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A bullet is fired from a gun. The force on the bullet is given by `F=600-2xx10^(5)` t, where F is in newtons and t in seconds. The force on the bullet becomes zero as soon as it leaves the barrel. What is the average impulse imparted to the bullet?A. 9NsB. zeroC. `0.9Ns`D. `1.8Ns` |
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Answer» Correct Answer - C `F=600-2xx10^(5)t=0` implies `t=3xx10^(-3)sec` impulse `I=int_(0)^(1)Fdt=int_(0)^(3xx10^(-3))(600-2xx10^(3)t)dt` `=[600t-10^(5)t^(2)]_(0)^(3xx10^(-3))=0.9Nxxsec` |
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