Question

At its peak, a tornado is 53.0 m in diameter and carries 465-km/h winds. What is its angular velocity in revolutions per second?

Answer 1

Answer:

0.7757 rev/s

Explanation:

d = Diameter of the tornado = 53 m

r = Radius of the tornado = 53/2 = 26.5 m

v = Velocity of wind = 465 km/h

Converting velocity to m/s

$465=465\times \frac{1000}{3600}=\frac{775}{6}$

Angular velocity

$\omega=\frac{v}{r}\\\Rightarrow \omega=\frac{\frac{775}{6}}{26.5}\\\Rightarrow \omega=4.87\ rad/s$

$\omega=4.87\frac{1}{2\pi}=0.7757\ rev/s$

∴ Angular velocity is 0.7757 rev/s