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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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A rocket moves upward, starting from rest with an acceleration of +30.0 m/s2 for 5.00 s. It runs out of fuel at the end of this
olchik [2.2K]

Answer:

The distance covered by the rocket after fuel ran out is 3442.04 m

Explanation:

Given that the rocket moves with an acceleration a=30m/s^2

time t=5 s

Since the rocket starts from rest initial velocity  u=0 s

The distance it travelled within this time is given by  s=ut+ \frac{1}{2} at^2                                                                                                  =0 \times 5+ \frac{1}{2} (30\times25)=375 m

Velocity at this point is given by v=u+at

v=0+30\times5=150m/s

Given that at this height it runs out of fuel but travels further. Here final velocity v=0(maximum height), initial velocityu=150 m/s  and time to zero velocity t=\frac{v}{g} = \frac{150}{9.8} =15.3 s.

Thus it travels 15.3 seconds more after fuel running out. The distance covered during this period is given

s= ut+\frac{1}{2} gt^2=150 \times 15.3+1/2 \times9.8 \times 15.3^2=3442.04 m

7 0
3 years ago
CAN AIR MAKE SHADOWS?
Tems11 [23]

Answer:

No because you cannot see air so therefore it cannot make shadows

Explanation:

3 0
3 years ago
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What is occurring when a light wave goes through a pane of glass in a window?
ycow [4]

Answer:

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Explanation:

5 0
2 years ago
A 4.00-m-long, 470 kg steel beam extends horizontally from the point where it has been bolted to the framework of a new building
adell [148]

Answer:

12164.4 Nm

Explanation:

CHECK THE ATTACHMENT

Given values are;

m1= 470 kg

x= 4m

m2= 75kg

Cm = center of mass

g= acceleration due to gravity= 9.82 m/s^2

The distance of centre of mass is x/2

Center of mass(1) = x/2

But x= 4 m

Then substitute, we have,

Center of mass(1) = 4/2 = 2m

We can find the total torque, through the summation of moments that comes from both the man and the beam.

τ = τ(1) + τ(2)

But

τ(1)= ( Center of m1 × m1 × g)= (2× 470× 9.81)

= 9221.4Nm

τ(2)= X * m2 * g = ( 4× 75 × 9.81)= 2943Nm

τ = τ(1) + τ(2)

= 9221.4Nm + 2943Nm

= 12164.4 Nm

Hence, the magnitude of the torque about the point where the beam is bolted into place is 12164.4 Nm

6 0
3 years ago
Directions: Using the T-chart below, compare balanced forces and unbalanced forces.
Salsk061 [2.6K]

Explanation:

unbalanced: a turning vehicle, apple falling on the ground, kicking a ball

balanced: floating on water, fruit hanging from tree, tug of war equally balanced teams

8 0
2 years ago
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