Two students hear the same sound and their eardrums receive the same power from the sound wave. The sound intensity at the eardrums of the first student is 0.58 W/m^2, while at the eardrums of the second student the sound intensity is 1.18 times greater. If the diameter of the second student’s eardrum is 1.1 cm, how much acoustic power, in microwatts, is striking each of his (and the other student’s) eardrums?

Acoustic power, in microwatt W is 65.04 μW

We know that;

I = W / a

In case 1;

I1 = 0.58 W / m²

So, I₂ = 1.18 I₁

I₂ = 1.18 × 0.58

I₂ = 0.6844 W / m²

Diameter of the size = 1.1 cm

So,

Radius r = 1.1/2 = 0.55 cm

Area A = πr²W = Iπr²

W = 0.6844(3.4)(0.55)²

W = 6.504 × 10⁻⁵ W

Acoustic power, in microwatt W = 65.04 μW

Learn more:

W = 65.04 μW

Explanation:

For this exercise we will use the definition of sound intensity which is the power per unit area

           I = W / A

In this case, they indicate that for the intensity of the first student is I1 = 0.58 W / m², and the intensity of the second student is

            I₂ = 1.18 I₁

            I₂ = 1.18 0.58

            I₂ = 0.6844 W / m²

Ask the power

              W = I A

the diameter of the size is 1.1 cm and the radius is r = 0.55 cm = 0.55 10⁻² m, the approximate eardrum area as a circle

             A = π r²

we substitute

             W = I π r²

let's calculate

             W = 0.6844 π (0.55 10⁻²)²

             W = 6.504 10⁻⁵ W

they ask us to reduce to microwatts

              W = 6.504 10⁻⁵ W (106 μW / 1W)

              W = 65.04 μW

the power is constant, what changes is the intensity that depends on the sensor area, therefore the two students review the same power

friction

explanation:

they us to move forward and backwards

could you add the photo?

explanation: