Answer:
(b) 2.40 x [tex]10^{3}[/tex] kg/s
Explanation:
Given that: Total mass of the rocket = 3.003 x [tex]10^{5}[/tex] kg
acceleration of the rocket = 36.0 m[tex]s^{-2}[/tex]
speed of the exhausted gases = 4.503 x [tex]10^{3}[/tex] m/s
Rate at which rocket was initially burning fuel = [tex]\frac{mass}{time}[/tex]
But,
time = [tex]\frac{velocity}{acceleration}[/tex]
= [tex]\frac{4.503*10^{3} }{36.0}[/tex]
= 125.0833 s
So that;
Rate at which rocket was initially burning fuel = [tex]\frac{3.003*10^{5} }{125.0833}[/tex]
= 2400.8001
= 2.40 x [tex]10^{3}[/tex] kg/s
Therefore, the initial rate at which the rocket burn fuel is 2.40 x [tex]10^{3}[/tex] kg/s.