sound 1 has an intensity of 38.0 w/m². sound 2 has an intensity level that is 2.5 db greater than the…

Explanation:

Step1: Recall the intensity - level formula

The intensity - level formula is $\beta = 10\log\left(\frac{I}{I_0}\right)$, where $\beta$ is the intensity level in decibels (dB), $I$ is the sound intensity, and $I_0 = 1\times10^{- 12}\ W/m^{2}$. Let $\beta_1$ be the intensity level of Sound 1 and $\beta_2$ be the intensity level of Sound 2. We know that $\beta_2=\beta_1 + 2.5$.

Step2: Express $\beta_1$ and $\beta_2$ in terms of intensities

$\beta_1 = 10\log\left(\frac{I_1}{I_0}\right)$ and $\beta_2 = 10\log\left(\frac{I_2}{I_0}\right)$. Since $\beta_2=\beta_1 + 2.5$, we have $10\log\left(\frac{I_2}{I_0}\right)=10\log\left(\frac{I_1}{I_0}\right)+2.5$.

Step3: Use logarithmic properties

Divide both sides of the equation by 10: $\log\left(\frac{I_2}{I_0}\right)=\log\left(\frac{I_1}{I_0}\right)+0.25$. Using the property of logarithms $\log a-\log b=\log\frac{a}{b}$ and $\log a+\log b = \log(ab)$, we get $\log\left(\frac{I_2}{I_0}\right)-\log\left(\frac{I_1}{I_0}\right)=0.25$, and $\log\left(\frac{\frac{I_2}{I_0}}{\frac{I_1}{I_0}}\right)=\log\left(\frac{I_2}{I_1}\right)=0.25$.

Step4: Solve for $I_2$

If $\log\left(\frac{I_2}{I_1}\right)=0.25$, then $\frac{I_2}{I_1}=10^{0.25}$ (since if $\log x = y$, then $x = 10^{y}$). Given $I_1 = 38.0\ W/m^{2}$, we have $I_2=I_1\times10^{0.25}$. $I_2=38.0\times10^{0.25}\approx38.0\times1.778 = 67.6\ W/m^{2}$

Answer:

C. $67.6\ W/m^{2}$