Use the formula $\mathrm { D } = 10.0 \log \left( \mathrm { S } _ { \mathrm { S } } \mathrm { S } \right)$, where the loudness of a sound in decibels is determined by $\mathrm { S }$, the number of watt/ $/ \mathrm { m } ^ { 2 }$ produced by the soundwave, and $\mathrm { S } _ { 0 } = 1.00 \times 10 ^ { - 12 } \mathrm { watt } / \mathrm { m } ^ { 2 }$. A certain noise produces $8.54 \times 10 ^ { - 5 }$ watt $/ \mathrm { m } ^ { 2 }$ of power. What is the decibel level of this noise? (Round to the nearest unit.)
A) 8 decibels
B) 69 decibels
C) 183 decibels
D) 79 decibels

Question

Quizplus · Accepted Answer

The formula for decibels is $ D = 10.0 \log \left( \frac{S}{S_0} \right) $. Substituting $ S = 8.54 \times 10^{-5} $ and $ S_0 = 1.00 \times 10^{-12} $, we get $ D = 10.0 \log \left( \frac{8.54 \times 10^{-5}}{1.00 \times 10^{-12}} \right) \approx 79 $ decibels.

Use the Formula D=10.0log⁡(SSS)\mathrm { D } = 10.0 \log \left( \mathrm { S } _ { \mathrm { S } } \mathrm { S } \right)D=10.0log(SS​S)

Use the Formula $\mathrm { D } = 10.0 \log \left( \mathrm { S } _ { \mathrm { S } } \mathrm { S } \right)$