Answer:
When two tectonic plates meet, we get a “plate boundary.” There are three major types of plate boundaries, each associated with the formation of a variety of geologic features.
Explanation:
Answer:

Explanation:
The magnitude of the net force exerted on q is known, we have the values and positions for
and q. So, making use of coulomb's law, we can calculate the magnitude of the force exerted by
on q. Then we can know the magnitude of the force exerted by
about q, finally this will allow us to know the magnitude of 
exerts a force on q in +y direction, and
exerts a force on q in -y direction.

The net force on q is:

Rewriting for
:

Explanation:
Solution,
- Mass(m)= 60 kg
- Force (F)= 20 N
- Acceleration (a)= ?
We know that,
- F=ma
- a=F/m
- a=20/60
- a=0.333 m/s²
So, her acceleration is 0.333 m/s².
Here when car in front of us applied brakes then it is slowing down due to frictional force on it
So here we can say that friction force on the car front of our car is given as

So the acceleration of car due to friction is given as



now it is given that


so here we have


so the car will accelerate due to brakes by a = - 8.52 m/s^2
Answer:
D
Explanation:
<em>The correct answer would be in the axle of the wheels while you ride your bicycle.</em>
Options A, B, and C requires that the forces of friction is increased in order to have more control.
However, option D requires that there is a minimal frictional force in the axle of the wheels of a bicycle while riding so that a little effort would be required to keep the bicycle moving.
<u>The lesser the friction, the lower the effort that would be needed to keep the bicycle moving and vice versa.</u>