Sound level: A sound source emits 20.0 W of acoustical power spread equally in all directions. The threshold of hearing is 1.0 × 10^-12 W/m². What is the sound intensity level 30.0 m from the source? A) 92.5 dB B) 81.5 dB C) 1.77 × 10^-3 dB D) 1.77 × 10^-3 W E) -27.5 dB

Question

Quizplus · Accepted Answer

The sound intensity level (in decibels, dB) can be calculated using the formula: $L = 10 \log_{10} \left( \frac{I}{I_0} \right)$, where $I$ is the sound intensity (in W/m$^2$) and $I_0$ is the threshold of hearing ($1.0 \times 10^{-12}$ W/m$^2$). At a distance of 30.0 m, the sound intensity $I$ can be found using the formula $I = \frac{P}{4\pi r^2}$, where $P = 20.0$ W is the power of the sound source and $r = 30.0$ m is the distance from the source. Substituting the values and calculating gives a sound intensity level of 92.5 dB.

Sound Level: a Sound Source Emits 20