To solve this problem it is necessary to apply the concepts related to the conservation of Energy. Mathematically the conservation of kinetic energy must be paid in the increase of potential energy or vice versa. This expressed in algebraic terms is equivalent to
Kinetic Energy = Potential Energy

Where
m = Mass
v = Velocity
g = Gravity
h = Height
As the mass is the same then we have to

Rearrange to find v,

Our values are given as


Therefore replacing we have


Hence the velocity at the moon would be 4.99m/s
The only direct affectation is that concerning the Resistance or drag force generated by a fluid - such as air in the ground - that can diminish / sharpen the direct effects of gravity. Disregarding the resistance of the air, as we can see in the equation previously given, there should be no affectation because the speed depends on the gravity and height.
<span>Cell membranes are selectively permeable because it allows some things to enter or leave the cell while keeping other things outside or inside the cell.</span>
True. There are forced acting on it, but as they're balanced it is unmoving
Answer:
A. F=6.65*10^{-10}N
B. south - north
Explanation:
A) We use the Lorentz force
F = qv X B
|F| = qvB
to calculate the magnitude of the force we need the speed of the of the ball.

and by replacing in the formula for the magnitude of the force we have (taking into account the excess of electrons)

B)
b. south - north (by the rigth hand rule)
I hope this is usefull for you
regards
You would need to use the equation a= (v-u)÷t
You need to substitute in the correct numbers.
a= (10-20)÷1
Your answer is -10m/s^2