Answer :
The speed of the winds corresponds to the tangential velocity of the tornado. Let's convert it into m/s:
[tex] v=500 km/h=500 \frac{km}{h} \frac{1000 m/km}{3600 s/h}=138.9 m/s [/tex]
The diameter of the tornado is [tex] d=60.0 m [/tex], so its radius is [tex] r=d/2=30.0 m [/tex]
The angular velocity is related to the tangential velocity by:
[tex] \omega=\frac{v}{r}=\frac{138.9 m/s}{30.0 m}=4.63 rad/s [/tex]
And keeping in mind that 1 revolution corresponds to [tex] 2 \pi [/tex] radians, we find the angular velocity in revolutions per second:
[tex] \omega=4.63 \frac{rad}{s} (\frac{1}{2 \pi} rev/rad )=0.74 rev/s [/tex]