The decibel level D of a sound is related to its intensity I by $\mathrm { D } = 10 \log \left( \frac { \mathrm { I } } { \mathrm { I } _ { 0 } } \right)$. If $\mathrm { I } _ { 0 }$ is $10 ^ { - 12 }$, then what is the intensity of a noise measured at 93 decibels? Express your answer in scientific notation, rounding to three significant digits, if necessary.
A) $9.3 \times 10 ^ { - 10 } \mathrm { watt } / \mathrm { m } ^ { 2 }$
B) $2 \times 10 ^ { - 3 } \mathrm { watt } / \mathrm { m } ^ { 2 }$
C) $1.09 \times 1016 _ { \mathrm { watt } } / \mathrm { m } ^ { 2 }$
D) $2 \times 10 ^ { - 2 } \mathrm { watt } / \mathrm { m } ^ { 2 }$

Question

Quizplus · Accepted Answer

To find the intensity I, use the formula $D = 10 \log \left( \frac {I}{I_0} \right)$. Given $D = 93$ and $I_0 = 10^{-12}$, solve for I:  $93 = 10 \log \left( \frac {I}{10^{-12}} \right)$  $9.3 = \log \left( \frac {I}{10^{-12}} \right)$  $10^{9.3} = \frac {I}{10^{-12}}$  $I = 10^{9.3} \times 10^{-12} = 10^{-2.7} = 2 \times 10^{-3}$ watts/m².

The Decibel Level D of a Sound Is Related to Its