To look for displacement, just draw a vector from your beginning stage to your last position and settle for the length of this line. So we begin by drawing a line to the north which is 30 ft, since it is north, the line is going up, then it move 5 ft to the south, so put a line going down, so we are in 25 ft, North so that would be the answer.
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:
Friction of the road on the motorcycle in the opposite direction
Explanation:
Khanacademy
Here, ball is released... and it is in free fall means with zero initial velocity.
We know, s = ut + 1/2 at²
Here, s = 1000 m
u = 0
a = 10 m/s2
Substitute their values,
1000 = 0 + 1/2 * 10 * t²
2000 = 10 * t²
t² = 2000 /10
t = √200
t = 14.14 s
In short, Your Answer would be 14.14 seconds
Hope this helps!
Answer:
weight at height = 100 N .
Explanation:
The problem relates to variation of weight due to change in height .
Let g₀ and g₁ be acceleration due to gravity , m is mass of the object .
At the surface :
Applying Newton's law of gravitation
mg₀ = G Mm / R²
At height h from centre
mg₁ = G Mm /h²
Given mg₀ = 400 N
400 = G Mm / R²
400 = G Mm / (6400 x 10³ )²
G Mm = 400 x (6400 x 10³ )²
At height h from centre
mg₁ = 400 x (6400 x 10³ )²/ ( 2 x 6400 x 10³)²
= 400 / 4
= 100 N .
weight at height = 100 N