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motikmotik
3 years ago
15

A textbook on a desk is experiencing a force of gravity of 1.56 Newtons.

Physics
1 answer:
disa [49]3 years ago
3 0

Answer:

m= 0.162kg

Explanation:

a) You would draw a square to represent the textbook, and draw the forces that are acting on it, which is force of gravity (Fg going down), and the normal force (Fn, going up)

b) To find its mass, you would use the equation Fg = mg, where m is mass in kg, and g is acceleration of gravity due to earth (9.8m/s^2), and Fg is force of gravity. We know what g is, and we know Fg (1.56 N), so just plug into the eqn and solve for m:

Fg = mg

1.59 = m (9.8)

m= 0.162kg

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Given,

Distance from the surface to the center of the earth, d=4000 miles

Distance from the center to you at a height of 8000 miles, a= 8000+4000=12000 miles

The gravitational force acting on a person at the surface is equal to his weight.

From Newton's Universal Law of Gravitation, the gravitational force is

F=\frac{G\times M\times m}{r^2}

Where G is the gravitational constant, M is the mass of the earth, m is the mass of the object/person, r is the distance between the center of the earth and the object/person

At the surface, this force is equal to the weight of the person, W=mg

i.e.

F_s=\frac{G\times M\times m}{d^2}=W

On substituting the of d,

W=\frac{\text{GMm}}{4000^2}

At a height of 8000 miles from the surface, the gravitational force is equal to,

F_a=\frac{GMm}{12000^2}

On dividing the above two equations,

\frac{F_a}{W}=\frac{4000^2^{}}{12000^2}=\frac{1}{9}

Therefore,

F_a=\frac{1}{9}W

Therefore at a height of 8000 miles above the surface of the earth, the force of gravity becomes 1/9 time your weight.

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Oduvanchick [21]

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A 45-gram object that has a volume of 9 mL has a density of
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Answer:

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Explanation:

Given data

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