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 1.90 10-6 W/m2 at a distance of 120 m from the explosion. Find the distance from the source at which the intensity is 0.80 10−6 W/m2.

If I1 is the intensity at a distance 150 m from the source and I2 is the intensity at a distance of d m from the source ,

I1 *150^2 = I2 *d^2

d^2 =[ I1 / I2 ]*150^2

[I1 / I2] = 1.9 /0.8= 2.375

d^2 = 2.375*150^2

d =231.17 m.

Well, this is quite a calculated explosion! Let's find the distance where the intensity drops to 0.80 x 10^(-6) W/m2.

Using the inverse square law for sound, we can say that the intensity of sound is inversely proportional to the square of the distance. That means we can set up the equation:

I1/I2 = (r2^2)/(r1^2)

Let's plug in the values we have:

1.9 x 10^(-6) / 0.8 x 10^(-6) = (r2^2) / (120^2)

Now, let's solve for r2:

r2^2 = (1.9 x 10^(-6) / 0.8 x 10^(-6)) * (120^2)

r2^2 = 570 * 120^2

r2^2 = 570 * 14400

r2^2 = 8208000

r2 ≈ 2862.4 m

So, the distance from the source at which the intensity is 0.80 x 10^(-6) W/m2 is approximately 2862.4 meters.

That's quite a distance! I hope no one's trying to have a conversation while standing there, or they'll have to shout really loudly!

To find the distance from the source at which the intensity is 0.80 × 10−6 W/m², we can use the inverse square law for sound intensity, which states that the intensity of a sound wave decreases as the square of the distance from the source.

The formula for the inverse square law is:

I1/I2 = (r2/r1)²

Where:
I1 is the initial intensity (1.90 × 10−6 W/m²),
I2 is the final intensity (0.80 × 10−6 W/m²),
r1 is the initial distance,
and r2 is the final distance (which we need to find).

Substituting the given values into the formula, we have:

(0.80 × 10−6)/(1.90 × 10−6) = (r2/120)²

Simplifying the equation, we get:

0.421/1.9 = (r2/120)²

0.221 = (r2/120)²

Taking the square root of both sides, we have:

√0.221 = r2/120

r2/120 = √0.221

r2 = √0.221 × 120

r2 ≈ 8.376 m

Therefore, the distance from the source at which the intensity is 0.80 × 10−6 W/m² is approximately 8.376 meters.

To find the distance from the source at which the intensity of the sound is 0.80 × 10^-6 W/m^2, we can use the inverse square law for sound propagation. According to this law, the intensity of sound decreases with the square of the distance from the source.

The formula for the inverse square law is:

I1 / I2 = (r2 / r1)^2

Where:
- I1 is the initial intensity of the sound (1.90 × 10^-6 W/m^2)
- I2 is the final intensity of the sound (0.80 × 10^-6 W/m^2)
- r1 is the initial distance from the source (120 m)
- r2 is the final distance from the source (unknown)

We can rearrange the formula to solve for r2:

r2 = sqrt((I2 * r1^2)/I1)

Plugging in the given values, we get:

r2 = sqrt((0.80 × 10^-6 * (120)^2) / (1.90 × 10^-6))

Simplifying the equation, we have:

r2 = sqrt((0.80 * 120^2) / 1.90)

Calculating this, we find:

r2 ≈ sqrt(38.4) ≈ 6.19 meters

Therefore, the distance from the source at which the intensity is 0.80 × 10^-6 W/m^2 is approximately 6.19 meters.