Fermi Questions C

Test your knowledge of various Science Olympiad events.
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Re: Fermi Questions C

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Riptide wrote:
PM2017 wrote: How many medals are won in Science Olympiad each season?
Assuming all 50 states participate in scioly, medals are on average given out to top 4 placings (often varies from 3 to 6 but 4 seems pretty average to me). 2 divisions, 23 events each, 184 medals per tournament. Let's assume 10 invitationals, 6 regionals, and obviously 1 state tournament. This gives us 3 medals won each season per state, so 5 total.
Not exactly sure where to find this information, so I'll let PM2017 give us the answer.

Since someone else just posted, check the post above for the newest question (Don't want to create 2 strands of questions lol)

Yeah, this was my bad. I didn't think about finding an actual solution. (although one could go through each state's website, and assume an average of 4.5 medals per comp, but really, who has the patience?)
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Re: Fermi Questions C

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PM2017 wrote:What is the mass of all the rivets in the wings of all the Boeing 747s ever made in terms of moles of botulinum toxin?
I'll estimate 2E4 Boeing 747's made. There are probably about E3 rivets per plane, each one being perhaps E2 grams, so 2E9 grams of rivets total rivets. 1 mole of a 1 Da substance being 1 gram, a molecule of botulinum toxin is probably around 2E2 Da - since it's likely a smallish organic molecule. This gives Fermi Answer: 7
There are around 1.5E3 Boeing 747s made ever, with 8E4 rivets per wing. Each rivet is closer to a gram or so, meaning that there is 1.2E8 grams of rivets. Unfortunately for me, botulinum toxin is 1.493E5 grams per mole, so Fermi Answer: 3
How many cones were produced by coastal redwoods in the US in 2015?
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Re: Fermi Questions C

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Unome wrote: How many cones were produced by coastal redwoods in the US in 2015?
pretty sure 13 trees in the world so 9 are coastal redwoods and maybe 8 exsists in US. I'll assume 3 cones drops per tree so 11
can't find anything about number of redwoods in US or cones dropped however 7 trees seems more accurate as thier only in certain parts of the US and thier population is decreasing. They are relatively large trees with smaller cones so probably 4 cones and  final answer 11 (probably)
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Assuming you gained the superpower of seeing infared and ultraviolet light how fast would you have to move so that the infared light with a wavelength of 1mcm your moving to appears to be UV with a wavelength of 100 nm(answer in meters per second)
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Re: Fermi Questions C

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Nobody did my last question so new question: what is the rotational inertia of the moon?
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Re: Fermi Questions C

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Name wrote:Nobody did my last question so new question: what is the rotational inertia of the moon?
Mass of the moon is a bit more than e23 kg. Radius is about e6 meters? Also going to assume that it's relatively uniform and spherical (in real life, a proportionally higher amount of mass is probably closer to the center since heavier elements would sink during the moon's formation?).

[math]I = \frac{2}{5} m r^{2}[/math]. Attempted answer is 35 (units of kg m^2)
Using the data from [url=https://nssdc.gsfc.nasa.gov/planetary/factsheet/moonfact.html]here[/url], it looks like 35 is indeed the correct answer (8.7 * 10^34). The mass is actually less than e23 kg and the radius is a bit over e6 m. 2/5 is very close to the actual value of 0.394
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Re: Fermi Questions C

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Adi1008 wrote:
Mass of the moon is a bit more than e23 kg. Radius is about e6 meters? Also going to assume that it's relatively uniform and spherical (in real life, a proportionally higher amount of mass is probably closer to the center since heavier elements would sink during the moon's formation?).

