Name wrote:
The number of moons that fit into earth can be determined by (radius earth/radius moon)^3
(6371/1737)^3 ends up to be 49.3... which would round to 1

Firstly, that is legitimately unfortunate...

Name wrote:
Question: according to a interesting video I was watching (by PBS spacetime, it's probably worth checking out if your bored) the maximum amount of information (in bits) that can be stored in a system is = surface area (in plank units)/4. Given that the actual information stored in the universe is closer to around 10^90 bits how many times larger is the radius of the observable universe then the radius of the volume of sphere the total information in the observable universe can be stored in?

Secondly, let's go Matt O'Dowd!

Thirdly:

e90 = pi*r^2 (the four from the SA formula cancels out with the other four)
3e89ish = r^2
r = 45 (rounded up, because of the three)
The radius of the observable universe is something like 40 billion light years, because of some reason that I don't remember.
so you have 10 fermi light years (almost 11, but since we're rounding down later, we're fine)
16 meters per light year, so 26 meters. (but almost 27)
one planck length is 2e-35m, so 5e34 planck lengths per meter.
so, we get 26+35 (cuz of rounding up) = 61 planck lengths.
61-45 = [b]16[/b]

Fourthly

so, actually doing the calculation yields a Bekenstein radius of 5.6 e44 planck lengths (uh oh, my rounding will cause issues later...)
observable universe is actually 46.6 billion light years, or 2.75e61 planck lengths
dividing 2.75e61 by 5.6e44 gives 4.9e16, which just [i]barely[/i] makes my answer of [b]16[/b] correct.

Fifthly:
If I made a "1-dimensional" cup tower, whose base stretched from Melbourne to Los Angeles, using 16 oz red Solo cups, how many cups would I need? Assume a flat earth (and confirm the Illuminati while you're at it...), though I'm not sure if it would make a difference.

Bonus question: how much would all these cups cost, in US dollars? (not necessary to answer; may be fun to add to your answer.)

Re: Fermi Questions C

Posted: October 13th, 2018, 7:11 pm

by Name

PM2017 wrote:
If I made a "1-dimensional" cup tower, whose base stretched from Melbourne to Los Angeles, using 16 oz red Solo cups, how many cups would I need? Assume a flat earth (and confirm the Illuminati while you're at it...), though I'm not sure if it would make a difference.

Bonus question: how much would all these cups cost, in US dollars? (not necessary to answer; may be fun to add to your answer.)

Distance from melbourne to LA maybe around 5e3 km? (im probably wrong im bad at distance between 2 places estimating). A cup is maybe a decimeter or so. Cup tower is when you have like 1 cup, 2 cups under it etc so the final number is 1+2+3+...n, and can be modeled as n(n-1)/2 or basically n^2/2. The number of cups in the bottom row is the same as the distance which is n, so about 5e7 cups. 5e7^2 is 3.5e15/2 is about 15 fermi. Its probably like 1 cent/cup so e13 dollars.

distance is like 1.3e4 km. Cups are probably slightly larger, but still fermi answer 15. Thier like $10 for 50 or 20 cents each according to walmart (that seems expensive lol) so cost is 14

Question: How many primes are there under 10^1000

Re: Fermi Questions C

Posted: October 13th, 2018, 8:38 pm

by PM2017

prime number theorem states that #p ~ x/logx, so (e1000)/1000, so fermi 1000-3 = 997. I don't see how to verify this, and I was correct about the prime number theorem, and my math seems to be correct, so...

The Lucas numbers follow the same rule are Fibonacci numbers, but are arguably "better" than Fibonacci numbers. They start with 2, then 1. Then the "Fibonacci rule" kicks in, os you have 2,1,3,4,7,11,18,29...
What is the 62nd Lucas number?

prime number theorem states that #p ~ x/logx, so (e1000)/1000, so fermi 1000-3 = 997. I don't see how to verify this, and I was correct about the prime number theorem, and my math seems to be correct, so...

