Motion in A Plane

Question: A proton of mass $1.6 \times 10^{–27}$ kg goes round in a circular orbit of radius 0.10 m under a centripetal force of $4 \times 10^{–13}$ N. then the frequency of revolution of the proton is about

1) $0.08 \times 10^8$ cycles per sec

2) $4 \times 10^8$ cycles per sec

3) $8 \times 10^8$ cycles per sec

4) $12 \times 10^8$ cycles per sec

Solution:

Explanation: Explanation

Rate this question:

★ ★ ★ ★ ★

Average rating: (0 votes)