Fermi Questions C

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Re: Fermi Questions C

Postby RJohnson » September 14th, 2017, 2:26 pm

NeilMehta wrote:
New question: what is price of single item in the displays at the NYC Apple Store converted to japanese yen?


Not entirely sure what you meant here, but here's my guess:

My guess is the new iPhone is what is currently in the displays at the apple store, price = $800 = 8e2. If I remember correctly, the yen floats around e2 yen to the dollar, so that takes the price in yen to 8e4, which rounds to fermi answer of 5.

Next Question: How many carbon atoms are in the average american?

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Re: Fermi Questions C

Postby whythelongface » September 14th, 2017, 2:46 pm

RJohnson wrote:
NeilMehta wrote:
New question: what is price of single item in the displays at the NYC Apple Store converted to japanese yen?


Not entirely sure what you meant here, but here's my guess:

My guess is the new iPhone is what is currently in the displays at the apple store, price = $800 = 8e2. If I remember correctly, the yen floats around e2 yen to the dollar, so that takes the price in yen to 8e4, which rounds to fermi answer of 5.

Next Question: How many carbon atoms are in the average american?


Attempt
The average American masses maybe 2E2kg. That's 2E5g.

Assuming the body is only about 25% non-water by mass, that's 5E4g of organic molecules and proteins. Estimating that organic molecules are around 80% carbon by mass, we get 4E4g of carbon. 6E23 atoms mass 12 grams, so the Fermi Answer would be 27.

Edit: Oh dearie me, I really mistook kilograms for pounds. The average human weighs maybe 70 kg. The answer, however, was 27, which means that another of my estimations was an order of magnitude off.


The Tsar Bomba was a thermonuclear test with a warhead yielding a total energy output equivalent to 50 megatons of TNT (already this is more information than I should be giving you, but I think this might be obscure). Using this information, calculate the number of simultaneous detonations needed to equal the energy output of the Sun throughout its lifetime.
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Re: Fermi Questions C

Postby NeilMehta » September 14th, 2017, 3:08 pm

RJohnson wrote:
NeilMehta wrote:
New question: what is price of single item in the displays at the NYC Apple Store converted to japanese yen?


Not entirely sure what you meant here, but here's my guess:

My guess is the new iPhone is what is currently in the displays at the apple store, price = $800 = 8e2. If I remember correctly, the yen floats around e2 yen to the dollar, so that takes the price in yen to 8e4, which rounds to fermi answer of 5.

Next Question: How many carbon atoms are in the average american?

Gah, sorry, I meant to say every single.. my fault
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Re: Fermi Questions C

Postby Torterra » September 15th, 2017, 12:01 pm

Answer
The luminosity of the sun is about 1E24W. I know this fluctuates, but I'm going to assume it's constant. Assume the sun has a lifetime of 10 billion, or 1E10 years. There are 3E7 seconds in a year, which gives a total time of 3E17s. P=E/t, so the total energy is 3E41J. 1 kilogram of TNT has about 1E6J. This means that one Tsar Bomba has about 5E16J. Divide 3E41 by 5E16 to get an answer of 25.



Question: If you stack every single order of playing cards(unique shuffling of a standard 52 card deck) on top of each other, how many times can you go from earth to the sun?

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Re: Fermi Questions C

Postby appleshake123 » September 17th, 2017, 4:47 am

Cards
FA:52
There's 52! orders possible. Each order of a deck is about half an inch. The light from the sun takes like 8.33 min to get from there to the earth and e1min *e2s/min * e8m/s = e11 metre
Distance of 52! orders = 52!* e-2m =
2*3 = 6
7*8 is about 50
22*23 is about 500
30*31 is about 1000
51*52 is a bit more than 2500
5/2ish e1s = e3
17/2ish e2s = e17
29/2ish e3s = e45
Total order amount is about e65.
e65 orders * e-2m size * 1 eart-sun amount/e11m = e52

Real thing
So the first part of the distance from the earth to teh sun is right at 1.5e11
52! = 8e67
Deck of card thickness = 5/8 inch oro 1.5e-2m
8e67 * 1.5e-2m/1 * 1d/1.5e11 = e55
I underestimated the factorial amount greatly which I should of expected as pairs between 32-52(10 pairs) are rounded down to e3 versus the 22-31(about 5 pairs) rounded up. I shouldve just added a few more exponents after the underestimation.

