Fermi Questions C

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

Postby Name » May 29th, 2018, 7:25 pm

TheChiScientist wrote:More or less the average word per tweet plus the theoretical "hot" air coming out at which rate it would take to make about 1 ml or water to reach 100C.
A morbid question....
How many nuclear ICBMs would it take to "nuke" the entire world. (Assume you're using UGM-133A Trident II D-5 W88/Mk5 missiles and blasting every single square cm of the world :twisted: )


Your 3 seems kinda low

Uh my solution to the cofveve assuming the same thing and water at room temp. Takes around 2.5E3 joules of energy for water to reach 100 degrees C and boil. One breathe is 500 ml, you can say cofveve around 10 times before breathing again so 50 ml. 50 ml weighs around 5E-5 kg. Specific heat of air is around 1 kj/kg (?) So 5E-2 joules in a cofveve. 2.5E3/5E-2 is 5E4. I'll round up because assuming normal breathing you probably would say cofveve less times so 5?

When doing Fermi questions here it'll be nice if you can explain the step by step process of how to obtain your answer so others can understand. Just saying a answer doesn't really do anything.

Attempt
There's about E8km of land on earth. I'll assume that's a nuke and could probably destroy a city which is E2 km of land so assuming ideal land destruction 6


Answer
so from some quick googling/Wikipedia I can't find the power of the wepon you specified so I'll assume a 1 megaton of tnt. That destroys 100 square miles not km. The surface area of land is around 1.5E8/256 is about 6E5 or 6


If you assumed 6E5 1 megaton wepons are used to destroy the world, assuming perfect energy conversation, what is the volume of a object that can be converted by fission into the amount of energy to destroy the world assuming a density of earth in cubic meters.
Last edited by Name on May 29th, 2018, 7:29 pm, edited 1 time in total.
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Re: Fermi Questions C

Postby Name » May 29th, 2018, 7:27 pm

Oops thought I was editing. Please delete. Sorry.
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Re: Fermi Questions C

Postby PM2017 » May 30th, 2018, 7:43 am

Name wrote:If you assumed 6E5 1 megaton wepons are used to destroy the world, assuming perfect energy conversation, what is the volume of a object that can be converted by fission into the amount of energy to destroy the world assuming a density of earth in cubic meters.


Attempt
Well, fission has a mass to energy efficiency of about 0.03%, so it takes about 3000 times as much mass (and therefore volume) than would be necessary. I'll come back to this fact later. 1 ton of tnt is 4e9 joules, one megaton would be 4e15, and times 6e5 would be 24e20, or 2.4e21 J. m = 3/c^2 ---> m = 2.4e21/(9e16). M ~2e4 kg. On my cheatsheet I had the density of the earth, which I remember was about e3.5 kg/m^3 (so 5500 kg/m^3). V, therefore is m/d = 20000/5500 which is about 4 (non fermi meters cubed). Now, times the 3000 from earlier, gives 12000, or fermi answer of 5.


I'll edit with the real solution later, when I have time.

Question: If, in a distant galaxy, there is a direct linear correlation between number of humanoids and surface area of the inhabited planet, and the Earth just so happens to fit the model perfectly, what is the gravitational force between a planet with 10 trillion humanoids, and a humanoid on the surafce of the planet, that has a mass of 100 kg? Give your answer in newtons. Assume all habitable planets have the same density as the earth.
Last edited by PM2017 on May 30th, 2018, 11:41 am, edited 1 time in total.
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Re: Fermi Questions C

Postby Name » May 30th, 2018, 8:28 am

PM2017 wrote:
Name wrote:If you assumed 6E5 1 megaton wepons are used to destroy the world, assuming perfect energy conversation, what is the volume of a object that can be converted by fission into the amount of energy to destroy the world assuming a density of earth in cubic meters.


Attempt
Well, fission has a mass to energy efficiency of about 0.03%, so it takes about 3000 times as much mass (and therefore volume) than would be necessary. I'll come back to this fact later. 1 ton of tnt is 4e9 joules, one megaton would be 4e15, and times 6e5 would be 24e20, or 2.4e21 J. m = 3/c^2 ---> m = 2.4e21/(9e16). M ~2e4 kg. On my cheatsheet I had the density of the earth, which I remember was about e3.5 kg/m^3 (so 5500 kg/m^3). V, therefore is m/d = 20000/5500 which is about 4 (non fermi meters cubed). Now, times the 3000 from earlier, gives 12000, or fermi answer of 5.


I'll edit with the real solution later, when I have time.


12000 is 4? Also I said assuming perfect energy conversation so no need to multiply by 3000, and the answer would be 0. I'll do the question later when I have time/or if someone else does it before me

Edit: in your question what's the distance between the planets?
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Re: Fermi Questions C

Postby PM2017 » May 30th, 2018, 11:41 am

Name wrote:
PM2017 wrote:
Name wrote:If you assumed 6E5 1 megaton wepons are used to destroy the world, assuming perfect energy conversation, what is the volume of a object that can be converted by fission into the amount of energy to destroy the world assuming a density of earth in cubic meters.


