A) \[19.6m{{s}^{-2}}\]
B) \[30.2m{{s}^{-2}}\]
C) \[40\,m{{s}^{-2}}\]
D) \[49.8m{{s}^{-2}}\]
Correct Answer: C
Solution :
To just lift of the rocket from the launching pad, thrust force, \[F={{v}_{r}}\left( \frac{-dm}{dt} \right)\] where\[{{v}_{r}},\]is exhaust speed and\[\left( \frac{-dm}{dt} \right)\]is the rate at which mass is ejecting. Also, \[F=ma\] \[\therefore \] \[ma={{v}_{r}}\left( \frac{-dm}{dt} \right)\] or \[a=acceleration\] \[=\frac{1}{m}\left( \frac{-dm}{dt} \right){{v}_{r}}\] Given, \[-\frac{dm}{dt}=\frac{1/60}{1}kg\,{{s}^{-1}},m=1\,kg,\] \[{{V}_{r}}=2400\text{ }m{{s}^{-1}}\] \[\therefore \] \[a=\frac{1}{1}\left( \frac{1}{60} \right)\times 2400=40\,m{{s}^{-2}}\]You need to login to perform this action.
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