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)
Amplitude: How dense the medium is in the compression part of the wave, and how empty the rarefied area is.
Frequency: The number of wavelengths that pass a position in 1 second.
loudness: The quality of the sound that is most closely linked to the amplitude of the sound wave.
Period: The amount of time that it takes one wavelength to pass by a position.
Pitch: The quality of the sound that is most closely linked to the frequency of the sound wave.
Answer:
The shortest braking distance is 35.8 m
Explanation:
To solve this problem we must use Newton's second law applied to the boxes, on the vertical axis we have the norm up and the weight vertically down
On the horizontal axis we fear the force of friction (fr) that opposes the movement and acceleration of the train, write the equation for each axis
Y axis
N- W = 0
N = W = mg
X axis
-Fr = m a
-μ N = m a
-μ mg = ma
a = μ g
a = - 0.32 9.8
a = - 3.14 m/s²
We calculate the distance using the kinematics equations
Vf² = Vo² + 2 a x
x = (Vf² - Vo²) / 2 a
When the train stops the speed is zero (Vf = 0)
Vo = 54 km/h (1000m/1km) (1 h/3600s)= 15 m/s
x = ( 0 - 15²) / 2 (-3.14)
x= 35.8 m
The shortest braking distance is 35.8 m
Answer: 117 kPa
Explanation:
For the liquid at depth 3 m, the gauge pressure is equal to = P₁=39 kPa
For the liquid at depth 9m, the gauge pressure is equal to= P₂
Now we are given the condition that the liquid is same. That must imply that the density must be same throughout the depth.
So, For finding gauge pressure we have formula P= ρ * g * h
Also gravity also remains same for both liquids
So taking ratio of their respective pressures we have
= 
So
= 
Or P₂= 39 * 3 = 117 kPa