Attempt
Well, fission has a mass to energy efficiency of about 0.03%, so it takes about 3000 times as much mass (and therefore volume) than would be necessary. I'll come back to this fact later. 1 ton of tnt is 4e9 joules, one megaton would be 4e15, and times 6e5 would be 24e20, or 2.4e21 J. m = 3/c^2 ---> m = 2.4e21/(9e16). M ~2e4 kg. On my cheatsheet I had the density of the earth, which I remember was about e3.5 kg/m^3 (so 5500 kg/m^3). V, therefore is m/d = 20000/5500 which is about 4 (non fermi meters cubed). Now, times the 3000 from earlier, gives 12000, or fermi answer of 5.


I'll edit with the real solution later, when I have time.


12000 is 4? Also I said assuming perfect energy conversation so no need to multiply by 3000, and the answer would be 0. I'll do the question later when I have time/or if someone else does it before me

Edit: in your question what's the distance between the planets?

right, I mistyped that.

also, if you re-read the question, its the force of gravitational between a humanoid (100kg) on the surface of the planet and the planet itself.
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Re: Fermi Questions C

Postby whythelongface » May 30th, 2018, 5:03 pm

A Trident D5 has a significantly larger yield than one megaton TNT equivalent. If I remember properly, it should be able to destroy land on the scale of E4 square kilometers
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Re: Fermi Questions C

Postby UTF-8 U+6211 U+662F » June 2nd, 2018, 5:24 pm

PM2017 wrote:Question: If, in a distant galaxy, there is a direct linear correlation between number of humanoids and surface area of the inhabited planet, and the Earth just so happens to fit the model perfectly, what is the gravitational force between a planet with 10 trillion humanoids, and a humanoid on the surafce of the planet, that has a mass of 100 kg? Give your answer in newtons. Assume all habitable planets have the same density as the earth.

Attempt
Gravitational acceleration is 9.8 m/s^2 or around 1E1. There are around 8 billion people on Earth. 10 trillion / 8 billion is around 1E3 (if we're using the short system for trillion). Ratio of surface areas is around 1E3. Ratio of radii is around 3E1. Ratio of volumes is around 27E3 or 3E4. Ratio of masses is then around 3E4. 1E1*100*3E4/3E1=1E6. Fermi answer: 6.

Edit: Need to divide by 3E1 again. Fermi answer: 4


Answer
9.8 m/s^2 * sqrt(10 trillion humanoids/7.442 billion humanoids) * 100 kg = 3.6E4.


How many drops of 0.1 M HCl would it take to turn Lake Victoria to a pH of 3?
Last edited by UTF-8 U+6211 U+662F on July 2nd, 2018, 7:32 pm, edited 3 times in total.

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Re: Fermi Questions C

Postby Name » June 2nd, 2018, 6:05 pm

UTF-8 U+6211 U+662F wrote:
Attempt
Gravitational acceleration is 9.8 m/s^2 or around 1E1. There are around 8 billion people on Earth. 10 trillion / 8 billion is around 1E3 (if we're using the short system for trillion). Ratio of surface areas is around 1E3. Ratio of radii is around 3E1. Ratio of volumes is around 27E3 or 3E4. Ratio of masses is then around 3E4. 1E1*100*3E4/3E1=1E6. Fermi answer: 6.


Radius is squared so 1E6 is divided by 3E1 again so final answer I think is 3E4 or 4
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Re: Fermi Questions C

Postby UTF-8 U+6211 U+662F » June 3rd, 2018, 6:45 am

Name wrote:
UTF-8 U+6211 U+662F wrote:
Attempt
Gravitational acceleration is 9.8 m/s^2 or around 1E1. There are around 8 billion people on Earth. 10 trillion / 8 billion is around 1E3 (if we're using the short system for trillion). Ratio of surface areas is around 1E3. Ratio of radii is around 3E1. Ratio of volumes is around 27E3 or 3E4. Ratio of masses is then around 3E4. 1E1*100*3E4/3E1=1E6. Fermi answer: 6.


Radius is squared so 1E6 is divided by 3E1 again so final answer I think is 3E4 or 4

Oh yeah, my bad (this does not bode well for my upcoming physics exam)

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Re: Fermi Questions C

Postby UTF-8 U+6211 U+662F » July 2nd, 2018, 7:25 pm

UTF-8 U+6211 U+662F wrote:How many drops of 0.1 M HCl would it take to turn Lake Victoria to a pH of 3?


Just reminding everyone this is here.

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Re: Fermi Questions C

Postby Name » August 30th, 2018, 7:18 pm

UTF-8 U+6211 U+662F wrote:
UTF-8 U+6211 U+662F wrote:How many drops of 0.1 M HCl would it take to turn Lake Victoria to a pH of 3?


Just reminding everyone this is here.


Revival time!

Attempt
Estimate lake victorias volume to be around eries so 4E2 cubic km or 4E14 liters. Im pretty sure that the current concentration doesn't matter. If one drop has .1 M of HCl 4E15 drops would give a pH of 1 (I think) and 4E13 for a pH of 3 so 13


answer
the actual volume is 2750 cubic km which doesn't affect the final value. However Wikipedia says 1 mole / liter is a pH of 0 not 1 so final answer 12???


Question: what is the density of the universe in kg/m cubed?
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mit- 1st code, 3rd fermi
cornell- 1st fermi
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