Fermi Questions C

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

Postby Unome » June 24th, 2017, 2:36 pm

Assuming this standard printer is an inkjet:
- For a fully covered page (i.e. there's no blank horizontal stretches - the speed isn't really affected very much by how much actual stuff is on the page), I'd estimate maybe E1 seconds to print. - The universe is 1.3E10 years old. - ~8.5E4 seconds per day, ~3E7 seconds per year. The universe is therefore around 4E17 seconds old. Dividing gives E16, but I've underestimated slightly so I'll go with E17 Fermi Answer: 17
Actual Answer
- The universe is 4.355E17 seconds old, based on an age of 13.8 billions years. The estimates are all within around .05 billion years of that. - It's hard to find data on inkjet printers in general (hence why specifying a model is ideal) but most sources seem to agree on around 10 seconds per page. Therefore, 4.355E16 Fermi Answer: 16
How many times more ink is contained in the printed words of King James Bible as compared to the Gettysburg Address? Assume the same font and font size.
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Re: Fermi Questions C

Postby NeilMehta » June 24th, 2017, 9:08 pm

How many times more ink is contained in the printed words of King James Bible as compared to the Gettysburg Address? Assume the same font and font size.
Attempt
I quite honestly have no idea how many pages the King James Bible has, but the book sounds long so I'll go with about 1000. The Gettysburg address was really short - half a page? My answer: E3
Actual answer
King James Bible: 783,137 words Gettysburg address: 272 words The former has just about 2879 times more words, and just that many times more ink, than the latter. Actual answer: E3
How many Wright Stuff models would it take to equal the weight of one Boeing 737
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Re: Fermi Questions C

Postby Adi1008 » June 24th, 2017, 9:43 pm

How many times more ink is contained in the printed words of King James Bible as compared to the Gettysburg Address? Assume the same font and font size.
Attempt
I quite honestly have no idea how many pages the King James Bible has, but the book sounds long so I'll go with about 1000. The Gettysburg address was really short - half a page? My answer: E3
Actual answer
King James Bible: 783,137 words Gettysburg address: 272 words The former has just about 2879 times more words, and just that many times more ink, than the latter. Actual answer: E3
How many Wright Stuff models would it take to equal the weight of one Boeing 737
My attempt
Wright Stuff is similar to Helicopters in that there is a minimum weight to the device. For helicopters, I think it was around 2.5 grams = 2.5E-3 kg. A Wright Stuff model seems much bigger. I'll guess that the minimum weight for a WS model is 10 grams = 0.010 kg so most models are around this mass. A Boeing 737 is really big, probably around 100 cars. I'll say a car is 10^3 kg, so a Boeing 737 is about 10^5 kg. 10^5 kg/10^-2 = 10^7 WS models, so a [b]Fermi Answer of 7[/b]
Actual Answer
A WS has a minimum mass of 7 grams = 0.007 kg. A Boeing 737 has an empty, operational mass of 4.1E4 kg. Dividing these gives 5.6E6, so a [b]Fermi Answer of 7[/b]
My question: What's 2^2017?
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Re: Fermi Questions C

Postby NeilMehta » June 24th, 2017, 10:04 pm

How many times more ink is contained in the printed words of King James Bible as compared to the Gettysburg Address? Assume the same font and font size.
Attempt
I quite honestly have no idea how many pages the King James Bible has, but the book sounds long so I'll go with about 1000. The Gettysburg address was really short - half a page? My answer: E3
Actual answer
King James Bible: 783,137 words Gettysburg address: 272 words The former has just about 2879 times more words, and just that many times more ink, than the latter. Actual answer: E3
How many Wright Stuff models would it take to equal the weight of one Boeing 737
My attempt
Wright Stuff is similar to Helicopters in that there is a minimum weight to the device. For helicopters, I think it was around 2.5 grams = 2.5E-3 kg. A Wright Stuff model seems much bigger. I'll guess that the minimum weight for a WS model is 10 grams = 0.010 kg so most models are around this mass. A Boeing 737 is really big, probably around 100 cars. I'll say a car is 10^3 kg, so a Boeing 737 is about 10^5 kg. 10^5 kg/10^-2 = 10^7 WS models, so a [b]Fermi Answer of 7[/b]
Actual Answer
A WS has a minimum mass of 7 grams = 0.007 kg. A Boeing 737 has an empty, operational mass of 4.1E4 kg. Dividing these gives 5.6E6, so a [b]Fermi Answer of 7[/b]
My question: What's 2^2017?
attempt
So 2^2017 is about (2^3.5)^500 which is about 10^500 Answer: 500
actual answer
So technically it should be (2^log2 10)^(2017/log2 10) Plug that into wolframalpha and you get just around 1.5*10^607 *At this point I realize I could've just plugged in the original problem* Actual: 607
How many Mars bars would it take lying end to end lengthwise to circle around the planet of mars?
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Re: Fermi Questions C

