Fermi Questions C

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Unome
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Fermi Questions C

Postby Unome » September 5th, 2017, 4:25 am

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How fast, in decimeters per hour squared, would a 2017 Volkswagen Beetle, which is currently orbiting the Earth at 700 km altitude, have to be accelerating in order to use as much power as the sun emits?
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Re: Fermi Questions C

Postby WhatScience? » September 7th, 2017, 5:02 pm

People are actually expected to figure this stuff out....

:lol: :lol: :lol: :lol:
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Re: Fermi Questions C

Postby whythelongface » September 7th, 2017, 8:38 pm

WhatScience? wrote:People are actually expected to figure this stuff out....

:lol: :lol: :lol: :lol:


From what I can tell, this is definitely from the more difficult end. Easier ones involve much simpler premises and conditions, i.e. what is the distance, in megaparsecs, that a snail travels in the amount of time it takes Pluto to make a full revolution around the sun or something like that.
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Re: Fermi Questions C

Postby jkang » September 7th, 2017, 9:35 pm

Unome wrote:And we return!

How fast, in decimeters per hour squared, would a 2017 Volkswagen Beetle, which is currently orbiting the Earth at 700 km altitude, have to be accelerating in order to use as much power as the sun emits?

Maybe
20?
We can say P*t = 1/2m*v^2.
Solving for v, v = sqrt(2P*t/m)
Differentiating wrt t, a = sqrt(P/2mt)
The luminosity of the sun is ~10^26 Watts, and I'd imagine a smaller car like the beetle would weigh a ton, or about 900 kg. Assuming we're trying to get the power output of the beetle to match that of the Sun, we set t = 1 second and find:
sqrt((10^26 kg*m^2/s^3)/(2*900 kg*1 s) ~ sqrt((10^26 kg*m^2/s^3)/(2*10^3 kg*s) ~ sqrt(10^22) ~ 11 m/s^2.
There's 10 meters in a decimeter and 3600 seconds in an hour, so 1 m/s^2 ~ 1e8 dm/s^2.
11 + 8 = 19, and counting all the rounding and numbers I've dropped, the answer might be closer to 20 so I'll go with that. That said, I'd imagine the car would explode long before it got anywhere near this level of acceleration. Also in this analysis I'm basically ignoring the altitude since that's basically the exosphere and unless you wanted me to talk about the car gaining even more potential energy from it falling back down or possibly decreased efficiency in the engine, but that's wayyy too much work beyond the simple algebra I'm willing to do
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Re: Fermi Questions C

Postby NeilMehta » September 8th, 2017, 1:01 pm

whythelongface wrote:
WhatScience? wrote:People are actually expected to figure this stuff out....

:lol: :lol: :lol: :lol:


From what I can tell, this is definitely from the more difficult end. Easier ones involve much simpler premises and conditions, i.e. what is the distance, in megaparsecs, that a snail travels in the amount of time it takes Pluto to make a full revolution around the sun or something like that.

Unome has high standards :P

Also, new question:
How many copies of Campbell's Biology would it take to cover the state of Texas?
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Re: Fermi Questions C

Postby whythelongface » September 8th, 2017, 4:28 pm

NeilMehta wrote:
whythelongface wrote:
WhatScience? wrote:People are actually expected to figure this stuff out....

:lol: :lol: :lol: :lol:


From what I can tell, this is definitely from the more difficult end. Easier ones involve much simpler premises and conditions, i.e. what is the distance, in megaparsecs, that a snail travels in the amount of time it takes Pluto to make a full revolution around the sun or something like that.

Unome has high standards :P

Also, new question:
How many copies of Campbell's Biology would it take to cover the state of Texas?


Let me take a stab:
Answer?
Campbell's Biology is slightly larger than an 8.5" x 11" sheet of paper, yielding an area of maybe about 5.5E2cm^2.
Assuming the state of Texas is the area of half a square 1000km wide, that's 5E5km^2 or 5E15cm^2.
Dividing, we get a Fermi answer of 13.
Google gives me 1.024E13, so it looks like I was correct.


New question:
What is the mass, in kilograms, of a stack of quarters stretching from the sun to the center of the Andromeda galaxy?
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Re: Fermi Questions C

Postby Raleway » September 8th, 2017, 7:05 pm

C'mon (Fox reference from SSBM weeb I know clarification)
Distance is approx let's say 10^25 m (Adi don't kill me I don't know Astro at all). Each quarter is let's say 1 mm thick, and approximately 6 grams. That's 10^28 quarters times 6 would round it up to 29 as a fermi answer.

Final answer: 29

I'm bad at math let's not try this again
Last edited by Raleway on September 8th, 2017, 7:06 pm, edited 2 times in total.
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Re: Fermi Questions C

Postby NeilMehta » September 8th, 2017, 7:05 pm

whythelongface wrote:
NeilMehta wrote:
whythelongface wrote:
From what I can tell, this is definitely from the more difficult end. Easier ones involve much simpler premises and conditions, i.e. what is the distance, in megaparsecs, that a snail travels in the amount of time it takes Pluto to make a full revolution around the sun or something like that.

