Suppose the sound is emitted uniformly in all directions by a public address system. The intensity at a location 22 m away from the sound source is 3.0 x 10-4 W/m2. What is the intensity at a spot that is 78 m away?

To find the intensity at a spot that is 78 m away from the sound source, we can use the inverse square law for sound propagation. According to the inverse square law, the intensity of sound decreases as the square of the distance from the source increases.

The formula for the inverse square law is:

I1 / I2 = (d2 / d1)²

Where:
I1 is the initial intensity (known)
I2 is the intensity at the second spot (to be determined)
d1 is the initial distance (known)
d2 is the distance at the second spot (known)

Using the given values, we can plug them into the formula:

I1 / I2 = (d2 / d1)²

I1 = 3.0 x 10^(-4) W/m²
d1 = 22 m
d2 = 78 m

Now we can calculate the intensity at the second spot (I2):

I2 = I1 * (d1 / d2)²

I2 = (3.0 x 10^(-4) W/m²) * (22 m / 78 m)²

Calculating the value:

I2 = (3.0 x 10^(-4) W/m²) * (0.28205128205)²

I2 ≈ 2.25 x 10^(-5) W/m²

Therefore, the intensity at a spot that is 78 m away from the sound source is approximately 2.25 x 10^(-5) W/m².

To find the intensity at a spot that is 78 m away, we can use the inverse square law of intensity, which states that the intensity of sound decreases as the square of the distance from the source increases.

The equation for the inverse square law is:

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

Where:
I1 = initial intensity at distance d1
I2 = intensity at distance d2

In this case, we know that:
I1 = 3.0 x 10^-4 W/m^2 (intensity at 22 m)
d1 = 22 m (initial distance)
d2 = 78 m (new distance)

Now we can substitute the values into the formula and calculate the intensity at the new distance:

I2 = (3.0 x 10^-4 W/m^2) * (22^2 / 78^2)

To simplify the calculation, let's first calculate the value of (22^2 / 78^2):

(22^2 / 78^2) ≈ 0.063

Now we can substitute this value back into the equation:

I2 ≈ (3.0 x 10^-4 W/m^2) * 0.063
≈ 1.89 x 10^-5 W/m^2

Therefore, the intensity at a spot that is 78 m away from the sound source is approximately 1.89 x 10^-5 W/m^2.