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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:

4 0
2 years ago
Solve -7= sqrt 2x-9​
Sauron [17]

Answer:

x = 2

Explanation:

if it was -7 = the square root of both 2x-9 together, it would be false.

if it was square root of just 2x in the equation, the answer is:

x = 2

°°°°°°°°°

-7 = √2x - 9

-√2x = -9 + 7

√-2x = -2

√2x = 2

2x = 4

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4 0
3 years ago
A comet is in an elliptical orbit around the sun. its closest approach to the sun is a distance of 4.5 1010 m (inside the orbit
barxatty [35]

r1 = 5*10^10 m , r2 = 6*10^12 m

v1 = 9*10^4 m/s

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K1 +U1 = K2 +U2

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0.5v1^2 - GM/r1 = 0.5v2^2 - GM/r2

M is mass of sun = 1.98*10^30 kg

G = 6.67*10^-11 N.m^2/kg^2

0.5*(9*10^4)^2 - (6.67*10^-11*1.98*10^30/(5*10^10)) = 0.5v2^2 - (6.67*10^-11*1.98*10^30/(6*10^12))

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An object is made of glass and has the shape of a cube 0.13 m on a side, according to an observer at rest relative to it. Howeve
Vlad1618 [11]

Answer:

v=0.9833\ c

Explanation:

The density changes means that the length in the direction of the motion is changed.

Therefore,

$\text{Density} = \frac{m}{lwh}$

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Mass, m = 3.3 kg

Density = 8100 kg/m^3

So,

$8100=\frac{3.3}{l \times 0.13 \times 0.13}$

$l=\frac{3.3}{8100 \times 0.13 \times 0.13}$

l = 0.024 m

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$l= l' \sqrt{1-\frac{v^2}{c^2}}$

$0.024= 0.13 \sqrt{1-\frac{v^2}{c^2}}$

$0.184=  \sqrt{1-\frac{v^2}{c^2}}$

$0.033=  1-\frac{v^2}{c^2}}$

$\frac{v^2}{c^2}= 0.967$

$\frac{v}{c}=0.9833$

v=0.9833\ c

Therefore, the speed of the observer relative to the cube is 0.9833 c (in the units of c).

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