a) A microphone has an area of 5cm^2. It receives during a 4.0 s time period a sound energy of 2.0*10^-11 J. What is the intensity of the sound?

b) using intensity from a) what is the variation in pressure in the sound wave, delta P ? Use T=293 K and density of air=1.2kg/m^3

To find the intensity of the sound, we can use the formula:

Intensity = Energy / Time

For part a):

Given:
Area (A) = 5 cm^2 = 5 * 10^-4 m^2 (since 1 cm^2 = 10^-4 m^2)
Energy (E) = 2.0 * 10^-11 J
Time (t) = 4.0 s

Plug in the values into the formula:

Intensity = Energy / Time
Intensity = (2.0 * 10^-11 J) / (4.0 s)
Intensity = 5.0 * 10^-12 W/m^2

Therefore, the intensity of the sound is 5.0 * 10^-12 W/m^2.

Moving on to part b):

To find the variation in pressure (delta P) in the sound wave, we can use the formula:

Intensity = (delta P)^2 / (2 * density * velocity)

where
Intensity = 5.0 * 10^-12 W/m^2 (from part a)
density of air = 1.2 kg/m^3
velocity = speed of sound = approximately 343 m/s at room temperature (T = 293 K)

Rearranging the formula to solve for delta P:

(delta P)^2 = Intensity * 2 * density * velocity
delta P = √(Intensity * 2 * density * velocity)

Plug in the values into the formula:

delta P = √((5.0 * 10^-12 W/m^2) * (2 * 1.2 kg/m^3) * (343 m/s))
delta P ≈ 3.73 * 10^-6 pa

Therefore, the variation in pressure in the sound wave (delta P) is approximately 3.73 * 10^-6 Pa.