Re: Fermi Questions C
Posted: May 28th, 2018, 6:40 pm
Fermi Questions C is for Cookie and Cookie is for me!
You should prob specify the amount of water lol and also current tempRaleway wrote:Time 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!
This is easy....NeilMehta wrote:You should prob specify the amount of water lol and also current tempRaleway wrote:Time 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!
What process did you use?TheChiScientist wrote:This is easy....NeilMehta wrote:You should prob specify the amount of water lol and also current tempRaleway wrote:Time 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)
Your 3 seems kinda lowTheChiScientist 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)
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
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.
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]
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 mePM2017 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.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]
right, I mistyped that.Name wrote: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 mePM2017 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.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?