The Lucas numbers follow the same rule are Fibonacci numbers, but are arguably "better" than Fibonacci numbers. They start with 2, then 1. Then the "Fibonacci rule" kicks in, os you have 2,1,3,4,7,11,18,29...
What is the 62nd Lucas number?

Hmm for some reason I remembered the prime number theorum as x/ln(x) but most sources say x/log(x).

well I remember my estimation of Fibonacci sequence to be approximately 2.5^(1/2n) while rounding up a bit for higher numbers. If we assume it's just fibbonacci we get .398 times 31 = 12.338. it's slightly higher cause my equation isn't perfect and with the slightly larger equation Ill say 13

well online it says the 62nd Lucas number is 9e12 so Fermi 13
Now the video. Hmm that was really interesting. So phi (about 1.62)^n rounded is always the nth lucas number. Log phi is about .2095(62)= 13. Also using log phi is probably a better equation then log2.5(1/2n) lol. That was just something I stumbed upon when trying to figure out how to do fibbonacci numbers.

Question: given the sequence 1,2,2,4,8,32 what is the 13th number of the sequence? As a hint (and to slightly reduce the math) the fibbonacci numbers are 1,1,2,3,5,8,13,21,34,55,89,144...

(don't look before solving) https://oeis.org/A000301

Re: Fermi Questions C

Posted: October 14th, 2018, 9:20 am

by PM2017

Name wrote:
Question: given the sequence 1,2,2,4,8,32 what is the 13th number of the sequence? As a hint (and to slightly reduce the math) the fibbonacci numbers are 1,1,2,3,5,8,13,21,34,55,89,144...

1,2,2,4,8,32
if I take log base 2 of each of these, I get
0,1,1,2,3,5
oh wow, it's the Fibonacci sequence!
I'm going to be wary of your version of the Fibonacci sequence because log base(2) of the sequence in question starts with 0,1,1... not 1,1,2...
(I'm assuming that's why you only gave twelve values)
so the 13th term is 2^144. The Fermi answer would then be 0.301*144, which works out to ~43.3, so Fermi answer [b]43[/b].

The integer series database provided gives the thirteenth term a value of 22300745198530623141535718272648361505980416, and taking the log of that gives 43.35, so Fermi answer [b]43[/b].

An indestructible 10-meter sphere is spinning extremely rapidly. In fact, no one knows the angular velocity of this sphere. Given that the speed of light is 3e8 m/s, what is the smallest amount of time, in microseconds, that this sphere might complete a rotation?

Re: Fermi Questions C

Posted: October 14th, 2018, 3:46 pm

by Name

PM2017 wrote:
An indestructible 10-meter sphere is spinning extremely rapidly. In fact, no one knows the angular velocity of this sphere. Given that the speed of light is 3e8 m/s, what is the smallest amount of time, in microseconds, that this sphere might complete a rotation?

assuming you mean 10 m radius 2pir is about 60 meters/3e8 is 2e-7 sec or 2e-4 ms or -4 fermi

Attempt should be correct unless the 10 meters refers to something else instead of radius

Question: In 1991 a single atomic nucleas (named the oh my god particle) was detected traveling at 99.99999999999999999999951% the speed of light. Due to relativistic effects, it has a Lorentz factor (the factor at which time/length/mass changes while moving) of 3.2×10^11. At what speed would you have to throw a baseball so that the kinetic energy of the oh my god particle = the kinetic energy of the baseball

Re: Fermi Questions C

Posted: October 14th, 2018, 4:07 pm

by PM2017

Name wrote:

PM2017 wrote:
An indestructible 10-meter sphere is spinning extremely rapidly. In fact, no one knows the angular velocity of this sphere. Given that the speed of light is 3e8 m/s, what is the smallest amount of time, in microseconds, that this sphere might complete a rotation?

