A6.0 c charge is placed at the origin and a second charge is placed on the x-axis at x 0.3m. if the resulting electric force on the second charge is 5.4 n in the positive x-direction, what is the value of its charge?

9 x 10^-12 c

Explanation:

q1 = 6 C

d = 0.3 m

q2 = ?

F = 5.4 N

As te force on the second charge is in the positive X axis direction so the charge on the second particle is positive in nature.

F = K q1 q2 / d^2

5.4 = ( 9 x 10^9 x 6 x q2) / (0.3 x 0.3)

5.4 x 0.3 x 0.3 = 54 x 10^9 x q2

q2 = 9 x 10^-12 C

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the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s