### Re: Fermi Questions C

Posted:

**May 28th, 2018, 6:40 pm**Fermi Questions

**C**is for**C**ookie and**C**ookie is for me!Science Olympiad Student Center

https://scioly.org/forums/

Page **17** of **18**

Posted: **May 28th, 2018, 6:40 pm**

Fermi Questions **C** is for **C**ookie and **C**ookie is for me!

Posted: **May 28th, 2018, 7:52 pm**

You should prob specify the amount of water lol and also current tempTime to restart this thread for now... b/c it'll be relevant next year!

With the hot air coming out of Donald Trump's mouth when he says cofveve, how many times would he need to repeat it in order to boil water? Whoever actually solves this (not you whythelongface) correctly gets a cookie!

Posted: **May 29th, 2018, 5:08 pm**

This is easy....You should prob specify the amount of water lol and also current tempTime to restart this thread for now... b/c it'll be relevant next year!

With the hot air coming out of Donald Trump's mouth when he says cofveve, how many times would he need to repeat it in order to boil water? Whoever actually solves this (not you whythelongface) correctly gets a cookie!

5x10^2 (He talks too much)

Posted: **May 29th, 2018, 6:01 pm**

What process did you use?This is easy....You should prob specify the amount of water lol and also current tempTime to restart this thread for now... b/c it'll be relevant next year!

With the hot air coming out of Donald Trump's mouth when he says cofveve, how many times would he need to repeat it in order to boil water? Whoever actually solves this (not you whythelongface) correctly gets a cookie!

5x10^2 (He talks too much)

Also, next question, please.

Posted: **May 29th, 2018, 6:37 pm**

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 )

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 )

Posted: **May 29th, 2018, 7:25 pm**

Your 3 seems kinda lowMore 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 )

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.

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

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

Posted: **May 29th, 2018, 7:27 pm**

Oops thought I was editing. Please delete. Sorry.

Posted: **May 30th, 2018, 7:43 am**

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.

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 [b]5.[/b]

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.

Posted: **May 30th, 2018, 8:28 am**

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 meIf 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.I'll edit with the real solution later, when I have time.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 [b]5.[/b]

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

Posted: **May 30th, 2018, 11:41 am**

right, I mistyped that.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 meIf 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.I'll edit with the real solution later, when I have time.

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

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.