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)
The potential across the capacitor at t = 1.0 seconds, 5.0 seconds, 20.0 seconds respectively is mathematically given as
- t=0.476v
- t=1.967v
- V2=4.323v
<h3>What is the potential across the capacitor?</h3>
Question Parameters:
A 1. 0 μf capacitor is being charged by a 9. 0 v battery through a 10 mω resistor.
at
- t = 1.0 seconds
- 5.0 seconds
- 20.0 seconds.
Generally, the equation for the Voltage is mathematically given as
v(t)=Vmax=(i-e^{-t/t})
Therefore
For t=1
V=5(i-e^{-1/10})
t=0.476v
For t=5s
V2=5(i-e^{-5/10})
t=1.967
For t=20s
V2=5(i-e^{-20/10})
V2=4.323v
Therefore, the values of voltages at the various times are
- t=0.476v
- t=1.967v
- V2=4.323v
Read more about Voltage
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Complete Question
A 1.0 μF capacitor is being charged by a 5.0 V battery through a 10 MΩ resistor.
Determine the potential across the capacitor when t = 1.0 seconds, 5.0 seconds, 20.0 seconds.
Answer:
HOPE THIS ANSWER WILL HELP YOU
The coefficient of linear expansion, given that the length of the pipe increased by 1.5 cm is 1.67×10¯⁵ /°F
<h3>How to determine the coefficient of linear expansion</h3>
From the question given above, the following data were obtained
- Original diameter (L₁) = 10 m
- Change in length (∆L) = 1.5 cm = 1.5 / 100 = 0.015 m
- Change in temperature (∆T) = 90 °F
- Coefficient of linear expansion (α) =?
The coefficient of linear expansion can be obtained as illustrated below:
α = ∆L / L₁∆T
α = 0.015 / (10 × 90)
α = 0.015 / 900
α = 1.67×10¯⁵ /°F
Thus, we can conclude that the coefficient of linear expansion is 1.67×10¯⁵ /°F
Learn more about coefficient of linear expansion:
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