Here is the full question
Suppose there are 10,000 civilizations in the Milky Way Galaxy. If the civilizations were randomly distributed throughout the disk of the galaxy, about how far (on average) would it be to the nearest civilization?
(Hint: Start by finding the area of the Milky Way's disk, assuming that it is circular and 100,000 light-years in diameter. Then find the average area per civilization, and use the distance across this area to estimate the distance between civilizations.)
Answer:
1000 light-years (ly)
Explanation:
If we go by the hint; The area of the disk can be expressed as:
where D = 100, 000 ly
Let's divide the Area by the number of civilization; if we do that ; we will be able to get 'n' disk that is randomly distributed; so ;
The distance between each disk is further calculated by finding the radius of the density which is shown as follows:
replacing d = in the equation above; we have:
The distance (s) between each civilization =
= 2 (500 ly)
= 1000 light-years (ly)
Answer:
Moreover, Boss says that even if Jupiter is proven to have a core, the planet still could have formed that core through disk instability. Enough dust could have collected and cemented together in the dense gas to form a core many times larger than the size of the Earth.
Explanation:
The same is true of most other objects in the solar system — except Jupiter. The gas giant is so big that it pulls the center of mass between it and the sun, also known as the barycenter, some 1.07 solar radii from the star's center — which is about 30,000 miles above the sun's surface.
69,911 km
69,911 kmJupiter/Radius
The minimum number of tickets that could admit all of them is six (6).
This thing is impossible to explain in words, so I shall attempt it with a diagram:
Here are the six ladies:
( A ) ( B )
| |
| |
( C ) ( D )
| |
| |
( E ) ( F )
-- 'E' and 'F' are the daughters of 'C' and 'D' .
-- 'C' and 'D' are the daughters of 'A' and 'B' .
So look what we have now:
-- 'A' and 'B' are the mothers of 'C' and 'D' .
There's 2 of the mothers.
-- 'C' and 'D' are the mothers of 'E' and 'F' .
There's the OTHER 2 mothers.
-- 'A' and 'B' are the grandmothers of 'E' and 'F' .
There's the 2 grandmothers.
-- 'E' and 'F' are the daughters of 'C' and 'D' .
There's 2 of the daughters.
-- 'C' and 'D' are the daughters of 'A' and 'B' .
There's the OTHER 2 daughters.
You want to know what ? !
The group is even bigger than THAT.
There are also 2 GRAND-daughters in the family ... 'E' and 'F' .
So now you have a list of 12 people ! ... 4 mothers, 2 grandmothers,
4 daughters, and 2 grand-daughters ... and they all get in to the
Christmas Market with only six tickets. Legally !
Such a deal !
Don't forget : Christmas this year is also the first day of Chanukah !
All for the same price !
Answer:
The answer would be B) The Same
Explanation:
Not gonna lie I checked my class notes but I figured this would help :)
Good luck!!!
A heavy truck moving a 30 mph. It has more mass.