Discover the answers you need at Westonci.ca, a dynamic Q&A platform where knowledge is shared freely by a community of experts. Get detailed and precise answers to your questions from a dedicated community of experts on our Q&A platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.
Sagot :
Sure, let's break down the solution to the problem step by step.
### Part (a): Finding the Decibel Rating
To find the decibel rating [tex]\( d \)[/tex] of a sound with an intensity [tex]\( 100,000,000,000 I_0 \)[/tex] (where [tex]\( I_0 \)[/tex] is the initial intensity), we use the formula:
[tex]\[ d = 10 \log \left( \frac{I}{I_0} \right) \][/tex]
Here:
- [tex]\( I \)[/tex] is the given intensity.
- [tex]\( I_0 \)[/tex] is the initial intensity, which we assume to be normalized to 1 for simplicity.
Given:
[tex]\[ I = 100,000,000,000 I_0 \][/tex]
[tex]\[ I_0 = 1 \][/tex]
Substitute the values into the formula:
[tex]\[ d = 10 \log \left( \frac{100,000,000,000 I_0}{I_0} \right) \][/tex]
[tex]\[ d = 10 \log (100,000,000,000) \][/tex]
Since [tex]\( \log (100,000,000,000) \)[/tex] can be computed as the log of [tex]\( 10^{11} \)[/tex]:
[tex]\[ d = 10 \log (10^{11}) \][/tex]
Using the logarithmic property that [tex]\( \log (10^x) = x \)[/tex]:
[tex]\[ d = 10 \times 11 \][/tex]
[tex]\[ d = 110 \][/tex]
Thus, the decibel rating [tex]\( d \)[/tex] is:
[tex]\[ \boxed{110.0 \text{ dB}} \][/tex]
### Part (b): Increase in Decibel Rating When Intensity is Doubled
If the intensity of the sound is doubled, the new intensity [tex]\( I_{\text{new}} \)[/tex] becomes:
[tex]\[ I_{\text{new}} = 2 \times 100,000,000,000 I_0 \][/tex]
[tex]\[ I_{\text{new}} = 200,000,000,000 I_0 \][/tex]
We need to find the new decibel rating [tex]\( d_{\text{new}} \)[/tex] using the same formula:
[tex]\[ d_{\text{new}} = 10 \log \left( \frac{I_{\text{new}}}{I_0} \right) \][/tex]
Substitute the new intensity [tex]\( I_{\text{new}} \)[/tex]:
[tex]\[ d_{\text{new}} = 10 \log \left( \frac{200,000,000,000 I_0}{I_0} \right) \][/tex]
[tex]\[ d_{\text{new}} = 10 \log (200,000,000,000) \][/tex]
Recognize that [tex]\( 200,000,000,000 \)[/tex] can be written as [tex]\( 2 \times 10^{11} \)[/tex]:
[tex]\[ d_{\text{new}} = 10 \log (2 \times 10^{11}) \][/tex]
Using the property of logarithms [tex]\( \log (ab) = \log (a) + \log (b) \)[/tex]:
[tex]\[ d_{\text{new}} = 10 (\log 2 + \log (10^{11})) \][/tex]
[tex]\[ d_{\text{new}} = 10 (\log 2 + 11) \][/tex]
Knowing [tex]\( \log 10^{11} = 11 \)[/tex]:
[tex]\[ d_{\text{new}} = 10 (\log 2 + 11) \][/tex]
Given that [tex]\( \log 2 \approx 0.3010 \)[/tex]:
[tex]\[ d_{\text{new}} = 10 (0.3010 + 11) \][/tex]
[tex]\[ d_{\text{new}} = 10 \times 11.3010 \][/tex]
[tex]\[ d_{\text{new}} = 113.010 \][/tex]
Thus, the new decibel rating [tex]\( d_{\text{new}} \)[/tex] is:
[tex]\[ \boxed{113.0103 \text{ dB}} \][/tex]
Finally, the increase in the decibel rating when the intensity is doubled is:
[tex]\[ \Delta d = d_{\text{new}} - d \][/tex]
[tex]\[ \Delta d = 113.0103 \text{ dB} - 110.0 \text{ dB} \][/tex]
[tex]\[ \Delta d = 3.0103 \text{ dB} \][/tex]
So, the increase in the decibel rating is:
[tex]\[ \boxed{3.0103 \text{ dB}} \][/tex]
To summarize:
- The decibel rating for an intensity of [tex]\( 100,000,000,000 I_0 \)[/tex] is [tex]\( 110.0 \text{ dB} \)[/tex].
- When the intensity is doubled, the decibel rating increases by [tex]\( 3.0103 \text{ dB} \)[/tex].
