Fermi Questions Marathon

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Re: Fermi Questions Marathon

Postby ryandm » Sun Oct 07, 2012 3:14 pm

haha you're absolutely right -- I should play it as it lays...

syo_astro wrote:How many animal units would it take to sustain a farm for a jiffy?


I'm going to think out loud for this one:

1. "Animal units" - don't think I've heard this before, but maybe it's a male-female pair of animals? Nah, usually there are more females than males on a farm (I think). Okay...I'm going to assume that an animal unit means one animal over the course of its lifetime. This makes sense to me in that a farmer could buy "animal units" to stock his or her farm.
2. "Jiffy" - I have heard this before, but I don't recall how much time it is. Maybe something like a tenth of a second? That's what I'll assume.
3. "sustain a farm" - I'm going to interpret this as "maintain the animal population levels on a farm."

Computation, guessing at several of the factors, desired units are: animalUnit/farm-jiffy

(1 animalUnit / 5 animal-year) * (1 year / 365 day) * (1 day / 9E3 sec) * (1 sec / 1E1 jiffy) ~= 1 animalUnit / 2E8 animal-jiffy
(1 animalUnit / 2E8 animal-jiffy) * (1E4 animal / farm) = 5E-5 animalUnit / farm-jiffy

I'd guess I slightly overestimated the number of animal/farm, so I'll go with Fermi Answer -5.

@syo_astro, sounds like you have a solution in mind? close at all? Time for me to go google animal unit & jiffy :D

How many seconds would it take a common snail to travel from Paris to Istanbul?
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Re: Fermi Questions Marathon

Postby syo_astro » Sun Oct 07, 2012 10:58 pm

OK, sorry if my question was confusing. What meant to say was:
1 animal unit is equal to the amount of mass needed to take care of a cow over the period of one day, but it can be applied to other animals as well. This is for about a 1000 pound (45 kg) cow, which needs about 26 pounds (12 kg) for the day. It can be applied to other animals, and it can account for other things, but this is the main fact, I'll pull up the animal units for other animals too (you can google animal unit calculator).
1 jiffy is in fact a unit of time. The original definition was in physics made by Gilbert Newton Lewis. It was the time it took light to travel one centimeter, about 30 picoseconds or 3E-11 seconds. Indeed a jiffy.
So, the main thing to estimate was how many animals on the farm, and how much food each one needs if you understood the jiffy. So, I am going to make a farm have about 30 chickens, 10-20 pigs, so 15 pigs, 20-100 cows, so 50 cows, 10 horses perhaps, 10 sheeps, and 30 turkeys. These are all estimates, and perhaps someone could research more into what a farm would have? But for now then the calculations for animal units:
1 cow requires 1 animal unit per day. This makes 50.
1 pig requires .3 animal units per day makes 4.5.
1 horse requires 1 makes 10.
1 sheep requires .1 makes 1.
1 turkey requires .0115 or just because we are estimate let's say .01 makes .3.
1 chicken requires .001 makes .03.
All together this makes about 66. This is for one day. Convert this for jiffys. So:
66 animal units/day * 1 day/86400 seconds * 3E-11 seconds/1 jiffy comes out to be E-14 or -14. Anyone want to check my work? Also, someone else can answer the next question. For a quicker estimate we could just say E2/E5*E-11=E-14
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Re: Fermi Questions Marathon

Postby space scientist » Mon Oct 08, 2012 12:18 am

ryandm wrote:How many seconds would it take a common snail to travel from Paris to Istanbul?


If I remember correctly, a snail travels at around 0.1 mph (E-1). Since Paris and Istanbul are both in the Eurasia region, distance traveled by ground is going to be roughly equivalent to straight line distance. I am not the best judge of distances, but I would estimate the distance between the 2 cities to round to 10,000 mi. (E4). 10,000 mi. at 0.1 mph is 100,000 hr. or 360,000,000 seconds. So my answer is 8.

