How many earths can fit on the moon
1 answer:
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
We hope to answer more of your questions and inquiries soon. Have a nice day ahead! :)
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
You might be interested in
The acceleration of the wagon is found by applying Newton's Second Law of motion.
1. The responses for question 1 are;
- x-component of the tension in the rope on the right is approximately <u>91.93 N</u>
- y-component of the tension in the rope on the right is approximately <u>71.135 N</u>
- x-component of the tension in the rope on the left is -80.0 N
- y-component of the tension in the rope on the left is 0
2. The net force in the x-direction is approximately <u>11.93 N</u>
3. The net acceleration of the wagon in the horizontal direction is approximately <u>0.1193 m/s²</u>.
Reasons:
The given parameters are;
Mass of the wagon, m = 100.0 kg
Angle of inclination to the horizontal of the rope to the right, θ = 40.0°
Tension in the rope on the right = 120.0 N
Direction in which the rope on the left is pulled = To the west
Tension in the rope on the left = 80.0 N
1. The <em>x</em> and <em>y</em> component of the tension in the rope on the right are;
x-component = 120.0 N × cos(40.0°) ≈ <u>91.93 N</u>
y-component = 120.0 N × sin(40.0°) ≈ <u>77.135 N</u>
The <em>x</em> and <em>y</em> component of the tension in the rope on the left are;
x-component = 80.0 N × cos(180°) = <u>-80.0 N</u>
y-component = 80.0 N × sin(180°) = <u>0.0 N</u>
2. The net force in the horizontal direction, Fₓ, is found as follows;
Fₓ = The x-component of the rope on the left + The x-component of the rope on the right
Which gives;
Fₓ = 91.93 N - 80.0 N = <u>11.93 N</u>
3. The net acceleration of the block is given as follows;
According to Newton's Second Law of motion, we have;
Force in the horizontal direction, Fₓ = Mass of wagon, m × Acceleration of the wagon in the horizontal direction, aₓ
Fₓ = m × aₓ
Therefore;
- The acceleration of the wagon in the horizontal direction, aₓ ≈ <u>0.1193 m/s²</u>.
Learn more here: