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-Dominant- [34]
2 years ago
13

If the sun is 400 times bigger than the moon, how couild the moon possibly cover the sun during a solar eclipse?​

Physics
2 answers:
Vilka [71]2 years ago
6 0

Answer:

It's all about perspective. The moon is far closer to the Earth than the Sun is, so they appear roughly the same size. If they were closer to each other, then obviously the moon wouldn't be large enough to cover a substantial amount of the sun's light. But given the huge distance both between them and the moon and the earth, to us they look relatively the same.

AlekseyPX2 years ago
3 0

Explanation:

the Moon passes between Earth and the Sun Even though the Moon is much smaller than the Sun, because it is just the right distance away from Earth, the Moon can fully block the Sun's light from Earth's perspective This completely blocks out the Sun's light

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Answer:

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Explanation:

In the early 1800's a system for naming geologic time periods was devised using four periods of geologic time. They were named using Latin root words, Primary, Secondary, Tertiary and Quaternary. ... Keep in mind that this chart is focused on geologic time periods. There are also geologic Eons, Eras, and epochs.

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galina1969 [7]

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It takes 6400 years for one gram of radium to decay away to only 1/16 (one-sixteenth) of a gram. The half-life of radium is
Vsevolod [243]
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4 0
2 years ago
An advertisement for an all-terrain vehicle (ATV) claims that the ATV can climb inclined slopes of 35°. What is the minimum coef
navik [9.2K]

An advertisement for an all-terrain vehicle (ATV) claims that the ATV can climb inclined slopes of 35°. The minimum coefficient of static friction needed for this claim to be possible is 0.7

In an inclined plane, the coefficient of static friction is the angle at which an object slide over another.  

As the angle rises, the gravitational force component surpasses the static friction force, as such, the object begins to slide.

Using the Newton second law;

\sum F_x = \sum F_y = 0

\mathbf{mg sin \theta -f_s= N-mgcos \theta = 0 }

  • So; On the L.H.S

\mathbf{mg sin \theta =f_s}

\mathbf{mg sin \theta =\mu_s N}

  • On the R.H.S

N = mg cos θ

Equating both force component together, we have:

\mathbf{mg sin \theta =\mu_s \ mg \ cos \theta}

\mathbf{sin \theta =\mu_s \ \ cos \theta}

\mathbf{\mu_s = \dfrac{sin \theta }{ cos \theta}}

From trigonometry rule:

\mathbf{tan \theta= \dfrac{sin \theta }{ cos \theta}}

∴

\mathbf{\mu_s =\tan \theta}}

\mathbf{\mu_s =\tan 35^0}}

\mathbf{\mu_s = 0.700}}

Therefore, we can conclude that the minimum coefficient of static friction needed for this claim to be possible is 0.7

Learn more about static friction here:

brainly.com/question/24882156?referrer=searchResults

8 0
2 years ago
A playground carousel is free to rotate about its center on frictionless bearings, and air resistance is negligible. The carouse
Sidana [21]

Answer:

m = 35.98 Kg ≈ 36 Kg

Explanation:

I₀ = 125 kg·m²

R₁ = 1.50 m

ωi = 0.600 rad/s

R₂ = 0.905 m

ωf = 0.800 rad/s

m = ?

We can apply The law of conservation of angular momentum as follows:

Linitial = Lfinal

⇒    Ii*ωi = If*ωf   <em>(I)</em>

where    

Ii = I₀ + m*R₁² = 125 + m*(1.50)² = 125 + 2.25*m

If = I₀ + m*R₂² = 125 + m*(0.905)² = 125 + 0.819025*m

Now, we using the equation <em>(I) </em>we have

(125 + 2.25*m)*0.600 = (125 + 0.819025*m)*0.800

⇒  m = 35.98 Kg ≈ 36 Kg

5 0
3 years ago
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