The block moves with constant velocity: for Newton's second law, this means that the resultant of the forces acting on the block is zero, because the acceleration is zero.
We are only concerned about the horizontal direction, and there are only two forces acting along this direction: the force F pushing the block and the frictional force

acting against the motion. Since their resultant must be zero, we have:

The frictional force is

where

is the coefficient of kinetic friction

is the weight of the block.
Substituting these values, we find the magnitude of the force F:
It is either A or C. i hope i could help. if not im really sorry
Answer:
<h2>Magnitude of the second charge is

</h2>
Explanation:
According to columbs law;
F = 
F is the attractive or repulsive force between the charges = 12N
q1 and q2 are the charges
let q1 = - 8.0 x 10^-6 C
q2=?
r is the distance between the charges = 0.050m
k is the coulumbs constant =9*10⁹ kg⋅m³⋅s⁻⁴⋅A⁻²
On substituting the given values
12 = 9*10⁹*( - 8.0 x 10^-6)q2/0.050²
Cross multiplying

Answer:
The square of the hypotenuse is equal to the sum of the squares of the other two lengths.
Explanation:
only one that made sense
Answer: <u>elastically</u> deformed or <u>non-permanently</u> deformed
Explanation:
According to classical mechanics, there are two types of deformations:
-Plastic deformation (also called irreversible or permanent deformation), in which the material does not return to its original form after removing the applied force, therefore it is said that the material was permanently deformed.
This is because the material undergoes irreversible thermodynamic changes while it is subjected to the applied forces.
-Elastic deformation (also called reversible or non-permanent deformation), in which the material returns to its original shape after removing the applied force that caused the deformation.
In this case t<u>he material also undergoes thermodynamic changes, but these are reversible, causing an increase in its internal energy by transforming it into elastic potential energy.</u>
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Therefore, the situation described in the question is related to elastic deformation.