Fermi Questions C

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talkingturtle101
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Re: Fermi Questions C

Postby talkingturtle101 » May 25th, 2017, 6:19 am

CHALLENGE: How many post-its would be required to fill the entire Pacific Ocean assuming it take the densest packing structure with no air gaps?
- A pad of like 200 post-its is 1 cm high.
- Using the thing that it would take 10^16 post-its to completely cover the Pacific.
- Assuming the Pacific is 5 km deep everywhere, and is a rectangular prism.
- 5 km = 500,000 cm = 500,000 * 200 post-its = 100,000,000 post-its = 10^8
- 10^8 times 10^16 equals
- Fermi Answer of 24
- On average the Pacific Ocean is 12,100 feet deep, or 3.688 km
- Apparently one post-it is 0.03 inches thick, or 0.0762 cm.
- This is 4.8 million post-its from the bottom of the ocean to the top, rounded to 10^6
- 10^6 times 10^16 equals
- Fermi Answer of 22
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NeilMehta
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Re: Fermi Questions C

Postby NeilMehta » May 25th, 2017, 7:03 am

Attempt: Post it is about .1m length and width, for a surface area of E-2
earth radius is 6 million meters, meaning surface area is about E15 m^2
pacific ocean is pretty big, about E14 at least
so E16 post its??

actual answer: Post-it area in m^2 = 0.00580644
Area of pacific ocean = 161.8 trillion m^2
161.8 trillion/0.00580644=2.7865611e+16
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Well a cube of post its has 100 post its
That cube has a volume of E-3
I'm going to take a hugeee guess and say that the ocean is about E3 deep, for a volume of E17
So we would need E20 (honestly this one is probably wrong)
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Re: Fermi Questions C

Postby Raleway » May 25th, 2017, 7:06 am

That's why I'm trying to get it now :(
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Re: Fermi Questions C

Postby Raleway » May 25th, 2017, 7:48 am

I'll keep a streak of questions :)

How many iPhone 7s need to be stacked (face up or face down, not length wise) to reach the combined height of the 10 tallest structures in the world?
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Re: Fermi Questions C

Postby Ashernoel » May 25th, 2017, 1:17 pm

I'll keep a streak of questions :)

How many iPhone 7s need to be stacked (face up or face down, not length wise) to reach the combined height of the 10 tallest structures in the world?
IF structure = everest i've never done fermi but this seems fun. so everest is like 24,000 feet idk so like 10,000m or 6 magnitudes. Then an iPhone is like 7->10 mm or -2 so need 8.
If structure = tall building = like a kilometer = 4 so then 6.
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Re: Fermi Questions C

Postby Raleway » May 25th, 2017, 2:06 pm

I'll keep a streak of questions :)

How many iPhone 7s need to be stacked (face up or face down, not length wise) to reach the combined height of the 10 tallest structures in the world?
IF structure = everest i've never done fermi but this seems fun. so everest is like 24,000 feet idk so like 10,000m or 6 magnitudes. Then an iPhone is like 7->10 mm or -2 so need 8.
If structure = tall building = like a kilometer = 4 so then 6.
Total height = 5772607.2 millimeters divided by 7.1 millimeters (height of iPhone 7) then you'd get 813043.268 iPhones or fermi answer of 6!
Congratulations! DING DING DING

Next Question: How many people would have to whisper to match the volume of a .357 magnum revolver firing? (Is this even possible- don't know if decibels from multiple sources are additive_
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Re: Fermi Questions C

Postby Unome » May 25th, 2017, 3:07 pm

Every ten decibels is one order of magnitude. I'll go with 30 for a whisper and 100 for a revolver.
Fermi Answer: 7
It's 20 and 160 apparently (should've known lol), so the true answer: 14
How many times could an average person throw a baseball at 60 mph using the energy content of a single day's worth of food?
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Re: Fermi Questions C

Postby Ashernoel » May 25th, 2017, 5:28 pm

Every ten decibels is one order of magnitude. I'll go with 30 for a whisper and 100 for a revolver.
Fermi Answer: 7
It's 20 and 160 apparently (should've known lol), so the true answer: 14
How many times could an average person throw a baseball at 60 mph using the energy content of a single day's worth of food?
day worth of food = 2,000,000 cal or 10,000,000 J so mag 7
base ball at 60 mph is uhhhhhhhhhh like a 1 meter with a hard pitch so humans can be strong and lift their weight which is like 1000N so they could throw that in ball so 1000J of force so 1000 J => 3, 7/3=>4?
lol in fermi when 7/3 = 4.

I would write an actual answer but idk how......
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Re: Fermi Questions C

Postby NeilMehta » May 25th, 2017, 6:55 pm

How many standard buckets of paint would it take to cover the half of the moon visible to us?
Bonus: If all of the 1-dollar bills used to pay for the paint were laid out, how much area, in m^2, would they take up (assume that paint costs 25 cents per liter)
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Re: Fermi Questions C

Postby Raleway » May 25th, 2017, 10:14 pm

How many standard buckets of paint would it take to cover the half of the moon visible to us?
Bonus: If all of the 1-dollar bills used to pay for the paint were laid out, how much area, in m^2, would they take up (assume that paint costs 25 cents per liter)
Hm, assuming the moon is 2/3 of the earth's SA (500 million sq km), then it should be about 175 million give or take. Assuming one bucket of paint covers 0.05 sq km, then it would be about 3 billion, or fermi answer of 9

Bonus: assuming a bucket of paint is I don't know I guess 5 liters??? then it would be 15 billion liters multiplied by 25 cents yields about 4 billion dollars. Assuming each dollar is around 80 cm^2, then it covers a total of 320 billion sq cm which converted to m^2 yields (divide by 10^5 as 100^2) which yields fermi answer of 7
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