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Chapter 23: Problem 96

A radar station is located at a height of \(24.0 \mathrm{m}\) above the shoreline. When the radar is aimed at a spot \(150.0 \mathrm{m}\) out to sea, it detects a whale at the bottom of the ocean. If it takes \(2.10 \mu \mathrm{s}\) for the radar to send out a beam and receive it again, how deep is the ocean where the whale is swimming?

Short Answer

Expert verified
Answer: The depth of the ocean where the whale is swimming is approximately \(269.37\,\text{m}\).

Step by step solution

01

Write down the given information

We are given: - Height of the radar station, \(h = 24.0\,\text{m}\) - Distance to the spot where the beam is aimed, \(d = 150.0\,\text{m}\) - Time for the radar to send and receive the beam, \(t = 2.10\,\text{µs}\)
02

Compute the total distance traveled by the radar beam

We can compute the total distance traveled by the radar beam using the speed of light, \(c = 3.0 \times 10^8\,\text{m/s}\). Since the radar beam travels back and forth, we must divide the total time it takes by 2 to get the time it takes for the beam to travel one way. So, we have: $$ T_{one\:way} = \frac{t}{2} = \frac{2.10\,\text{µs}}{2} = 1.05\, \text{µs} $$ Now, we can find the distance traveled by the radar beam using the speed of light and the one-way time: $$ dist_{beam} = c \times T_{one\:way} = 3.0 \times 10^8\,\text{m/s} \times 1.05\,\text{µs} = 315\,\text{m} $$
03

Calculate the depth of the ocean using the Pythagorean theorem

Let's call the length of the ocean floor from the radar station to the spot where the beam is aimed \(a\). Let's call the depth of the ocean at that spot \(b\). The distance traveled by the radar beam is the hypotenuse of the right triangle formed by \(a\), \(b\), and \(dist_{beam}\). Using the Pythagorean theorem: $$ dist_{beam}^2 = a^2 + b^2 $$ We know that \(a = 150\,\text{m} - 24\,\text{m} = 126\,\text{m}\), and \(dist_{beam} =315\,\text{m}\). Thus, we can solve for \(b\): $$ b^2 = dist_{beam}^2 - a^2 $$ Substitute the known values: $$ b^2 = 315^2 - 126^2 $$ $$ b^2 = 72561 $$ Now, find the square root: $$ b = \sqrt{72561} = 269.37\,\text{m} $$
04

Interpret the result

The depth of the ocean where the whale is swimming is approximately \(269.37\,\text{m}\).

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