Answer:
Explanation:
Potential energy on the surface of the earth
= - GMm/ R
Potential at height h
= - GMm/ (R+h)
Potential difference
= GMm/ R - GMm/ (R+h)
= GMm ( 1/R - 1/ R+h )
= GMmh / R (R +h)
This will be the energy needed to launch an object from the surface of Earth to a height h above the surface.
Extra energy is needed to get the same object into orbit at height h
= Kinetic energy of the orbiting object at height h
= 1/2 x potential energy at height h
= 1/2 x GMm / ( R + h)
a. Independent variable: <em>TIME
</em>
b. Dependent variable: <em>DISTANCE</em>
To solve this problem it is necessary to apply the concepts related to the conservation of the Momentum describing the inelastic collision of two bodies. By definition the collision between the two bodies is given as:
Where,
= Mass of each object
= Initial Velocity of Each object
= Final Velocity
Our values are given as
Replacing we have that
Therefore the the velocity of the 3220 kg car before the collision was 0.8224m/s
V=IR
Potential Difference (v)= Current (A) * Resistance (Ω)
As V increases, R also increases.
