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:
3.63 hours or 3 and 37.5 minutes
Explanation:
200/55
Hope this helps :)
Answer:
B. The elastic portion of a straight-line, downward-sloping demand curve corresponds to the segment above the midpoint.
Explanation:
Elasticity measures the sensitivity of one variable to another. Specifically it is a figure that indicates the percentage variation that a variable will experience in response to a variation of another one percent.
The elasticity of demand measures the reaction of demand when one of the factors that affects it varies.
<u>Elasticity - Price of demand.</u>
easure the sensitivity of the quantity demanded to price variations. It indicates the percentage variation that the quantity demanded of a good will experience if its price rises by 1 percent.
<u>
Elastic Demand
</u>
The demand quantity is relatively sensitive to price variations, so the total expenditure on the product decreases when the price rises, the price elasticity takes value greater than -∞ but less than -1
Answer:
v = 6t² + t + 2, s = 2t³ + ½ t² + 2t
59 m/s, 64.5 m
Explanation:
a = 12t + 1
v = ∫ a dt
v = 6t² + t + C
At t = 0, v = 2.
2 = 6(0)² + (0) + C
2 = C
Therefore, v = 6t² + t + 2.
s = ∫ v dt
s = 2t³ + ½ t² + 2t + C
At t = 0, s = 0.
0 = 2(0)³ + ½ (0)² + 2(0) + C
0 = C
Therefore, s = 2t³ + ½ t² + 2t.
At t = 3:
v = 6(3)² + (3) + 2 = 59
s = 2(3)³ + ½ (3)² + 2(3) = 64.5
