sound intensity model: $l = 10\\log(\\frac{i}{i_0})$ \n$l$ = loudness, in decibels (db); $i$ = sound…

Explanation:

Step1: Rearrange the sound - intensity formula for $I$

Given $L = 10\log(\frac{I}{I_0})$, first divide both sides by 10: $\frac{L}{10}=\log(\frac{I}{I_0})$. Then, rewrite in exponential form: $10^{\frac{L}{10}}=\frac{I}{I_0}$, and $I = I_0\times10^{\frac{L}{10}}$.

Step2: Calculate the sound intensity of the jack - hammer

Since $L = 96$ dB and $I_0=10^{- 12}\text{ watts/m}^2$, substitute into the formula $I = I_0\times10^{\frac{L}{10}}$. So $I = 10^{-12}\times10^{\frac{96}{10}}$. Using the rule of exponents $a^m\times a^n=a^{m + n}$, we have $I = 10^{-12}\times10^{9.6}=10^{-12 + 9.6}=10^{-2.4}\text{ watts/m}^2\approx3.98\times10^{-3}\text{ watts/m}^2$.

Step3: Calculate the sound intensity of the compactor

Since $L = 94$ dB and $I_0 = 10^{-12}\text{ watts/m}^2$, substitute into the formula $I = I_0\times10^{\frac{L}{10}}$. So $I=10^{-12}\times10^{\frac{94}{10}}=10^{-12}\times10^{9.4}=10^{-12 + 9.4}=10^{-2.6}\text{ watts/m}^2\approx2.51\times10^{-3}\text{ watts/m}^2$.

Answer:

The sound intensity of the jack - hammer is about $3.98\times10^{-3}\text{ watts/m}^2$. The sound intensity of the compactor is about $2.51\times10^{-3}\text{ watts/m}^2$.