To research the correct answer:
speed of a snail: 0.002 mph
straight line distance between Paris and Istanbul: 1,400 mi.
1,400/0.002=700,000 hours= 2,520,000,000 sec.= 2.52*10^9
correct answer: 9

If all of the thermal energy received by the Earth in the form of sunlight over the course of 1 year was put into 1 gram of water at 0 degrees Celsius, what would be the resulting temperature in degrees Delisle, assuming no state changes?

The Answer
amount of solar energy received by Earth: 1,368 W/m^2
surface area of Earth: 510,072,000 km^2= 510,072,000,000,000 m^2
number of seconds in a year: 31,536,000
specific heat capacity of water: 4.1813 J/g*K
Celsius to Delisle conversion: 150-(C*3/2)
1,368*510,072,000,000,000*31,536,000=22,005,142,649,856,000,000,000,000 (or 2.2 E25) Joules
22,005,142,649,856,000,000,000,000/4.1813*1=5.26275145286 E24 degrees Kelvin= 5.26275145286 E24 degrees Celsius
150-(5.26275145286 E18*3/2)=-7.89412717929 E24 degrees Delisle
answer: 25


Edit: I changed my question so that I can find the information to calculate the correct answer and adding the correct answer.
Edit 2: fixing my calculations
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Re: Fermi Questions Marathon

Postby hmcginny » Wed Oct 24, 2012 9:09 pm

Alright, so sunlight hits the earth at a rate of about 100 W per square foot, a foot is 3E1 cm, a square foot is 9E2 cm^2 or 9E-2 m^2. The earth's radius is 6E6 m, 4*pi*(6E6)^2 is ~4E14 m^2. If we say half the earth is receiving direct sunlight at a given time then 2E14 m^2 receive sunlight, and they would receive 2E14/9E-2 * 1E2 or 2E17 W. Thats a unit of energy per time, there are 3E7 seconds in a year so 6E24 Joules of energy. Q=mcT so the change in temperature is 1.5E24 degrees celsius. I'm going to assume Delisle is a unit that follows similarly, the way that E24 celsius would be E24 fahrenheit, so lets go with 24.

I just looked at your answer, you didn't account for not all of the earth being hit with sunlight at a given time.

Since we're on the topic of thermo, how many joules of energy would it take to turn all of the water on earth into steam?
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Re: Fermi Questions Marathon

Postby OldSpice » Sat Nov 03, 2012 10:09 pm

hmcginny wrote:Alright, so sunlight hits the earth at a rate of about 100 W per square foot, a foot is 3E1 cm, a square foot is 9E2 cm^2 or 9E-2 m^2. The earth's radius is 6E6 m, 4*pi*(6E6)^2 is ~4E14 m^2. If we say half the earth is receiving direct sunlight at a given time then 2E14 m^2 receive sunlight, and they would receive 2E14/9E-2 * 1E2 or 2E17 W. Thats a unit of energy per time, there are 3E7 seconds in a year so 6E24 Joules of energy. Q=mcT so the change in temperature is 1.5E24 degrees celsius. I'm going to assume Delisle is a unit that follows similarly, the way that E24 celsius would be E24 fahrenheit, so lets go with 24.

I just looked at your answer, you didn't account for not all of the earth being hit with sunlight at a given time.

Since we're on the topic of thermo, how many joules of energy would it take to turn all of the water on earth into steam?


Hey everybody.
Okay so I haven't studied in a while but I think there's something around E21 liters of water on the earth, which means there's E24 grams. If I remember correctly it takes 4.18J to heat a gram of water up 1 degree C, and we'll assume that all the water is about 20C and the energy needed to change phases is negligible (I hope) so that means it takes 4E0*8E2*E24 J to get boil all the water in the world into steam. Answer: E27.
Okay it turns out I was wrong, it takes 2E3 joules of energy to vaporize a gram of water at 100C. I was correct on everything else, so the answer would be E30.

