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 3.14 x 10-6 W/m2 at a distance of 121 m from the explosion. Find the distance (in m) from the source at which the intensity is 1.80 x 10-6 W/m2.
See the answer to a similar question below under related questions
(121^2/d^2)*3.14*10^-6 = 1.8*10^-6.
0.046/d^2 = 1.8*10^-6.
1.8*10^-6*d^2 = 0.046.
d^2 = 0.02556*10^6 = 25,560.
d = 160 m.
To find the distance from the source at which the intensity is 1.80 x 10-6 W/m2, we can use the inverse square law for sound intensity. According to the inverse square law, the intensity of sound decreases as the square of the distance.
The inverse square law equation for sound intensity is:
I1/I2 = (r2/r1)^2
Where I1 and I2 are the intensities at distances r1 and r2 respectively.
Let's substitute the given values into this equation. Given that:
I1 = 3.14 x 10^-6 W/m^2
I2 = 1.80 x 10^-6 W/m^2
r1 = 121 m
We can plug these values into the equation and solve for r2, the distance we are trying to find.
(3.14 x 10^-6 / 1.80 x 10^-6) = (r2 / 121)^2
To simplify and solve, we can cross-multiply:
(3.14 x 10^-6) * (121)^2 = (1.80 x 10^-6) * (r2)^2
Simplifying further:
(3.14 x 121^2) / (1.80) = (r2)^2
(3.14 * 121 * 121) / 1.80 = (r2)^2
Now, let's calculate this expression:
(r2)^2 = 10276.44
Taking the square root on both sides:
r2 = sqrt(10276.44)
r2 ≈ 101.375
Therefore, the distance from the source at which the intensity is 1.80 x 10^-6 W/m^2 is approximately 101.375 meters.