I'm assuming you mean a femto decimeter by meter by foot (correct me if I'm wrong because volumexlengthxlength doesn't make sense) So the volume of a decimeter foot meter is 3E-2 or 3E43 femto DMF in a meter cubed. Pretty sure alveolus are tiny air sacks. Pretty sure thier tiny (gonna assume volume of 100 mcm or about 12 would fit in a meter cubed. 43-12 is 31
diameter equals 200 mcm radius .1 mm volume is 4E-3 mm 4E-12 meter or 2E11 per meter which is 32
Question: In a theoretical universe, a beam of light has a energy level of 100 joules per photon. Assuming it has a wavelength of the height of Angel Falls what is the speed of light in m/s
I have no idea how to do that question honestly
How far does an air molecule (assume at STP) travel on average before it hits another air molecule, in meters?
Do you mean O2 or the "air" compound?
Re: Fermi Questions C
Posted: April 25th, 2018, 3:35 pm
by PM2017
TheChiScientist wrote:
Adi1008 wrote:
Name wrote:
I'm assuming you mean a femto decimeter by meter by foot (correct me if I'm wrong because volumexlengthxlength doesn't make sense) So the volume of a decimeter foot meter is 3E-2 or 3E43 femto DMF in a meter cubed. Pretty sure alveolus are tiny air sacks. Pretty sure thier tiny (gonna assume volume of 100 mcm or about 12 would fit in a meter cubed. 43-12 is 31
diameter equals 200 mcm radius .1 mm volume is 4E-3 mm 4E-12 meter or 2E11 per meter which is 32
Question: In a theoretical universe, a beam of light has a energy level of 100 joules per photon. Assuming it has a wavelength of the height of Angel Falls what is the speed of light in m/s
I have no idea how to do that question honestly
How far does an air molecule (assume at STP) travel on average before it hits another air molecule, in meters?
Do you mean O2 or the "air" compound?
He said:
Adi1008 wrote:
on average
So I would assume the average value between all gas in the atmosphere.
Re: Fermi Questions C
Posted: April 25th, 2018, 3:35 pm
by Name
Adi1008 wrote:
Name wrote:
I'm assuming you mean a femto decimeter by meter by foot (correct me if I'm wrong because volumexlengthxlength doesn't make sense) So the volume of a decimeter foot meter is 3E-2 or 3E43 femto DMF in a meter cubed. Pretty sure alveolus are tiny air sacks. Pretty sure thier tiny (gonna assume volume of 100 mcm or about 12 would fit in a meter cubed. 43-12 is 31
diameter equals 200 mcm radius .1 mm volume is 4E-3 mm 4E-12 meter or 2E11 per meter which is 32
Question: In a theoretical universe, a beam of light has a energy level of 100 joules per photon. Assuming it has a wavelength of the height of Angel Falls what is the speed of light in m/s
I have no idea how to do that question honestly
The frequency of light is speed/wavelength so X/E3 meters. Frequency times plancks constant equals energy per photon. So X/E3 times 6E34 equals E2. Divide both sides by 6E-34 (or multiply by about 2E33) X/E3 equals 2E35. Multiply both sides by E3, X equals 2E38 or 38
Re: Fermi Questions C
Posted: April 25th, 2018, 3:49 pm
by Name
Adi1008 wrote:
How far does an air molecule (assume at STP) travel on average before it hits another air molecule, in meters?
22 liters in a mole 6E23/22 is 3E22. Assuming a 1 dimentional movement the cube root of 3E22 is about 7E7. Because a liter is 1/10 of a meter, 7E8 air molecules in a straight line 1 meter. 1/7E8 is around E-9 so -9
I saw both mentiones of -9 and -7 as a answer (3 nm and 68). I think it might be 68 but I can't understand the math behind it. Can you confirm what the answer is?
Re: Fermi Questions C
Posted: April 25th, 2018, 4:55 pm
by Adi1008
Name wrote:
Adi1008 wrote:
How far does an air molecule (assume at STP) travel on average before it hits another air molecule, in meters?
