rocket is using 200 kg of fuel per second for its flight. Gas produced during combustion is ejected at a velocity of 6000 m`cdot s^(-1)`. What is the force acting on the rocket?

Explanation

Force acting on the rocket, F = - `(dm)/(dt)` u.
Given, rate of combustion of fuel, `(dm)/(dt) = 200 "kg"cdot s^(-1)`, the velocity of ejection of gas, u = 6000 `m cdot s^(-1)`.
`therefore" " F = -200 xx 6000 "kg" cdot m cdot s^(-2) = - 1.2 xx 10^(6)`N
Negative sign indicates that the force acts in the direction opposite to that of the ejected gas.