Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 20 - Traveling Waves - Conceptual Questions: 11

Answer

The sound intensity level would be 72 decibels.

Work Step by Step

We can find the intensity of one professor talking. $I_1= I_0~10^{\frac{\beta_1}{10}}$ $I_1= (10^{-12}~W/m^2)~10^{\frac{52}{10}}$ $I_1 = 1.585\times 10^{-7}~W/m^2$ We can find the intensity of 100 professors talking. $I_2 = 100~I_1$ $I_2 = (100)(1.585\times 10^{-7}~W/m^2)$ $I_2 = 1.585\times 10^{-5}~W/m^2$ We can find the sound intensity level in decibels. $I_2= I_0~10^{\frac{\beta_2}{10}}$ $10^{\frac{\beta_2}{10}} = \frac{I_2}{I_0}$ $log(10^{\frac{\beta_2}{10}}) = log(\frac{I_2}{I_0})$ $\frac{\beta_2}{10} = log(\frac{I_2}{I_0})$ $\beta_2 = 10~log(\frac{I_2}{I_0})$ $\beta_2 = 10~log(\frac{1.585\times 10^{-5}~W/m^2}{10^{-12}~W/m^2})$ $\beta_2 = 72~dB$ The sound intensity level would be 72 decibels.
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