Answer:13.9 on edgy I just did it
Explanation:
I guess the problem is asking for the distance between Earth and Mars in Astronomical Units (AU).
Since

and the distance between the two planets is

we can convert this distance into AU by using the following proportion:

from which we find
There's no physics or electronics to this question.
It's just arithmetic.
The last part of the question TELLS you how many 'coulombs per electron'.
If you just flip that fraction (divide ' 1 ' by it, take its reciprocal), then
you'll have 'electrons per coulomb', and 5 of those will answer the question.
I got 31,250,000,000,000,000,000 . I could be wrong. You should check it.
That all depends on the planet toward which the mass is falling.
If this happens to be taking place near the Earth, then the object accelerates
at the rate of about 9.8 meters per second every second.
Furthermore, if there is truly no air resistance, then it makes no difference whether
the object is a feather, a mass of 30 kg, or a school-bus. All objects accelerate at
the same rate regardless of their mass.
(a) the initial kinetic energy of the projectile is equal to:

The projectile is fired straight up, so at the top of its trajectory, its velocity is zero; this means that it has no kinetic energy left, so for the law of conservation of energy, all its energy has converted into potential energy, which is equal to

b) If the projectile is fired with an angle of

, its velocity has 2 components, one in the x-direction and one in the y-direction:


This means that at the top of its trajectory, only the vertical velocity will be zero (because the horizontal velocity is constant, since the motion on the x-axis is a uniform motion). Therefore, at the top of the trajectory, the projectile will have some kinetic energy left:

For the conservation of energy, the initial energy mechanical energy must be equal to the mechanical energy at the highest point:

the initial kinetic energy is the same as point a), so we can re-arrange this equation to find the new potential energy at the top of the trajectory: