medium through which it travels, not on its Frequency, wave- length, or intensity (until the intensity is so great that a shock wave results.) 002 10.0 points A stereo speaker represented by P in the fg- ure emits sound waves with a power output oF 102.3 W. P x 13 . 2m What is the intensity oF the sound waves at point x , 13.2 m away? Correct answer: 0 . 0467216 W / m 2 . Explanation: Let : P = 102 . 3 W and r = 13 . 2m . I = P 4 π r 2 = 102 . 3W 4 π (13 . 2 m) 2 = 0 . 0467216 W / m 2 . 003 10.0 points What is the time lapse between seeing a lightning strike and hearing the thunder iF the lightning ±ash is 31 km away? The speed oF lightwaves in air is 3 × 10 8 m / s and the speed oF sound waves in air is 333 m / s. Correct answer: 93 . 093 s. Explanation: Let : c =3 × 10 8 m / s , v = 333 m / s , and d = 31 km . The time For the light to travel 31 km is t 1 = d c and the time For the sound to travel 31 km is t 2 = d v 2 ,
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strong (ams4684) – HW13 – ditmire – (58216) 2 so the time lapse is Δ t = t 2 - t 1 = d ± 1 v 2 - 1 c ² = (31 km) ± 1 333 m / s - 1 3 × 10 8 m / s ² = 93 . 093 s . 004 10.0 points At a maximum level of loudness, the power output of a 75-piece orchestra radiated as sound is 72.0 W. What is the intensity of these sound waves to a listener who is sitting 28.8 m from the orchestra? Correct answer: 0 . 00690777 W / m 2 . Explanation: Let : P = 72 . 0 W and r = 28 . 8m . I = P 4 π r 2 = 72 W 4 π (28 . 8 m) 2 = 0 . 00690777 W / m 2 . 005 10.0 points The speed of sound in water is 1432 m / s. A sonar signal is sent from a ship at a point just below the water surface and 2 . 25 s later the reFected signal is detected. How deep is the ocean beneath the ship? Correct answer: 1611 m. Explanation: Let : v = 1432 m / s and t =2 . 25 s . The sound has to travel to the ocean Foor and back, so 2 d = vt d = vt 2 = (1432 m / s) (2 . 25 s) 2 = 1611 m . 006
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This note was uploaded on 11/03/2010 for the course PHYSICS 303 taught by Professor Shih during the Spring '10 term at University of Texas at Austin.

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hw13 - strong (ams4684) HW13 ditmire (58216) This print-out...

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