A 4.55 Times 10^-5-kg positive charged ball with charge q = 4 90 Times 10^4 C is

Question

A 4.55 Times 10^-5-kg positive charged ball with charge q = 4 90 Times 10^4 C is rising on a flat. frictionless horizontal surface For a time of I = 0 195 s a constant electric field of magnitude E = 735 N/C is directed vertical to the ball which makes the ball rise to a height of d After this time the electric field is turned off and the ball returns to the surface Find the height in meters which the ball is able to be lifted off the surface during the time t. Draw a tree body diagram of the ball end sum the forces acting on it Next, after finding the acceleration. Use an equation used to relate position and acceleration for 1-D motion.

Explanation / Answer

q = charge on the ball = 4.90 x 10-6 C

m = mass of the ball = 4.55 x 10-5 kg

E = electric field = 735 N/C

a = acceleration of the ball due to electric force by electric field = qE/m = (4.90 x 10-6 ) (735)/(4.55 x 10-5 )

a = 79.2 m/s2

Vi = initial velocity = 0

t = time of travel = 0.195 s

d = height gained

using the equation

d = Vi t + (0.5) a t2

d = 0 (0.195) + (0.5) (79.2) (0.195)2

d = 1.51 m

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