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3 years ago

Is an object on the surface of Earth sitting still relative to outer space?

Physics

1 answer:

DIA [1.3K]3 years ago

4 0

Answer:

The object is not still as it observe from the outer space.

Explanation:

Our earth is not stationary.

It rotates about its own axis with the period of 24 hours and it revolves around the sun with the period of 1 year.

An object is said to be at rest if the position of the object with respect to time changes.

An object is said to be in motion when the position of the object changes with respect to time.

Both the terms rest and motion are relative to each other.

If an observer observe an object on the earth from the outer space, he observe that the object s not still it is also moving as the earth is moving.

So, the object is not still as it observe from the outer space.

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How do people grow what makes them grow

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Answer:

people grow by the year sometimes it even by months

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3 years ago

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A uniform marble rolls down a symmetric bowl, starting from rest atthe top of the left side. The top of each side is a distanceh

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Answer:

A. 5/7h

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3 years ago

A man-made satellite orbits the earth in a circular orbit that has a radius of 32000 km. The mass of the Earth is 5.97e 24 kg. W

Gwar [14]

Answer:

= 3521m/s

The tangential speed is approximately 3500 m/s.

Explanation:

F = m * v² ÷ r

Fg = (G * M * m) ÷ r²

(m v²) / r = (G * M * m) / r²

v² = (G * M) / r

v = √( G * M ÷ r)

G * M = 6.67 * 10⁻¹¹ * 5.97 * 10²⁴ = 3.98199 * 10¹⁴

r = 32000km = 32 * 10⁶ meters

G * M / r = 3.98199 * 10¹⁴ ÷ 32 * 10⁶

v = √1.24 * 10⁷

v = 3521.36m/s

The tangential speed is approximately 3500 m/s.

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3 years ago

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Is it possible for a population with a high birth rate to decrease in size.

matrenka [14]

Sure, if the mortality (death) rate is even higher than the birth rate.

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3 years ago

A person who weighs 800N on the earth's surface will weigh 200N at what height above the earth

Marina86 [1]

Answer: 6,400 km

Explanation:

The weight of a person is given by:

$W=mg$

where m is the mass of the person and g is the acceleration due to gravity. While the mass does not depend on the height above the surface, the value of g does, following the formula:

$g=\frac{GM}{r^2}$

where

G is the gravitational constant

M is the Earth's mass

r is the distance of the person from the Earth's center

The problem says that the person weighs 800 N at the Earth's surface, so when r=R (Earth's radius):

$800 N= W=mg=m \frac{GM}{R^2}$ (1)

Now we want to find the height h above the surface at which the weight of the man is 200 N:

$200 N = W' = mg' = m \frac{GM}{(R+h)^2}$ (2)

If we divide eq.(1) by eq.(2), we get

$\frac{800 N}{200 N}=\frac{W}{W'}=\frac{(R+h)^2}{R^2}$

$4=\frac{(R+h)^2}{R^2}$

By solving the equation, we find:

$4R^2 = (R+h)^2=R^2+2Rh+h^2\\h^2 +2Rh-3R^2 =0$

which has two solutions:

$h=-3R$ --> negative solution, we can ignore it

$h=R$ --> this is our solution

Since the Earth's radius is $R=6.4\cdot 10^6 m$ , the person should be at $h=R=6.4\cdot 10^6 m=6400 km$ above Earth's surface.

5 0

4 years ago