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A 10.1 g bullet leaves the muzzle of a rifle with a speed of 425 m/s. What constant force is exerted on the bullet while it is traveling down the 0.6 m length of the barrel of the rifle?

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Answer

Force will be 1520.2604 N

Explanation:

We have given mass of the bullet m = 10.1 gram = 0.0101 kg

Distance S= 0.6 m

Final velocity v = 425 m/sec

Initial velocity u will be 0 m/sec

From third equation of motion [tex]v^2=u^2+2as[/tex]

[tex]425^2=0^2+2\times a\times 0.6[/tex]

[tex]a=150520.833m/sec^2[/tex]

We know that force is given by F = ma

So force = 0.0101×150520.833 = 1520.2604 N

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