Fermi Questions Marathon

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jacobbigham
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Re: Fermi Questions Marathon

Post by jacobbigham »

Let's say it's 20m high and 500m long. (Actual dimensions are about 20m by 270m, so not off by too much, especially given that it's curved, contributing to higher surface area)

That means:
20m high
times 500m long
times 2 sides
times say 1 um = E-6 m thick of gold = 2E-2 cubic meters of gold (apparently, actual gold plating is .5 um)

2E-2 cubic meters is equal to 2E2 cubic centimeters:
2E2 cubic centimeters
times 20 grams per cc
times 450 grams per lb
times 1/16 ~ 6E-1 ounce per lb
= 1.08 E 6 ounces. --> [6]

How long would it take the raise the entire ocean's temperature by 13 K?
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Re: Fermi Questions Marathon

Post by hmcginny »

ok well q=mcdeltat, there are 1E21 liters of water on the earth, at a density of 1g/ml thats 1E24 g of water, specific heat is 4E0, so you need 6E25 joules of energy to raise the temperature. Since the only way it is warming is through the heat of the sun, which sends E21 joules of energy per day, meaning we need 6E4 days to gain or 2E2 years.

How many sheets of paper would it take to cover the surface of the land on earth?
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Re: Fermi Questions Marathon

Post by quizbowl »

hmcginny wrote:ok well q=mcdeltat, there are 1E21 liters of water on the earth, at a density of 1g/ml thats 1E24 g of water, specific heat is 4E0, so you need 6E25 joules of energy to raise the temperature. Since the only way it is warming is through the heat of the sun, which sends E21 joules of energy per day, meaning we need 6E4 days to gain or 2E2 years.

How many sheets of paper would it take to cover the surface of the land on earth?
So earth's radius is 6E3km, or 6E6m. SA = 4pi*r^2 (12pi^2) but since earth is 70% ocean and 30% land thats 4 r^2. = 1E14 meters squared
sheet of paper: 8 1/2 by 11 inches. Thats 22 by 28 cm, which is .22m by .28m. = 6E-2 meters squared
1E14/6E-2 = 2E15

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Re: Fermi Questions Marathon

Post by jacobbigham »

That would just be (1/365)^3, right? I mean, there are other factors that go in as well, because one's birthday isn't totally random, but that sounds right:

So I'll do (1/400)^3 = (.0025)^3 = (2.5E-3)^3 = ~14E-9 = 1.4E-8 so [-8].

What fraction of his or her life does a human spend waiting for red lights to turn green?
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Re: Fermi Questions Marathon

Post by schylarshae »

I'd say I hit about -E1 lights a day with -E2 days a year and -E1 year lifetime meaning I wait at -E4 lights during my life. There are about -E5 total minutes in a lifetime and if every redlight lasts approximately 2 minutes, the average person would spend -E4/-E5 (1/10) of their life waiting at stopligts.

How many metal track cleats are used in the U.S. each year.
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Re: Fermi Questions Marathon

Post by OldSpice »

schylarshae wrote:I'd say I hit about -E1 lights a day with -E2 days a year and -E1 year lifetime meaning I wait at -E4 lights during my life. There are about -E5 total minutes in a lifetime and if every redlight lasts approximately 2 minutes, the average person would spend -E4/-E5 (1/10) of their life waiting at stopligts.

How many metal track cleats are used in the U.S. each year.
That's a tricky one.
The average track team has E1 members, There are E5 high schools and E4 colleges.
I'm probably really off but I'd say around E7.

If you take every book sold in the world in the past ten years and stack them vertically, how high would the stack be in terameters?
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Re: Fermi Questions Marathon

Post by hmcginny »

well there are 7E9 people in the world, and each year each person probably buys E1 books, but we'll round it up to a low E2 because of schools and libraries. so thats E12 books per year, and E13 for 10 years. Each book is probably around an inch thick, so E13 inches tall, which is 2E13 centimeters or 2E-1 terameters so E-1 is my answer.

How many pages of paper will be used to print the midterm exams for every class in every US high school?
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Re: Fermi Questions Marathon

Post by A123456789 »

There are about 20 million high school students in the U.S. Suppose that each student takes 4 exams on average and that the average midterm uses 20 pages of paper. 2E7X4X20=160E7=9.

How many grains of sand would it take to fill an Olympic swimming pool?
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Re: Fermi Questions Marathon

Post by ndclscienceolympiad »

So I'll say a grain of sand has a rough volume of 2E-2 cm^3 which converts to 2E-8 m^3. An olympic swimming pool is 2,500m^3 in volume, so 2.5E3. 2.5E3/2E-8 = 1.25E11------>[11]


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Re: Fermi Questions Marathon

Post by Schrodingerscat »

I will simply guess 10^3, as I am guessing around 1000lbs of beef, although it could round up to 10^4.

How many atoms of helium are in the average classroom?
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