A) \[19.6\text{ }m/{{s}^{2}}\]
B) \[30.2\text{ }m/{{s}^{2}}\]
C) \[40\text{ }m/{{s}^{2}}\]
D) \[49.8\text{ }m/{{s}^{2}}\]
Correct Answer: C
Solution :
To just lift of the rocket from the launching pad, Trust 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 \[F=ma\] \[ma={{v}_{r}}\left( \frac{-dm}{dr} \right)\] or\[a=\]acceleration \[=\frac{1}{m}\left( \frac{-dm}{dt} \right){{v}_{r}}\] Given, \[\frac{-dm}{dt}=\frac{\frac{1}{60}}{1}kg{{s}^{-1}},m=1\,kg\] \[{{V}_{r}}=2400\text{ }m{{s}^{-1}}\] \[a=\frac{1}{1}\left( \frac{1}{60} \right)\times 2400=4\,m{{s}^{-2}}\]
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