Postby Unome » June 25th, 2017, 7:07 am

Attempt
- Mars is around 1/2 length of the Earth. Thanks to an early post in this thread I've memorized the circumference of Earth, 4E7 meters, so Mars is 2E7 meters. - A Mars bar is somewhat less than E1 centimeters which is equal to somewhat less than E-1 meters, dividing gives E8, but I can probably round that to E9. Fermi Answer: 9
Actual Answer
- A Mars bar is 9.8 centimeters. Mars has a circumference of 2134400000 centimeters. Clearly I shouldn't have rounded that extra bit. Fermi Answer: 8
How many tennis balls are produced each year in the entire world?
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Re: Fermi Questions C

Postby Justin72835 » June 25th, 2017, 7:34 am

Attempt
- Mars is around 1/2 length of the Earth. Thanks to an early post in this thread I've memorized the circumference of Earth, 4E7 meters, so Mars is 2E7 meters. - A Mars bar is somewhat less than E1 centimeters which is equal to somewhat less than E-1 meters, dividing gives E8, but I can probably round that to E9. Fermi Answer: 9
Actual Answer
- A Mars bar is 9.8 centimeters. Mars has a circumference of 2134400000 centimeters. Clearly I shouldn't have rounded that extra bit. Fermi Answer: 8
How many tennis balls are produced each year in the entire world?
Attempt
- I'd say that for every tennis ball factory, a tennis ball is produced every minute. So that's about 1440 a day at each factory - There are probably about 1000 tennis ball factories all over the world so that's 1 440 000 being produced each day - Multiply that with 365 for each day of the year to get an answer of 525 600 000 which gives E9 Fermi Answer: 9
Actual Answer
- Wikipedia says around 325 million tennis balls around produced each year, which gives a result of E9. Fermi Answer: 9
How many times larger is free fall acceleration on the Sun than the acceleration of a bullet in the barrel of a gun?
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Re: Fermi Questions C

