Fermi Questions Marathon

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3nv1r0nm3ntal ch3m
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Re: Fermi Questions Marathon

Postby 3nv1r0nm3ntal ch3m » April 18th, 2012, 12:58 pm

Wouldn't it depend on the shape of the dice?

anepictimelord
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Re: Fermi Questions Marathon

Postby anepictimelord » April 18th, 2012, 1:07 pm

Wouldn't it depend on the shape of the dice?
thats why I said D20 all d20's are icosahedrons, the 20 sided regular 3d solid. all the platonic solids have the golden ration in some way. http://en.wikipedia.org/wiki/Icosahedron
2012 Regionals:
Technical Problem Solving -1
Thermodynamics -1
Fermi Questions -3
Experimental Design -3
Write It Do IT -6

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Re: Fermi Questions Marathon

Postby nejanimb » April 20th, 2012, 8:16 am

How many pounds of fries does McDonald's sell each day?
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Re: Fermi Questions Marathon

Postby A123456789 » April 20th, 2012, 1:31 pm

Suppose that (very roughly) McDonald's sells meals to 10^8 people per day. The average person perhaps orders 3 ounces of fries. The answer is thus 7.

How many websites are there on the internet that have been viewed by at least 1 million different computers?

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Re: Fermi Questions Marathon

Postby Herr Doktor » May 2nd, 2012, 3:44 pm

Hmmm, that's a rough one. From what I know, very obscure websites can have millions of visitors so long as they have a very precise concentration (like birding in certain states, for example). Simply from past experience, I know that there are roughly 6E8 websites on the Internet to begin with. But, only a small proportion of these are capable of garnering one million views. Taking into account personal webpages and other useless pages, perhaps 1 in 1000 are actually capable of breaking the one million view barrier. That could be an optimistic estimate; anybody want to back me up? We arrive at 6E8/1E3=6E5=1E6. So I answer 6. Anybody have any other methods?

This next one I came up with on my own during one long Tennessee day. Enjoy.

Calculate the number of Whoppers that could be purchased using one year's healthcare costs due to obesity in the number of average American households powered by the rest energy of the amount of glucose produced through photosynthesis by the volume of water that can be boiled away by the rest energy of the number of cells that completely fill the universe, assuming excess carbon dioxide, full sunlight and that the water is already at 100C.

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Re: Fermi Questions Marathon

Postby Schrodingerscat » May 16th, 2012, 6:38 am

This is going to be long and probably very inaccurate. We will assume that since the cells are filled with water, they have a similar density of 1000kg/m^3. One lightyear is 1E15 ly, and if I assume you are discussing the observable universe with its expanded size to this day, I will assume it has a radius of around 4E10 ly, so thus its 4E25 meters in radius. The volume of a sphere is roughly 4 times the radius cubed, so its 16E75 or about 2E76 m^3 and 2E79 kg. E=MC^2, so 9E18J/kg or 18e97 or 2E98J. We will say a kilogram of water can be boiled by something in the order of 1E4 Joules, so we Have 2E94 kg of water. A kg of water is roughly 50 moles, so we have 1E96 moles of water, which can support 1E95 moles of glucose. I would guess something like 1E5 Joules to react a mole of glucose, so we have 1E100 Joules to power the homes. I would guess a home consumes about 3000W*3600*24=3E3*4E3*2E1=24E7= 2E8J. I will also assume for consistency this is year of energy, so more like 8E11J per year. Thus we have 1E88 american households. Of course some have significantly more and significantly less, but I would guess the average is something in the order of $2000. per person, so we have 1E91 dollars. Thus I would guess 90 as my Fermi answer for the number of whoppers. (plus or minus many powers, I am sure)

How many meters of film would a professional photographer use in his or her career assuming that he or she uses only film and never digital?

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Re: Fermi Questions Marathon

Postby BYHscioly » June 29th, 2012, 11:08 am

Well, it depends on the photographer and varies from person to person :P

Let's assume professional photographers probably take about 250 pictures every work day, which is about 10 meters of film. The photographer works 300 days a year for 30 years. That's roughly 10^4 days. Multiply that by 10 meters and you get 10^5 or 5. Probably inaccurate though.

On average, how many of Joules of gravitational potential energy has the CN Tower in Toronto (tallest freestanding structure until Burj Khalifa came along) gained since it was built due to isostatic rebounding?
1st Fermi (2013), 2nd Astro (2014), 3rd DP (2014), 4th DP (2012)


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