Suppose that the cyclist begins his journey from the rest from the top of a wedge with a slope of a degree above the horizontal.
At point A (where it starts its journey), the energy is:
Ea = m * g * h
In other words, energy is only potential.
At point B (located at the bottom of the wedge), the energy is:
Eb = (1/2) * (m) * (v ^ 2)
In other words, the energy is only kinetic.
For energy conservation we have:
Ea = Eb
That is, we have that all potential energy is transformed into kinetic energy.
Which means that the cyclist has less kinetic energy at point A because that's where he has more potential energy.
answer:
the cyclist has less kinetic energy at point A because that's where he has more potential energy.
Answer:

Explanation:
The difference between top and bottom surfaces is computed by the following hydrostatic equivalence:



Momentum = mass x velocity so if velocity is kept same and mass is increased then momentum will increase.
Answer:

Explanation:
Here we can use energy conservation
As per energy conservation conditions we know that work done by external source is converted into kinetic energy of the disc
Now we have

now we know that work done is product of force and displacement
so here we have


now for moment of inertia of the disc we will have



now from above equation we will have


This happens<span> at the boiling </span>temperature<span> of every substance that can vaporize. At the boiling </span>temperature<span>, adding </span>heat<span> energy converts the liquid into a gas WITHOUT RAISING THE </span>TEMPERATURE<span>. Adding </span>heat<span> to a boiling liquid is an important exception to general rule that more </span>heat<span> makes a higher </span>temperature<span>.</span>