Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 14 - Waves and Sound - Problems and Conceptual Exercises - Page 496: 98

Answer

$108$

Work Step by Step

The intensity level is given as $\beta=10log(\frac{I}{I_{\circ}}) $...eq(1) The intensity level for 120 machines is given as $\beta_{120}=10log (\frac{120I}{I_{\circ}})$.....eq(2) The intensity level for $nI$ machines is $\beta_{nI}=10log(\frac{nI}{I_{\circ}})$....eq(3) $\implies 92dB=10log(\frac{120I}{I_{\circ}})$ $\implies 92dB=10log120+10log\frac{I}{I_{\circ}}$.....eq(4) Now we substitute $82db$for $\beta_{nI}$ in eq(3) to obtain: $82dB=10logn+10log(\frac{I}{I_{\circ}})$...eq(5) Subtracting eq(4) from eq(5), we obtain: $10dB=10log120-10logn$ $\implies 10dB=10log\frac{120}{n}$ This simplifies to: (number of machines that will produce 82 db) $n=12$ Thus, turned off machines = total machines - number of machines that produce 82 db $=120-12=108$
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