noise in a quiet room is 500 times as intense as the threshold of sound. what is the decibel measurement for…

Explanation:

Step1: Recall decibel - intensity formula

The formula for the decibel level $D$ of a sound is $D = 10\log\left(\frac{I}{I_0}\right)$, where $I$ is the intensity of the sound and $I_0$ is the threshold of sound. Given that $I = 500I_0$.

Step2: Substitute $I = 500I_0$ into the formula

$D=10\log\left(\frac{500I_0}{I_0}\right)$. Since $\frac{500I_0}{I_0}=500$, the formula becomes $D = 10\log(500)$.

Step3: Calculate $\log(500)$

We know that $\log(500)=\log(5\times100)=\log(5)+\log(100)$. Since $\log(100) = 2$ and $\log(5)\approx0.699$, then $\log(500)\approx2 + 0.699=2.699$.

Step4: Calculate the decibel level

$D = 10\times2.699\approx27\approx28$ decibels.

Answer:

28 decibels