Fermi Questions Marathon

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Fermi Questions Marathon

Postby Jim_R » October 3rd, 2012, 5:55 am

This is a thread to post fermi questions and then answer them.
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Re: Fermi Questions Marathon

Postby Schrodingerscat » October 3rd, 2012, 6:30 am

At a given instant, how many photons are in a room of your choice (with electric lighting turned on)?

*Also consider only the photons emitted by the electric lights, ignoring thermal radiation of other objects, virtual photons, etc.

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Re: Fermi Questions Marathon

Postby Slarik » October 3rd, 2012, 7:56 am

At a given instant, how many photons are in a room of your choice (with electric lighting turned on)?

*Also consider only the photons emitted by the electric lights, ignoring thermal radiation of other objects, virtual photons, etc.
After someone answers this, can you please explain it? I guess you're going to have to try to figure out how many photons a roomful of typical lightbulb emits during the time it would take a photon to be absorbed... is that along the right lines? But I have no idea how to calculate it. Would you learn about that in physics?
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Re: Fermi Questions Marathon

Postby Schrodingerscat » October 3rd, 2012, 9:08 am

At a given instant, how many photons are in a room of your choice (with electric lighting turned on)?

*Also consider only the photons emitted by the electric lights, ignoring thermal radiation of other objects, virtual photons, etc.
After someone answers this, can you please explain it? I guess you're going to have to try to figure out how many photons a roomful of typical lightbulb emits during the time it would take a photon to be absorbed... is that along the right lines? But I have no idea how to calculate it. Would you learn about that in physics?
I meant that if one were to freeze time and count every photon traveling from the lights to somewhere in the room, how many photons would one count.

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Re: Fermi Questions Marathon

Postby SeemsLegit » October 5th, 2012, 1:40 pm

Here's a semi-guess solution which may be close. Assume a room is 25 meters*10 meters=250 meters. Assume a photo is around 1/10^5 of a meter, so therefore we get FA=7. This may be very off, since a photon is not around 1/10^5 of a meter, but it's an OK estimate.

New question: Find the number of fermi questions made anywhere up to this point.
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Re: Fermi Questions Marathon

Postby Schrodingerscat » October 5th, 2012, 2:26 pm

Here's a semi-guess solution which may be close. Assume a room is 25 meters*10 meters=250 meters. Assume a photo is around 1/10^5 of a meter, so therefore we get FA=7. This may be very off, since a photon is not around 1/10^5 of a meter, but it's an OK estimate.

New question: Find the number of fermi questions made anywhere up to this point.
I will make the assumption that this is referring to specifically Fermi questions, as opposed to solving problems with the same methodology, which would be extremely difficult to calculate. Based on the number of tournaments and the average length of tests (20 questions from my experience), I would guess there may be as many as E4 questions from Science Olympiad alone, plus when they are used separately for others purposes, I will answer with 4.

Also, for my solution, I will pick a fairly small room with 3 100W incandescent light bulbs. I will guess an average wavelength of 600nm to find the photon energy. E=Hc/λ=6E-34Js*3E8s-1/6E-7m=3E-19J/photon. I will then invert this to get 3E18 photons per joule. 300W is 300 joules per second, so we have 9E20 photons per second emitted into the room. However, these photons will only take t=d/v=3m/3E8m/s=1E-8 s to travel across the room. Thus you have 9E12, or E13 photons in the room at a particular instant.

How many people are driving on 30-mile stretch of 2-lane interstate with "bumper to bumper" traffic?

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Re: Fermi Questions Marathon

Postby SeemsLegit » October 5th, 2012, 2:39 pm

Let's say that one car is around 10 feet. Therefore, we want to find 2*(5000)*30/10=10^3*30=3*10^4... assume there is 2 people in a car so it's 6*10^4 or FA=5.

