Apositively charged particle moves through an electric field. as part of a complicated trajectory, the particle passes through the points a and b. the electric potential of the two points is such that vb > va. the electric force is the only force acting on the particle. what can be deduced about the speed, v, of the particle at the two points?

(a) the speed is equal at the two points.
(b) the speed is larger at a than at b.
(c) the speed is larger at b than at a.

(B) The speed is larger at A than at B.

Explanation:

Point B, the final point of the trajectory, has higher electric potential than point A, the initial point of the trajectory, so the electric potential energy of the charged particle increases, which means that its kinetic energy must be decreasing, thus the speed at B must be lower than the speed at A.

the tires are rotating at 0.434 rpmlinear speed=45 m/sdiameter=0.550 mradius=0.275 mnow v= rωω= v/rω=45/0.275ω=163.6 rad/sω=163.6 rad/s (1rev/2π rad) (60s/1m)ω=1563 rpm