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The problem provided asks to calculate the sound intensity \( I \) given a sound level \( \beta = 130 \, \text{dB} \), using the formula:

\[
\beta = 10 \log \left( \frac{I}{I_0} \right)
\]

Where:
- \( \beta = 130 \, \text{dB} \)
- \( I_0 = 10^{-12} \, \text{watts/meter}^2 \) (the reference sound intensity)

### Step-by-Step Solution:

1. Start with the formula:
   \[
   \beta = 10 \log \left( \frac{I}{I_0} \right)
   \]
   
2. Substitute \( \beta = 130 \, \text{dB} \) and \( I_0 = 10^{-12} \, \text{watts/meter}^2 \):
   \[
   130 = 10 \log \left( \frac{I}{10^{-12}} \right)
   \]

3. Divide both sides by 10:
   \[
   13 = \log \left( \frac{I}{10^{-12}} \right)
   \]

4. Convert from logarithmic form to exponential form:
   \[
   10^{13} = \frac{I}{10^{-12}}
   \]

5. Solve for \( I \):
   \[
   I = 10^{13} \times 10^{-12} = 10^1 = 10 \, \text{watts/meter}^2
   \]

### Final Answer:
The sound intensity is \( 10 \, \text{watts/meter}^2 \).

Would you like more details or have any questions?

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Here are 5 related questions:
1. What would the sound intensity be for a sound level of 100 dB?
2. How does increasing the decibel level affect the sound intensity?
3. What is the reference intensity \( I_0 \) based on in terms of human hearing?
4. How can we calculate sound levels for noises louder than 130 dB?
5. How does the logarithmic scale help in measuring sound intensities?

**Tip:** When using logarithmic formulas, always remember to isolate the logarithmic term before converting it back to exponential form.

What is the sound intensity of a noise that is 130 dB? The formula for sound intensity is β = 10 log (I/I₀), where I₀ = 10^-12 watts/meter².

Learn how to calculate the sound intensity for a noise that is 130 dB using the formula β = 10 log (I/I₀). This example uses the reference intensity I₀ = 10^-12 watts/meter² and shows step-by-step how to solve for I.

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