A) Calculate the kinetic energy in joules of a 1200-kg automobile moving at 18 m/s
B) convert this energy into calories?
c) what happens to this energy when the automobile brakes to a stop
A= kinetic energy= (1/2)(m)(v^2)= 1/2*1200*18= 10,800J
B= kinetic energy J into calories=
1 cal= 4.1868J
?= 10,800J10,800*1/4.1868=2,579.54 caloriesC. The kinetic energy is dissipated
C. The kinetic energy is dissipated
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