assuming you mean 10 m radius 2pir is about 60 meters/3e8 is 2e-7 sec or 2e-4 ms or -4 fermi

Attempt should be correct unless the 10 meters refers to something else instead of radius

Question: In 1991 a single atomic nucleas (named the oh my god particle) was detected traveling at 99.99999999999999999999951% the speed of light. Due to relativistic effects, it has a Lorentz factor (the factor at which time/length/mass changes while moving) of 3.2×10^11. At what speed would you have to throw a baseball so that the kinetic energy of the oh my god particle = the kinetic energy of the baseball

Wow, you really like PBS Spacetime...
So, I remember that the neutron had 45 J.
I'm assuming that a baseball is 0.1 kg, and e = 1/2*m*v^2
so, 90 = 0.1*v^2
900 = v^2
v = 30 (m/s)
I'm going to assume that you wanted meters/second...
fermi answer = [b]1[/b]

The wikipedia page for the oh-my-god particle [url=https://en.wikipedia.org/wiki/Oh-My-God_particle](article)[/url] says it would be 26m/s, so my answer of [b]1[/b] is correct.

"See You Again" by Wiz Khalifa, featuring Charlie Puth is one of the most viewed videos on youtube. Assuming that each view means that someone watched the whole video, on average, since it released, how many people are viewing the video at any given time.

(Can you we get someone other than just me and Name involved in this? I mean, I don't mind, but...)

Re: Fermi Questions C

Posted: October 16th, 2018, 9:43 pm

by acidbeaker

PM2017 wrote:
"See You Again" by Wiz Khalifa, featuring Charlie Puth is one of the most viewed videos on youtube. Assuming that each view means that someone watched the whole video, on average, since it released, how many people are viewing the video at any given time.

I've never done this type of problem so I'm BSing this at this point. I know that Despacito has 5 billion views or so, so I'm going to estimate See You Again has about 3 billion. It's been about 3 years, or 1.3 * 10^3 days since the music video released. Assuming linear view rate (which it's not) we can say there's about 10^6 views per day, or 1.4 * 10^3 views a minute. The song duration is about 3 minutes, so we get 1.4*3 * 10^3 views or 5.2*10^3. My final answer is 4.

Coachella is an annual music event in the Coachella Valley of South California. How many milliliters of water did those who attended Coachella drink during Coachella 2018?

Re: Fermi Questions C

Posted: October 17th, 2018, 2:33 pm

by PM2017

I'm going to assume that e5 people attended.
I know nothing about Coachella, so I'll assume it was 5 days long.
I drink about a liter to 1.5 liters of water per day, so 1.5e3 mililiters per person per day, but they're not going to be at the convention all day, so just e3 milliliter per person per day, so e8 milliliter per day. I'll round up because of the 5 days,
so fermi answer [b]9[/b]

So the combined total attendance was 750,000 people. Apparently, the average person drinks about 2000 ml a day. Multiply the two together, and take the log, and you get 9.17, so Fermi answer [b]9[/b]. (Even if I divided the 75000*2000 by 2, assuming that they drink half the water while not in Coachella, I still get 8.8 something, so 9 is still correct.)

Over the course of a regular NFL season, how many total passing yards are accumulated by all the teams combined?

Re: Fermi Questions C

Posted: October 17th, 2018, 5:02 pm

by Name

PM2017 wrote:
Over the course of a regular NFL season, how many total passing yards are accumulated by all the teams combined?

*sighs as I realize not everything in fermi is scientific facts and having a life is probably a good idea*

ik NBA has 30 teams so lets assume the same thing. theres maybe 15 games so 4.5e2 games. A quarterback might throw 100 yards so 4.5e4 yards. However two teams are playing so 9e4 yards or 5 fermi yards

In 2017 the lowest passing yards accumilated by a team was 2811 yards and the highest was 4431 yards. Assuming the average of the two accuratly represents all 30 teams (which it appears to do so) the final passing yards is about 1.1e5 yards or 5 fermi