Answer:
The "solid force"? ... The direction of the force always seems to be coming out of the solid surface. A direction which is perpendicular to the plane of a surface is said to be normal. The force that a solid surface exerts on anything in the normal direction is called the normal force.
Explanation:
i think i hope this helps
Answer:
I can't see the picture
Explanation:
Im sorry can you right whats on the picture down in the comments plz.
The force of the racket affects the ball's motion because it changes the momentum of the ball.
<h3>Impulse received by the ball</h3>
The impulse received by the ball through the racket affects the motion because it changes the momentum of the ball.
The ball which is initially at rest, will gain momentum after been hit with the racket.
J = ΔP = Ft
where;
- J is the impulse received by the ball
- ΔP is change in momentum of the ball
- F is the applied force
- t is the time of action
Thus, the force of the racket affects the ball's motion because it changes the momentum of the ball.
Learn more about impulse here: brainly.com/question/25700778
Answer:

Explanation:
The force on the point charge q exerted by the rod can be found by Coulomb's Law.

Unfortunately, Coulomb's Law is valid for points charges only, and the rod is not a point charge.
In this case, we have to choose an infinitesimal portion on the rod, which is basically a point, and calculate the force exerted by this point, then integrate this small force (dF) over the entire rod.
We will choose an infinitesimal portion from a distance 'x' from the origin, and the length of this portion will be denoted as 'dx'. The charge of this small portion will be 'dq'.
Applying Coulomb's Law:

The direction of the force on 'q' is to the right, since both charges are positive, and they repel each other.
Now, we have to write 'dq' in term of the known quantities.

Now, substitute this into 'dF':

Now we can integrate dF over the rod.
