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3 years ago

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The value of the gravitational acceleration g decreases with elevation from 9.807 m/s2 at the sea level to 9.767 m/s2 at an alti

tude of 13000 m, where large passenger planes cruise. Determine the percentage reduction in the weight of an airplane cruising at 13000 m relative to its weight at sea level.

1 answer:

alexira [117]3 years ago

6 0

Answer: 0.41%

Explanation:

If we want to calculate the percentage decrease (the reduction) in weight of an airplane, we can use the given values of gravity, because the weight is directly proportional to the gravity acceleration.

Now the formula for the percentage decrease for ths case is:

$\% weightdecrease=\frac{g_{sea}-g_{13000m}}{g_{sea}}(100)$

Where:

$g_{sea}=9.807 m/s^{2}$ is the gravity at sea level

$g_{13000m}=9.767 m/s^{2}$ is the gravity at 13000 m

$\% weightdecrease=\frac{9.807 m/s^{2} - 9.767 m/s^{2}}{9.807 m/s^{2}}(100)$

$\% weightdecrease=0.00407 (100)$

$\% weightdecrease=0.407 \% \approx 0.41 \%$ This is the percentage reduction in the weight of an airplane

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Answer:

1.0 x 10^-3 m

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To teach you how to find the parameters characterizing an object in a circular orbit around a much heavier body like the earth.

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Answer:

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Explanation:

1) for this exercise we will use Newton's second law

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in this case the dependency is the inverse of the root of the distance

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U =

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Answer:

$\frac{1}{10}$ M

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As the <u>asteroid is travelling directly towards the center of the Earth</u>, after impact ,it <u>does not impose any torque on earth's rotation,</u> So angular momentum of earth is conserved

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where $M_{1}$ is mass of asteroid

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