A 1600-kg automobile traveling at 55 mph brakes smoothly to a stop. Assume 40.0% of the heat generated in stopping the car is dissipated in the front steel brake disks. Each front disk has a mass of 3.0 kg. What is the temperature rise of the front brake disks during the stop?
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Answer:
v² = v₀² + 2as
a = (v² - v₀²) / (2s)
= [(0 m/s)² - (15 m/s)²] / [2(40 m)] = -2,8125 m/s²
W = Fs
= mas
= (1600 kg)(-2,8125 m/s²)(40 m)
= -180000 J = -180 kJ
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