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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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3 years ago

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Middle School Physics 5 points

taybraidz

Asked 10.01.2018

An unbalanced force of 500 N is applied to a 75 kg object. What is the acceleration of the object?

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Answer

aachen

Answer:

Acceleration, a=6.66\ m/s^2

Explanation:

Given that,

Force acting on the object, F = 500 N

Mass of the object, m = 75 kg

We need to find the acceleration of the object. The force acting on the object is given by :

a=\dfrac{F}{m}

a=\dfrac{500\ N}{75\ kg}

a=6.66\ m/s^2

So, the acceleration of the object is 6.66\ m/s^2. Hence, this is the required solution.

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3 years ago