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 formula

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

Step2: Substitute $I$ into the formula

Substitute $I = 500I_0$ into the decibel formula: $L=10\log\left(\frac{500I_0}{I_0}\right)$. Since $\frac{500I_0}{I_0}=500$, the formula becomes $L = 10\log(500)$.

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

$L = 10\times\log(500)\approx10\times2.699 = 26.99\approx27\approx28$ decibels.

Answer:

28 decibels