Answer:

Explanation:
Given:
- mass of the object on a horizontal surface,

- coefficient of static friction,

- coefficient of kinetic friction,

- horizontal force on the object,

<u>Now the value of limiting frictional force offered by the contact surface tending to have a relative motion under the effect of force:</u>

where:
normal force of reaction acting on the body= weight of the body


As we know that the frictional force acting on the body is always in the opposite direction:
So, the frictional force will not be at its maximum and will be equal in magnitude to the applied external force and hence the body will not move.
so, the frictional force will be:

Answer:
The surface of Mercury has landforms that indicate its crust may have contracted. They are long, sinuous cliffs called lobate scarps. These scarps appear to be the surface expression of thrust faults, where the crust is broken along an inclined plane and pushed upward.
Explanation:
I hope this helps a little bit.
a.) Plants that thrive in the shade are often able to hold on to sunlight for extensive periods of time; they're in a sense like the camels of the plaNt WoRld.
b.) Though artificial lights are not nearly as beneficial as the sun, one could invest in one of them plant growing light thingies, but sun-loving plants might be sad if u do this instead of letting them soak in ePic rays from the sun.
The given question is incomplete. The complete question is as follows.
In a nuclear physics experiment, a proton (mass
kg, charge +e =
C) is fired directly at a target nucleus of unknown charge. (You can treat both objects as point charges, and assume that the nucleus remains at rest.) When it is far from its target, the proton has speed
m/s. The proton comes momentarily to rest at a distance
m from the center of the target nucleus, then flies back in the direction from which it came. What is the electric potential energy of the proton and nucleus when they are
m apart?
Explanation:
The given data is as follows.
Mass of proton =
kg
Charge of proton = 
Speed of proton = 
Distance traveled = 
We will calculate the electric potential energy of the proton and the nucleus by conservation of energy as follows.
=

where, 
U = 
Putting the given values into the above formula as follows.
U = 
= 
= 
Therefore, we can conclude that the electric potential energy of the proton and nucleus is
.