Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 12 - Sound - Problems - Page 354: 14

Answer

(a) The eardrum receives $1.58\times 10^{-11}~joules$ of energy per second. (b) It would take 2010 years to receive 1.0 J of energy.

Work Step by Step

(a) We can calculate the intensity of the sound wave. $\beta = 10~log(\frac{I}{I_0})$ $log(\frac{I}{I_0})= \frac{\beta}{10}$ $10^{log(\frac{I}{I_0})}= 10^{\frac{\beta}{10}}$ $I= I_0~10^{\frac{\beta}{10}}$ $I= (10^{-12}~W/m^2)~10^{\frac{55}{10}}$ $I = 3.16\times 10^{-7}~W/m^2$ We can find the power received by the eardrum each second. $P = I~A$ $P = (3.16\times 10^{-7}~W/m^2)(5.0\times 10^{-5}~m^2)$ $P = 1.58\times 10^{-11}~W$ The eardrum receives $1.58\times 10^{-11}~joules$ of energy per second. (b) We can calculate the time to receive 1.0 J of energy. $t = \frac{1.0~J}{1.58\times 10^{-11}~J/s}$ $t = 6.33\times 10^{10}~s$ We can convert the time to units of years. $t = (6.33\times 10^{10}~s)(\frac{1~hr}{3600~s})(\frac{1~day}{24~hr})(\frac{1~yr}{365~days})$ $t = 2010~years$ It would take 2010 years to receive 1.0 J of energy.
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