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3 years ago

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]3 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.

AlekseyPX3 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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Over a time interval of 1.99 years, the velocity of a planet orbiting a distant star reverses direction, changing from +20.7 km/

madam [21]

Answer:

(a) - 42700 m/s

(b) - 6.8 x 10^-4 m/s^2

Explanation:

initial velocity of star, u = 20.7 km/s

Final velocity of star, v = - 22 km/s

time, t = 1.99 years

Convert velocities into m/s and time into second

So, u = 20700 m / s

v = - 22000 m/s

t = 1.99 x 365.25 x 24 x 3600 = 62799624 second

(a) Change in planet's velocity = final velocity - initial velocity

= - 22000 - 20700 = - 42700 m/s

(b) Accelerate is defined as the rate of change of velocity.

Acceleration = change in velocity / time

= ( - 42700 ) / (62799624) = - 6.8 x 10^-4 m/s^2

8 0

3 years ago

A small block with mass 0.0400 kg slides in a vertical circle of radius 0.600 m on the inside of a circular track. During one of

Maurinko [17]

Answer:

Explanation:

Given that

The mass of the body is 0.04kg

M=0.04kg

The radius of the paths is 0.6m

r=0.6m

The normal force exerted at A is 3.9N

Fa=3.9N

The normal force exerted at B is 0.69N

Fb=0.69N

Then work done by friction from point A to B will be the change in K.E

W=∆K.E+P.E

So we need to know the velocity at both point A and B

Then since the centripetal force is given as

Ft=mv²/r

Then,

For point A

Fa=mv²/r

3.9=0.04v²/0.6

3.9=0.0667v²

v²=3.9/0.0667

v²=58.5

v=√58.5

v=7.65m/s

Va=7.65m/s

Now at point B

Fb=mv²/r

0.69=0.04v²/0.6

0.69=0.0667v²

v²=0.69/0.0667

v²=10.35

v=√10.35

v=3.22m/s

Vb=3.22m/s

Then, the work done is

W=∆K.E+P.E

P.E is given as mgh

The height will be 2R =1.2m

P.E=mgh

P.E=0.04×9.81×1.2

P.E=0.471J

Final kinetic energy at B minus initial kinetic energy at A

W=K.Eb-K.Ea

K.E is given as 1/2mv²

W=1/2m(Vb²-Va²) +P.E

W=0.5×0.04(3.22²-7.65²) +0.471

W=0.5×0.04×(-48.1541) +0.471

W=-0.96+0.471

W=-0.49J

work was done on the block by friction during the motion of the block from point A to point B is 0.49J.

Friction opposes motions and that is why the work done is negative

3 0

3 years ago

According to the law of conservation of energy, the total amount of energy in the universe does not change, but remains constant

Evgen [1.6K]

<span>The statement : According to the law of conservation of energy, the total amount of energy in the universe does not change, but remains constant - is TRUE. The main point which led me to decide is that our universe is the example of a closed system which means there's nothing inside. Hope you will agree with my opinion.
</span>

6 0

3 years ago

Read 2 more answers

Some planetary scientists have suggested that the planet Mars has an electric field somewhat similar to that of the earth, produ

aalyn [17]

Answer:

324795 C

252.637820565 N/C

$2.235844712\times 10^{-9}\ C/m^2$

Explanation:

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

R = Radius of Mars = $3.4\times 10^6\ m$

A = Area = $4\pi R^2$

$\phi$ = Electric flux = $3.67\times 10^{16}\ Nm^2/C$

Electric flux is given by

$\phi=\dfrac{q}{\epsilon_0}\\\Rightarrow q=\phi\epsilon_0\\\Rightarrow q=3.67\times 10^{16}\times 8.85\times 10^{-12}\\\Rightarrow q=324795\ C$

The charge is 324795 C

Electric field is given by

$E=\dfrac{\phi}{A}\\\Rightarrow E=\dfrac{3.67\times 10^{16}}{4\pi (3.4\times 10^6)^2}\\\Rightarrow E=252.637820565\ N/C$

The electric field is 252.637820565 N/C

Surface charge density is given by

$\sigma=\dfrac{q}{4\pi R^2}\\\Rightarrow \sigma=\dfrac{324795}{4\pi (3.4\times 10^6)^2}\\\Rightarrow \sigma=2.235844712\times 10^{-9}\ C/m^2$

The surface charge density is $2.235844712\times 10^{-9}\ C/m^2$

6 0

3 years ago

An airplane is flying with a velocity of 100 m/s at an angle of 25° above the horizontal. When the plane is 114 m directly above

alex41 [277]

Answer:

The suitcase will land 976.447m from the dog.

Explanation:

The velocity in its component in the X and Y axis is decomposed:

Vx= 100m/s × cos(25°)= 90.63m/s

Vy= 100m/s × sen(25°)= 42.26m/s

Time it takes for the suitcase to reach maximum height, the final speed on the axis and at the point of maximum height is zero whereby:

VhmaxY= Voy- 9.81(m/s^2) × t ⇒ t= (42.26 m/s) / (9.81(m/s^2)) = 4.308s

The space traveled on the axis and from the moment the suitcase is thrown until it reaches its maximum height will be:

Dyhmax= Voy × t - (1/2) × 9.81(m/s^2) × (t^2) =

= 42.26m/s × 4.308s - 4.9 (m/s^2) × (4.308s)^2 =

=182.056m - 90.938m= 91.118m

The time from the maximum height to touching the ground is:

Dtotal y= 114m + 91.118m = (1/2) × 9.81(m/s^2) × (t^2) =

= 205.118m = 4.9 (m/s^2) × (t^2) ⇒ t= (41.818 s^2) ^ (1/2)= 6.466s

The total time of the bag in its rise and fall will be:

t= 4.308s + 6.466s = 10.774s

With this time and the initial velocity at x which is constant I can obtain the distance traveled by the suitcase on the x-axis:

Dx= 90.63 (m/s) × 10.774s = 976.447m

8 0

4 years ago

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

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