To determine the sound level for a noise with an intensity of [tex]\( 7.6 \times 10^{-4} \)[/tex] watts/m², you can use the following formula for the sound level in decibels (dB):
[tex]\[
\beta = 10 \log \left( \frac{I}{I_0} \right)
\][/tex]
where
- [tex]\(\beta\)[/tex] is the sound level in decibels,
- [tex]\(I\)[/tex] is the intensity of the sound, and
- [tex]\(I_0\)[/tex] is the reference intensity, typically [tex]\( 1 \times 10^{-12} \)[/tex] watts/m².
Let's plug in the values:
- [tex]\(I = 7.6 \times 10^{-4}\)[/tex] watts/m²,
- [tex]\(I_0 = 1 \times 10^{-12}\)[/tex] watts/m².
The formula becomes:
[tex]\[
\beta = 10 \log \left( \frac{7.6 \times 10^{-4}}{1 \times 10^{-12}} \right)
\][/tex]
First, calculate the ratio [tex]\(\frac{I}{I_0}\)[/tex]:
[tex]\[
\frac{7.6 \times 10^{-4}}{1 \times 10^{-12}} = 7.6 \times 10^8
\][/tex]
Next, calculate the logarithm base 10 of the ratio:
[tex]\[
\log(7.6 \times 10^8)
\][/tex]
The logarithm of a product can be broken down into the sum of logarithms:
[tex]\[
\log(7.6 \times 10^8) = \log(7.6) + \log(10^8)
\][/tex]
Since [tex]\(\log(10^8) = 8\)[/tex]:
[tex]\[
\log(7.6 \times 10^8) = \log(7.6) + 8
\][/tex]
Now, we calculate [tex]\(\log(7.6)\)[/tex]:
[tex]\[
\log(7.6) \approx 0.880
\][/tex]
So:
[tex]\[
\log(7.6 \times 10^8) \approx 0.880 + 8 = 8.880
\][/tex]
Finally, we multiply by 10 to get the sound level in decibels:
[tex]\[
\beta = 10 \times 8.880 = 88.80 \, \text{dB}
\][/tex]
The nearest answer from the given choices is:
C. 88.81 dB
