A proton of mass `1.6xx10^-(27)` kg, revolves in a circular path of radius 0.1 m. Calculate the angular velocity of the proton if it is acted upon by a centripetal force of `2.5xx10^(-12)` N.

Explanation

`F=(mv^2)/r=mrw^2`
`therefore w^2=F/(mr)=(4xx10^(-13))/(1.6xx10^(-27)xx10^(-1)`
`w^2 =25xx10^14 therefore w=5xx10^7 (rad)/s`