About how many times more intense will the normal ear perceiver a sound of `10 ^(6) W //m^(2)` than one of `10^(9) W//m^(2)` ?

Explanation

To solve the problem of how many times more intense a sound of \(10^6 \, \text{W/m}^2\) is perceived compared to a sound of \(10^9 \, \text{W/m}^2\), we will follow these steps:

### Step 1: Understand the relationship between intensity and loudness
Loudness (in decibels, dB) is related to intensity (I) by the formula:
\[ L = 10 \log \left( \frac{I}{I_0} \right) \]
where \(I_0\) is the reference intensity, typically taken as \(10^{-12} \, \text{W/m}^2\).
...