Respuesta :
Explanation:
First, we will calculate fuel consumption is as follows.
[tex]372 L/min \times 1000 ml/L \times 0.730 g/ml \times \frac{1}{60} min/s[/tex]
= 4526 g/s
Now, we will calculate the power as follows.
Power = Fuel consumption rate × -enthalpy of combustion
= [tex]4526 g/s \times -26.5 kJ/g[/tex]
= [tex]1.19 \times 10^{5}[/tex] kW
Thus, we can conclude that maximum power (in units of kilowatts) that can be produced by this spacecraft is [tex]1.19 \times 10^{5}[/tex] kW.
