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2 years ago

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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?(unit=W)PLEASE H

ELP

1 answer:

Aleks04 [339]2 years ago

3 0

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 .

Explanation:

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$V_{ft}= 317 cm/s$

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Where:

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Changed to cm/s, we get:

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So, the initial K is:

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And the final K is:

$K_f = \frac{1}{2}M_cV_{fc}^2+\frac{1}{2}M_tV_{ft}^2$

$K_f = \frac{1}{2}(1.1)(1.8)^2+\frac{1}{2}(3.55)(3.17)^2$

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