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Rudik [331]
3 years ago
9

How does being on the moon effect gravitational field strength

Physics
1 answer:
SVETLANKA909090 [29]3 years ago
5 0

Your being on the moon has no effect on the moon's
gravitational field strength, or on the Earth's for that
matter.

However, YOU notice a change on YOU when YOU move
from one to the other, because of the effect of the gravitational
field strength on you and your internal organs.

If you could stand on the moon, you would experience an incredible
sense of lightness, since the forces of attraction between the moon
and anything else are only 16% as great as the same forces are on
Earth.

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What is the formula for calcium chloride
Naddik [55]

The correct answer is - CaCl2

The calcium chloride is a salt, an inorganic compound. Its formula is CaCl2, with Ca being calcium, Cl being chloride, and the number 2 representing the number of chloride molecules.

The calcium chloride is a white colored crystalline solid when it is at room temperature, and it is highly soluble in water, acetone, and acetic acid. It has a molar mass of 110.98 g/mol, density of 2.15 g/cm³, and melting point at 772 °C.

6 0
3 years ago
Read 2 more answers
A(n) _____ is an organic compound that changes color in acid or bases.
Romashka-Z-Leto [24]
The answer is indicator.
6 0
3 years ago
PLS HELP
RoseWind [281]

Answer:

A and C are correct.

4 0
3 years ago
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The period of a satellite circling planet Nutron is observed to be 84 s when it is in a circular orbit with a radius of 8.0 x 10
antoniya [11.8K]

Answer:

4.3 * 10^28 kg

Explanation:

Given:

Period, T = 84s

Radius of satellite orbit, r = 8*10^6

Using the relation :

M = 4π²r³ / GT²

Where G = Gravitational constant, 6.67 * 10^-11

M = 4*π^2*(8*10^6)^3 / 6.67 * 10^-11 * 84^2

M = (20218.191872 * 10^18) / 47063.52 * 10^-11

M = 0.4295937 * 10^18 - (-11)

M = 0.4295937 * 10^29

M = 4.295937 * 10^28 kg

M = 4.3 * 10^28 kg

Mass of planet Nutron = 4.3 * 10^28 kg

8 0
2 years ago
A total charge of 9.0 mC passes through a cross-sectional area of a nichrome wire in 3.6s. The number of electrons passing throu
Setler79 [48]
<h2>Given :</h2>

  • total charge = 9.0 mC = 0.009 C

Each electron has a charge of :

1.6 \times 10 {}^{ - 19} \:  C

For producing 1 Cuolomb charge we need :

  • \mathrm{\dfrac{1}{1.6 \times 10 {}^{ - 19} } }

  • \dfrac{10 {}^{19} }{1.6}

  • \dfrac{10\times 10 {}^{19} }{16}

  • \dfrac{100 \times 10 {}^{18} }{16}

  • \mathrm{6.24 \times 10 {}^{18}  \:  \: electrons}

Now, for producing 0.009 C of charge, the number of electrons required is :

  • 0.009 \times 6.24 \times  {10}^{18}

  • 0.05616 \times 10 {}^{18}

  • \mathrm{5.616 \times 10 {}^{16}  \:  \: electons}

_____________________________

So, Number of electrons passing through the cross section in 3.6 seconds is :

\mathrm{5.616 \times 10 {}^{16} \:  \: electrons}

Number of electrons passing through it in 1 Second is :

  • \dfrac{5.616 \times  {10}^{16} }{3.6}

  • \mathrm{1.56 \times 10 {}^{16}  \:  \: electrons}

Now, in 10 seconds the number of electrons passing through it is :

  • 10 \times  \mathrm{1.56 \times 10 {}^{16}  \:  \: }

  • \mathrm{1.56 \times 10 {}^{17}  \:  \: electrons}

_____________________________

\mathrm{ \#TeeNForeveR}

6 0
2 years ago
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