Unome wrote:
What was the average rate of text messages sent, in messages per second, during 2016?
Assuming 4 texts per person per day --> ~[strikethrough]30 billion[/strikethrough]3x10^10 texts per day --> divided by ~[strikethrough]100,000[/strikethrough]10^5 seconds per day --> ~3x10^5 texts per second -->10^5 --> FA=5
Using the Google machine yields the figure of 23 billion per day in 2015 (I wasn't able to find a figure for 2016 from the couple sites I clicked on, so I'll just use 2015). Divide this number by 86400 seconds per day to get 266,203 = 2.7x10^5. FA=5
Source: https://teckst.com/19-text-messaging-stats-that-will-blow-your-mind/
How many cans of diet coke would be required to fill crater lake? Sponsored by Coca-Cola
- I have no idea how big crater lake is, nor can I even remember what it is. I'll guess 500 km^2 and 200 meters deep, which gives a total volume of E2 cubic kilometers which equals E11 cubic meters.
- This is equal to E17 milliliters.
- A diet coke is probably like E2 milliliters, so E15
Fermi Answer: 15
The volume of Crater Lake (which is apparently in Oregon) is 18.7 cubic kilometers (from Wikipedia - numbers vary slightly between sources). The volume of a diet coke can is 12 fluid ounces, which is 354.88 milliliters. Dividing these gives 5.27E13
True Answer: 13
If you used a single standard size #2 pencil to draw a line (assuming an infinite amount of standard printer paper), and used up the pencil entirely, how many times could that line cross the English channel (at the channel's narrowest point)?
- I remember reading somewhere that you could draw a line for 40 miles-ish with one pencil.
- The English Channel might be 20 miles wide at its narrowest.
So dividing gives 2, Fermi Answer of 0?
- Google says you could draw a line for 35 miles.
- At it's narrowest, it's 20.9 miles.
- So Fermi Answer of 0
Sticking with the topic of bodies of water, how many dictionaries do you need to print out to have the amount of ink used to be equal to the volume of the Pacific Ocean?
- Most printers print somewhere in the range of E2-E3 pages on one cartridge, so I'll go with 5E2.
- A cartiridge is like 5E1 mL, dividing gives E1 pages per mL. I'll underestimate for simplicity and say it's E-1 dictionaries per mL.
- The Earth is ~E4 km radius, so ~ 4E8 km^2 surface area (can't remember the exact formula sadly), ~E8 km^2 Pacific Ocean, ~E18 cm^2.
- The depth is ~E1 km so ~E6 cm. Multiplying gives E24 mL of water. I should really memorize the volume of the Pacific Ocean at some point.
E24 mL multiplied by E-1 dictionaries per mL gives E23 dictionaries.
Fermi Answer: 23
- It appears that the ratio is 2E1 pages per mL of ink. Sicne you didn't specify I'll assume a dictionary size of 1000 pages - this gives 2E-2 dictionaries per mL.
- The volume of the Pacific Ocean is around 7.1E23 mL.
- Multiplying gives 1.42E22.
Fermi Answer: 22
How many times could a full Boeing 747 fly around the Earth if it had begun when the Earth was formed? Assume infinite fuel, continuous top speed, and an atmosphere identical to the present one.
- Earth formed 4.6 billion years ago, or 5E9
- It could probably go 1000 kph, or 1E3.
- The Earth is around 40000 km in circumference, because light can go around it 7 times per second. This is 4E4.
- There's around 8,000 hours in a year, so the earth is 4E13 hours old.
- So it would take 40 hours for a 747 to go around the Earth once, or 4E1.
- Dividing gives a nice Fermi Answer of 12.
- The Earth formed 4.03 E13 hours ago.
- Top speed of a 747 is 9.2 E2 kilometers per hour.
- Earth is almost exactly 40,000 km in circumference, or 4 E4.
- So it would take 43.48 hours for a 747 to go around the Earth once.
- Dividing 4.03 E13 by 43.48 gives us 9.27 E11, which gives us
- Fermi Answer of 12
How many pages could a standard printer print off if it started printing when the universe formed? Assume infinite ink.
- 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
- 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.
Unome wrote: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.
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
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
i can't feel my arms wtf i think i'm turning into a lamp
voted least likely to sleep 2018, most likely to sleep in class 2017+2018, biggest procrastinator 2018
Unome wrote: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.
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
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
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]
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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Unome wrote: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.
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
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
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]
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?
So 2^2017 is about (2^3.5)^500 which is about 10^500
Answer: 500
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?
i can't feel my arms wtf i think i'm turning into a lamp
voted least likely to sleep 2018, most likely to sleep in class 2017+2018, biggest procrastinator 2018
- 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
- 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?
