# 2-2 - Create assignment 00224 Homework 2 Jan 24 at 10:21 am...

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Unformatted text preview: Create assignment, 00224, Homework 2, Jan 24 at 10:21 am 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. Hikers Hear an Echo 17:02, trigonometry, numeric, > 1 min, nor- mal. 001 A group of hikers hear an echo 3 s after they shout. The temperature is 22 ◦ C. How far away is the mountain that reflected the sound wave? At this altitude, the speed of sound in air at 273 ◦ C is 329 m / s. Correct answer: 512 . 999 m. Explanation: Let : t = 3 s , T = 22 ◦ C , T = 273 ◦ C , and v = 329 m / s . The velocity at 22 ◦ C is v = v r 1 + T T v = (329 m / s) r 1 + 22 ◦ C 273 ◦ C = 342 m / s . If d is the distance to the object, the total distance traveled by the sound (there and back) is 2 d. Thus, 2 d = v t d = v t 2 = (342 m / s) (3 s) 2 = 512 . 999 m . keywords: Serway CP 14 20 17:05, trigonometry, numeric, < 1 min, nor- mal. 002 A train at rest emits a sound at a frequency of 1000 Hz. An observer in a car travels away from the sound source at a speed of 30 m / s. What is the frequency heard by the ob- server? Assume the speed of sound in air to be 343 m / s. Correct answer: 912 . 536 Hz. Explanation: Given : f = 1000 Hz , v = 30 m / s , and v = 343 m / s . f = f µ v ± v v ∓ v s ¶ = f µ v- v v ¶ since the source is stationary and the observer is moving away from the source. Then f = (1000 Hz) µ 343 m / s- 30 m / s 343 m / s ¶ = 912 . 536 Hz ....
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2-2 - Create assignment 00224 Homework 2 Jan 24 at 10:21 am...

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