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Veseljchak [2.6K]
3 years ago
9

Two airplanes leave an airport at the same

Physics
1 answer:
PSYCHO15rus [73]3 years ago
7 0

Answer:

1753.67 m

Explanation:

Using the law of cosine

c=\sqrt {(a^{2}+ b^{2}-2abcos\theta_{c})}

After 2.9 hours, the first plane has moved 730*2.9=2117  m, let this be a

The second plane has moved 550*2.9=1595 m, let this be b

The angle between them is 111-56.8=54.2, let this be \theta_{c}

c=\sqrt {(2117^{2}+ 1595^{2}-2abcos 54.2)}

c=1753.67 m

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Now  we can evaluate the distance of the center of mass to the rear axle as follows

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          W_{front \  axle} *  D  -  W_{rear \  axle} *  k  = 0

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substituting values

             \frac{W_{front \  axle}}{W_{rear \  axle} }       =    \frac{4}{2.2}

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