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Temka [501]
3 years ago
11

The gravitational force experienced by Earth due to the Moon is ________ the gravitational force experienced by the Moon due to

Earth. (Note that the Moon is about 1% of the mass of Earth.)
Physics
1 answer:
Vsevolod [243]3 years ago
5 0

The gravitational force experienced by Earth due to the Moon is <u>equal to </u>the gravitational force experienced by the Moon due to Earth.

<u>Explanation</u>:

The force that attracts any two objects/bodies with mass towards each other is defined as gravitational force. Generally the gravitational force is attractive, as it always pulls the masses together and never pushes them apart.

The gravitational force can be calculated effectively using the following formula: F=GMmr^2  

where “G” is the gravitational constant.

Though gravity has the ability to pull the masses together, it is the weakest force in the nature.

The mass of the Earth and moon varies, but still the gravitational force felt by the Earth and Moon are alike.

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Saturn is 890,700,000 miles from the Sun. What is the distance in meters?
Maslowich

Answer:

1.4936 trillion meters

Explanation:

From an average distance of 886 million miles Saturn is 9.5 astronomical units away from the sun

6 0
3 years ago
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PLEASE HELP!!!! 15 point!!!!!
Nadusha1986 [10]
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3 0
4 years ago
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The magnitude E of an electric field depends on the radial distance r according to E = A/r4, where A is a constant with unit vol
Lesechka [4]

Answer:

\Delta V = 0.053 A

Explanation:

Electric field in a given region is given by equation

E = \frac{A}{r^4}

as we know the relation between electric field and potential difference is given as

\Delta V = -\int E. dr

so here we have

\Delta V = - \int (\frac{A}{r^4}) .dr

\Delta V = \frac{A}{3r_1^3} - \frac{A}{3r_2^3}

here we know that

r_1 = 1.71 m  and r_2 = 2.89 m

so we will have

\Delta V = \frac{A}{3}(\frac{1}{1.71^3} - \frac{1}{2.89^3})

so we will have

\Delta V = 0.053 A

8 0
3 years ago
Which best describes accuracy?
Likurg_2 [28]
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7 0
3 years ago
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A satellite moves in a circular orbit at a constant speed v0 around Earth at a distance R from its center. The force exerted on
postnew [5]

Answer:

vT = v0/3

Explanation:

The gravitational force on the satellite with speed v0 at distance R is F = GMm/R². This is also equal to the centripetal force on the satellite F' = m(v0)²/R

Since F = F0 = F'

GMm/R² = m(v0)²/R

GM = (v0)²R  (1)

Also, he gravitational force on the satellite with speed vT at distance 3R is F1 = GMm/(3R)² = GMm/27R². This is also equal to the centripetal force on the satellite at 3R is F" = m(vT)²/3R

Since F1 = F'

GMm/27R² = m(vT)²/3R

GM = 27(vT)²R/3

GM = 9(vT)²R (2)

Equating (1) and (2),

(v0)²R = 9(vT)²R

dividing through by R, we have

9(vT)² = (v0)²

dividing through by 9, we have

(vT)² = (v0)²/9

taking square-root of both sides,

vT = v0/3

6 0
3 years ago
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