Michelle is enjoying a picnic across the valley from a cliff. She is playing music on her radio (assume it to be an isotropic source) and notices an echo from the cliff. She claps her hands and the echo takes 2.6 s to return.

(a) Given that the speed of sound in air is 343 m/s on that day, how far away is the cliff? m

(b) If the intensity of the music 1.0 m from the radio is 1.0 10-5 W/m2, what is the intensity of the music arriving at the cliff? W/m2

2646,8

To answer both parts of this question, we need to know the relationship between distance, speed, and time for sound waves.

(a) Let's start with the first part of the question. We know that the speed of sound in air is 343 m/s, and the echo takes 2.6 s to return. We can use the formula:

Distance = Speed × Time

In this case, the distance is the distance between Michelle and the cliff. Plugging in the values we know:

Distance = 343 m/s × 2.6 s = 891.8 m

Therefore, the distance between Michelle and the cliff is 891.8 m.

(b) Now, let's move on to the second part. We know the intensity of the music 1.0 m from the radio is 1.0 × 10^-5 W/m^2. We need to find the intensity of the music arriving at the cliff.

The intensity of a sound wave decreases with distance according to the inverse square law, which states that the intensity is inversely proportional to the square of the distance. The formula is:

Intensity2 = Intensity1 × (Distance1^2 / Distance2^2)

In this case, we can use the given intensity of the music 1.0 m from the radio as Intensity1 and the distance between Michelle and the cliff (891.8 m) as Distance1. We want to find the intensity at the cliff, so let Intensity2 be the unknown and Distance2 is 891.8 m.

Plugging in the values we know:

Intensity2 = (1.0 × 10^-5 W/m^2) × (1.0 m / 891.8 m)^2

Simplifying:

Intensity2 = (1.0 × 10^-5 W/m^2) × (1/891.8)^2

Intensity2 = (1.0 × 10^-5 W/m^2) × (1/795026.24)

Intensity2 ≈ 1.26 × 10^-11 W/m^2

Therefore, the intensity of the music arriving at the cliff is approximately 1.26 × 10^-11 W/m^2.