contestada

Consider a category five hurricane that has a maximum wind speed of 160 mph at the eye wall, 10 miles form the center of the hurricane. if the flow in the hurricane outside of the hurricane's eye is approximate as a free vortex, determine the wind speeds at the location 20 mi, 30 mi, and 40 mi from the center of the storm.

Respuesta :

mkasblog

Answer:

See explanation for step by step answer.

Explanation:

Correct answer approximates the hurricane as a free vortex as the problem requests.

The correct equation to use is

V = (-M/(2xpixr))

where V is the tangential velocity, M is the circulation of the vortex, and r is the radius.

so -M=(160 mph)x(2 x pi x 10) at r=10 miles so -M=10,053.096

From here its just plug and chug with the radi of 20, 30, and 40 miles.

so 20 miles= 80 mph

30 miles=53.333 mph

and 40 miles=40 mph

These answers make much more sense and also model a true huricane more closely, as you increse distance from the eye of the storm wind speeds decrese.

batolisis

Answer:

  • at 20 miles = 80 mph
  • at 30 miles = 53.3 mph
  • at 40 miles = 40 mph

Explanation:

maximum wind speed ( V ) = 160 mph ( tangential velocity )

distance from the center of the hurricane ( r ) = 10, 20, 30, 40

M = circulation of the Vortex

since the hurricane outside of the hurricane's eye is approximate as a free vortex the equation for calculating the wind speeds at different ( r ) will be

V = [tex](\frac{-M}{(2\pi*r) })[/tex]  equation 1

therefore - M = V * 2[tex]\pi[/tex]r

at r = 10 miles

- M = 160 * 2 * [tex]\pi[/tex] * 10 = 160 * 2 * 3.14 * 10 = 10048

wind speed at r = 20 miles

- M = 10048

insert into equation 1 : V = [tex]\frac{10048}{2 * 314 * 20}[/tex] = 10048 / 125.6 = 80 mph

wind speed at r = 30 miles

-M = 10048

insert into equation 1 : V = 10048 / 2 * 3.14 * 30 = 10048 / 188.4 = 53.3 mph

wind speed at r = 40 miles

-M = 10048

insert back into equation 1 : V = 10048 / 2 * 3.14 * 40 = 10048 / 251.2 = 40 mph

ACCESS MORE
ACCESS MORE
ACCESS MORE
ACCESS MORE

Otras preguntas