Answer:
impulse, J = 200,000 kg-m/s
Step-by-step explanation:
It is given that,
The momentum of the rocket is increased by 200,000 kg-m/s during the first 60 seconds. The impulse applied to the rocket during this time is given by the change in momentum i.e.
[tex]\Delta p=p_f-p_i[/tex]
Another definition of impulse is given by the force applied and the small interval of time i.e.
[tex]J=\Delta p=F\Delta t[/tex]
So, in this case impulse applied to the rocket during 60 seconds is equivalent to the change in momentum i.e. 200,000 kg-m/s. Hence, the correct option is (d).
