PHYIIHW2Solutions

# PHYIIHW2Solutions - hopkins(tlh982 – HW02 – criss...

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Unformatted text preview: hopkins (tlh982) – HW02 – criss – (4908) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A skyrocket explodes 126 m above the ground. Three observers are spaced 107 m apart, with observer A directly under the point of the explosion. A B C 126 m 107 m 107 m Find the ratio of the sound intensity heard by observer A to that heard by observer B. Correct answer: 1 . 72115. Explanation: Given : h = 126 m and d = 107 m . 107 m 107 m A B C Source r B r C r A =126m b b b b The intensity at a distance r from the source is I = P 4 π r 2 , and distances from the source to points A and B are r A = h and r B = radicalbig h 2 + d 2 , so I A I B = r 2 B r 2 A = h 2 + d 2 h 2 = (126 m) 2 + (107 m) 2 (126 m) 2 = 1 . 72115 . 002 (part 2 of 2) 10.0 points Find the ratio of the intensity heard by ob- server A to that heard by observer C. Correct answer: 3 . 88461. Explanation: The distance from the source to C is r C = radicalBig h 2 + (2 d ) 2 , so I A I C = r 2 C r 2 A = h 2 + 4 d 2 h 2 = (126 m) 2 + 4 (107 m) 2 (126 m) 2 = 3 . 88461 . 003 10.0 points A siren creates a uniform sound level of 94 . 6 dB at 704 m away. The siren is pow- ered by a battery that delivers a total energy of 0 . 621 kJ. Assuming that the efficiency of the siren is 44 . 8 percent ( i.e. , 44 . 8 percent of the supplied energy is transformed into sound energy), determine the total time the siren can sound. Correct answer: 0 . 0154887 s. Explanation: The sound intensity of the siren is I = I 10 β/ 10 = 0 . 00288403 W / m 2 . Then, using E = P t and P = 4 π L 2 I , we obtain for the length of time it can sound with energy E t = ( n/ 100) 4 π L 2 E I hopkins (tlh982) – HW02 – criss – (4908) 2 = (44 . 8 percent / 100) 4 (3 . 14159) (704 m) 2 × (0 . 621 kJ) (1000 J / kJ) (0 . 00288403 W / m 2 ) = . 0154887 s . 004 10.0 points The sound level produced by one singer is 64 . 8 dB. What would be the sound level produced by a chorus of 19 such singers (all singing at the same intensity at approximately the same distance as the original singer)? Correct answer: 77 . 5875 dB. Explanation: The total sound intensity is the sum of the sound intensities produced by each individual singer. Note: This is NOT true of sound levels measured on the decibel scale. Therefore I n = nI 1 . The resulting sound level in decibels is β n = 10 log I n I = 10 log nI 1 I = 10 log I 1 I + 10 log n = β 1 + Δ β = (64 . 8 dB) + (12 . 7875 dB) = 77 . 5875 dB . 005 10.0 points Everyone at a party is talking equally loudly. If only one person were talking, the sound level would be 19 dB. Find the sound level when all 42 people are talking....
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PHYIIHW2Solutions - hopkins(tlh982 – HW02 – criss...

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