F=ma
F=QE = 1.602e-19C*700N/C = 1.1214e-16N
1.1214e-16N = ma = 1.6726e-27kg * a
a = 6.702e10 m/s² along the direction of the field line
Answer:
56.86153 N
Explanation:
t =Time taken
F = Force
Power
Work done
The magnitude of the force that is exerted on the handle is 56.86153 N
Answer:
u = - 38.85 m/s^-1
Explanation:
given data:
acceleration = 2.10*10^4 m/s^2
time = 1.85*10^{-3} s
final velocity = 0 m/s
from equation of motion we have following relation
v = u +at
0 = u + 2.10*10^4 *1.85*10^{-3}
0 = u + (21 *1.85)
0 = u + 38.85
u = - 38.85 m/s^-1
negative sign indicate that the ball bounce in opposite directon