### Part (a): Finding the Decibel Rating
To find the decibel rating [tex]\( d \)[/tex] of a sound with an intensity [tex]\( 100,000,000,000 I_0 \)[/tex] (where [tex]\( I_0 \)[/tex] is the initial intensity), we use the formula:
[tex]\[ d = 10 \log \left( \frac{I}{I_0} \right) \][/tex]
Here:
- [tex]\( I \)[/tex] is the given intensity.
- [tex]\( I_0 \)[/tex] is the initial intensity, which we assume to be normalized to 1 for simplicity.
Given:
[tex]\[ I = 100,000,000,000 I_0 \][/tex]
[tex]\[ I_0 = 1 \][/tex]
Substitute the values into the formula:
[tex]\[ d = 10 \log \left( \frac{100,000,000,000 I_0}{I_0} \right) \][/tex]
[tex]\[ d = 10 \log (100,000,000,000) \][/tex]
Since [tex]\( \log (100,000,000,000) \)[/tex] can be computed as the log of [tex]\( 10^{11} \)[/tex]:
[tex]\[ d = 10 \log (10^{11}) \][/tex]
Using the logarithmic property that [tex]\( \log (10^x) = x \)[/tex]:
[tex]\[ d = 10 \times 11 \][/tex]
[tex]\[ d = 110 \][/tex]
Thus, the decibel rating [tex]\( d \)[/tex] is:
[tex]\[ \boxed{110.0 \text{ dB}} \][/tex]
### Part (b): Increase in Decibel Rating When Intensity is Doubled
If the intensity of the sound is doubled, the new intensity [tex]\( I_{\text{new}} \)[/tex] becomes:
[tex]\[ I_{\text{new}} = 2 \times 100,000,000,000 I_0 \][/tex]
[tex]\[ I_{\text{new}} = 200,000,000,000 I_0 \][/tex]
We need to find the new decibel rating [tex]\( d_{\text{new}} \)[/tex] using the same formula:
[tex]\[ d_{\text{new}} = 10 \log \left( \frac{I_{\text{new}}}{I_0} \right) \][/tex]
Substitute the new intensity [tex]\( I_{\text{new}} \)[/tex]:
[tex]\[ d_{\text{new}} = 10 \log \left( \frac{200,000,000,000 I_0}{I_0} \right) \][/tex]
[tex]\[ d_{\text{new}} = 10 \log (200,000,000,000) \][/tex]
Recognize that [tex]\( 200,000,000,000 \)[/tex] can be written as [tex]\( 2 \times 10^{11} \)[/tex]:
[tex]\[ d_{\text{new}} = 10 \log (2 \times 10^{11}) \][/tex]
Using the property of logarithms [tex]\( \log (ab) = \log (a) + \log (b) \)[/tex]:
[tex]\[ d_{\text{new}} = 10 (\log 2 + \log (10^{11})) \][/tex]
[tex]\[ d_{\text{new}} = 10 (\log 2 + 11) \][/tex]
Knowing [tex]\( \log 10^{11} = 11 \)[/tex]:
[tex]\[ d_{\text{new}} = 10 (\log 2 + 11) \][/tex]
Given that [tex]\( \log 2 \approx 0.3010 \)[/tex]:
[tex]\[ d_{\text{new}} = 10 (0.3010 + 11) \][/tex]
[tex]\[ d_{\text{new}} = 10 \times 11.3010 \][/tex]
[tex]\[ d_{\text{new}} = 113.010 \][/tex]
Thus, the new decibel rating [tex]\( d_{\text{new}} \)[/tex] is:
[tex]\[ \boxed{113.0103 \text{ dB}} \][/tex]
Finally, the increase in the decibel rating when the intensity is doubled is:
[tex]\[ \Delta d = d_{\text{new}} - d \][/tex]
[tex]\[ \Delta d = 113.0103 \text{ dB} - 110.0 \text{ dB} \][/tex]
[tex]\[ \Delta d = 3.0103 \text{ dB} \][/tex]
So, the increase in the decibel rating is:
[tex]\[ \boxed{3.0103 \text{ dB}} \][/tex]
To summarize:
- The decibel rating for an intensity of [tex]\( 100,000,000,000 I_0 \)[/tex] is [tex]\( 110.0 \text{ dB} \)[/tex].
- When the intensity is doubled, the decibel rating increases by [tex]\( 3.0103 \text{ dB} \)[/tex].
Visit us again for up-to-date and reliable answers. We're always ready to assist you with your informational needs. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Find reliable answers at Westonci.ca. Visit us again for the latest updates and expert advice.
