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1 year ago

What is the gravitational field strength at a distance of 60.0 km above the surface of the earth

Physics

1 answer:

zloy xaker [14]1 year ago

6 0

Answer:

g=9.64m/s^2.

Explanation:

Gravitational field strength (in other words, gravitational acceleration) is given as follows:g=GMR2g=R2GMwhere G=6.674×10−11m3kg⋅s2G=6.674×10−11kg⋅s2m3 is the gravitational constant, M=5.972×1024kgM=5.972×1024kg is the mass of the Earth, and R=6.371×106m+0.06×106m=6.431×106mR=6.371×106m+0.06×106m=6.431×106m is the distance from the center of the Earth to the required point above the surface (radius plus 60 km).

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Answer:

a. 11 m/s at 76° with respect to the original direction of the lighter car.

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$\theta = cos^{-1}(\frac{v_{fx} }{v_{f}} )\\\theta = cos^{-1}(\frac{2.6}{10.7} )\\\theta = 76^{o}$

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