Answer:
Explanation:
Given,
- Work done by the rope 900 m/s.
- Angle of inclination of the slope =
- Initial speed of the skier = v = 1.0 m/s
- Length of the inclined surface = d = 8.0 m
part (a)
The rope is doing the work against the gravity on the skier to uplift up to the inclined surface. Therefore the work done by the rope is equal to the work done on the skier due to the gravity
In both cases the height attained by the skier is equal. and the work done by gravity does not depend upon the speed of the skier.
part (b)
- Initial speed of the skier = v = 1.0 m/s.
Rate of the work done by the rope is power of the rope.
Part (c)
- Initial speed of the skier = v = 2.0 m/s.
Rate of the work done by the rope is power of the rope.
Answer:
A vector is a quantity that has both magnitude and direction. DEF: A scalar is a quantity that has magnitude but NO direction.
Explanation:
Answer: 0.8
Explanation:
The efficiency is defined as the ratio of total imput energy that is actually utilized to the end of the device.
So if we have a total transfer of 1500j of energy, and 1200j are used to heat the thing inside the kettle, 300j are not used to the actual function of the kettle.
So the efficiency is n = 1200j/1500j = 0.8
This means that a 80% of the energy imput is actually used to heat the kettle.
Answer:
w = w₀ M / (M + 2m)
Explanation:
This exercise can be solved using the concepts of conservation of angular momentum
L = I w
Let's write in angular momentum in two points
Initial. Before impact
L₀ = I w₀
Final. After the rock has stuck
= I w + (m r²) w
The system is formed by the disk and the rock, so that the forces and moments during the crash are internal and the angular momentum is preserved
L₀ =
I w₀ = (I + m r²) w
w = w₀ I / (I + m r²)
The roundabout is a disk so its moment of inertia is
I = ½ M r²
w = w₀ ½ Mr² / (½ M r² + mr²)
w = w₀ ½ M / (½ M + m)
w = w₀ ½ M2 / (M + 2m)
w = w₀ M / (M + 2m)
That ratio is the efficiency of whatever the power went into and came out of. There's no reason the efficiency has to be expressed as a percentage.