In a simple model of the wind speed associated with hurricane Emily, we assume there is calm eye 10.0 km in radius. The winds, which extend to a height of 5550 m, begin with a speed of 208.0 km/hr at the eye wall and decrease linearly with radial distance down to 0 km/hr at a distance of 150.0 km from the center. Assume the average density of the air from sea level to an altitude of 5550 m is 0.891 kg/m3. Calculate the total kinetic energy of the winds. Note: To appreciate the hurricane's KE, compare your answer to the Hiroshima atomic bomb which had an energy equivalent to about 15,000 tons of TNT, representing an energy of about 6.00e+13 J.
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To calculate the total kinetic energy of the winds associated with Hurricane Emily, we can break it down into smaller radial sections and calculate the kinetic energy for each section.
First, let's calculate the kinetic energy for the section from the eye wall to a distance of 150.0 km from the center. We know that the wind speed at the eye wall is 208.0 km/hr, and it decreases linearly to 0 km/hr at 150.0 km.
To calculate the kinetic energy, we use the formula:
KE = (1/2) * mass * velocity^2
Let's calculate the mass of air in this section. We assume that the wind extends to a height of 5550 m. So, the volume of air in this section is:
V = pi * radius^2 * height
V = pi * (10.0 km)^2 * 5550 m
Since the density of air is given as 0.891 kg/m^3, we can calculate the mass of air:
mass = density * volume
Now, let's calculate the kinetic energy for this section:
KE = (1/2) * mass * velocity^2
Repeat this calculation for various radial sections, up to the radius of the eye, and sum up all the kinetic energies to get the total kinetic energy of the winds for Hurricane Emily.
Once we have this total kinetic energy, we can compare it to the energy equivalent of the Hiroshima atomic bomb, which is about 6.00e+13 J, to appreciate the scale of the hurricane's kinetic energy.