FIRED per SECOND by the GUN \[\frac{dm}{dt}\]= Mass of bullet (mB) × Bullets fired per sec (N) Maximum force that man can exert \[F=u\ \left( \frac{dm}{dt} \right)\] \[\therefore \] \[F=u\TIMES {{m}_{B}}\times N\] Þ \[N=\frac{F}{{{m}_{B}}\times u}=\frac{144}{40\times {{10}^{-3}}\times 1200}=3\]

"> FIRED per SECOND by the GUN \[\frac{dm}{dt}\]= Mass of bullet (mB) × Bullets fired per sec (N) Maximum force that man can exert \[F=u\ \left( \frac{dm}{dt} \right)\] \[\therefore \] \[F=u\TIMES {{m}_{B}}\times N\] Þ \[N=\frac{F}{{{m}_{B}}\times u}=\frac{144}{40\times {{10}^{-3}}\times 1200}=3\]

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A machine gun fires a bullet of mass 40 g with a velocity \[1200\,\,m{{s}^{-1}}.\] The man holding it can exert a maximum force of 144 N on the gun. How many bullets can he fire per second at the most                        [AIEEE 2004]

Joint Entrance Exam - Main (JEE Main) Physics in Joint Entrance Exam - Main (JEE Main) . 10 months ago

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u = velocity of bullet \[\frac{dm}{dt}=\]Mass FIRED per SECOND by the GUN \[\frac{dm}{dt}\]= Mass of bullet (mB) × Bullets fired per sec (N) Maximum force that man can exert \[F=u\ \left( \frac{dm}{dt} \right)\] \[\therefore \] \[F=u\TIMES {{m}_{B}}\times N\] Þ \[N=\frac{F}{{{m}_{B}}\times u}=\frac{144}{40\times {{10}^{-3}}\times 1200}=3\]

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Posted on 20 Aug 2024, this text provides information on Joint Entrance Exam - Main (JEE Main) related to Physics in Joint Entrance Exam - Main (JEE Main). Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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