homework 34-solutions - klare(alk736 – homework 34 –...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: klare (alk736) – homework 34 – Turner – (58220) 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A flowerpot is knocked o ff a balcony 18 . 8 m above the sidewalk and falls toward an un- suspecting 1 . 71 m tall man who is standing below. How close to the side walk can the flower pot fall before it is too late for a warning shouted from the balcony to reach the man in time? Assume that the man below requires . 203 s to respond to the warning, and the velocity of sound in air to be 344 m / s. The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 6 . 02171 m. Explanation: Let : v s = 344 m / s , H = 18 . 8 m , h m = 1 . 71 m , and t r = 0 . 203 s . h 2 = gt 2 2 /2 t 2 = t- t 1 h = H- h m t = (2 h / g ) 1/2 h 1 = h- h 2 t 1 = t r + t s h m h m H H 1 The distance from the balcony to the man’s head is h = H- h m and the time for a warning to travel this dis- tance is t s = h v s = H- h m v s . The total time needed to receive the warn- ing and react is t 1 = t s + t r = H- h m v s + t r , and the time for the pot to fall this distance (starting from rest) is t = 2 h g = 2 ( H- h m ) g . Thus the latest the warning can be sent is at t 2 = t- t 1 = 2 ( H- h m ) g- H- h m v s- t r = 2 (18 . 8 m- 1 . 71 m) 9 . 8 m / s 2- 18 . 8 m- 1 . 71 m 344 m / s- . 203 s = 1 . 61487 s into the fall. In this time the pot has fallen a distance of h 2 = 1 2 g t 2 2 and the corresponding height above the side- walk is H 1 = H- h 2 = H- 1 2 g t 2 2 = 18 . 8 m- 1 2 ( 9 . 8 m / s 2 ) (1 . 61487 s) 2 = 6 . 02171 m . 002 10.0 points Sound waves travel through a liquid of density 783 kg / m 3 at a speed of 4450 m / s. What is the bulk modulus of this liquid? Correct answer: 1 . 55054 × 10 10 Pa. Explanation: klare (alk736) – homework 34 – Turner – (58220) 2 Let : ρ = 783 kg / m 3 and v sound = 4450 m / s . In fluids, sound waves are pressure waves, and their speed depends on the fluid’s density and bulk modulus: v sound = B ρ B = ρ v 2 sound = (783 kg / m 3 ) (4450 m / s) 2 = 1 . 55054 × 10 10 Pa ....
View Full Document

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.

Page1 / 6

homework 34-solutions - klare(alk736 – homework 34 –...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online