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 1.5 s to return. (a) Given that the speed of sound in air is \(343 \mathrm{m} / \mathrm{s}\) on that day, how far away is the cliff? (b) If the intensity of the music $1.0 \mathrm{m}\( from the radio is \)1.0 \times 10^{-5} \mathrm{W} / \mathrm{m}^{2},$ what is the intensity of the music arriving at the cliff?

To find the distance to the cliff, we first should calculate the total time taken for the sound to travel from Michelle to the cliff and back, which is given as 1.5 seconds. The speed of sound is given as \(343 \mathrm{m} / \mathrm{s}\). Let d be the distance to the cliff. Since the sound takes the same time to travel to the cliff and back, we can divide the total time by 2 to find the time it takes to travel one way: time_one_way = total_time / 2 = 1.5 s / 2 = 0.75 s Now, we can use the formula distance = speed × time to find the distance to the cliff: distance = speed × time_one_way d = 343 m/s × 0.75 s d = 257.25 m The distance to the cliff is 257.25 meters.

We are given the intensity of the music at a distance of 1.0 m from the radio, which is \(1.0 \times 10^{-5} \mathrm{W} / \mathrm{m}^{2}\). Since the radio is an isotropic source, we can use the formula for intensity at a distance r to find the intensity at the cliff: intensity_cliff = (power) / (4π × (distance_to_cliff)^2) We first need to find the power of the radio. We can use the given initial intensity and distance to calculate this: power = intensity_initial × (4π × r_initial^2) power = \((1.0 \times 10^{-5} \mathrm{W} / \mathrm{m}^{2}) (4π × (1.0 \mathrm{m})^2)\) power = \(4π \times 10^{-5} \mathrm{W}\) Now we can find the intensity of the music at the cliff: intensity_cliff = \((4π \times 10^{-5} \mathrm{W})\) / \((4π × (257.25 \mathrm{m})^2)\) We can now simplify and calculate the intensity_cliff: intensity_cliff = \(10^{-5} \mathrm{W} / \mathrm{m}^{2} \times 1.0 \mathrm{m}^{2}\) / \((257.25 \mathrm{m})^2\) intensity_cliff = \(10^{-5} \mathrm{W} / \mathrm{m}^{2} \times \frac{1}{(257.25)^2}\) intensity_cliff = \(1.515 \times 10^{-10} \mathrm{W} / \mathrm{m}^{2}\) The intensity of the music at the cliff is \(1.515 \times 10^{-10} \mathrm{W} / \mathrm{m}^{2}\).