Which Astronomer estimated the relative size and distance of the moon and sun?

Suppose there are 10,000 civilizations in the Milky Way Galaxy. If the civilizations were randomly distributed throughout the di

vekshin1

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:

$A = \pi (\frac{D}{2})^2$

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 ;

$d= \frac{A}{N} =\frac{\pi (\frac{D}{2})^2 }{10, 000}$

The distance between each disk is further calculated by finding the radius of the density which is shown as follows:

$d = \pi r^2 e$

$r^2_e= \frac{d}{\pi}$

$r_e = \sqrt{\frac{d}{\pi} }$

replacing d = $\frac{\pi (\frac{D}{2})^2 }{10, 000}$ in the equation above; we have:

$r_e = \sqrt{\frac{\frac{\pi (\frac{D}{2})^2 }{10, 000}}{\pi} }$

$r_e = \sqrt{\frac{(\frac{D}{2})^2 }{10, 000}}$

$r_e = \sqrt{\frac{(\frac{100,000}{2})^2 }{10, 000}}$

$r_e = 500 ly$

The distance (s) between each civilization = $2(r_e)$

= 2 (500 ly)

= 1000 light-years (ly)

4 0

3 years ago

When you stand outside your are standing in the, Stratosphere. Troposphere. Mesosphere no

AVprozaik [17]

Uhhh I’m not really sure of the answer i think it’s stratosphere

8 0

3 years ago

A perpetual motion machine can never be built because it is not possible to eliminate

d1i1m1o1n [39]

Answer:

D. Friction

Explanation:

Friction is a force that opposes motion. So a perpetual motion machine can never be built because it is impossible to eliminate frictional force. It can only be reduced

8 0

2 years ago

What is the net force acting on the piano

CaHeK987 [17]

answer: 500 net force

5 0

3 years ago

Example: A wooden crate with mass 100kg is at rest on a stone floor. You know that the coefficients of kinetic and static fricti

alexgriva [62]

Answer

Any force greater 490N

Explanation

The force required just to make an object slide over a rough horizontal surface is any force greater that the static friction which given by;

$F=\mu_s mg.............(1)$

Given;

$\mu_s=0.5\\m=100kg\\g=9.8m/s^2$

Hence;

F = 0.5 x 100 x 9.8

F = 490N.

We will only need the coefficient of kinetic friction if we were asked to find the force required to keep the object moving uniformly. Usually, the force needed to keep an object moving uniformly over a rough surface is lesser that which is needed to start its motion.

In this problem, we were only asked to find the minimum force required to make the object move which we have done.

7 0

2 years ago