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algol13
3 years ago
15

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?

Physics
1 answer:
iogann1982 [59]3 years ago
4 0
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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Explanation:

Given;

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heat transferred to the system, q₁  30 kJ

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heat gained by the system, Q = q₁ - q₂ = 30 kJ - 5kJ = 25 kJ

work done on the system, W = 500 J = 0.5 kJ

Apply first law of thermodynamic,

ΔU = Q + W

where;

ΔU  is change in internal energy

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W is work done on the system

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For t2:

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The answer is 3.19 : 2.25
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