Answer:
The angular velocity of the tornado at its peak is 0.478 revolutions per second.
Explanation:
Let consider that tornado rotates at constant speed. From Rotation Physics we get the following relationship between linear ([tex]v[/tex]) and angular speeds ([tex]\omega[/tex]), measured in meter per second and radians per second, respectively.
[tex]\omega = \frac{v}{R}[/tex] (Eq. 1)
Where [tex]R[/tex] is the radius of the tornado, measured in meters.
If we know that [tex]v = 109.722\,\frac{m}{s}[/tex] and [tex]R = 36.5\,m[/tex], then the angular speed of the tornado at its peak is:
[tex]\omega = \frac{109.722\,\frac{m}{s} }{36.5\,m}[/tex]
[tex]\omega = 3.006\,\frac{rad}{s}[/tex] [tex]\left(0.478\,\frac{rev}{s} \right)[/tex]
The angular velocity of the tornado at its peak is 0.478 revolutions per second.
