Answer: Option (b) is the correct answer.
Explanation:
It is given that a positive test charge q is released from rest at a distance r away from a charge of +Q and a distance 2r which is away from a charge of +2Q.
Then test charge to the right immediately after being released.
Therefore, the net force will be as follows.
F = 
= 
= 
F = > 0
Thus, we can conclude that the test charge move to the right immediately after being released.
Answer:
Explanation:
For an electric force, F the formula:
F = kQq/r^2
Given:
r2 = 1/2 × r1
F1 × r1 = k
F1 × r1 = F2 × r2
F2 = (F1 × r1^2)/(0.5 × r1)^2
= (F1 × r1^2)/0.25r1^2
= 4 × F1.
The speed of light is......299 792 458 m / s
Answer:
incorrect its 987 for exact
Answer: The free - body diagrams for blocks A and B. frictionless surface by a constant horizontal force F = 100 N. Find the tension in the cord between the 5 kg and 10 kg blocks. The string that attaches it to the block of mass M2 passes over a frictionless pulley of negligible mass. The coefficient of kinetic friction Hk between M.
Explanation: Hope this helped :)
