Newton's first and second laws of motion both do, but I think the one you're looking for is: <em>The First Law of Motion</em>. That description is a little more direct.
It says that if an object is not acted on by a net external force, then it continues in "constant, uniform motion".
Answer:
The options are not shown, so let's derive the relationship.
For an object that is at a height H above the ground, and is not moving, the potential energy will be:
U = m*g*H
where m is the mass of the object, and g is the gravitational acceleration.
Now, the kinetic energy of an object can be written as:
K = (1/2)*m*v^2
where v is the velocity.
Now, when we drop the object, the potential energy begins to transform into kinetic energy, and by the conservation of the energy, by the moment that H is equal to zero (So the potential energy is zero) all the initial potential energy must now be converted into kinetic energy.
Uinitial = Kfinal.
m*g*H = (1/2)*m*v^2
v^2 = 2*g*H
v = √(2*g*H)
So we expressed the final velocity (the velocity at which the object impacts the ground) in terms of the height, H.
Answer: 9.9%
Explanation: efficiency = (work output /work input) × 100
Note that, 1 kilocalorie = 4184 joules, hence 22kcal = 22× 4184 = 92048 joules.
Work output = 9200 j and work input = 92048 j
Efficiency = (9200/92048) × 100 = 0.099 × 100 = 9.9%
I think you can just sub the values in? unless the qn is asking for smth else?
