## Fermi Questions C

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### Re: Fermi Questions C

Unome wrote:By the way, with my neutrino question I was going for the number of times a neutrino actually interacts with matter in the Earth, rather than the number of neutrinos that pass through.
Unome wrote:How long would it take for the volume of the Sun (of water) to pour down Niagara Falls? (both the Canadian and American sides combined)
$r_{sun}\approx 7 \times 10^8 m$. The volume of the Sun would be $V \approx 4 \times 243 \times 10^{24} m^3 \approx 10^{27} m^3 = 10^{30} L$

I'll take a wild guess and say $10^6 L$ of water falls every second. $10^{30}/10^{6} = 10^{24} s$.
$\framebox{FA = 24}$
[url=https://www.wolframalpha.com/input/?i=((4%2F3)*pi*(6.955*10%5E8)%5E3)%2F(168000%2F60)]Wolfram Alpha[/url] just barely gives $\framebox{FA = 24}$
How many complete orbits around its parent star would HAT-P-11b, a "Hot Neptune", make in the time it takes for the average human to run 5 miles without stopping?
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### Re: Fermi Questions C

Adi1008 wrote: How many complete orbits around its parent star would HAT-P-11b, a "Hot Neptune", make in the time it takes for the average human to run 5 miles without stopping?
- I've never heard of it, but it seems like planets of that size wouldn't be too close to their star. So say it takes 50 Earth years to orbit (5 E1)
- Average mile time is probably 7-8 minutes, so 5 times that plus some extra because of getting tired would be around 45 minutes, or 0.75 hours, so 7.5 E-1.
- There's around E4 hours in a year, so it takes 5 E5 hours for the planet to orbit.
- Dividing gives us less than one, but in your question it said complete orbits, so I'm assuming that means round down, so it would be zero, with no Fermi Answer?
- If it wasn't complete orbits, the Fermi Answer would be -6.
- The planet orbits once every 5 days
- The average 5 mile time is 40-45 minutes, but lets go with 45.
- The planet takes 120 hours to orbit, and 5 miles takes 0.75 hours, so dividing gives us 0.00625, which rounds up to 0.01, so actual Fermi Answer of -2.
- However, if it was complete orbits, there would be no Fermi Answer, because there were no complete orbits.
How many times larger than the mass of all Apple devices across the world is the mass of all apple trees across the world?
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### Re: Fermi Questions C

talkingturtle101 wrote:How many times larger than the mass of all Apple devices across the world is the mass of all apple trees across the world?
- Most of the mass probably comes from heavy computers. These come out at around E1 kg.
- An apple tree is around E3 or E4 kg.
- There's probably E1 apple trees per device, since apple trees aren't super common. Dividing,
Fermi Answer: 4
- A single apple tree produces around 200 pounds of apples each year, and the worldwide production of apples per year in 2014 was 84.6 billion kg, so 9.33E8 commercial apple trees in the world. The number of wild apple trees is probably negligible. An average apple tree seems to weight upwards of 1 or 2 million grams, so perhaps 2E15 total grams of apple tree mass.
- The average iPhone mass is ~150 grams, with 700 million currently active iPhones. The average iPad weighs around 500 grams, with 360 million total iPads sold (probably 250 million of these are currently active, based on the ratio for iPhones). The total mass of Apple computers is probably similar to that of iPads - they are heavier, but sell less.This gives a total of around 4E11 grams.
Fermi Answer: 4 (wow, didn't think I would be that close)
How many centimeters of railway tracks existed in the US at the start of the Civil War?
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### Re: Fermi Questions C

Unome wrote:
talkingturtle101 wrote:How many times larger than the mass of all Apple devices across the world is the mass of all apple trees across the world?
- Most of the mass probably comes from heavy computers. These come out at around E1 kg.
- An apple tree is around E3 or E4 kg.
- There's probably E1 apple trees per device, since apple trees aren't super common. Dividing,
Fermi Answer: 4
- A single apple tree produces around 200 pounds of apples each year, and the worldwide production of apples per year in 2014 was 84.6 billion kg, so 9.33E8 commercial apple trees in the world. The number of wild apple trees is probably negligible. An average apple tree seems to weight upwards of 1 or 2 million grams, so perhaps 2E15 total grams of apple tree mass.
- The average iPhone mass is ~150 grams, with 700 million currently active iPhones. The average iPad weighs around 500 grams, with 360 million total iPads sold (probably 250 million of these are currently active, based on the ratio for iPhones). The total mass of Apple computers is probably similar to that of iPads - they are heavier, but sell less.This gives a total of around 4E11 grams.
Fermi Answer: 4 (wow, didn't think I would be that close)
How many centimeters of railway tracks existed in the US at the start of the Civil War?
I'm approximating that there were closer to 100,000 km of railroad tracks existing at the time of the Civil War. Using dimensional analysis, you find that there are 1e10 cm in 100,000 km, which gives a Fermi Answer of 10. This is a little more difficult to approximate though.
Fermi Answer: 10
According to https://www.civilwar.org/learn/articles/railroads-confederacy, there were 22,000 miles of railroad in the North and 9,500 miles of railroad in the South, which gives 31,500 miles or about 51,000 km. Using dimensional analysis there are 5.1e9 cm in 51,000 km.

