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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. |
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Answer» Solution :`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` |
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