Answer:
Option (A) , (b) and (d) are correct option
Explanation:
According to Coulomb's law electric force between two charges is given by

From the relation we can say that force is directly proportional to magnitude of charges and inversely proportional to distance between them '
So if we increase the distance then force will decrease
Increase if any of the charge get larger
If force is attractive then both the charge will be of different sign and is force is repulsive then both the charges of same sign
From above conclusion we can say that (a), (b) and (d) are correct option
Kinetic energy = (1/2) (mass) (speed)²
The rock's kinetic energy is not
(1/2) (4 kg) (10 m/s)²
= (1/2) (4 kg) (100 m²/s²)
= 200 Joules .
It may be more, or it may be less. The only thing
we can be sure of is that it is not 200 Joules.
There is many activities you can do like running, sit-ups, push-ups, and squats.
Answer:

Explanation:
From the free-body diagram for the car, we have that the normal force has a vertical component and a horizontal component, and this component act as the centripetal force on the car:

Solving N from (2) and replacing in (1):

The centripetal acceleration is given by:

Replacing and solving for v:

<span> Displacement of the medium perpendicular to the direction of propagation of the wave. that would be your answer</span>