Apositively charged particle initially at rest on the ground accelerates upward to 140 m/s in 2.10 s. the particle has a charge-to-mass ratio of 0.100 c/kg and the electric field in this region is constant and uniform. what are the magnitude and direction of the electric field?

764.7 N/C, the direction is upward

Explanation:

E electric field of  the electron = F / charge

F force of the electron = Eq

acceleration of the body = change in velocity / time = 140 m / 2.10 s = 66.67 m/s² since the electron is from rest.

F upward = F of electron - force of gravity

F upward = Eq - mg

ma = Eq -mg

divide through with m

a = E (q/m) - g

(a + g) / (q/m) = E

(66.67 + 9.8) / 0.100 = 764.7 N/C, the direction is upward

g and the electric field in this region is constant and uniform.What are the magnitude and direction of the electric field?



the moon affects the earth gravitation ally, causing certain motions of that body and also producing tides in its ocean and atmosphere. the tidal wave in our atmosphere is, however, far too minute to be an effective cause in any known meteorological phenomenon.

explanation: