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The decibel level of sound is [tex]$50 \, \text{dB}$[/tex] greater on a busy street than in a quiet room where the intensity of sound is [tex]$10^{-10} \, \text{watt/m}^2$[/tex].

The level of sound in the quiet room is [tex]$\square \, \text{dB}$[/tex], and the intensity of sound on the busy street is [tex]$\square \, \text{watt/m}^2$[/tex].

Use the formula [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 sound, and [tex]$I_0$[/tex] is the smallest sound intensity that can be heard by the human ear (roughly equal to [tex]$1 \times 10^{-12} \, \text{watt/m}^2$[/tex]).



Answer :

GinnyAnswer
To solve this problem, let's follow a step-by-step approach using the provided formulas and given data:

1. Identify the given values:
- Intensity in a quiet room ([tex]\( I_{\text{quiet room}} \)[/tex]): [tex]\( 10^{-10} \)[/tex] watts/m²
- Threshold of hearing intensity ([tex]\( I_0 \)[/tex]): [tex]\( 10^{-12} \)[/tex] watts/m²
- The sound level increase on a busy street is 50 dB greater than in a quiet room.

2. Calculate the sound level in the quiet room ([tex]\( \beta_{\text{quiet room}} \)[/tex]):

The formula to calculate the decibel level ([tex]\( \beta \)[/tex]) is:
[tex]\[ \beta = 10 \log \left(\frac{I}{I_0}\right) \][/tex]

Substitute [tex]\( I_{\text{quiet room}} \)[/tex] and [tex]\( I_0 \)[/tex] into the formula:
[tex]\[ \beta_{\text{quiet room}} = 10 \log \left(\frac{10^{-10}}{10^{-12}}\right) \][/tex]

Calculate the ratio inside the logarithm:
[tex]\[ \frac{10^{-10}}{10^{-12}} = 10^2 = 100 \][/tex]

Now calculate the logarithm:
[tex]\[ \log(100) = 2 \][/tex]

Therefore:
[tex]\[ \beta_{\text{quiet room}} = 10 \times 2 = 20 \text{ dB} \][/tex]

3. Calculate the sound level on a busy street ([tex]\( \beta_{\text{busy street}} \)[/tex]):

Given that the sound level on a busy street is 50 dB greater than in a quiet room:
[tex]\[ \beta_{\text{busy street}} = \beta_{\text{quiet room}} + 50 = 20 \text{ dB} + 50 \text{ dB} = 70 \text{ dB} \][/tex]

4. Calculate the intensity of sound on the busy street ([tex]\( I_{\text{busy street}} \)[/tex]):

We use the inverse formula to find [tex]\( I \)[/tex] from [tex]\( \beta \)[/tex]:
[tex]\[ I = I_0 \times 10^{\frac{\beta}{10}} \][/tex]

Substitute [tex]\( \beta_{\text{busy street}} \)[/tex] and [tex]\( I_0 \)[/tex] into the formula:
[tex]\[ I_{\text{busy street}} = 10^{-12} \times 10^{\frac{70}{10}} \][/tex]

Simplify the exponent:
[tex]\[ 10^{\frac{70}{10}} = 10^7 \][/tex]

Therefore:
[tex]\[ I_{\text{busy street}} = 10^{-12} \times 10^7 = 10^{-5} \text{ watts/m}^2 \][/tex]

5. Conclusively:

- The level of sound in the quiet room is [tex]\( 20 \)[/tex] dB.
- The intensity of sound on the busy street is [tex]\( 10^{-5} \)[/tex] watts/m².

Thus, the completed sentences with the correct values are:

- The level of sound in the quiet room is 20 dB.
- The intensity of sound in the busy street is [tex]\( 10^{-5} \)[/tex] watt/m².

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