22 liters in a mole 6E23/22 is 3E22. Assuming a 1 dimentional movement the cube root of 3E22 is about 7E7. Because a liter is 1/10 of a meter, 7E8 air molecules in a straight line 1 meter. 1/7E8 is around E-9 so -9
I saw both mentiones of -9 and -7 as a answer (3 nm and 68). I think it might be 68 but I can't understand the math behind it. Can you confirm what the answer is?
I believe that it would be , where is the number density and is the cross sectional area of an air molecule (going to make the rough assumption that diatomic oxygen and nitrogen have relatively similar sizes).
is about (you can calculate this using the Ideal Gas Law if you don't have it memorized). The cross sectional area of diatomic oxygen or nitrogen is about .
Doing the calculation would give a fermi answer of about -7.
Re: Fermi Questions C
Posted: April 25th, 2018, 5:33 pm
by Name
Adi1008 wrote:
Name wrote:
Adi1008 wrote:
How far does an air molecule (assume at STP) travel on average before it hits another air molecule, in meters?
22 liters in a mole 6E23/22 is 3E22. Assuming a 1 dimentional movement the cube root of 3E22 is about 7E7. Because a liter is 1/10 of a meter, 7E8 air molecules in a straight line 1 meter. 1/7E8 is around E-9 so -9
I saw both mentiones of -9 and -7 as a answer (3 nm and 68). I think it might be 68 but I can't understand the math behind it. Can you confirm what the answer is?
I believe that it would be , where is the number density and is the cross sectional area of an air molecule (going to make the rough assumption that diatomic oxygen and nitrogen have relatively similar sizes).
is about (you can calculate this using the Ideal Gas Law if you don't have it memorized). The cross sectional area of diatomic oxygen or nitrogen is about .
Doing the calculation would give a fermi answer of about -7.
Ok thanks! Also oops forgot to set up a question
Lets say you gained the ability to harvest salt from the ocean on a mass scale. If you take the salt and compacted it into a 1 meter radius sphere, what is it's escape velocity
Re: Fermi Questions C
Posted: April 28th, 2018, 1:12 pm
by Adi1008
Name wrote:
Adi1008 wrote:
Name wrote:
22 liters in a mole 6E23/22 is 3E22. Assuming a 1 dimentional movement the cube root of 3E22 is about 7E7. Because a liter is 1/10 of a meter, 7E8 air molecules in a straight line 1 meter. 1/7E8 is around E-9 so -9
I saw both mentiones of -9 and -7 as a answer (3 nm and 68). I think it might be 68 but I can't understand the math behind it. Can you confirm what the answer is?
I believe that it would be , where is the number density and is the cross sectional area of an air molecule (going to make the rough assumption that diatomic oxygen and nitrogen have relatively similar sizes).
is about (you can calculate this using the Ideal Gas Law if you don't have it memorized). The cross sectional area of diatomic oxygen or nitrogen is about .
Doing the calculation would give a fermi answer of about -7.
Ok thanks! Also oops forgot to set up a question
Lets say you gained the ability to harvest salt from the ocean on a mass scale. If you take the salt and compacted it into a 1 meter radius sphere, what is it's escape velocity
Volume of the oceans is about 18, so the mass is about 21. Guess that 1% of the ocean's mass is salt, so the salt mass is 19. Escape velocity is sqrt(2GM/r). G is -10, M is 19, r is 1. Escape velocity would be 4.
According to [url=https://www.reddit.com/r/askscience/comments/1cyizs/if_all_the_salt_in_the_oceans_was_removed_and/]this reddit post[/url], there's about 4.9 x 10^22 grams of salt = 4.9 x 10^19 kg of salt. [url=http://www.wolframalpha.com/input/?i=sqrt(2*6.67*10%5E(-11)*4.9*10%5E(19)%2F1)]Wolfram Alpha[/url] gives an answer of 5 overall.
How many complete rotations has the fastest pulsar made in the time between the extinction of the dinosaurs and the birth of Newton?
