Fermi Questions Marathon

[syo_astro]

Uh, I think we should check eachother and explain calculations and all, but I think we are also supposed to answer the question to make it a real marathon .
So just for that I'll do the fedora one. Let's see, 7E9 people on Earth. Not all of them have hats I would say, and specifically the fedora. I would personally estimate it at 7, but then again it's in the world and perhaps that would be more yearly.
I'll try out a calculation. The number of people who have watched Indiana Jones I'd say is E8 to E9, so I'll go with 5E8. Let's say that's for like 10 years (obviously an overestimate, but I feel like maybe I should add on some more for fedoras in the past). This would make it 5E9 or 10. One last way to estimate it I can think of is that it's probably existed for 100 years. Let's say about 10 million or 1E7 were produced per year thinking that both my estimate earlier and that probably less were made earlier and more were made later. That comes to E9. I guess with this my final answer is 9 (probably got this so wrong, but whatever).

NEW QUESTION!!!!! Hopefully this is phrased OK:
How many animal units would it take to sustain a farm for a jiffy?

[ryandm]

@sye_astro, sorry! didn't mean to rain on the parade...ahem, marathon

my background -- did fq for two years, medaled 1st and 2nd at states, and developed a deep respect for the event

mentioning this only so that you might have reason to believe me when I say that the only way to get better is to research the answers! Otherwise it's like playing darts without ever looking at where the dart lands. That's all I meant to communicate/contribute with my earlier comment. Absolutely no intention to be rude, or otherwise do anything but be positive about the fun of solving fq.

To make it up, I'll be happy to answer your question after a small clarification...what is an animal unit? (I assume that "jiffy" refers to a unit of time.)

[syo_astro]

@sye_astro, sorry! didn't mean to rain on the parade...ahem, marathon

my background -- did fq for two years, medaled 1st and 2nd at states, and developed a deep respect for the event

mentioning this only so that you might have reason to believe me when I say that the only way to get better is to research the answers! Otherwise it's like playing darts without ever looking at where the dart lands. That's all I meant to communicate/contribute with my earlier comment. Absolutely no intention to be rude, or otherwise do anything but be positive about the fun of solving fq.

To make it up, I'll be happy to answer your question after a small clarification...what is an animal unit? (I assume that "jiffy" refers to a unit of time.)

Ah, no not at all! I wasn't meaning to say that you were being bad in any way. I completely agree that one must research answers, develop processes/methods to remember, and practice like this. I just meant that you should provide research for an answer AND a new question/answer. I think that's understood, though. As for the two measures...um, well, on your tests if you were given random units were you usually told what they would be? If not then just estimate, and I'll show you the answer/whatever needed.

[Mathdino]

If, I remember correctly, a jiffy has had many different definitions (the one I remember is about E-1 seconds), and an animal unit is a cow. I feel like these units are incompatible...

[syo_astro]

If, I remember correctly, a jiffy has had many different definitions (the one I remember is about E-1 seconds), and an animal unit is a cow. I feel like these units are incompatible...

You are right in that the jiffy has multiple definitions, one is 1/50 or 1/60, so not exactly E-1. So, just for that I'll say use the original definition relating to physics (that one is used for electronics). Sorry for not clarifying that. But I am fairly sure the animal unit works.

[ryandm]

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

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

[syo_astro]

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

[space scientist]

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 (Click to Show Hidden Text)

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

[hmcginny]

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?

[OldSpice]

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?