# A spacecraft weighing 10 ^ 6 kg begins to rise vertically upward. The thrust of its engine is 2.94 · 10

August 22, 2021 | education

| **A spacecraft weighing 10 ^ 6 kg begins to rise vertically upward. The thrust of its engine is 2.94 · 10 (in the seventh power) N. Determine the acceleration of the ship.**

m = 10 ^ 6 kg.

g = 9.8 N / kg.

Ft = 2.94 * 107 N.

a -?

Let us write Newton’s 2 law for a spacecraft during takeoff: m * a = Fт – F, where m is the mass of the spacecraft, a is the acceleration of its motion, Ft is the thrust force of the engines, F is the gravity of the spacecraft.

a = (Fт – F) / m.

The force of gravity of the spacecraft F is expressed by the formula: F = m * g.

a = (Fт – m * g) / m.

a = (2.94 * 10 ^ 7 N – 10 ^ 6 kg * 9.8 N / kg) / 10 ^ 6 kg = 19.6 m / s2.

Answer: during takeoff, the spacecraft moves with acceleration a = 19.6 m / s2.

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