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A 3.0 g bullet (c=.0305 cal/g and Degree=128 Degree C) moving at 180 m/s enters a bag of sand and stops. By what amount does the temp of bullet change if all its KE becomes heat energy added to bullet

Nicole366

by Nicole366 at April 13, 2011

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Find the kinetic energy 1/2mv^2 Equate to mc(change in temperature)=1/2mv^2 c *Change in tem=v^2/2 solve for change in tem and find the tem

chitra003 chitra003 April 13, 2011

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KE = 1/2 mv^2 = 48.6 Jq = mC(delta T) 48.6 J = 3 g * .127612 J/g*deg * (delta T) delta T = +127 degrees C

Seth_Warncke Seth_Warncke June 07, 2011

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