Answer: The initial force is reduced a factor 1/4 when the separation between charge is doubled
Explanation: As it well known the electric force between two charges is given by:
Finitial=k*q1*q2/d^2 where d is the distance between charges and k is a constant
if the distance is doubled this means 2*dinitial thus the new force is equal to F initial* 1/4
Answer:
Explanation:
Component of force perpendicular to stick
= F Sin 60°
=√3 / 2 F.
Taking torque about the other end
= √3 / 2 F x 1 Nm
Weight of stick = 60 gm
= 60 x 10⁻³ kg
= 60 x 10⁻³ x 9.8 N
= .588 N
This weight will act from the middle point of stick so torque about the
other end
= .588 x 1 Nm
Balancing these two torques we have
.588 = √3 /2 F

F = 0.679 N
The answer is D
Hope this helps
In this problem we have the electric field intensity E:
E = 6.5 ×
newtons/coulomb
We have the magnitude of the load:
q = 6.4 ×
coulombs
We also have the distance d that the load moved in a direction parallel to the field 1.2 ×
meters.
We know that the electric potential energy (PE) is:
PE = qEd
So:
PE = (6.4 ×
)(6.5 ×
)(1.2 ×
)
PE = 5.0 x
joules
None of the options shown is correct.
Answer:
t = 4.1 seconds
Explanation:
It is given that,
Width of road which is to be crossed by a man is 8.25 m, it means it is distance to be covered.
Speed of man is 2.01 m/s
We need to find the time taken by the man to cross the road. It is a concept of speed. Speed of a person is given by total distance covered divided by time taken. So,

t is time taken

So, the time taken by the man to cross the road is 4.1 seconds.