[math]I = \frac{2}{5} m r^{2}[/math]. Attempted answer is 35 (units of kg m^2)
Using the data from [url=https://nssdc.gsfc.nasa.gov/planetary/factsheet/moonfact.html]here[/url], it looks like 35 is indeed the correct answer (8.7 * 10^34). The mass is actually less than e23 kg and the radius is a bit over e6 m. 2/5 is very close to the actual value of 0.394
Yup, your turn to make a question
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Re: Fermi Questions C

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Name wrote:
Adi1008 wrote:
Mass of the moon is a bit more than e23 kg. Radius is about e6 meters? Also going to assume that it's relatively uniform and spherical (in real life, a proportionally higher amount of mass is probably closer to the center since heavier elements would sink during the moon's formation?).

[math]I = \frac{2}{5} m r^{2}[/math]. Attempted answer is 35 (units of kg m^2)
Using the data from [url=https://nssdc.gsfc.nasa.gov/planetary/factsheet/moonfact.html]here[/url], it looks like 35 is indeed the correct answer (8.7 * 10^34). The mass is actually less than e23 kg and the radius is a bit over e6 m. 2/5 is very close to the actual value of 0.394
Yup, your turn to make a question
How many neutrinos would be produced by nuclear reactions in the Sun in the time since arXiv was launched?
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Re: Fermi Questions C

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Adi1008 wrote:How many neutrinos would be produced by nuclear reactions in the Sun in the time since arXiv was launched?
Neutrino flux at Earth is something like E16 m^-2 - I unfortunately can't remember the exact value. Multiplying by pi(6E6)^2 =
 1.3E14 m^2 gives 1.3E30 neutrinos per second. Multiplying this by 3E7 seconds per year gives 4E37 neutrinos per year. arXiv has probably been around for 3E1 years, so 1.2E39 - Fermi Answer: 39
Neutrino flux at Earth appears to be around 7E14 particles per square meter. Most of my estimates were fine - arXiv was launched 26.5 years ago. Using the exact values for the number of seconds in a year and cross sectional area of the Earth yields 7.5E37,
 Fermi Answer: 38
On average, how many golf balls were produced worldwide in 2016 each second?
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Re: Fermi Questions C

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Unome wrote:
Adi1008 wrote:How many neutrinos would be produced by nuclear reactions in the Sun in the time since arXiv was launched?
Neutrino flux at Earth is something like E16 m^-2 - I unfortunately can't remember the exact value. Multiplying by pi(6E6)^2 =
 1.3E14 m^2 gives 1.3E30 neutrinos per second. Multiplying this by 3E7 seconds per year gives 4E37 neutrinos per year. arXiv has probably been around for 3E1 years, so 1.2E39 - Fermi Answer: 39
Neutrino flux at Earth appears to be around 7E14 particles per square meter. Most of my estimates were fine - arXiv was launched 26.5 years ago. Using the exact values for the number of seconds in a year and cross sectional area of the Earth yields 7.5E37,
 Fermi Answer: 38
On average, how many golf balls were produced worldwide in 2016 each second?
I'm going to say that out of 1000 people, maybe 1-2 are golfers (0.1%-0.2%). This, times the 8 -9 billion people yields 8 million golfers. I'll assume that an active golfer plays once a month, and each golfer loses 3 balls each time he or she plays, so monthly, the collective loss is approx 24 million balls per month. Now, there are 86,400 seconds in a day, and approx 30 days per month,so ~ 2,500,000 seconds per month. 24 million/ 2.5 million is approx 10, and now I'm getting the ridiculously high answer of 10 golf balls per second, or fermi answer [b]1[/b]
SOOOO, that number was actually not ridiculous. There are 840 million golf balls produced yearly, and there are 31,600,000 seconds in a year. With some dimensional analysis (840 mil gb/yr *(1yr/31.6 mil sec)) you get approx 26 golf balls per second, so fermi answer [b]1[/b]
Although I did well on this one, does anyone have a good method to solve the archetype production questions? (like the following, which I guess will be the next question.)

How many tons of rocks and minerals are produced in an average fortnight?
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Re: Fermi Questions C

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Would anyone like to revive this thread?
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