Answer: 24 kN
Explanation:
Given
The rocket burns propellant at the rate of
[tex]\dfrac{dm}{dt}=3\ kg/s[/tex]
Relative ejection of gases [tex]v=8000\ m/s[/tex]
The magnitude of thrust force is given by
[tex]F_t=v\dfrac{dm}{dt}\\\\F_t=8000\times 3=24,000\ N\ or\ 24\ kN[/tex]
