# An advertisement shows a 1239 kg car slowly pulling at constant speed a large passenger airplane [Solved] An commercial reveals a 1239 kg automobile slowly pulling at fixed pace a big passenger airplane to reveal the ability of its newly-designed 58.7-hp (horsepower) engine. Through the pull, the automobile handed two landmarks spaced 21.0 meters aside in 15.3 seconds. The passenger airplane being towed is a Boeing 707, with a complete weight of 62.5 tons. Calculate the power that opposes the movement. Assume that the effectivity of the automobile is such that 18.7% of the engine energy is obtainable to propel the automobile ahead.

The web power f = MA = P F = 0; the place F is the power opposing the pull P.
As there isnt any acceleration, we are able to say is that P = F; in order that A = 0 is the acceleration (none).
If we assume all the ability out there is definitely used, then 58.7*.187 = 10.9769 HP is used when dragging the airplane that 21 m. 745.7 J/s = 1 HP; so 8185.5 J/s = 11 HP and work performed in that 15.3 seconds is QE = 15.3*8185.5 = 125238.15 Joules. And over that 21 m F = QE/S = 125238.15/21 = 5963.7 N is the power opposing the tow automobile. ANS.
And from earlier, QE = 125238.15 Joules is the work performed by the tow automobile. ANS.
The airplanes drag/friction works to offset the vitality enter by the tow automobile. So it is equal to minus QE. ANS. Wordeach QEs are performed on or by your entire system, not simply the automobile or the airplane, however each, the full system. That is apparent should you simply do a thought experiment and put the brakes on for the airplane. Each the automobile and the airplane will decelerate, each are acted on by what the airplane does.

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