Answer:
Mehar cant ride down the slope
Explanation:
She does not has a bicycle
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, 

Well the heat that is needed to raise the temperature of 10g of water by 17oC is 7
Explanation:
Given:
v₀ = 0 m/s
a = 3 m/s²
t = 4 s
Find: Δx and v
Δx = v₀ t + ½ at²
Δx = (0 m/s) (4 s) + ½ (3 m/s²) (4 s)²
Δx = 24 m
v = at + v₀
v = (3 m/s²) (4 s) + 0 m/s
v = 12 m/s