phyhw18 - valdez(vv689 Homework 18 orin(58140 This...

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valdez (vv689) – Homework 18 – florin – (58140) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Sound waves travel through a liquid of density 3090 kg / m 3 at a speed of 6840 m / s. What is the bulk modulus of this liquid? Correct answer: 1 . 44568 × 10 11 Pa. Explanation: Let : ρ = 3090 kg / m 3 and v sound = 6840 m / s . In fluids, sound waves are pressure waves, and their speed depends on the fluid’s density and bulk modulus: v sound = radicalBigg B ρ B = ρ v 2 sound = (3090 kg / m 3 ) (6840 m / s) 2 = 1 . 44568 × 10 11 Pa . 002 (part 1 of 3) 10.0 points An outside loudspeaker (considered a small source) emits sound waves with a power out- put of 73 W. Find the intensity 17 . 2 m from the source. Correct answer: 0 . 0196361 W / m 2 . Explanation: Let : P = 73 W and r = 17 . 2 m . The wave intensity is I = P 4 π r 2 = 73 W 4 π (17 . 2 m) 2 = 0 . 0196361 W / m 2 . 003 (part 2 of 3) 10.0 points Find the intensity level in decibels at this distance. Correct answer: 102 . 931 dB. Explanation: Let : I 0 = 1 × 10 12 W / m 2 . The intensity level is β = 10 log parenleftbigg I I 0 parenrightbigg = 10 log parenleftbigg 0 . 0196361 W / m 2 1 × 10 12 W / m 2 parenrightbigg = 102 . 931 dB . 004 (part 3 of 3) 10.0 points At what distance would you experience the sound at the threshold of pain, 120 dB ? Correct answer: 2 . 41022. Explanation: Let : β = 120 dB . β = 10 log parenleftbigg I I 0 parenrightbigg I = I 0 10 β/ 10 = ( 1 × 10 12 W / m 2 ) 10 12 = 1 W / m 2 , so the wave intensity is I = P 4 π r 2 r = radicalbigg P 4 π I = radicalBigg 73 W 4 π (1 W / m 2 ) = 2 . 41022 m . 005 10.0 points The sound level produced by one singer is 85 . 6 dB. What would be the sound level produced by a chorus of 21 such singers (all singing at
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valdez (vv689) – Homework 18 – florin – (58140) 2 the same intensity at approximately the same distance as the original singer)? Correct answer: 98 . 8222 dB. Explanation: Let : β = 85 . 6 dB and n = 21 . The total sound intensity is the sum of the sound intensities produced by each individual singer: I n = n I 1 , so β n = 10 log parenleftbigg I n I 0 parenrightbigg = 10 log parenleftbigg n I 1 I 0 parenrightbigg = 10 log parenleftbigg I 1 I 0 parenrightbigg + 10 log n = β 1 + 10 log n = 85 . 6 dB + 10 log 21 = 98 . 8222 dB . 006 10.0 points A copper rod is given a sharp compressional blow at one end. The sound of the blow, traveling through air at 4 . 77 C, reaches the opposite end of the rod 3 . 32 ms later than the sound transmitted through the rod. What is the length of the rod? The speed of sound in copper is 3050 m / s and the speed of sound in air at 4 . 77 C is 331 m / s. Correct answer: 1 . 2327 m. Explanation: Let : v air = 331 m / s , t = 0 . 00332 s , and v Cu = 3050 m / s . The difference in the times for sound trav- eling through the copper rod and the air is t = L parenleftbigg 1 v air 1 v Cu parenrightbigg , where L is the distance traveled ( i.e. , the length of the rod). Then L = t bracketleftbigg v air v Cu v Cu v air bracketrightbigg = (0 . 00332 s) · (331 m / s) (3050 m / s) 3050 m / s 331 m / s = 1 . 2327 m .
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