If every student only ate cookies for the rest of their student career as replacement for their daily meals(in terms of Caloric intake), how many hours of jump pullups amongst every student would they need to burn 60% of their total cookie consumption?
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Re: Fermi Questions C

Postby whythelongface » September 24th, 2017, 1:09 pm

Answer
Since the question is a bit vague, I will make a few assumptions. I will take "student" to mean K-12. Also, judging from your question, wouldn't any food item do? You're not asking for anything based off the number of cookies, you're asking total Caloric intake, which should stay constant whether you're eating cookies or pizzas. In other words, unless I'm misinterpreting something, "cookies" didn't have to be specified. I probably didn't understand the question fully, but here goes.

The growing child needs on an average about 1800 Calories a day. (I am basing from K-12). 60% of that is 1000 C/day. That's 3.7E5 Calories a year, or about 4.5E6 Calories K-12.

Assume one hour of high-intensity exercise burns 100 Calories, That gives a Fermi Answer of 4.

If you meant "entire academic career" like undergrad, grad school, I'd say that jacks it up by about a magnitude, yielding 5.


If you allowed an entire grocery store's stock of 12-oz coke to go flat, what is the volume the carbon dioxide would occupy in one atmosphere?
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Re: Fermi Questions C

Postby NeilMehta » October 7th, 2017, 6:14 pm

whythelongface wrote:
Answer
Since the question is a bit vague, I will make a few assumptions. I will take "student" to mean K-12. Also, judging from your question, wouldn't any food item do? You're not asking for anything based off the number of cookies, you're asking total Caloric intake, which should stay constant whether you're eating cookies or pizzas. In other words, unless I'm misinterpreting something, "cookies" didn't have to be specified. I probably didn't understand the question fully, but here goes.

The growing child needs on an average about 1800 Calories a day. (I am basing from K-12). 60% of that is 1000 C/day. That's 3.7E5 Calories a year, or about 4.5E6 Calories K-12.

Assume one hour of high-intensity exercise burns 100 Calories, That gives a Fermi Answer of 4.

If you meant "entire academic career" like undergrad, grad school, I'd say that jacks it up by about a magnitude, yielding 5.


If you allowed an entire grocery store's stock of 12-oz coke to go flat, what is the volume the carbon dioxide would occupy in one atmosphere?

gonna completely guess, but, let’s say each bottle has E-1 m^3 of CO2. Assuming the store has about E2 packs of 36 each, there should be E1 m^3

Solution:
apparently each can has about E-3 of CO2...
True answer: E-1

New question: how many base pairs of DNA are there in a human?
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Re: Fermi Questions C

Postby whythelongface » October 7th, 2017, 6:43 pm

Answer
Eukaryotic DNA has about E6 bps per molecule. If the human body has E10 cells (purely guessing right now, likely a very large underestimation), that means E16 bps, or E13 kbps.

Actual answer: REALLY off. For starters, the human body has about E13 cells, and each molecule of eukaryotic DNA has about E9 bp (I must've meant E6 kbp.) So it's actually E22 bps
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Re: Fermi Questions C

Postby Torterra » October 19th, 2017, 10:24 am

If the entire population of the earth was converted to energy and used to power a MacBook Air, how far would the chain of all the electrons that pass through the MacBook in that time stretch in light seconds?

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Re: Fermi Questions C

Postby Unome » October 19th, 2017, 11:59 am

Torterra wrote:converted to energy

I will assume that this is via the energy-mass equivalence.

Attempt
- 7E9 people, ~7E1 kg each, multiply to get 5E11 kg of human biomass.
- Converting to energy in joules by multiplying by 3E8 meters per second twice gives 1.5E28 joules.
- Joules to electron volts - ~1.5E19 (if I remember correctly), so ~3E47 eV - multiply by ~100 volts (random guess) to get ~E49 electrons.
- An electron is probably ~E-20 meters, so ~E29 meters.
- Convert to light second by dividing by 3E8 meters, gives ~E21.
Fermi Answer: 21

Actual Answer
- 2.87E11 total biomass.
- Convert to energy gives 2.583E28 Joules.
- It looks like the MacBook Air takes 110 volts, so that's 1.77E49 electrons.
- The classical electron diameter is 5.64E-15 meters (darn...) which gives almost exactly E35 meters.
- Converting gives 3.34E26 light-seconds.
Actual Fermi Answer: 26

You drop a golf ball from the top of the Empire State Building - when it hits the (perfectly elastic) ground, it causes a vibration. How many golf balls would have to simultaneously hit the ground at the same point to produce a seismic wave as powerful as the 1906 San Francisco earthquake?
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Re: Fermi Questions C

Postby Alex-RCHS » November 23rd, 2017, 7:11 am

Because of thanksgiving:

How many Joules of turkey energy are consumed by Americans on thanksgiving?

