Fermi Questions Marathon

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Re: Fermi Questions Marathon

Postby ryandm » October 7th, 2012, 8:14 am

haha you're absolutely right -- I should play it as it lays...
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 » October 7th, 2012, 3:58 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 » October 7th, 2012, 5:18 pm

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?
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 » October 24th, 2012, 2:09 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 » November 3rd, 2012, 3:09 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?
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 » November 3rd, 2012, 3:31 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?
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?
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 » November 3rd, 2012, 6:31 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?
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?
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 » November 18th, 2012, 1:53 pm

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 » November 24th, 2012, 12:05 pm

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 » November 24th, 2012, 4:34 pm

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.

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