A 5.0 g bullet leaves the muzzle of a riffle with a speedof 320 m/s. What force

Question

A 5.0 g bullet leaves the muzzle of a riffle with a speedof 320 m/s. What force (assumed constant) is exerted on the bulletwhile it is traveling down the 0.82 m long barrel of theriffle? I need to see step by step A 5.0 g bullet leaves the muzzle of a riffle with a speedof 320 m/s. What force (assumed constant) is exerted on the bulletwhile it is traveling down the 0.82 m long barrel of theriffle? I need to see step by step

Explanation / Answer

We know that        Woekdone = KE    ==> F.S = 0.5mV2    ==> F = mV2 / 2S              = 0.005 * 102400 / 2*0.82             = 312.19 N

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