JEE Main Physics question
A diwali rocket is ejecting 0.05 kg of gases per second at a velocity of 400 m/sec. The accelerating force on the rocket is
Answer: (B)
Solution
Force = (rate of gas ejection) × (velocity) = 0.05 × 400 = 20 N.
A diwali rocket is ejecting 0.05 kg of gases per second at a velocity of 400 m/sec. The accelerating force on the rocket is
Force = (rate of gas ejection) × (velocity) = 0.05 × 400 = 20 N.