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 and $I_0 = 1.0 \times 10^{-12} \, \text{W/m}^2$ is the threshold of hearing. The sound intensity $I$ at a distance $r$ from a point source is given by $I = \frac{P}{4\pi r^2}$, where $P = 20.0 \, \text{W}$ is the power of the source. At $r = 30.0 \, \text{m}$, $I = \frac{20.0}{4\pi (30.0)^2}$. Substituting $I$ and $I_0$ into the formula for $L$ gives a result in decibels, which matches choice A, 92.5 dB.

A Sound Source Emits 20