Let's solve the problem where we need to calculate the sound intensities given the loudness in decibels for a jack hammer and a compactor.
We'll use the formula that relates loudness [tex]\(L\)[/tex] in decibels to the sound intensity [tex]\(I\)[/tex] in watts per square meter:
[tex]\[ L = 10 \cdot \log_{10}\left(\frac{I}{I_0}\right) \][/tex]
where [tex]\( I_0 = 10^{-12} \,\text{W/m}^2 \)[/tex] is the reference sound intensity.
First, we solve for the sound intensity [tex]\(I\)[/tex]. Rearranging the formula gives:
[tex]\[ \frac{I}{I_0} = 10^{\frac{L}{10}} \][/tex]
[tex]\[ I = I_0 \cdot 10^{\frac{L}{10}} \][/tex]
### For the Jack Hammer
Given:
- Loudness [tex]\( L = 96 \, \text{dB} \)[/tex]
Substituting into the intensity formula:
[tex]\[ I_{\text{jack\_hammer}} = 10^{-12} \cdot 10^{\frac{96}{10}} \][/tex]
[tex]\[ I_{\text{jack\_hammer}} = 10^{-12} \cdot 10^{9.6} \][/tex]
[tex]\[ I_{\text{jack\_hammer}} \approx 0.003981071705534969 \, \text{W/m}^2 \][/tex]
So, the sound intensity of the jack hammer is approximately [tex]\( 0.003981071705534969 \, \text{W/m}^2 \)[/tex].
### For the Compactor
Given:
- Loudness [tex]\( L = 94 \, \text{dB} \)[/tex]
Substituting into the intensity formula:
[tex]\[ I_{\text{compactor}} = 10^{-12} \cdot 10^{\frac{94}{10}} \][/tex]
[tex]\[ I_{\text{compactor}} = 10^{-12} \cdot 10^{9.4} \][/tex]
[tex]\[ I_{\text{compactor}} \approx 0.002511886431509582 \, \text{W/m}^2 \][/tex]
So, the sound intensity of the compactor is approximately [tex]\( 0.002511886431509582 \, \text{W/m}^2 \)[/tex].
### Summary
- The sound intensity of the jack hammer is about [tex]\( 0.003981071705534969 \, \text{W/m}^2 \)[/tex].
- The sound intensity of the compactor is about [tex]\( 0.002511886431509582 \, \text{W/m}^2 \)[/tex].
