IDNLearn.com is committed to providing high-quality answers to your questions. Get the information you need from our community of experts who provide accurate and thorough answers to all your questions.
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.
We appreciate your contributions to this forum. Don't forget to check back for the latest answers. Keep asking, answering, and sharing useful information. Thank you for trusting IDNLearn.com with your questions. Visit us again for clear, concise, and accurate answers.