I assume you mean that the car's motor is not running ... the car is just
sitting there.
If that's so, then the car's mechanical energy is just like the mechanical
energy of any other object. It has potential energy if it's in a high place
from which it can roll or fall, and it has kinetic energy if it's moving.
-- If you make the car move by pushing it, then you gave it kinetic energy
that it didn't have while it was just sitting there.
-- If it's already moving slowly, and you're able to make it move faster by
pushing, then you increased its kinetic energy.
-- If you're able to push it up a hill, no matter how small the hill is but just
to any higher place, then you gave it more gravitational potential energy
than it had before you came along.
In all of these cases, if you exert a force and keep exerting it through some
distance while the car moves, then you have done "work", which is just
another name for mechanical energy, and your work adds to the mechanical
energy of the car.
But if you didn't move the car, then no matter how hard you pushed, no work
was done, and the car's mechanical energy didn't change.
Answer:
73.67 m
Explanation:
If projected straight up, we can work in 1 dimension, and we can use the following kinematic equations:
,
Where 
its our initial height, 
our initial speed, a the acceleration and t the time that has passed.
For our problem, the initial height its 0 meters, our initial speed its 38.0 m/s, the acceleration its the gravitational one ( g = 9.8 m/s^2), and the time its uknown.
We can plug this values in our equations, to obtain:
note that the acceleration point downwards, hence the minus sign.
Now, in the highest point, velocity must be zero, so, we can grab our second equation, and write:
and obtain:
Plugin this time on our first equation we find:
Answer:
formed by the process of glaciation.
Explanation:
efficiency = (useful energy transferred ÷ energy supplied) × 100
It's easy to use this formula, but we have to know both the useful energy and the energy supplied. The drawing doesn't tell us the useful energy, so we have to find a clever way to figure it out. I see two ways to do it:
<u>Way #1:</u>
We all know about the law of conservation of energy. So we know that the total energy coming out must be 250J, because that's how much energy is going in. The wasted energy is 75J, so the rest of the 250J must be the useful energy . . . (250J - 75J) = 175J useful energy.
(useful energy) / (energy supplied) = (175J) / (250J) = <em>70% efficiency</em>
================================
<u>Way #2: </u>
How much of the energy is wasted ? . . . 75J wasted
What percentage of the Input is that 75J ? . . . 75/250 = 30% wasted
30% of the input energy is wasted. That leaves the other <em>70%</em> to be useful energy.
Using the formula A squared plus B squared equals C squared, we can find the solution by substituting 5 for A and 12 for B.
By squaring 5, we get 25, and by squaring 12, we get 144. Adding these, we get 169. The square root of this is 13.