New problem: Find the number of weapons used in world war 2.
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Re: Fermi Questions Marathon

Postby Mathdino » October 5th, 2012, 2:46 pm

I'm going to assume that a weapon means something an individual soldier can hold, so disregarding artillery, tanks, planes, etc. About E8 soldiers participated in WW2, and accounting for broken and lost weapons, probably 2 weapons per soldier, so E8+, or FA 8.

Accounting for tanks, artillery, planes, etc, the total amount would probably push it over 5E8, so FA 9.

New Question: How many grains of rice could fit in the number of shoes manufactured in China in one year?
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Re: Fermi Questions Marathon

Postby SeemsLegit » October 5th, 2012, 2:54 pm

Let's say an average shoe is 3 inches*12 inches=36 inches (looking at my own foot, which may be a bit large, I'm size 11. So let's change it to 25 inches. Let's also say they make the number of shoes for EVERYBODY in China which is a LOT of people (1 billion). Therefore, we get 10^9 people, with 25 inches per shoe. And now, a grain of rice takes up 1 inch*1/10 inch=1/10 inch. Therefore, we get 10^9*25/(1/10) or FA=10 yay.

Calculate the number of fedoras in the world.
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Re: Fermi Questions Marathon

Postby ryandm » October 5th, 2012, 10:44 pm

Let's say an average shoe is 3 inches*12 inches=36 inches (looking at my own foot, which may be a bit large, I'm size 11. So let's change it to 25 inches. Let's also say they make the number of shoes for EVERYBODY in China which is a LOT of people (1 billion). Therefore, we get 10^9 people, with 25 inches per shoe. And now, a grain of rice takes up 1 inch*1/10 inch=1/10 inch. Therefore, we get 10^9*25/(1/10) or FA=10 yay.

Calculate the number of fedoras in the world.
SeemsLegit, I agree with your assumptions that (1) shoe production in China is dominated by domestic rather than exports (2) the average Chinese person uses 1 pair of shoes per year (3) we should neglect the fact that space is lost when packing rice into a container.

However I would've calculated the volume of a shoe, rather than its area, and guessed the population of China to be higher.

My computation would look something like:

(5 wide * 12 long * 4 high) (inch^3/shoe) * 2 (shoes/person-year) * (1.5E9 people / China) * 16 cm^3/in^3 ~= 1.2E13 (cm^3 of shoe space / China-year)

I'd have approximated the volume of rice as 1mm x 1mm x 10mm = 1E1 mm^3 = 1E-2 cm^3, so my overall answer would be 1.2E13 / 1E-2 = 15.

Now for the hard/annoying part of Fermi Questions...researching the answer so I'll do better next time:
http://www.unc.edu/~andrewsr/ints092/clark.html says that global shoe production was 10 billion pairs in 1996, and China produced 1/3.
https://www.google.com/search?q=population+of+china says that the population of China is 1.34 billion
http://www.wolframalpha.com/input/?i=grain+of+rice says that the serving volume of a grain of rice is 0.0025 fluid ounces = 0.0739 cm^3
I measured the volume of my own shoe (by stuffing it, then measuring the stuffing). It came out to 43.3 in^3, but I'm a size 8.5

Let's use the assumptions that (1) since 1996 China's share of shoe production has stayed the same, (2) since 1996 global shoe production has grown at the same rate as global population, and (3) my shoe is the average size of shoes produced -- in expectation it is :)

Additional source:
http://www.google.com/publicdata/explor ... population
says that world population in 1996 was 5.8 billion, 2011 was 7.0 billion.

Then the "exact" computation (still neglects the fact that space is lost when packing rice into a container) is:
1E10 (pairs/year) * 2 (shoes/pair) * (7.0 / 5.8) * (1/3) * 43.3 (in^3/shoe) * 16.387064 (cm^3/in^3) / (0.0739 cm^3/grain) ~= 7.7E13

It could be a bit bigger since my shoe size may be below average, or a bit smaller because space is lost when packing rice into a container, but overall I'm convinced that it's probably Fermi Answer 14, and almost certainly in the range 13-15.
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