In 1993 the radius of Hurricane Emily was about 350 km. The wind speed near the

Question

In 1993 the radius of Hurricane Emily was about 350 km. The wind speed near the center ("eye") of the hurricane, whose radius was about 30 km, reached about 200 km/h. As air swirled in from the rim of the hurricane toward the eye, its angular momentum remained roughly constant.

Estimate the wind speed at the rim of the hurricane.

Express your answer using two significant figures.

Estimate the pressure difference at the earth's surface between the eye and the rim. (Hint: The density of air (1 atm, 20 C) is 1.20 kg/m3.)

Express your answer using two significant figures.

Where is the pressure greater?

Where is the pressure greater?

at the rim

If the kinetic energy of the swirling air in the eye could be converted completely to gravitational potential energy, how high would the air go?

Express your answer using two significant figures.

Explanation / Answer

angular momentum L = m*v*r

given that L1 = L2

m*v1*r1 = m*v2*r2

200*30 = 350*v2

v2 = 6000/350 = 17.14 km/hr is the wind speed at the rim of the hurricane.

-------------------------------------------

Pressure difference is P2-P1 = rho*g*h = 1.2*9.8.1*(330*10^3) = 3884760 Pa

since the air swirled in from the rim of the hurricane toward the eye

then at the rim the pressure is greater than eye

So at rim pressure is greater

at eye the pressure is smaller

KE of the air at the eye is = 0.5*m*v^2 = 0.5*m*(200*5/18)^2

but this is equal to gravitational PE = m*g*h = m*9.81*h

then 0.5*m*(200*5/18)^2 = m*g*h

h = 157.3 m = 0.1573 km

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.