Three point charges are arranged on a line. Charge q3 = +5.00 nC and is at the origin. Charge q2 = -3.00 nC and is at x = 3.50 cm . Charge q1 is at x = 1.00 cm .What is q1 (magnitude and sign) if the net force on q3 is zero?

Now using

K*q3q1/(1cm)² = kq3q2/(3.5cm)²

So

q1/1²= q2/3.5²

But q2= -3nC

So q1= (1/3.5)²* 3*10-9

q1= 0.24nC

q1 will be positive because to become the net force on the q3 charge due to the remaining two charges, the force due to each charge on the third force should be equal and opposite thus q1 is positive

okay i have so many questions. but the big one here is: um wt..f is this mess? ?

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