Question

sound intensity varies inversely as the square of the distance from the sound source. if you are in a movie theater and you change your seat to one that is eight times as far from the speakers, how does the new sound intensity compare with that of your original seat? the sound intensity is of what it was originally. (type an integer or a fraction.)

Explanation:

Step1: Define the inverse - square relationship

Let $I_1$ be the initial sound intensity, $d_1$ be the initial distance from the sound source, $I_2$ be the new sound intensity, and $d_2$ be the new distance from the sound source. The relationship is $I_1d_1^{2}=I_2d_2^{2}$.

Step2: Substitute the given distance relationship

We know that $d_2 = 8d_1$. Substituting this into the equation $I_1d_1^{2}=I_2d_2^{2}$, we get $I_1d_1^{2}=I_2(8d_1)^{2}$.

Step3: Solve for the ratio of intensities

Expand the right - hand side: $I_1d_1^{2}=I_2\times64d_1^{2}$. Divide both sides of the equation by $d_1^{2}$ (since $d_1
eq0$), we have $I_1 = 64I_2$. Then, solve for $\frac{I_2}{I_1}$, we get $\frac{I_2}{I_1}=\frac{1}{64}$.

Answer:

$\frac{1}{64}$