# A rocket in a fireworks display explodes high in the air. The sound spreads out uniformly in all directions. The intensity of the sound is 2.89 x 10-6 W/m2 at a distance of 127 m from the explosion. Find the distance (in m) from the source at which the intensity is 1.56 x 10-6 W/m2.

Isn't the intensity inversely prop to the square of distance?

I1/d1^2 = I2/d2^2

Solve for d2

i tried that equation, but it turned out to be incorrect.
can u plz retry to help me out plz.
I appreciate it. Thank You!

8 months ago

## Yes, you are correct that the intensity (I) of a sound wave is inversely proportional to the square of the distance (d) from the source. So, the equation you wrote, I1/d1^2 = I2/d2^2, is the correct formula to use. Let's solve it step by step.

Given:
I1 = 2.89 x 10^-6 W/m^2 (intensity at distance d1)
I2 = 1.56 x 10^-6 W/m^2 (desired intensity)
d1 = 127 m (distance from the explosion at intensity I1)

Now, substitute the given values into the equation:

I1 / d1^2 = I2 / d2^2

Solving for d2, we can rearrange the equation as:

d2^2 = (I2 * d1^2) / I1

Now plug in the given values and calculate:

d2^2 = (1.56 x 10^-6 W/m^2 * (127 m)^2) / (2.89 x 10^-6 W/m^2)

Simplifying:

d2^2 = (1.56 x 10^-6 W/m^2 * 16129 m^2) / (2.89 x 10^-6 W/m^2)

d2^2 = 2.50144 x 10^-2 m^2

To find d2, take the square root of both sides:

d2 = âˆš(2.50144 x 10^-2 m^2)

d2 â‰ˆ 0.158 m

Therefore, the distance from the source at which the intensity is 1.56 x 10^-6 W/m^2 is approximately 0.158 meters.

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