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Sagot :
To determine how many times more intense the sound level of a rock concert is compared to a hot water pump, we can use the formula for sound intensity in decibels:
[tex]\[ d = 10 \log \left(\frac{I}{I_0}\right) \][/tex]
where:
- [tex]\( d \)[/tex] is the sound level in decibels,
- [tex]\( I \)[/tex] is the intensity of the sound,
- [tex]\( I_0 \)[/tex] is the reference intensity, usually the threshold of hearing.
Given:
- The noise rating of the hot water pump ([tex]\( d_{\text{pump}} \)[/tex]) is 58 decibels.
- The noise rating of the rock concert ([tex]\( d_{\text{concert}} \)[/tex]) is 124 decibels.
We need to find how many times more intense the sound level of the rock concert is compared to the hot water pump. Here's the step-by-step solution:
Step 1: Calculate the intensity ratio for the hot water pump.
Using the formula:
[tex]\[ d_{\text{pump}} = 10 \log \left(\frac{I_{\text{pump}}}{I_0}\right) \][/tex]
Rearranging for [tex]\( \frac{I_{\text{pump}}}{I_0} \)[/tex]:
[tex]\[ \frac{I_{\text{pump}}}{I_0} = 10^{\frac{d_{\text{pump}}}{10}} \][/tex]
Substituting [tex]\( d_{\text{pump}} = 58 \)[/tex]:
[tex]\[ \frac{I_{\text{pump}}}{I_0} = 10^{\frac{58}{10}} = 10^{5.8} \approx 630957.34 \][/tex]
Step 2: Calculate the intensity ratio for the rock concert.
Using the same formula:
[tex]\[ d_{\text{concert}} = 10 \log \left(\frac{I_{\text{concert}}}{I_0}\right) \][/tex]
Rearranging for [tex]\( \frac{I_{\text{concert}}}{I_0} \)[/tex]:
[tex]\[ \frac{I_{\text{concert}}}{I_0} = 10^{\frac{d_{\text{concert}}}{10}} \][/tex]
Substituting [tex]\( d_{\text{concert}} = 124 \)[/tex]:
[tex]\[ \frac{I_{\text{concert}}}{I_0} = 10^{\frac{124}{10}} = 10^{12.4} \approx 2511886431509.58 \][/tex]
Step 3: Find the ratio of the intensities.
Now we need to find how many times more intense the rock concert is compared to the hot water pump:
[tex]\[ \text{Intensity Ratio} = \frac{\frac{I_{\text{concert}}}{I_0}}{\frac{I_{\text{pump}}}{I_0}} = \frac{10^{12.4}}{10^{5.8}} = 10^{(12.4 - 5.8)} = 10^{6.6} \approx 3981071.71 \][/tex]
Therefore, the rock concert sound level is approximately [tex]\( 3981071.71 \)[/tex] times as intense as the hot water pump sound level, rounded to two decimal places.
[tex]\[ d = 10 \log \left(\frac{I}{I_0}\right) \][/tex]
where:
- [tex]\( d \)[/tex] is the sound level in decibels,
- [tex]\( I \)[/tex] is the intensity of the sound,
- [tex]\( I_0 \)[/tex] is the reference intensity, usually the threshold of hearing.
Given:
- The noise rating of the hot water pump ([tex]\( d_{\text{pump}} \)[/tex]) is 58 decibels.
- The noise rating of the rock concert ([tex]\( d_{\text{concert}} \)[/tex]) is 124 decibels.
We need to find how many times more intense the sound level of the rock concert is compared to the hot water pump. Here's the step-by-step solution:
Step 1: Calculate the intensity ratio for the hot water pump.
Using the formula:
[tex]\[ d_{\text{pump}} = 10 \log \left(\frac{I_{\text{pump}}}{I_0}\right) \][/tex]
Rearranging for [tex]\( \frac{I_{\text{pump}}}{I_0} \)[/tex]:
[tex]\[ \frac{I_{\text{pump}}}{I_0} = 10^{\frac{d_{\text{pump}}}{10}} \][/tex]
Substituting [tex]\( d_{\text{pump}} = 58 \)[/tex]:
[tex]\[ \frac{I_{\text{pump}}}{I_0} = 10^{\frac{58}{10}} = 10^{5.8} \approx 630957.34 \][/tex]
Step 2: Calculate the intensity ratio for the rock concert.
Using the same formula:
[tex]\[ d_{\text{concert}} = 10 \log \left(\frac{I_{\text{concert}}}{I_0}\right) \][/tex]
Rearranging for [tex]\( \frac{I_{\text{concert}}}{I_0} \)[/tex]:
[tex]\[ \frac{I_{\text{concert}}}{I_0} = 10^{\frac{d_{\text{concert}}}{10}} \][/tex]
Substituting [tex]\( d_{\text{concert}} = 124 \)[/tex]:
[tex]\[ \frac{I_{\text{concert}}}{I_0} = 10^{\frac{124}{10}} = 10^{12.4} \approx 2511886431509.58 \][/tex]
Step 3: Find the ratio of the intensities.
Now we need to find how many times more intense the rock concert is compared to the hot water pump:
[tex]\[ \text{Intensity Ratio} = \frac{\frac{I_{\text{concert}}}{I_0}}{\frac{I_{\text{pump}}}{I_0}} = \frac{10^{12.4}}{10^{5.8}} = 10^{(12.4 - 5.8)} = 10^{6.6} \approx 3981071.71 \][/tex]
Therefore, the rock concert sound level is approximately [tex]\( 3981071.71 \)[/tex] times as intense as the hot water pump sound level, rounded to two decimal places.
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