I think its c sorry if I’m wrong
Answer:
D) 25 m/s
Explanation:
In order to solve this problem we must use the following kinematics equation.

where:
Vf = final speed [m/s]
Vi = initial speed = 0
a = acceleration = 5[m/s^2]
t = time = 5[s]
After 5 seconds the acceleration is equal to 5 [m/s^2]
Now replacing the values in the equation:
Vf = 0 + (5*5)
Vf = 25[m/s]
Complete Question:
A 10 kg block is pulled across a horizontal surface by a rope that is oriented at 60° relative to the horizontal surface.
The tension in the rope is constant and equal to 40 N as the block is pulled. What is the instantaneous power (in W) supplied by the tension in the rope if the block when the block is 5 m away from its starting point? The coefficient of kinetic friction between the block and the floor is 0.2 and you may assume that the block starting at rest.
Answer:
Power = 54.07 W
Explanation:
Mass of the block = 10 kg
Angle made with the horizontal, θ = 60°
Distance covered, d = 5 m
Tension in the rope, T = 40 N
Coefficient of kinetic friction, 
Let the Normal reaction = N
The weight of the block acting downwards = mg
The vertical resolution of the 40 N force, 





Power, 

Answer:
16250 kgm/s due south
Explanation:
Applying,
M = mv................. Equation 1
Where M = momentum, m = mass, v = velocity.
From the car,
Given: m = 1000 kg, v = 6.5 m/s
Substitute these values into equation 1
M = 1000(6.5)
M = 6500 kgm/s
For the truck,
Given: m = 3500 kg, v = 6.5 m/s
Substitute these values into equation 1
M' = 3500(6.5)
M' = 22750 kgm/s.
Assuming South to be negative direction,
From the question,
Total momentum of the two vehicles = (6500-22750)
Total momentum of the two vehicles = -16250 kgm/s
Hence the total momentum of the two vehicles is 16250 kgm/s due south