log(5.1e9) = 9.7 which rounds to 10

Fermi Answer: 10
How long would it take for kitchen sink to fill a volume equivalent to Lake Michigan, assuming that the flow of water from the sink is constant and the answer is measured in seconds?
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### Re: Fermi Questions C

talkingturtle101 wrote:
- I've never heard of it, but it seems like planets of that size wouldn't be too close to their star. So say it takes 50 Earth years to orbit (5 E1)
- Average mile time is probably 7-8 minutes, so 5 times that plus some extra because of getting tired would be around 45 minutes, or 0.75 hours, so 7.5 E-1.
- There's around E4 hours in a year, so it takes 5 E5 hours for the planet to orbit.
- Dividing gives us less than one, but in your question it said complete orbits, so I'm assuming that means round down, so it would be zero, with no Fermi Answer?
- If it wasn't complete orbits, the Fermi Answer would be -6.
- The planet orbits once every 5 days
- The average 5 mile time is 40-45 minutes, but lets go with 45.
- The planet takes 120 hours to orbit, and 5 miles takes 0.75 hours, so dividing gives us 0.00625, which rounds up to 0.01, so actual Fermi Answer of -2.
- However, if it was complete orbits, there would be no Fermi Answer, because there were no complete orbits.
You're right, I worded that very poorly. I meant for it to be like fractions of a complete orbit but I wasn't clear at all
Justin72835 wrote:
Unome wrote:
talkingturtle101 wrote:How many times larger than the mass of all Apple devices across the world is the mass of all apple trees across the world?
- Most of the mass probably comes from heavy computers. These come out at around E1 kg.
- An apple tree is around E3 or E4 kg.
- There's probably E1 apple trees per device, since apple trees aren't super common. Dividing,
Fermi Answer: 4
- A single apple tree produces around 200 pounds of apples each year, and the worldwide production of apples per year in 2014 was 84.6 billion kg, so 9.33E8 commercial apple trees in the world. The number of wild apple trees is probably negligible. An average apple tree seems to weight upwards of 1 or 2 million grams, so perhaps 2E15 total grams of apple tree mass.
- The average iPhone mass is ~150 grams, with 700 million currently active iPhones. The average iPad weighs around 500 grams, with 360 million total iPads sold (probably 250 million of these are currently active, based on the ratio for iPhones). The total mass of Apple computers is probably similar to that of iPads - they are heavier, but sell less.This gives a total of around 4E11 grams.
Fermi Answer: 4 (wow, didn't think I would be that close)
How many centimeters of railway tracks existed in the US at the start of the Civil War?
I'm approximating that there were closer to 100,000 km of railroad tracks existing at the time of the Civil War. Using dimensional analysis, you find that there are 1e10 cm in 100,000 km, which gives a Fermi Answer of 10. This is a little more difficult to approximate though.
Fermi Answer: 10
According to https://www.civilwar.org/learn/articles/railroads-confederacy, there were 22,000 miles of railroad in the North and 9,500 miles of railroad in the South, which gives 31,500 miles or about 51,000 km. Using dimensional analysis there are 5.1e9 cm in 51,000 km.

log(5.1e9) = 9.7 which rounds to 10

Fermi Answer: 10
How long would it take for kitchen sink to fill a volume equivalent to Lake Michigan, assuming that the flow of water from the sink is constant and the answer is measured in seconds?
I'll assume that water comes from a faucet is about 10L/min and that the volume of Lake Michigan 10^12 L. 10^11 minutes * 60 = 6*10^12, rounds to 10^13. Fermi Answer is 13.
[url=https://www.wolframalpha.com/input/?i=(4.918+%C3%97+10%5E15)%2F(8.3%2F60)]Wolfram Alpha[/url] says that the Fermi Answer is 16
How many times would a ceiling fan spin in the time between the discovery of Europa and the discovery of the structure of DNA?
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### Re: Fermi Questions C

