Answer:
KE (5) = 1/2 M V^2 = 25/2 M
KE (10) = 1/2 M V^2 = 100/2 M
KE (10) - KE (5) = M/2 (100 - 25) = 75/2 / M
An object traveling at 10 m/s has 4 times the kinetic energy as an object traveling at 5 m/s
Total work would depend on the mass being accelerated
A geostationary orbit can be achieved only at an altitude very close to 35,786 km (22,236 mi) and directly above the equator. This equates to an orbital velocity of 3.07 km/s (1.91 mi/s) and an orbital period of 1,436 minutes, which equates to almost exactly one sidereal day (23.934461223 hours).
Answer:
OPTIMISATION
Explanation:
Optimisation is a mathematical theory for developing strategies that maximize gains and minimize losses while adhering to a given set of rules and constraints.
The theory has a target function to be maximised or minimised, dependent on its explanatory variable(s), with respect to which the function has to be maximised or minimised. It also has constraints which might be binding factors to maximisation / minimisation.
Eg : Revenue optimising output is found by maximising profit function with respect to constraint function in forms of cost etc.
Answer is 4,400,000 kg • m/s