Here's a question taken from the Athen's Invite last year:
How many Barns are there in an Attometer?
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Re: Fermi Questions Marathon

Postby syo_astro » Sat Nov 03, 2012 10:31 pm

OldSpice wrote:
hmcginny wrote:Alright, so sunlight hits the earth at a rate of about 100 W per square foot, a foot is 3E1 cm, a square foot is 9E2 cm^2 or 9E-2 m^2. The earth's radius is 6E6 m, 4*pi*(6E6)^2 is ~4E14 m^2. If we say half the earth is receiving direct sunlight at a given time then 2E14 m^2 receive sunlight, and they would receive 2E14/9E-2 * 1E2 or 2E17 W. Thats a unit of energy per time, there are 3E7 seconds in a year so 6E24 Joules of energy. Q=mcT so the change in temperature is 1.5E24 degrees celsius. I'm going to assume Delisle is a unit that follows similarly, the way that E24 celsius would be E24 fahrenheit, so lets go with 24.

I just looked at your answer, you didn't account for not all of the earth being hit with sunlight at a given time.

Since we're on the topic of thermo, how many joules of energy would it take to turn all of the water on earth into steam?


Hey everybody.
Okay so I haven't studied in a while but I think there's something around E21 liters of water on the earth, which means there's E24 grams. If I remember correctly it takes 4.18J to heat a gram of water up 1 degree C, and we'll assume that all the water is about 20C and the energy needed to change phases is negligible (I hope) so that means it takes 4E0*8E2*E24 J to get boil all the water in the world into steam. Answer: E27.
Okay it turns out I was wrong, it takes 2E3 joules of energy to vaporize a gram of water at 100C. I was correct on everything else, so the answer would be E30.

Here's a question taken from the Athen's Invite last year:
How many Barns are there in an Attometer?

Did you mean square attometer? That would make more sense >.>. Also, to explain. If I recall, a barn is E-28 m^2. So just convert units:
1 attometer=1E-18 m, square to get 1E-36 m^2 divide by E-28 m^2. We get -8 to be the answer.

Onto the next question! Well, we just had a hurricane. So, how about:
Let's say we had the amount of money that Hurricane Sandy costed (at least so far approximately) in pennies. What would be the weight of that in troy ounces?
Answer
The cost was about $20-60 billion. So, I'll just say about 40 billion. To pennies we multiply by 100 to get 4 trillion pennies. A penny weights about 2.5 grams. This equals 1E13 grams. A troy ounce is about 31 grams. The answer is 3.1x10^14 grams or 14.
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Re: Fermi Questions Marathon

Postby OldSpice » Sun Nov 04, 2012 1:31 am

syo_astro wrote:
OldSpice wrote:
hmcginny wrote:Alright, so sunlight hits the earth at a rate of about 100 W per square foot, a foot is 3E1 cm, a square foot is 9E2 cm^2 or 9E-2 m^2. The earth's radius is 6E6 m, 4*pi*(6E6)^2 is ~4E14 m^2. If we say half the earth is receiving direct sunlight at a given time then 2E14 m^2 receive sunlight, and they would receive 2E14/9E-2 * 1E2 or 2E17 W. Thats a unit of energy per time, there are 3E7 seconds in a year so 6E24 Joules of energy. Q=mcT so the change in temperature is 1.5E24 degrees celsius. I'm going to assume Delisle is a unit that follows similarly, the way that E24 celsius would be E24 fahrenheit, so lets go with 24.

I just looked at your answer, you didn't account for not all of the earth being hit with sunlight at a given time.

Since we're on the topic of thermo, how many joules of energy would it take to turn all of the water on earth into steam?


Hey everybody.
Okay so I haven't studied in a while but I think there's something around E21 liters of water on the earth, which means there's E24 grams. If I remember correctly it takes 4.18J to heat a gram of water up 1 degree C, and we'll assume that all the water is about 20C and the energy needed to change phases is negligible (I hope) so that means it takes 4E0*8E2*E24 J to get boil all the water in the world into steam. Answer: E27.
Okay it turns out I was wrong, it takes 2E3 joules of energy to vaporize a gram of water at 100C. I was correct on everything else, so the answer would be E30.