Postby Adi1008 » June 25th, 2017, 10:14 am

Attempt
- Mars is around 1/2 length of the Earth. Thanks to an early post in this thread I've memorized the circumference of Earth, 4E7 meters, so Mars is 2E7 meters. - A Mars bar is somewhat less than E1 centimeters which is equal to somewhat less than E-1 meters, dividing gives E8, but I can probably round that to E9. Fermi Answer: 9
Actual Answer
- A Mars bar is 9.8 centimeters. Mars has a circumference of 2134400000 centimeters. Clearly I shouldn't have rounded that extra bit. Fermi Answer: 8
How many tennis balls are produced each year in the entire world?
Attempt
- I'd say that for every tennis ball factory, a tennis ball is produced every minute. So that's about 1440 a day at each factory - There are probably about 1000 tennis ball factories all over the world so that's 1 440 000 being produced each day - Multiply that with 365 for each day of the year to get an answer of 525 600 000 which gives E9 Fermi Answer: 9
Actual Answer
- Wikipedia says around 325 million tennis balls around produced each year, which gives a result of E9. Fermi Answer: 9
How many times larger is free fall acceleration on the Sun than the acceleration of a bullet in the barrel of a gun?
My Attempt
The ratio we're looking for here is [math]a_{sun}/a_{gun}[/math] The acceleration on the sun can be found using [math]\frac{GM_{sun}}{r_{sun}^2}[/math]. Let [math]G \approx 6.67 \times 10^{-11}, M \approx 1.98 \times 10^{30}, r_{sun} \approx 6.955 \times 10^8[/math]. From this, I get [math]a_{sun} \approx 2 \times 10^2[/math] Using [math]v_f^2 - v_0^2 = 2as[/math] and assuming [math]s = 0.10, v_f = 400[/math], I get [math]a_{gun} \approx 8 \times 10^5[/math] The ratio of these two is [math]2.5 \times 10^{-4} \therefore FA = -4[/math]
Actual answer
[url=http://hypertextbook.com/facts/2003/MichaelTse.shtml]Hypertextbook[/url] gives [math]a = 4.4 \times 10^5[/math]. Plugging it into [url=https://www.wolframalpha.com/input/?i=(((6.67*10%5E-11)*(1.98*10%5E30))%2F((6.955*10%5E8)%5E2))%2F(4.4*10%5E5)]Wolfram Alpha[/url] gives [math]FA = -3[/math]
My question: What's 13^81?
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Re: Fermi Questions C

Postby Justin72835 » June 25th, 2017, 12:46 pm

Attempt
- Mars is around 1/2 length of the Earth. Thanks to an early post in this thread I've memorized the circumference of Earth, 4E7 meters, so Mars is 2E7 meters. - A Mars bar is somewhat less than E1 centimeters which is equal to somewhat less than E-1 meters, dividing gives E8, but I can probably round that to E9. Fermi Answer: 9
Actual Answer
- A Mars bar is 9.8 centimeters. Mars has a circumference of 2134400000 centimeters. Clearly I shouldn't have rounded that extra bit. Fermi Answer: 8
How many tennis balls are produced each year in the entire world?
Attempt
- I'd say that for every tennis ball factory, a tennis ball is produced every minute. So that's about 1440 a day at each factory - There are probably about 1000 tennis ball factories all over the world so that's 1 440 000 being produced each day - Multiply that with 365 for each day of the year to get an answer of 525 600 000 which gives E9 Fermi Answer: 9
Actual Answer
- Wikipedia says around 325 million tennis balls around produced each year, which gives a result of E9. Fermi Answer: 9
How many times larger is free fall acceleration on the Sun than the acceleration of a bullet in the barrel of a gun?
My Attempt
The ratio we're looking for here is [math]a_{sun}/a_{gun}[/math] The acceleration on the sun can be found using [math]\frac{GM_{sun}}{r_{sun}^2}[/math]. Let [math]G \approx 6.67 \times 10^{-11}, M \approx 1.98 \times 10^{30}, r_{sun} \approx 6.955 \times 10^8[/math]. From this, I get [math]a_{sun} \approx 2 \times 10^2[/math] Using [math]v_f^2 - v_0^2 = 2as[/math] and assuming [math]s = 0.10, v_f = 400[/math], I get [math]a_{gun} \approx 8 \times 10^5[/math] The ratio of these two is [math]2.5 \times 10^{-4} \therefore FA = -4[/math]
Actual answer
[url=http://hypertextbook.com/facts/2003/MichaelTse.shtml]Hypertextbook[/url] gives [math]a = 4.4 \times 10^5[/math]. Plugging it into [url=https://www.wolframalpha.com/input/?i=(((6.67*10%5E-11)*(1.98*10%5E30))%2F((6.955*10%5E8)%5E2))%2F(4.4*10%5E5)]Wolfram Alpha[/url] gives [math]FA = -3[/math]
My question: What's 13^81?
Attempt
- http://imgur.com/gallery/k4Erf (this is a picture of my written response) Fermi Answer: 91
Actual Answer
- 13^81 = 1.6959441E+90, which gives a Fermi Answer of 90. Fermi Answer: 90
How many dollar bills would it take to stretch from the Sun to Pluto?
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Re: Fermi Questions C

