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If the satellite doesn't have little rocket engines or other thrusters on it, AND it stays far enough from Earth that it doesn't have to plow through any air molecules, AND no pieces break off of it and drift away, AND there are no hamsters inside it running on treadmills connected to external thrusters, then there's no way for it to gain or lose energy, and its total energy remains constant.
Some of its energy is always changing, either from potential to kinetic or from kinetic to potential, as its distance from Earth changes. But the total stays constant.
Answer: E = 5.80*10^-13 J
Explanation:
Given
We use the law of conservation of momentum to solve this
Momentum before breakup = momentum after breakup
0 = m1v1 + m2v2
0 = 238m * -2.2*10^5 + 4m * v2
0 = -523.6m m/s + 4m * v2
v2 * 4m = 523.6m m/s
v2 = 523.6 m m/s / 4m
v2 = 130.9*10^5 m/s
v2 = 1.31*10^7 m/s
Using this speed in the energy equation, we have
E = 1/2m1v1² + 1/2m2v2²
E = 1/2 * (238 * 1.66*10^-27) * -2.2*10^5² + 1/2 * (4 * 1.66*10^-27) * 1.31*10^7²
E = [1/2 * 3.95*10^-25 * 4.84*10^10] + [1/2 * 6.64*10^-27 * 1.716*10^14]
E = (1/2 * 1.911*10^-14) + (1/2 * 1.139*10^-12)
E = 9.56*10^-15 + 5.7*10^-13
E = 5.80*10^-13 J
Answer:
a point representing the mean position of the matter in a body or system.
Explanation:
The body of a porter carrying load on head on a level road uses more energy than is required when the porter is resting completely. In this way the porter is doing some work.
However, when we think of theoretical definition of work, there is no force resisting the horizontal movement of the load when carried on a level road, and therefore there is no work done when a porter carrying a load on his head walk along a level road.
When the porter is going down the stairs the gravitational energy acting on the load is assisting the downward movement of the load. In this way energy is released by downward movement of the load. Thus the porter is doing negative work on the load.