Here's a question taken from the Athen's Invite last year:
How many Barns are there in an Attometer?

Did you mean square attometer? That would make more sense >.>. Also, to explain. If I recall, a barn is E-28 m^2. So just convert units:
1 attometer=1E-18 m, square to get 1E-36 m^2 divide by E-28 m^2. We get -8 to be the answer.

Onto the next question! Well, we just had a hurricane. So, how about:
Let's say we had the amount of money that Hurricane Sandy costed (at least so far approximately) in pennies. What would be the weight of that in troy ounces?
Answer
The cost was about $20-60 billion. So, I'll just say about 40 billion. To pennies we multiply by 100 to get 4 trillion pennies. A penny weights about 2.5 grams. This equals 1E13 grams. A troy ounce is about 31 grams. The answer is 3.1x10^14 grams or 14.


Yeah I meant square attometer, silly me.
I think I heard the damages were around E10 dollars, in pennies that'd be E12. The mass of a penny is probably E1 grams, so the mass is E13. I think a troy ounce is like E1 ounces. So I think it's E14.

How many carbon atoms are in a single printed tittle in your standard newspaper headline?
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Re: Fermi Questions Marathon

Postby The Eviscerator » Sun Nov 18, 2012 9:53 pm

OldSpice wrote:How many carbon atoms are in a single printed tittle in your standard newspaper headline?


I don't actually do Fermi, but I'd like to give this a shot. Average length of a title in a standard newspaper headline: 5 words, each 7 letters. so 3.5E1; number carbon atoms per letter: E29?, so E30 in all... This is probably wrong.

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Re: Fermi Questions Marathon

Postby OldSpice » Sat Nov 24, 2012 8:05 pm

The Eviscerator wrote:
OldSpice wrote:How many carbon atoms are in a single printed tittle in your standard newspaper headline?


I don't actually do Fermi, but I'd like to give this a shot. Average length of a title in a standard newspaper headline: 5 words, each 7 letters. so 3.5E1; number carbon atoms per letter: E29?, so E30 in all... This is probably wrong.

A tittle is the dot on an i. (http://en.wikipedia.org/wiki/Tittle) But I can see how you might've taken that as a typo. Regardless, your carbon atoms/letter estimate is very high. 6E23 is the amount of carbon atoms per 12 grams, so you estimated that a letter's mass in carbon is ~E4 grams, or 10 kilograms per letter. So unless you're reading a giant newspaper you're a bit off.

I approached this problem by figuring out how much ink goes into printing a single piece of paper full of words. A standard 14mL HP cartridge can print ~300 pages. So 1mL of ink = 2E1 pages. A page = 2E3 letters. A mL = 4E4 letters. Therefore a letter is E-4 mL of ink. Which is E-4 grams, which is E18 carbon atoms. Since the question was how many per tittle, we'll assume that a tittle is about a tenth of a letter so the answer is E17.

Before we lock in our answer, we must consider that carbon is not the sole ingredient in the ink. Newspaper ink is primarily mineral oil which is composed of C15-C40 alkanes, or carbon chains 15 to 40 carbons long. Each of these chains has 32 to 82 hydrogen atoms attached. Therefore the mass of the chains are ~26% hydrogen by mass. This makes carbon 74% of the solution by mass. This does not change our answer this time, but it was definitely worth considering.

Next question:
How many meters of railroad tracks are there in the United States?
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Re: Fermi Questions Marathon

Postby space scientist » Sun Nov 25, 2012 12:34 am

I'm a little uncertain about how to approach this problem, so please correct my logic if there are any mistakes.

The United States is about 3,000 mi. (about 4.5*10^6 meters.) wide. There are at least 2 transcontinental railroads, so there is at least 9*10^6 meters of transcontinental railroad tracks. However, the question asks about how many meters of railroad tracks there are, and there are smaller, regional railroad lines that need to be taken into consideration. Dividing the length of the transcontinental railroad tracks by the fraction of the total railroad tracks that are part of a transcontinental railroad yields the total length of the railroad tracks in the United States. I would estimate that between a third and a tenth of all the railroad tracks in the United States are part of a transcontinental railroad.