Re: Fermi Questions C
Posted: April 28th, 2018, 2:59 pm
by whythelongface
Adi1008 wrote:How many complete rotations has the fastest pulsar made in the time between the extinction of the dinosaurs and the birth of Newton?
The number of years between Newton's death and the present is negligible compared to the K-T extinction event, 65 mya (I think: this could just be one of those numbers the public loves to misquote). 7E7 years = what, 1 to 2E15 seconds?
I have no idea how many rotations a fast pulsar makes per second, but I think 40 rps was a number I encountered somewhere for pulsars? So my final answer is [b]17[/b].
PSR J1748-2446ad spins at [b]716 hz[/b]. My K-T extinction estimate was correct, so my answer is off by a magnitude from the real answer, [b]18.[/b]
RuBisCO is the most abundant enzyme in the entire biosphere. RuBisCO fixes carbon dioxide into ribulose bisphosphate, but can also fix diatomic oxygen into RuBP in a wasteful process known as photorespiration. How many kilograms of O2 are metabolized by RuBisCO within the entire biosphere in one second?
Re: Fermi Questions C
Posted: April 28th, 2018, 3:36 pm
by Name
whythelongface wrote: RuBisCO is the most abundant enzyme in the entire biosphere. RuBisCO fixes carbon dioxide into ribulose bisphosphate, but can also fix diatomic oxygen into RuBP in a wasteful process known as photorespiration. How many kilograms of O2 are metabolized by RuBisCO within the entire biosphere in one second?
RuBisCO I think is in the leaves. Trees make up the majority of leaves. There are about E12 trees. I have no idea how many RuBisCO are in a leaf. I'm gonna assume a tree uses 100 calories a day (thier large but don't spend energy moving and whatnot) so E14 calories of energy are used per day. There are 5 Cal of energy in a gram of glucose so 2E13 grams of glucose have to be produced. About 200 grams per mole 6E34 glucose molecules. I think it takes 6 cycles for 1 molecule (I suck at bio) so 6 o2 will be fixed or 3E35. Or 5E11 moles of 02 or about 2E10 grams of o2 or 2E7 kg per day or 2 per second.
Can't find anything on the answer, and my answer seems off. Can you confirm what the answer is?
At the current rate this thread is moving at (starting from the first post) how many posts will there be in the time it takes for sound to travel from here to Sagittarius A*
Re: Fermi Questions C
Posted: April 28th, 2018, 4:36 pm
by whythelongface
Name wrote:
whythelongface wrote: RuBisCO is the most abundant enzyme in the entire biosphere. RuBisCO fixes carbon dioxide into ribulose bisphosphate, but can also fix diatomic oxygen into RuBP in a wasteful process known as photorespiration. How many kilograms of O2 are metabolized by RuBisCO within the entire biosphere in one second?
RuBisCO I think is in the leaves. Trees make up the majority of leaves. There are about E12 trees. I have no idea how many RuBisCO are in a leaf. I'm gonna assume a tree uses 100 calories a day (thier large but don't spend energy moving and whatnot) so E14 calories of energy are used per day. There are 5 Cal of energy in a gram of glucose so 2E13 grams of glucose have to be produced. About 200 grams per mole 6E34 glucose molecules. I think it takes 6 cycles for 1 molecule (I suck at bio) so 6 o2 will be fixed or 3E35. Or 5E11 moles of 02 or about 2E10 grams of o2 or 2E7 kg per day or 2 per second.
Can't find anything on the answer, and my answer seems off. Can you confirm what the answer is?
I cannot confirm because I have not calculated it myself and am far too lazy to do so, but I'll give you some numbers:
"For every person on Earth, there are around five kilograms of RuBisCO in the biosphere."
"At ambient levels of carbon dioxide and oxygen, the ratio of the reactions is about 4 to 1, which results in a net carbon dioxide fixation of only 3.5."
I can't find the data for rate of carbon fixation, but if you really wanted to, you probably could.