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Chapter 16: Problem 12

Let's imagine waves hitting the entire \(2,000 \mathrm{~km}\) Pacific coastline of the U.S. that are different from those evaluated in the text. This time, the waves have \(50 \mathrm{~m}\) wavelength, arriving every 10 seconds, and \(2 \mathrm{~m}\) crest-to-trough amplitude. How much power does the coast receive under these more active conditions?

Short Answer

Expert verified
Under these more active conditions, the coast receives approximately \(5.09825\times 10^9 \mathrm{~W}\) of power.

Step by step solution

01

Convert length of coastline to meters

To proceed with our calculations, we need to convert the length of the coastline to meters. $$L = 2000 \mathrm{~km} * 1000 \mathrm{~m/km} = 2,000,000 \mathrm{~m}$$
02

Calculate wave speed

We can calculate the wave speed (v) by dividing the wavelength (λ) by the period (T): $$v = \frac{\lambda}{T} = \frac{50 \mathrm{~m}}{10 \mathrm{~s}} = 5 \mathrm{~m/s}$$
03

Calculate wave frequency

Next, we can calculate the frequency (f) of the waves by taking the inverse of the period (T): $$f = \frac{1}{T} = \frac{1}{10 \mathrm{~s}} = 0.1 \mathrm{~Hz}$$
04

Calculate wave energy density

We will use the wave energy density formula to calculate the energy density (E) of the waves: $$E = \frac{1}{8}*\rho*g*A^2$$ Where ρ is the density of seawater, g is the acceleration due to gravity, and A is the amplitude of the waves. Assuming the density of seawater is 1025 kg/m³ and the acceleration due to gravity is 9.81 m/s², we can find the wave energy density: $$E = \frac{1}{8}\times 1025 \mathrm{~kg/m^3}\times 9.81 \mathrm{~m/s^2}\times (2 \mathrm{~m})^2 = 5098.25 \mathrm{~J/m^2}$$
05

Calculate the power received by the coast

Now, we can calculate the power received by the coast (P) using the wave energy density (E), the wave speed (v), the length of the coastline (L), and the frequency (f): $$P = E \times v \times L \times f = 5098.25 \mathrm{~J/m^2} \times 5 \mathrm{~m/s} \times 2,000,000 \mathrm{~m} \times 0.1 \mathrm{~Hz}$$ $$P = 5.09825\times 10^9 \mathrm{~W}$$ Therefore, under these more active conditions, the coast receives approximately \(5.09825\times 10^9 \mathrm{~W}\) of power.

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