A) \[137.5\,\,kg/s\]
B) \[185.5\,\,kg/s\]
C) \[127.5\,\,kg/s\]
D) \[187.5\,\,kg/s\]
Correct Answer: D
Solution :
Key Idea: The product of the force and the time interval is called the impulse of the force. If a constant force \[\overset{\to }{\mathop{\mathbf{F}}}\,\] is applied on a body for a short interval of time \[\Delta t\] then the impulse of this force is\[\overset{\to }{\mathop{\mathbf{F}}}\,\times \Delta t\]. Let \[m\] be the mass of the body. On applying a constant force\[\overset{\to }{\mathop{\mathbf{F}}}\,\], for a time interval\[\Delta t\], body suffers a velocity charge\[\Delta \overset{\to }{\mathop{\mathbf{v}}}\,\]. Then, according to Newton's second law, we have \[\overset{\to }{\mathop{\mathbf{F}}}\,=m\overset{\to }{\mathop{\mathbf{a}}}\,=m\frac{\Delta \overset{\to }{\mathop{\mathbf{v}}}\,}{\Delta t}\] \[\overset{\to }{\mathop{\mathbf{F}}}\,\Delta t=m\Delta \overset{\to }{\mathop{\mathbf{v}}}\,=-\overset{\to }{\mathop{\mathbf{v}}}\,(\Delta m)\] But\[m\Delta \overset{\to }{\mathop{\mathbf{v}}}\,=\Delta \overset{\to }{\mathop{\mathbf{p}}}\,\](change in momentum) Mass of rocket\[m=5000\,\,kg\] Exhaust speed \[v=800\,\,m/s\] Acceleration \[a=20\,\,m/{{s}^{2}}\] Putting these values in' above equation, we have \[\frac{\Delta m}{\Delta t}=m(a+g)\] \[\frac{\Delta m}{\Delta t}\times 800=5000(10+20)\] \[\frac{\Delta m}{\Delta t}=\frac{5000\times 30}{800}=187.5\,\,kg/s\]You need to login to perform this action.
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