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A bullet has a mass of 0.0042kg. The muzzle velocity of the bullet coming
out of the barrel of the rifle is 993m/s. What is the KE of the bullet as it
exits the gun barrel?

Sagot :

dawitbrhane438

Answer:

KE= 1/2mv^2= 1/2*0.0042kg*993m/s= 2.0853joule

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