The taken off plane, rising to an altitude of 11 km, picks up speed of 900 km / h. Compare the kinetic and potential energies acquired by the plane: which of them is greater and how many times?
Task data: h (the height to which the indicated aircraft climbed) = 11 km = 11 * 10 ^ 3 m; V (gained speed) = 900 km / h = 250 m / s.
Constants: g (acceleration due to gravity) ≈ 10 m / s2.
1) Potential energy: Ep = m * g * h = m * 10 * 11 * 10 ^ 3 = 110m * 10 ^ 3 J = 110m kJ.
2) Kinetic energy: Ek = m * V ^ 2/2 = m * 250 ^ 2/2 = 31250m J = 31.25m kJ.
3) Energy comparison: Ep (110m kJ)> Ek (31.25m kJ), the potential energy is greater.
4) Energy ratio: n = Ep / Ek = 110m / 31.25m = 3.52 p.
Answer: The potential energy of the indicated aircraft is 3.52 times greater than the kinetic one.
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