The plane flies at an altitude of 4 km from the earth’s surface at a speed of 720 km / h. What is the ratio of the aircraft’s

| September 20, 2021 | education

The plane flies at an altitude of 4 km from the earth’s surface at a speed of 720 km / h. What is the ratio of the aircraft’s potential energy to the kinetic energy? The potential energy on the surface of the earth is assumed to be zero.

h = 4 km = 4000 m.

g = 9.8 m / s2.

V = 720 km / h = 200 m / s.

Ek / Ep -?

The kinetic energy of the body Ek is determined by the formula: Ek = m * V2 / 2, where m is the mass of the body, V is the speed of movement of the body.

The potential energy of a body Ep of mass m, raised to a height h above the earth’s surface, is determined by the formula: Ep = m * g * h, where g is the acceleration of gravity.

Ek / En = m * V ^ 2/2 * m * g * h = V ^ 2/2 * g * h.

Ek / Ep = (200 m / s) ^ 2/2 * 9.8 m / s ^ 2 * 4000 m = 0.5.

Answer: the potential energy of the plane Ep is 2 times its kinetic energy Ek: Ek / Ep = 0.5.



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