Question

At its peak, a tornado is 60.0 m in diameter and carries 500 km/h winds. What is its angular velocity in revolutions per second?

a. 0.065 rev/s
b. 0.075 rev/s
c. 0.085 rev/s
d. 0.095 rev/s

Answer 1

The angular velocity of the tornado is 0.736 rev/s.

Step-by-step explanation:

To find the angular velocity of the tornado, we need to first calculate its angular speed. Angular speed is defined as the angle through which an object rotates in a given time period. We can find the angular speed by dividing the linear speed of the tornado by its radius. The tornado's linear speed can be converted from km/h to m/s by dividing it by 3.6. The radius of the tornado is half of its diameter.

Angular Speed = Linear Speed / Radius

Angular Velocity = Angular Speed / (2π)

Given: Diameter = 60.0 m, Linear Speed = 500 km/h

Radius = Diameter / 2 = 60.0 m / 2 = 30.0 m

Linear Speed = 500 km/h = 500 km/h * (1000 m/km) / (3600 s/h) = 138.9 m/s

Angular Speed = 138.9 m/s / 30.0 m = 4.63 rad/s

Angular Velocity = 4.63 rad/s / (2π) = 0.736 rev/s