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:
find the sum of the inital and final velocitys and divide by 2 to find the average
The net force on the barge is 8000 N
Explanation:
In order to find the net force on the badge, we have to use the rules of vector addition, since force is a vector quantity.
In this problem, we have two forces:
- The force of tugboat A,
, acting in a certain direction - The force of tugboat B,
, also acting in the same direction
Since the two forces act in the same direction, this means that we can simply add their magnitudes to find the net combined force on the barge. Therefore, we get

and the direction is the same as the direction of the two forces.
Learn more about forces:
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Answer:
1. it explains what's happening im the universe now
Answer:
a) 37.70 m/s
b)710.6 m/s²
Explanation:
Given that ;
Mass of object = 2 kg
Radius of the motion = 2m
Frequency of motion = 3 rev/s
The formula to apply is;
v= 2πrf where v is linear speed
v = 2×π×2×3 =12π = 37.70 m/s
Centripetal acceleration is given as;
a= 4×π²×r×f²
a= 4×π²×2×3²
a=710.6 m/s²