a 1430 kg car speeds up from 7.50 m/s to 11.0 m/s in 9.30 s. Ignoring friction, how much power did that require?
1 answer:
Kinetic energy = (1/2) (mass) (speed²) Original KE = (1/2) (1430 kg) (7.5 m/s)² = 40,218.75 joules Final KE = (1/2) (1430 kg) (11.0 m/s)² = 86,515 joules Work done during the acceleration = (40218.75 - 86515) = 46,296.25 joules Power = work/time = 46,296.25 joules / 9.3 sec = 4,978.1 watts . _______________________________ One interesting thing to notice while you're (I'm) solving this problem is the fact that although the car's speed only increased by about 47%, its kinetic energy more than doubled ... increased by about 115%. That's because kinetic energy is proportional to the SQUARE of the speed.
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thank you so much for the schlatt
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On 23rd September
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Answer:
W = Fd = KE =1/2mv²
Explanation:
not sure if that's what your looking for but i'm pretty sure this is it.
Explanation:
because it doesn't depend upon other unit like kg meter and second