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3 years ago

How far (in meters) above the earth's surface will the acceleration of gravity be 85.0 % of what it is on the surface?

Physics

1 answer:

suter [353]3 years ago

4 0

Answer:

X = 6910319.7 m

Explanation:

let X be the distance where the acceleration of gravity is 85% of what it is on the surface and g1 be the acceleration of gravity at the surface and g2 be the acceleration of gravity at some distance X above the surface.

on the surface of the earth, the gravitational acceleration is given by:

g1 = GM/(r^2) = [(6.67408×10^-11)(5.972×10^24)]/[(6371×10^3)^2] = 9.82 m/s^2

at X meters above the earth's surface, g2 = 85/100(9.82) = 8.35m/s^2

then:

g2 = GM/(X^2)

X^2 = GM/g2

X = \sqrt{GM/g2}

= \sqrt{(6.67408×10^-11)(5.972×10^24)/ 8.35

= 6910319.7 m

Therefore, the acceleration of gravity becomes 85% of what it is on the surface at 6910319.7 m .

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Answer:

4. both blocks will both have the same amount of kinetic energy.

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4 years ago

A mountain climber of mass 60.0 kg slips and falls a distance of 4.00 m, at which time he reaches the end of his elastic safety

Sever21 [200]

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Ep=Er

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3 years ago

A large pendulum with a 200-lb gold-plated bob 12 inches in diameter is on display in the lobby of the United Nations building.

Tpy6a [65]

Answer:

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Explanation:

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3 years ago

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Answer

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