Answer:
a)Taking into consideration Newton’s second law, we know that
Net_Force = mass * acceleration
Since the box is pulled at constant speed, the acceleration is equal to zero.
This means that
Net_Force = 0 N
b) Force of friction
The net force is equal to the sum of all forces,
Net_Force = Force_applied - Friction
We found that Net_Force = 0, which means
Friction = Force_applied = 48 N
c) If the box comes to a stop. And the applied force becomes zero, the friction force becomes also zero.
Friction = 0 N
Answer:
B or A Try your best to choose Between those 2 if not I'll help u
The answer is A. Hope this helps I tried my best!
The final speed is 
Why?
To calculate the final speed, we need to know that we are working with an inelastic collision, it means that both bodies stick together after the colission. We can calculate the final speed of the bodies (if exists) using the following equation:

We are given the following information:

So, substituting and calculating, we have:

Hence, the final speed is 1.09 m/s.
Have a nice day!
Answer:
* if the two plates have the same charge sign F_net = 0
*if one plate is positive and the other is negative F_net = 59.4 N
Explanation:
The electric field created by a parallel plate is
E = 
where sigma is the charge density
σ = Q / A
we substitute
E =
E =
E = 4.487 10⁶ N / C
the electric force is
F = E / q
in this force it is a vector, so if the charges are of the same sign they repel and if they are of the opposite sign they attract. In this case, the test load is between the two plates that have the same load sign, so the forces are in the opposite direction.
* if the two plates have the same charge sign
F_net = F₁ - F₂
F_net = q (E₁ -E₂)
since the electric field does not depend on the distance
E₁ = E₂
in consecuense
F_net = 0
In a more interesting case
*if one plate is positive and the other is negative
F_net = F₁ + F₂
F_net = q (E₁ + E₂) = 2 q E₁
F_net = 2 6.62 10⁻⁶ 4.487 10⁶
F_net = 59.4 N
net force goes from positive to negative plate