(9*10^6)/(1/3)=2.7*10^7 and (9*10^6)/(1/10)=9*10^7

That establishes that the answer is either 7 or 8. I am leaning towards an answer of 8 since it seems a little unreasonable for only 2/3 of the railroad tracks in the United States to belong to regional lines. Therefore my answer is going to be 8.

To look up the correct answer: According to http://en.wikipedia.org/wiki/List_of_co ... twork_size, the United States has 226,427 km. of railroad tracks, which is 2E8 meters.

Question: How much energy in British thermal units was released by the atomic bombs dropped on Hiroshima and Nagasaki?
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Re: Fermi Questions Marathon

Postby Schrodingerscat » Sun Nov 25, 2012 12:54 am

space scientist wrote:Question: How much energy in British thermal units was released by the atomic bombs dropped on Hiroshima and Nagasaki?

I believe their combine yields are on the order of 10 kilotons of tnt. I also know that a gram of tnt is defined as 1 kCal or 4kJ. You have then E4tons=E7kg=E10grams. Thus you have 4E10kJ. I believe a BTU is reasonably close to a kJ, so I will say 10.

Answer
The actual answer would be 11 due to the yield being 36kT


If one were to collect all the rain that fell on the islands of Hawaii in an hour with a constant flux of one aluminum can height per half life of a neutron as drinking water, for how many seconds could one provide drinking water for all of humanity?

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Re: Fermi Questions Marathon

Postby OldSpice » Mon Nov 26, 2012 11:51 pm

Schrodingerscat wrote:If one were to collect all the rain that fell on the islands of Hawaii in an hour with a constant flux of one aluminum can height per half life of a neutron as drinking water, for how many seconds could one provide drinking water for all of humanity?


This question is ridiculous. But I'll bite.
Okay so I know the surface area of a body of water close in size to Hawaii, so from there I'm guessing the surface area of Hawaii is probably about E10 m^2. A half life of a neutron is about 15 minutes, so four aluminum can heights fall, or about two feet. For simplicity's sake, since I'm doing this in my head, I'll round that up to a meter. Now we have E10 cubic meters of rain water. That translates to E13 liters of water. I know it's recommended that we drink 8 glasses of water a day or about a liter of water a day. There's 7E9 people in the world. That should give humanity water for about E3 days. With 8E4 seconds in a day I'd estimate that with all that water you'd be able to give humanity drinking water for E8 seconds.
Fermi Answer: 8

Approximately how many electrons flowed through my laptop as I typed out this question?
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Re: Fermi Questions Marathon

Postby Schrodingerscat » Thu Dec 13, 2012 12:21 am

OldSpice wrote:
Schrodingerscat wrote:If one were to collect all the rain that fell on the islands of Hawaii in an hour with a constant flux of one aluminum can height per half life of a neutron as drinking water, for how many seconds could one provide drinking water for all of humanity?


This question is ridiculous. But I'll bite.
Okay so I know the surface area of a body of water close in size to Hawaii, so from there I'm guessing the surface area of Hawaii is probably about E10 m^2. A half life of a neutron is about 15 minutes, so four aluminum can heights fall, or about two feet. For simplicity's sake, since I'm doing this in my head, I'll round that up to a meter. Now we have E10 cubic meters of rain water. That translates to E13 liters of water. I know it's recommended that we drink 8 glasses of water a day or about a liter of water a day. There's 7E9 people in the world. That should give humanity water for about E3 days. With 8E4 seconds in a day I'd estimate that with all that water you'd be able to give humanity drinking water for E8 seconds.
Fermi Answer: 8

Approximately how many electrons flowed through my laptop as I typed out this question?


I will guess that it took you about 500s to type out the question. A laptop probably draws something in the order of an ampere from the battery, so you have 500C of charge or about Fermi answer 21 elementary charges. However, to be technical, not that many electrons actually drift through the laptop, but I am unsure how to calculate the actual flow rate of electrons.

