The solution for this problem is computed by through this formula, F = kQq / d²Plugging in the given values above, we can now compute for the answer.
F = 8.98755e9N·m²/C² * -(7e-6C)² / (0.03m)² = -489N, the negative sign denotes attraction.
J can get answer on this way:
Ek=m*V*V/2= (24kg*2m/s*2m/s)/2=48 Ј
Acceleration = v-u/t
= (45-30)/15
= 1 km/h
Distance = ut + 1/2 at^2
s = [30 x 15] + 1/2 x 1 x 15^2
= 562.5 km
Hope this helps
What we call a "year" is the time a body takes to complete one orbital revolution
in its path around the sun. The way gravity works, the farther a planet is from the
sun, the slower it moves, and the longer it takes to complete that trip. So, farther
out from the sun means a longer "year".
Everybody knows that if you want to get more warmth, then you have to stand closer
to the fire, and it's the same with planets. The farther a planet is from the sun, the less
heat it gets from the sun, and in most cases, that means its average temperature is
lower. (The planet's average temperature is affected by other things besides its distance
from the sun, such as how much heat comes up from inside, and how much heat its
atmosphere traps.)
The farther a planet's rotation axis is tilted from being perpendicular to the plane
of its orbit, the more seasonal variation there can be in the temperature at any one
place on its surface. Of course, this is kind of irrelevant if the planet has no surface.