Unome has high standards :P

Also, new question:
How many copies of Campbell's Biology would it take to cover the state of Texas?


Let me take a stab:
Answer?
Campbell's Biology is slightly larger than an 8.5" x 11" sheet of paper, yielding an area of maybe about 5.5E2cm^2.
Assuming the state of Texas is the area of half a square 1000km wide, that's 5E5km^2 or 5E15cm^2.
Dividing, we get a Fermi answer of 13.
Google gives me 1.024E13, so it looks like I was correct.


New question:
What is the mass, in kilograms, of a stack of quarters stretching from the sun to the center of the Andromeda galaxy?

attempt
I thankfully remember that Andromeda is about 10 million light years away (E7)
I also remember that a light year is E13 km
Quarter in width is just about 1mm, so E6 quarters to make a km and each quarter is E1 grams
Altogether that makes E26

oops I messed up not rly
Apparently Andromeda is actually E6 away but quarters are actually E2 grams so they cancel out


New question:
How many paper airplanes woudm it take to equal the mass of one airplane? (use Boeing 777 as an example)

EDIT: raleway posted before me but didn't put a solution or anything so I'll leave my post here for solution plus next question
EDIT 2: spelling
Last edited by NeilMehta on September 8th, 2017, 7:13 pm, edited 2 times in total.
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Re: Fermi Questions C

Postby Raleway » September 8th, 2017, 7:12 pm

For Neil (Lemme not fail again at units)

Assume one plane is like 1 million pounds. I read on a study of paper that 100 sheets of standard A4 paper weigh exactly 1 pound... so the fermi answer would be 8?

Final Answer: 8
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Re: Fermi Questions C

Postby Raleway » September 8th, 2017, 7:17 pm

How many times would a violet light (assume perfect light) wrap around the universe if it traveled for Googleplex to the Googleplex lunar months?
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Re: Fermi Questions C

Postby whythelongface » September 9th, 2017, 7:37 am

If I am interpreting this properly, you want the distance light travels in Googolplex^Googolplex lunar months, and how many times that would wrap around an arbitrary equator of the visible universe.

If that's the case, I really can't see how you'd express that as a Fermi answer, unless you accept 10^(100+10^100)-12 .

More likely than not though I'm just reading this problem wrong.
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Re: Fermi Questions C

Postby NeilMehta » September 9th, 2017, 8:27 am

Raleway wrote:How many times would a violet light (assume perfect light) wrap around the universe if it traveled for Googleplex to the Googleplex lunar months?

https://www.wolframalpha.com/input/?i=g ... googolplex
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Re: Fermi Questions C

Postby Alex-RCHS » September 9th, 2017, 2:30 pm

Raleway wrote:How many times would a violet light (assume perfect light) wrap around the universe if it traveled for Googleplex to the Googleplex lunar months?

I'll admit, I had to look up the radius of the observable universe. I got:
this
googleplex to the googleplex, as a fermi number, would be 1000googleplex. That's how many months light would be traveling. Divide by 12 to get the distance in light years and you get 999googleplex as a fermi number. Divide by the radius of the observable universe (46 billion light years) and you get 989googleplex as your fermi answer.


Not 100% sure I did that right.

A horse walks into a bar. How fast would he have to be walking into the bar in order to have the equivalent amount of (kinetic) energy as 10 million chickens crossing the road at normal chicken speed?
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Re: Fermi Questions C

Postby whythelongface » September 9th, 2017, 7:14 pm

Alex-RCHS wrote:
Raleway wrote:How many times would a violet light (assume perfect light) wrap around the universe if it traveled for Googleplex to the Googleplex lunar months?

I'll admit, I had to look up the radius of the observable universe. I got:
this
googleplex to the googleplex, as a fermi number, would be 1000googleplex. That's how many months light would be traveling. Divide by 12 to get the distance in light years and you get 999googleplex as a fermi number. Divide by the radius of the observable universe (46 billion light years) and you get 989googleplex as your fermi answer.


Not 100% sure I did that right.

A horse walks into a bar. How fast would he have to be walking into the bar in order to have the equivalent amount of (kinetic) energy as 10 million chickens crossing the road at normal chicken speed?


I don't think that's right, but whatever.

Answer
Let's assume a horse masses about E3 kg, and that a chicken massess around E1 kg. Ten million chickens would therefore mass E8 kg. KE = 0.5mv^2, Assume chickens move at E-1 m/s. The kinetic energy of ten million chickens would therefore be about 5E5. This means the the horse's velocity squared would be about E3 (m/s)^2, and so the Fermi answer would be 2, if we are measuring in meters per second. If we're lucky, maybe we can have supersonic horses by the end of this century.


Supposing pencils can be made out of any wood (not just cedar), if I cut down the entire Amazon Rain Forest and used the wood to solely manufacture pencils, and used them to draw one, straight, unbroken line, how long would the line be, in Angstroms?
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Re: Fermi Questions C

Postby NeilMehta » September 11th, 2017, 5:14 am

I doubt I could ever solve this without info, so I'll post the solution for anyone who wants to check their work:

BREAK GLASS IN CASE OF EMERGENCY
Solution: 31
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