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Igoryamba
3 years ago
12

How many earths can fit on the moon

Physics
1 answer:
liberstina [14]3 years ago
8 0
Hello My Dear Friend!

Do you mean Sun or actually Moon?

If Moon is what you're looking for...then that is absolutely IMPOSSIBLE, due to that the Earth is WAY bigger than the Moon, therefore, not even HALF of the Earth would fit into the Moon since it's too small XD.

But if it's Sun...well then, t<span>he Answer is that it would take </span>1.3 million Earths<span> to fill up the Sun. That's a lot of Earths LOL XD

</span>I Hope my answer has come to your Help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead! :)

(Ps. Mark As Brainliest IF Helped!)

-TheOneAboveAll :D
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We don't know the change in velocity, so can't answer.

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3 years ago
"In a Young’s double-slit experiment, the separation between slits is d and the screen is a distance D from the slits. D is much
san4es73 [151]

Answer:

The number of bright fringes per unit width on the screen is, x=\dfrac{\lambda D}{d}      

Explanation:

If d is the separation between slits, D is the distance between the slit and the screen and \lambda is the wavelength of the light. Let x is the  number of bright fringes per unit width on the screen is given by :

x=\dfrac{n\lambda D}{d}

\lambda is the wavelength

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If n = 1,

x=\dfrac{\lambda D}{d}

So, the the number of bright fringes per unit width on the screen is \dfrac{\lambda D}{d}. Hence, the correct option is (B).

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3 years ago
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horrorfan [7]

Answer:

12 units

Explanation:

This problem can be solved if we take into account the equation for a sphere

x^{2}+y^{2}+z^{2}=r^{2}\\x^{2}+y^{2}+z^{2}=(\frac{26}{2})^{2}=13^{2}

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x^{2}+y^{2}+(5)^{5}=13^{2}\\x^{2}+y^{2}=144=(12)^{2}

where we have taken x2 +y2 because if the equation of a circunference.

In this case the intersection is made when we take z=5, for this value the sphere and the plane coincides in values.

Hence, the radius is 12 units

I hope this is useful for you

regards

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