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Lelu [443]
3 years ago
5

What is the weight of a feather (mass = 0.0001 kg) that floats through earth's and the moon's atmospheres?

Physics
1 answer:
olya-2409 [2.1K]3 years ago
6 0

Weight = (mass) x (acceleration of gravity)

Acceleration of gravity = 9.81 m/s² on Earth, 1.62 m/s² on the Moon.

The feather's weight is . . .

On Earth:  (0.0001 kg) x (9.81 m/s²) = <em>0.000981 Newton </em>

On the Moon:  (0.0001 kg) x (1.62 m/s²) = <em>0.000162 N</em>

The presence or absence of atmosphere makes no difference.  In fact, the numbers would be the same if the feather were sealed in a jar, or spinning wildly in a tornado, or hanging by a thread, or floating in a bowl of water or chicken soup.  Weight is just the force of gravity between the feather and the Earth.  It's not affected by what's around the feather, or what's happening to it.

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Explain whether a particle moving in a straight line with constant speed does or does not have an acceleration. b) Explain wheth
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Answer:

A: No because it is nor changing speed or direction

B: Yes because it changes direction even though the speed is constant

Please Give Brainliest

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2 years ago
Which of the following is a circuit component whose function includes changing an open circuit to a closed circuit?
blondinia [14]

Answer:

<h2>B. Switch</h2>

Explanation:

<u>A device designed to open or close a circuit under controlled conditions is called a switch. The terms “open” and “closed” refer to switches as well as entire circuits. An open switch is one without continuity: current cannot flow through it.</u>

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3 years ago
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A fully loaded, slow-moving freight elevator has a cab with a total mass of 1500 kg, which is required to travel upward 57 m in
Kazeer [188]

Answer:

Explanation:

it equals16 2211

8 0
3 years ago
A man-made satellite of mass 6105 kg is in orbit around the earth, making one revolution in 430 minutes. What is the magnitude o
blondinia [14]

Answer:

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

Explanation:

At first we assume that Earth is represented by an uniform sphere, such that the man-made satellite rotates in a circular orbit around the planet. Hence, the following condition must be satisfied:

\left(\frac{4\pi^{2}}{T^{2}} \right)\cdot r = \frac{G\cdot M}{r^{2}} (1)

Where:

T - Period of rotation of the satellite, measured in seconds.

r - Distance of the satellite with respect to the center of the planet, measured in meters.

G - Gravitational constant, measured in newton-square meters per square kilogram.

M - Mass of the Earth, measured in kilograms.

Now we clear the distance of the satellite with respect to the center of the planet:

r^{3} = \frac{G\cdot M\cdot T^{2}}{4\pi^{2}}

r = \sqrt[3]{\frac{G\cdot M\cdot T^{2}}{4\pi^{2}} } (2)

If we know that G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}, M = 6.0\times 10^{24}\,kg and T = 25800\,s, then the distance of the satellite is:

r = \sqrt[3]{\frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6.0\times 10^{24}\,kg)\cdot (25800\,s)^{2}}{4\pi^{2}} }

r \approx 18.897\times 10^{6}\,m

The gravitational force exerted on the satellite by the Earth is determined by the Newton's Law of Gravitation:

F = \frac{G\cdot m\cdot M}{r^{2}} (3)

Where:

m - Mass of the satellite, measured in kilograms.

F - Force exerted on the satellite by the Earth, measured in newtons.

If we know that G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}, M = 6.0\times 10^{24}\,kg, m = 6105\,kg and r \approx 18.897\times 10^{6}\,m, then the gravitational force is:

F = \frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6105\,kg)\cdot (6\times 10^{24}\,kg)}{(18.897\times 10^{6}\,m)^{2}}

F = 6841.905\,N

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

4 0
3 years ago
A rocket takes off from Earth. It travels 825km in 75 seconds. What is the
Naya [18.7K]

Answer:

11 km/s

Explanation:

v=s/t

v=825km/75s

v=11km/s

4 0
3 years ago
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