I'm not sure if the fan is spinning slow or fast, so I'm guessing close to top speed. I think that a ceiling fan spins at about 5 spins per second. I'm not completely sure when Europa was discovered so I'm going to estimate that it was about 400 years ago (it was during the early 1600s). The discovery of the structure of DNA was fairly recent, having been discovered in the early 1960s, so I'm going to use 350 years as the time between the two events. Using dimensional analysis:

5 spins/second * 3600 seconds/hr * 24 hr/day * 365 hr/year * 350 years = 5.52e10

Fermi Answer is 11.
A normal ceiling can spin at rpm of 60 to 240 rpm, which is about 1 to 5 spins per second. Assuming top speed, I'm going to use 5 spins per second. The discovery of Europa was in 1610 by Galileo, while the structure of DNA was discovered in 1953 by Watson and Crick (so I was a decade off). That's 343 years apart. Using dimensional analysis:

5 spins/second * 3600 seconds/hr * 24 hr/day * 365 hr/year * 343 years = 5.41e10

Fermi Answer is 11
How many Olympic-size swimming pools could you fit in the Pacific Ocean?
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### Re: Fermi Questions C

Justin72835 wrote:How many Olympic-size swimming pools could you fit in the Pacific Ocean?
Olympic swimming pool = E3 m^3 of water which is about E-6 km^3
huge overestimation that the pacific ocean is 50% of earth surface area for easy math
using that estimate the pacific ocean has E^7 km^2 times 10km depth so E^8 km^3
E14?
Olympic swimming pool = 2,500m^3 of water = 2.5e-6
Pacific ocean = 710,000,000 km^3 of water = 7.1e8
Actual answer: E14
How many E Coli bacteria would it take to fill a water bottle?
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### Re: Fermi Questions C

NeilMehta wrote:How many E Coli bacteria would it take to fill a water bottle?
- An E. coli bacterium is probably around 100 nm x 10 nm x 10 nm, so that's E4 nm^3 which is E-17 mL.
- A tall water bottle is around E3 mL. Dividing,
Fermi Answer: 20
- I underestimated the size of E. coli, it's actually around 5E-13 mL.
- A tall water bottle is 5E2 mL. Dividing,
Fermi Answer: 15
How many liters of water precipitated on Earth in 2016?
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### Re: Fermi Questions C

Unome wrote:
NeilMehta wrote:How many E Coli bacteria would it take to fill a water bottle?
- An E. coli bacterium is probably around 100 nm x 10 nm x 10 nm, so that's E4 nm^3 which is E-17 mL.
- A tall water bottle is around E3 mL. Dividing,
Fermi Answer: 20
- I underestimated the size of E. coli, it's actually around 5E-13 mL.
- A tall water bottle is 5E2 mL. Dividing,
Fermi Answer: 15
How many liters of water precipitated on Earth in 2016?
Earth SA = about 10*E3^2 = E7 km^2
going to guess that average rainfall around the earth is E1 cm which is E-4 km
that would be a total of E3 km^3 of rain
That equals E12 m^3,which is E15 liters
Answer: E15
Apparently average rainfall is 100cm so E2 cm
Earth surface area is E9 (oops)
In liters, multiplying those gives 5E17, or E18
Actual answer: E18
Lets say every SAT test taker used some number of SAT prep books, that if stacked (horizontally/shortest side for height), would equal their height.
How many SAT prep books were used overall?
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### Re: Fermi Questions C

NeilMehta wrote:
Unome wrote:
NeilMehta wrote:How many E Coli bacteria would it take to fill a water bottle?
- An E. coli bacterium is probably around 100 nm x 10 nm x 10 nm, so that's E4 nm^3 which is E-17 mL.
- A tall water bottle is around E3 mL. Dividing,
Fermi Answer: 20
- I underestimated the size of E. coli, it's actually around 5E-13 mL.
- A tall water bottle is 5E2 mL. Dividing,
Fermi Answer: 15
How many liters of water precipitated on Earth in 2016?
Earth SA = about 10*E3^2 = E7 km^2
going to guess that average rainfall around the earth is E1 cm which is E-4 km
that would be a total of E3 km^3 of rain
That equals E12 m^3,which is E15 liters
Answer: E15
Apparently average rainfall is 100cm so E2 cm
Earth surface area is E9 (oops)
In liters, multiplying those gives 5E17, or E18
Actual answer: E18
Lets say every SAT test taker in a year used some number of SAT prep books, that if stacked (horizontally/shortest side for height), would equal their height.
How many SAT prep books were used overall that year?

EDIT: sorry for double post, was trying to edit something but accidentally quoted
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