Also, how many (completely empty) Empire State buildings could you fill with those turkeys?
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Re: Fermi Questions C

Postby whythelongface » December 13th, 2017, 3:31 pm

I'll ask a new question: How many cubic meters of sweat does Valery Gergiev extrude while conducting Stravinsky's Firebird? :?: :?:
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Re: Fermi Questions C

Postby AnikJain3pro5un00bs » January 1st, 2018, 9:09 pm

Ha! A funny question! I do not know if the man made many recordings of Firebird, but I'll choose the famous one, since the others are probably the same!

Ok! Here we go!

ok so, Firebird is about 45 minutes long if my memory serves me correct, not including audience claps.
Now Valery does quite a bit of working out, haha, and we might equate it to a normal workout...! In fact, we shall do so!
I do not know how much I sweat when I work out, but I do know how much water I usually drink! About half a litre, I think. And maybe I sweat as much.
Now we must examine, if Valery is really doing so much of a workout! I think not. Perhaps.
However, we must look that he is conducting in a hot suit, and waving a stick around for 45 minutes! Perhaps this makes you sweat a lot!
Furthermore, he has much energy! I remember that he does sweat a lot even at the beginning, before the finale, et cetera...so I would have to go with 1 litre, which comes out to be -3, I think.


What the actual answer is, I do not think we shall ever know! But we can look at the actual numbers, and it says you sweat around a liter every hour when you exercise! So I think I may be right.

Fun problem I have for you! So fun, I make into a poem. Or a song, you can sing if you want :D
I went to tower of Hanoi one day,
and there I saw them stacking rings!
In my head maths began to play,
And I wondered how many ways to stack those things.
How many rings, I don't know, but let us say there were 88.
On three stands, were these rings,
Each stacked on another with a radius more great.
All at first were on one stand,
A move is to take the top ring of a stand and put it on another stand, fun!
But the catch here really is quite grand!
You cannot put it on a smaller one.
So I sat there and I ask to me,
If all start on one stand and we make only legal move,
At any time, how many legal positions can there be?
If solve this you can, a Fermi master you will prove!
:D hoping to enjoy another year of the science olympiad!!

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Re: Fermi Questions C

Postby Name » January 21st, 2018, 10:26 pm

AnikJain3pro5un00bs wrote:Ha! A funny question! I do not know if the man made many recordings of Firebird, but I'll choose the famous one, since the others are probably the same!

Ok! Here we go!

ok so, Firebird is about 45 minutes long if my memory serves me correct, not including audience claps.
Now Valery does quite a bit of working out, haha, and we might equate it to a normal workout...! In fact, we shall do so!
I do not know how much I sweat when I work out, but I do know how much water I usually drink! About half a litre, I think. And maybe I sweat as much.
Now we must examine, if Valery is really doing so much of a workout! I think not. Perhaps.
However, we must look that he is conducting in a hot suit, and waving a stick around for 45 minutes! Perhaps this makes you sweat a lot!
Furthermore, he has much energy! I remember that he does sweat a lot even at the beginning, before the finale, et cetera...so I would have to go with 1 litre, which comes out to be -3, I think.


What the actual answer is, I do not think we shall ever know! But we can look at the actual numbers, and it says you sweat around a liter every hour when you exercise! So I think I may be right.

Fun problem I have for you! So fun, I make into a poem. Or a song, you can sing if you want :D
I went to tower of Hanoi one day,
and there I saw them stacking rings!
In my head maths began to play,
And I wondered how many ways to stack those things.
How many rings, I don't know, but let us say there were 88.
On three stands, were these rings,
Each stacked on another with a radius more great.
All at first were on one stand,
A move is to take the top ring of a stand and put it on another stand, fun!
But the catch here really is quite grand!
You cannot put it on a smaller one.
So I sat there and I ask to me,
If all start on one stand and we make only legal move,
At any time, how many legal positions can there be?
If solve this you can, a Fermi master you will prove!

Hmmm 1 ring 3 combination 2 ring is 9 and 3 is 27 I think which means equation is 3^n. Log(3 is .477 times 88 is 42

Question: what is the gravitational acceleration In m/s^2 of 2 identical balls weighing the mass of the sun separated by the plancks length

Question 2: If the balls both gained 1 electron how many newtons of force would the balls initially experience
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Re: Fermi Questions C

Postby NeilMehta » February 8th, 2018, 6:28 am

Been well over two weeks, will try to start this again:
if you took every blood vessel in your body and laid it end to end, how many times could that length (ignoring the width!) wrap around a neutron?
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