If one were to print the entire internet for a Science Olympiad binder on A4 paper, how thick would the binder be?

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Re: Fermi Questions Marathon

Postby iwonder » Thu Dec 13, 2012 1:26 am

Interesting... google crawls something like a few petabytes of data a day, I believe, they crawl the whole internet in one week, so we can assume that the internet has around 2e16 bytes of data in it(images included) so knowing that a large amount of that will be images, lets assume that it's 50% images(images take up more space than text, but there are fewer of them, presumably) so it's 1e16 characters, and 1e16 bytes of images. Most images are in jpeg format, a 3000x2000 image in jpeg format takes up about 2e6 bytes, so that's 5e9 3000x2000 images(this is probably an underestimation, jpeg is one of the more compressed formats, I'll underestimate later) so if 2 of those pictures take up about the same area as a 11x17 sheet of paper(that's A4, I believe), that's about 2e9 pages of images(underestimate on the division). Now, add that to the 1e16 charecters, since an 8.5x11 page can hold approximately 4e2 words, that's 2e3 letters, an A4 page could hold 4e3 letters, so that's 2.5e14 pages of A4 paper, total so far we've got just about 3e14 pages of A4, but this is SciOly, we're going to put 4 standard pages to a sheet, so now we're right around 7.5e13, this seems a little high(on pages of text, honestly) so I'll just round down and say 5e13 pages. Since a page is 2e-3 inches thick, that's 2.5e10 inches thick, since you didn't specify units, I'll answer in inches and put down 10.

Researched answer...

So the internet contains something like 46 billion web pages(http://www.worldwidewebsize.com/), 5e9 pages, using this site (http://httparchive.org/interesting.php) we can figure that 7.5e5 bytes of images(I'll focus on images first) are going to be requested per page. That's 3.75e15 bytes of images in the entire web. Now, there's on average 5.3e4 bytes of HTML per page, since there's not stats on how much of this is visible text, I'll estimate that 30% of html is actually visible text(disclaimer: I do web development for the company I work for and admin our SciOly site). This translates to 8e13 bytes of text on the entire internet. By opening up word and selecting size 8pt font(what scioly sheets seem to use) I know that 1e4 characters can fit on a 8.5x11 sheet of paper, so about the same on A4(which, it turns out, is 8.25x11.75, whoops). Since a byte of ascii is a character, that's 8e13 chars in the internet that would actually print, since I assume this is double sided, that's 4e9 pages of text. Now the images are much harder... taking the weighted average of the image sizes(from the HTTP Archive) we can find that every picture on the internet would take up 1.44e17 pixles, average printer dpi is 600, so that's 3.6e5 pixels per square inch, thus 4e11 square inches of images on the internet. A sheet of A4 paper is 9.69e1 square inches per side, so 2e9 pages of images. Total, we get 6e9 pages of information on the internet, since a sheet of paper is .003" (I measured it), that's 1.8e7 inches thick. Thus a fermi answer of 7.

Wow that was difficult... Here's an easier one from a recent test I took. If one planted grape vines 5 feet apart, how many grape vines would cover central park?
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Re: Fermi Questions Marathon

Postby DGing94 » Sat Dec 15, 2012 2:31 pm

Central Park is about 1 mi^2. So for simplicity's sake we'll convert that to feet squared (1 mi^2)= (5280 ft* 5280ft)= 27,878,400 ft^2. If the grapevines are 5 feet apart they can be divided into sqaure units to make a grid within central park with each cell being 25 ft^2. So take (27,878,400/25) which gives you an answer of 1,115,136 cells. Assuming that the grapevines sit in the exact corners we can assume that 1 cell has all four grapevines and the rest have only two because two are already counted from a previous cell. So (1,115,135 * 2)= 2,230,270 + 4= 2,230,274. Giving an exponent of 6.

I'm a little new at this so let me try a question.
What is the molecular weight of water in an average 150lb man?


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