16homework13 - Husain, Zeena Homework 13 Due: Nov 25 2003,...

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Husain, Zeena – Homework 13 – Due: Nov 25 2003, 4:00 am – Inst: H L Berk 1 This print-out should have 19 questions, check that it is complete. Multiple-choice questions may continue on the next column or page: fnd all choices beFore making your selection. The due time is Central time. 001 (part 1 oF 1) 10 points Sound with a Frequency oF 437 Hz travels through steel. A wavelength oF 11 . 44 m is measured. What is the speed oF the sound in steel? Correct answer: 4999 . 28 m / s. Explanation: v = = (437 Hz)(11 . 44 m) = 4999 . 28 m / s . 002 (part 1 oF 1) 10 points A bat can detect small objects such as an insect whose size is approximately equal to one wavelength oF the sound the bat makes. IF bats emit a chirp at a Frequency oF 25 kHz, and iF the speed oF sound in air is 328 m / s, what is the smallest insect a bat can detect? Correct answer: 13 . 12 mm. Explanation: ±rom λ = v f , we get λ = (328 m / s) (25000 Hz) = 13 . 12 mm . 003 (part 1 oF 1) 10 points Karen shouts across a canyon and hears an echo 6 . 2 s later. Assume: The speed oF sound is 343 m / s. How wide is the canyon? Correct answer: 1063 . 3 m. Explanation: Karen hears the sound aFter it had traveled across the canyon and back, so 2 d = v t d = v t 2 = (343 m / s)(6 . 2 s) 2 = 1063 . 3 m . 004 (part 1 oF 1) 10 points The speed oF sound in air is v s = s γ P ρ the constant γ = 1 . 4, P is the air pressure, and ρ is the density oF air. Calculate the speed oF sound For P = 0 . 651 atm and ρ = 1 . 12 kg / m 3 . Correct answer: 287 . 111 m / s. Explanation: Applying the Formula v = s γP ρ = s (1 . 4)(0 . 651 atm)(101300 Pa / atm) 1 . 12 kg / m 3 = 287 . 111 m / s . 005 (part 1 oF 1) 10 points A copper rod is given a sharp compressional blow at one end. The sound oF the blow, traveling through air at 7 . 72 C, reaches the opposite end oF the rod 9 . 32 ms later than the sound transmitted through the rod. The speed oF sound in copper is 3500 m / s and the speed oF sound in air at 7 . 72 C is 331 m / s. What is the length oF the rod? Correct answer: 3 . 40714 m. Explanation: The di²erence in the times oF travel oF sound through the copper rod and the air is t = L µ 1 v air - 1 v Cu , where L is the distance traveled ( i.e. , the length oF the rod). Then L = t v air v Cu v Cu - v air = (0 . 00932 s) × (331 m / s)(3500 m / s) (3500 m / s) - (331 m / s) = 3 . 40714 m .
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Husain, Zeena – Homework 13 – Due: Nov 25 2003, 4:00 am – Inst: H L Berk 2 006 (part 1 of 1) 10 points An experimenter wishes to generate in air a sound wave that has a displacement ampli- tude equal to 3 × 10 - 6 m. The pressure am- plitude is to be limited to 0 . 267 Pa. Use the equilibrium density of air to be 1 . 29 kg / m 3 and the speed of sound in air to be 343 m / s. What
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This note was uploaded on 01/27/2009 for the course PHY 303K taught by Professor Turner during the Fall '08 term at University of Texas at Austin.

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16homework13 - Husain, Zeena Homework 13 Due: Nov 25 2003,...

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