Choose the one alternative that best completes the statement or answers the question.
-Use the formula $\mathrm { D } = 10.0 \log \left( \mathrm { S } / \mathrm { S } _ { 0 } \right)$, 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 $\mathrm { S } _ { 0 } = 1.00 \times 10 ^ { - 12 } \mathrm { watt } / \mathrm { m } ^ { 2 }$. What is the intensity in watt $/ \mathrm { m } ^ { 2 }$ of a noise measured at 82 decibels? (Round to the nearest tenth.)
A) $8.2 \times 10 ^ { - 10 }$ watt $/ \mathrm { m } ^ { 2 }$
B) $1.6 \times 10 ^ { - 4 }$ watt $/ \mathrm { m } ^ { 2 }$
C) $1.6 \times 10 ^ { - 3 }$ watt $/ \mathrm { m } ^ { 2 }$
D) $3.6 \times 10 ^ { 15 }$ watt $/ \mathrm { m } ^ { 2 }$

Question

Quizplus · Accepted Answer

To find the intensity $S$ in watts/$m^2$ for a noise measured at 82 decibels, we rearrange the formula to solve for $S$: $S = S_0 \times 10^{(D/10)}$, where $D = 82$ decibels and $S_0 = 1.00 \times 10^{-12}$ watt/$m^2$. Plugging in the values gives $S = 1.00 \times 10^{-12} \times 10^{(82/10)} = 1.6 \times 10^{-4}$ watt/$m^2$.

Choose the One Alternative That Best Completes the Statement or Answers