Postby Unome » June 25th, 2017, 2:17 pm

How many dollar bills would it take to stretch from the Sun to Pluto?
Attempt
- Assuming end to end, a dollar bill is around 1.5E-1 meters. - Pluto orbits at like 40 AU (can't remember), 1 AU is around 1.5E13 meters, so that's around 6E14 meters. Dividing: Fermi Answer: 15
Actual Answer
- Pluto is 5.9E12 meters from the Sun on average. For some reason I thought a million was E8 and based my numbers of off that... - A US dollar bill is 15.61 cm long. - Dividing gives 3.78E13 meters. Fermi Answer: 13
Over the course of one week, how many neutrinos interact with matter on Earth?
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Re: Fermi Questions C

Postby Adi1008 » June 25th, 2017, 6:30 pm

How many dollar bills would it take to stretch from the Sun to Pluto?
Attempt
- Assuming end to end, a dollar bill is around 1.5E-1 meters. - Pluto orbits at like 40 AU (can't remember), 1 AU is around 1.5E13 meters, so that's around 6E14 meters. Dividing: Fermi Answer: 15
Actual Answer
- Pluto is 5.9E12 meters from the Sun on average. For some reason I thought a million was E8 and based my numbers of off that... - A US dollar bill is 15.61 cm long. - Dividing gives 3.78E13 meters. Fermi Answer: 13
Over the course of one week, how many neutrinos interact with matter on Earth?
My attempt
Unfortunately, I don't have the solar neutrino flux memorized. Luckily, however, there's a way to figure it out theoretically. The proton-proton chain in the Sun is the dominant nuclear reaction. In the Sun, the majority of the energy is released through PP1 (as opposed to PP2 or PP3), which releases 26.2 MeV. [math]26.2 MeV \approx 4 \times 10^{-12} J[/math]. This is the amount of "energy" released during one reaction. I know from prior Astronomy knowledge that 2 neutrinos are released per reaction and the total amount of energy the Sun emits per second (luminosity). I can set these equal to each other and then solve for the number of reactions, and then multiply by two. Then, using the inverse-square law, I can approximate the flux of neutrinos at Earth. [math]n_{reactions}\times 4 \times 10^{-12} = 4 \times 10^{26}[/math] so [math]n_{reactions} \approx 10^{38}[/math]. Therefore, the amount of neutrinos will be [math]n_{\nu} \approx 2 \times 10^{38}[/math] Now, using the inverse square law: the flux at Earth will be about [math]\frac{n_{\nu}}{4\pi\times (1.5 \times 10^{11})^2}[/math]. [math]4\pi\times 2.25\approx 25[/math]. [math]\frac{2\times 10^{38}}{25\times 10^{22}} = 8\times 10^{14}[/math]. This is the solar neutrino flux, in meters, at the distance from Earth Now, I multiply this by the cross sectional area of Earth. I know that the radius of the Earth is ~6370000 m. [math]\pi \times (10^7)^2 \approx 3 \times 10^{14}[/math]. The product is approximately [math]10^{29}[/math] There are roughly [math]10^6[/math] seconds in a week. [math]10^6 \times 10^{29} = 10^{35} \therefore FA = 35[/math]
Actual answer
The flux is about [math]7 \times 10^14 \nu \times m^{-2} \times s^{-1}[/math], so the Fermi Answer is still 35
How many 2x2 LEGO bricks would you need to cover all of Texas?
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Re: Fermi Questions C

Postby NeilMehta » June 25th, 2017, 9:01 pm

attempt
A 2x2 Lego brick is about 1cm x 1cm, or 1 cm^3 One km^2 of this has e10 bricks I'm taking a guess and saying Texas is just about 400x400km, so e5 km^2 That means we have about e15 Answer: 15
solution
Brick area = 1.2324 cm^2 Texas area: 268597 mi^2=6.957*10^15 cm^2 Dividing gives 5.645*10^15 Actual answer=16
How many grains of sand would it take to fill the Gulf of Mexico?
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Re: Fermi Questions C

