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Sonbull [250]
4 years ago
13

The atoms in the answer choices have at least 2 electron shells each. Which atom would tend to lose 1 valence electron to anothe

r atom in order to become stable?
A) C
B) O
C) Na
D) Ar
Physics
1 answer:
Afina-wow [57]4 years ago
5 0

Your answer is D) Ar

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A shopper walks westward 5.4 meters and then eastward 7.8 meters
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Answer:

13.2 meters

Explanation:

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Can you please help quick
Mademuasel [1]

Answer:

1. the difference between a series circuit and parallel are if a bulb goes out in a series circuit all go out and in a parallel if one goes out all the others stay on.

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magine an astronaut on an extrasolar planet, standing on a sheer cliff 50.0 m high. She is so happy to be on a different planet,
Mama L [17]

Answer:

\Delta t=(\frac{20}{g'}+\sqrt{\frac{400}{g'^2}+\frac{100}{g'}  }  )-(\frac{20}{g}+\sqrt{\frac{400}{g^2}+\frac{100}{g}  }  )

Explanation:

Given:

height above which the rock is thrown up, \Delta h=50\ m

initial velocity of projection, u=20\ m.s^{-1}

let the gravity on the other planet be g'

The time taken by the rock to reach the top height on the exoplanet:

v=u+g'.t'

where:

v= final velocity at the top height = 0 m.s^{-1}

0=20-g'.t' (-ve sign to indicate that acceleration acts opposite to the velocity)

t'=\frac{20}{g'}\ s

The time taken by the rock to reach the top height on the earth:

v=u+g.t

0=20-g.t

t=\frac{20}{g} \ s

Height reached by the rock above the point of throwing on the exoplanet:

v^2=u^2+2g'.h'

where:

v= final velocity at the top height = 0 m.s^{-1}

0^2=20^2-2\times g'.h'

h'=\frac{200}{g'}\ m

Height reached by the rock above the point of throwing on the earth:

v^2=u^2+2g.h

0^2=20^2-2g.h

h=\frac{200}{g}\ m

The time taken by the rock to fall from the highest point to the ground on the exoplanet:

(50+h')=u.t_f'+\frac{1}{2} g'.t_f'^2 (during falling it falls below the cliff)

here:

u= initial velocity= 0 m.s^{-1}

\frac{200}{g'}+50 =0+\frac{1}{2} g'.t_f'^2

t_f'^2=\frac{400}{g'^2}+\frac{100}{g'}

t_f'=\sqrt{\frac{400}{g'^2}+\frac{100}{g'}  }

Similarly on earth:

t_f=\sqrt{\frac{400}{g^2}+\frac{100}{g}  }

Now the required time difference:

\Delta t=(t'+t_f')-(t+t_f)

\Delta t=(\frac{20}{g'}+\sqrt{\frac{400}{g'^2}+\frac{100}{g'}  }  )-(\frac{20}{g}+\sqrt{\frac{400}{g^2}+\frac{100}{g}  }  )

3 0
3 years ago
Al2(SO4)3<br> Name of this?
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The answer to your question is Aluminum sulfate  also plz mark brainliest

5 0
3 years ago
Earth pulls on the Moon with a gravitational force, keeping it in its nearly circular, 27‑day orbit. According to Newton's third
arlik [135]

Answer:

Centripetal Force

Explanation:

The reaction to the earth's pull on the moon is the centripetal force

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3 years ago
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