Yes that looks correct to me. good luck!!
If the distance around the equator is reduced by half, then the radius is also reduced by half.
Since the acceleration due to gravity is proportional to 1/(radius²),
the acceleration changes by a factor of 1/(1/2)² = 1/(1/4) = <em>4 </em>.
The acceleration due to gravity ... and also the weight of everything on Earth ...
becomes <em>4 times what it is now</em>.
If the distance between two charges is halved, the electrical force between them increases by a factor 4.
In fact, the magnitude of the electric force between two charges is given by:
where
k is the Coulomb's constant
q1 and q2 are the two charges
r is the separation between the two charges
We see that the magnitude of the force F is inversely proportional to the square of the distance r. Therefore, if the radius is halved:
the magnitude of the force changes as follows:
so, the force increases by a factor 4.
I think the answer will be A
Is the component perpendicular to the surface on contact of the contact force <span />