Postby Justin72835 » June 25th, 2017, 9:04 pm

Attempt
To fill a liter of sand you would need approximately 1,000,000 grains of sand. So that would mean that a cubic meter holds 1e9 grains of sand. The length of the Gulf of Mexico is about 1,000 km, which would give an area of 1,000,000 square kilometers. The depth of the Gulf is around 2 or 3 kilometers so I'll round that to 1 kilometer. This makes the volume of the Gulf of Mexico 1,000,000 cubic kilometers. Using dimensional analysis: 1e9 (grains in a cubic meter) * 1e9 (cubic meters in a cubic kilometer) * 1e6 (volume of the Gulf of Mexico) = 1e24 Fermi Answer: 24
Actual Answer
The actual volume of the Gulf of Mexico is 2.5 million cubic kilometers, while the number of grains of sand in a cubic meter is 8 billion, depending on the size of the grains. Using dimensional analysis: 8e9 (grains in a cubic meter) * 1e9 (cubic meters in a cubic kilometer) * 2.5e6 (volume of the Gulf of Mexico) = 2e25 Fermi Answer: 25
How long would it take for an ant to travel from Paris, France to Beijing, China?

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Re: Fermi Questions C

Postby Unome » June 26th, 2017, 7:17 am

Attempt, assuming the ant doesn't stop and rest for several hours or something. Also, you really should have specified the units
- Estimating ant speed at 5E-2 meters per second - The distance from Paris to Beijing (I'll assume great circle distance, since you didn't specify) is probably around E7 meters. Dividing to get the time in seconds, Fermi Answer: 9
Actual Answer
- Looks like 5 cm per second is a good estimate (based on [url=http://www.theincredibleant.com/ant-how/how-fast-are-ants]this source[/url]). - The distance between the Charles deg Gaulle and Beijing International airports is 8212 km. - This works out to around 1.6E8 seconds. Fermi Answer: 8
How many posts on scioly.org, as of the June 2008 forums, contain the word "microcontroller"?
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Re: Fermi Questions C

Postby Adi1008 » June 26th, 2017, 10:41 am

Attempt, assuming the ant doesn't stop and rest for several hours or something. Also, you really should have specified the units
- Estimating ant speed at 5E-2 meters per second - The distance from Paris to Beijing (I'll assume great circle distance, since you didn't specify) is probably around E7 meters. Dividing to get the time in seconds, Fermi Answer: 9
Actual Answer
- Looks like 5 cm per second is a good estimate (based on [url=http://www.theincredibleant.com/ant-how/how-fast-are-ants]this source[/url]). - The distance between the Charles deg Gaulle and Beijing International airports is 8212 km. - This works out to around 1.6E8 seconds. Fermi Answer: 8
How many posts on scioly.org, as of the June 2008 forums, contain the word "microcontroller"?
My attempt
I have no idea, honestly. Microcontrollers are an important part of events involving electronics, such as Electric Vehicle. I feel like there should be more than 50 posts with that word, but less than 500. Fermi answer of 2
Actual answer
Doing searches on the forums, I find: 2009: 16 posts 2010: 2 posts 2011: 2 posts 2012: 8 posts 2013: 2 posts 2014: 4 posts 2015: 1 post 2016: 52 posts (!!!) probably solely because of Electric Vehicle 2017: 4 posts [b]Total: 91[/b] Fermi answer: 2!!!
How far could a cheetah run (at top speed, assuming it doesn't get tired), in picometers, in the time since the Big Bang?
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Re: Fermi Questions C

Postby Unome » June 26th, 2017, 12:07 pm

How far could a cheetah run (at top speed, assuming it doesn't get tired), in picometers, in the time since the Big Bang?
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.
Attempt
- E2 km/hour, which is E17 picometers per hour. - The universe is 1.3E10 years old. - 365*24 is a little less than E3 hours per year, so that makes E13 hours (yay convenient canceling!) Fermi Answer: 30
Actual Answer
- A cheetah runs at around 115 km/hour - Using the age of the universe, 13.8 billions years (converted to hours), we get 1.39E16 kilometers. Converting to picometers, that's 1.39E31. Fermi Answer: 31
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)
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