Answer in Physics for jay #254810
January 23rd, 2023
Estimate the average power of a water wave when it hits the chest of an adult male standing in the water
at the seashore. Assume that the amplitude of the wave is 0.50 m, the wavelength is 2.5 m, and the
period is 4.0 s.
The power of the wave is given as follows (see http://large.stanford.edu/courses/2010/ph240/bonifacio1/):
“P = dfrac{rho g^2 T H^2}{32pi}”
where “rho approx 1000kg/m^3” is the density of water, “g= 9.81m/s^2” is the gravitational acceleration, “T = 4.0s” is the period, and “H = 0.50m” is the amplitude. Thus, obtain:
“P = dfrac{1000cdot 9.81^2 cdot 4cdot 0.5^2}{32pi} approx 9.6times 10^2W”
Answer. “9.6times 10^2W”.
