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irga5000 [103]
3 years ago
10

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?

Physics
1 answer:
expeople1 [14]3 years ago
8 0

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

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