If the force on a rocket moving with a velocity of 300 m/s is 210 N, then the rate of combustion of fuel is:
Correct answer: A. 0.7 kg/s
- A. 0.7 kg/s
- B. 1.4 kg/s
- C. 7 kg/s
- D. 10.7 kg/s
Explanation
The correct answer is 0.7 kg/s. The thrust force exerted by the rocket engine is given by the formula F = rate of fuel combustion × velocity of exhaust gases. Here, F = 210 N and velocity = 300 m/s. Solving for the rate of fuel combustion gives 210 N ÷ 300 m/s = 0.7 kg/s. Other options result from misapplying or miscalculating this formula.
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Displacement, velocity and acceleration describe motion, including the horizontal and vertical components of projectile motion. The chapter connects these quantities with Newton's laws, momentum and impulse, then applies conservation of momentum to collisions, distinguishing elastic collisions from inelastic ones.
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