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To calculate the decibel level \( N \) of a sound given its intensity \( I \), we use the formula:

\[
N = 10 \log_{10} \left( \frac{I}{I_0} \right)
\]

where:
- \( I \) is the intensity of the sound (given as \( 3.5 \times 10^{-11} \, \text{watts/cm}^2 \)),
- \( I_0 \) is the reference intensity level, usually taken as \( 10^{-16} \, \text{watts/cm}^2 \) for air.

### Step-by-Step Calculation

1. **Insert the values**:
   \[
   N = 10 \log_{10} \left( \frac{3.5 \times 10^{-11}}{10^{-16}} \right)
   \]

2. **Simplify inside the logarithm**:
   \[
   \frac{3.5 \times 10^{-11}}{10^{-16}} = 3.5 \times 10^{5}
   \]

3. **Calculate the logarithm**:
   \[
   N = 10 \log_{10}(3.5 \times 10^{5})
   \]

4. **Split the logarithm**:
   \[
   N = 10 \left( \log_{10}(3.5) + \log_{10}(10^{5}) \right)
   \]

5. **Solve**:
   \[
   \log_{10}(3.5) \approx 0.544
   \]
   \[
   \log_{10}(10^{5}) = 5
   \]
   \[
   N = 10 \times (0.544 + 5) = 10 \times 5.544 = 55.4 \, \text{dB}
   \]

### Answer
The decibel level \( N \) is approximately **55.4 dB**.

Would you like more details, or do you have questions?

---

1. What is the formula for calculating decibel levels from intensity?
2. How does changing the reference intensity \( I_0 \) affect the decibel calculation?
3. What are common reference intensities in different media?
4. How does intensity in watts/cm² relate to watts/m²?
5. What does a change in decibel level represent perceptually in sound?

**Tip:** Remember that every 10 dB increase represents a tenfold increase in intensity.

What is the decibel level of a sound whose intensity is 3.5 × 10^-11 watts/cm²?

Learn how to calculate the decibel level of a sound with an intensity of 3.5 × 10^-11 watts/cm². This problem involves logarithmic calculations and is ideal for students in Grades 11-12.

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