Fermi Questions C

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Re: Fermi Questions C

Postby Unome » February 8th, 2018, 7:13 am

Neutron radius is something like E-13 meters (guesstimating since atoms are around E-10 and electrons are around E-15).
 Total length of blood vessels is dominated by capillaries of course. It's hard to say, but I'd take the volume of a human (7E-2 cubic meters) and multiply it by perhaps 1 cm of capillary per cubic micrometer - E16 meters per cubic meter, therefore E13 meters of blood vessels. Hence, Fermi Answer: 26
Neutron radius is E-15 meters. Length of blood vessels is less than I had thought, about 1.6E8 meters, giving a Fermi Answer: 23
How much energy would be required to boil all of the world's oceans, in units of the amount of calories consumed in the form of potato chips annually in the US?
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Re: Fermi Questions C

Postby NeilMehta » February 8th, 2018, 8:09 am

Neutron radius is something like E-13 meters (guesstimating since atoms are around E-10 and electrons are around E-15).
 Total length of blood vessels is dominated by capillaries of course. It's hard to say, but I'd take the volume of a human (7E-2 cubic meters) and multiply it by perhaps 1 cm of capillary per cubic micrometer - E16 meters per cubic meter, therefore E13 meters of blood vessels. Hence, Fermi Answer: 26
Neutron radius is E-15 meters. Length of blood vessels is less than I had thought, about 1.6E8 meters, giving a Fermi Answer: 23
How much energy would be required to boil all of the world's oceans, in units of the amount of calories consumed in the form of potato chips annually in the US?
A serving of potato chips has about 2E2 kilocalories, which is 2E5 calories, or about E6 joules of energy.
q=mct
we need to raise the temperature by a bit under 100 degrees C
the earth has (guess  :( ) E16 L of water? which should be E19 mL of water, which should have a mass of E19 g
so E19*E2=E21 bags of chips
in the united states, the average person probably consumes about E2 bags of chips (or equivalent) per year. with a population of E8, that leaves us with E10 bags eaten per year, or a final answer of E11 units.
1292000000 pounds of potato chips eaten per year, or 3E12 calories
E21 liters of water, so E24 mL.
we need to raise by 83 degrees celsius
>>realized i made a mistake in calculations, C of water is in calories not joules
E24 grams * 83 degrees C is E25 calories
final answer= E13
distance from attempt=2
new question: how many µL of ink would it take to print one copy of Shakespeare's Romeo and Juliet?
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Re: Fermi Questions C

Postby Name » February 8th, 2018, 11:24 am

new question: how many µL of ink would it take to print one copy of Shakespeare's Romeo and Juliet?
I'm gonna assume that Romeo and Juliet is around 300 pages and I doubt a page takes that much ink. 1 ML seems too much and I'll assume 3 microliters which is Fermi answer 3.
It seems like a page uses around .05 microliter and the copy of Romeo and Juliet I have is just over 300 pages so Fermi answer 1
New question
Assuming everyone in the world weighs 100 kg, if everyone in the world was converted to pure energy, how long will it take for the sun to create the same amount of energy
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Re: Fermi Questions C

Postby Unome » February 8th, 2018, 12:08 pm

E2 kg*7E9 = E11 kg, E11*3E8*3E8=9E27. The sun outputs either E24 or E26 Watts - can't remember, so I'll go with E26 to account for the coefficient - assuming you wanted the answer in seconds (since you didn't specify) , Fermi Answer: 2
There are approximately 7.58 billion people in the world currently (I underestimated). Energy produced by people = 6.822E28. The sun produces 3.85E26 Watts. Hence, Fermi Answer: 2
How many pixels are on the screens of all iPhone 5's (standard version only) ever produced?
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Re: Fermi Questions C

Postby Name » February 8th, 2018, 3:57 pm

How many pixels are on the screens of all iPhone 5's (standard version only) ever produced?
gonna assume a phone is around a decimeter by decimeter. I'll go with E3 pixels in a decimeter so E6 pixel/phone and I'll assume maybe 5E7 iPhone 5 have been purchased so 14
I'm pretty sure there is 6E5 pixels in a iPhone 5 and E7 has been sold so 13
How many moles of oxegen molecules will one RBC carry throughout it's lifetime?
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Re: Fermi Questions C

Postby Riptide » February 10th, 2018, 5:28 pm

How many pixels are on the screens of all iPhone 5's (standard version only) ever produced?
gonna assume a phone is around a decimeter by decimeter. I'll go with E3 pixels in a decimeter so E6 pixel/phone and I'll assume maybe 5E7 iPhone 5 have been purchased so 14
I'm pretty sure there is 6E5 pixels in a iPhone 5 and E7 has been sold so 13
How many moles of oxegen molecules will one RBC carry throughout it's lifetime?
(I really hope RBC stands for red blood cells xD) Lets assume the life span of a red blood cell to be one month. Assuming a red blood cell can hold 1 billion oxygen molecules at a time, let's say it takes the red blood cell 10 seconds to transport the oxygen molecules. 30 days = 2E6 seconds, divided by 10 gives us 2E5 total transportations in the lifespan. Assuming maximum capacity everytime, this would give us 2E5*E9 = 14 oxygen molecules.
Not exactly sure if I'm finding accurate information so please correct me if I'm wrong Name. A red blood cell can in fact hold 1 billion molecules at a time, but its lifespan is much longer than i estimated, clocking in at 120 days. One circulation takes approximately a minute, so there are 172800 circulations in its lifespan. E9 oxygen molecules in every circulation gives us Fermi Answer 14. My incorrect estimations worked out in the end I guess lol. There was some contrasting information on the web so yeah don't take this as fact.
How many rotations are there of all the carbon-carbon single bonds in one mole of natural gas at room temperature in 1 hour?
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Re: Fermi Questions C

Postby Name » February 10th, 2018, 6:08 pm

(I really hope RBC stands for red blood cells xD) Lets assume the life span of a red blood cell to be one month. Assuming a red blood cell can hold 1 billion oxygen molecules at a time, let's say it takes the red blood cell 10 seconds to transport the oxygen molecules. 30 days = 2E6 seconds, divided by 10 gives us 2E5 total transportations in the lifespan. Assuming maximum capacity everytime, this would give us 2E5*E9 = 14 oxygen molecules.
Not exactly sure if I'm finding accurate information so please correct me if I'm wrong Name. A red blood cell can in fact hold 1 billion molecules at a time, but its lifespan is much longer than i estimated, clocking in at 120 days. One circulation takes approximately a minute, so there are 172800 circulations in its lifespan. E9 oxygen molecules in every circulation gives us Fermi Answer 14. My incorrect estimations worked out in the end I guess lol. There was some contrasting information on the web so yeah don't take this as fact.
How many rotations are there of all the carbon-carbon single bonds in one mole of natural gas at room temperature in 1 hour?
Looks correct but I asked for moles -10
Uh I'll assume natrual gas is ethane so 6E23 single bonds. Idk how often a rotation happens tbh but I'll assume 1/10 of a second and over an hour that would give 26
I saw something (idk if it's right,
there wasn't much of rate of rotation) that said 43 picoseconds so 37 (wow I was off) tho I'm likely wrong Edit: I found a test with this question and it is indeed 37
Question: If the earth became the density of a marshmallow what would the diameter (In meters) have to be to have the same gravitational attraction
Last edited by Name on February 21st, 2018, 8:28 pm, edited 1 time in total.
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Re: Fermi Questions C

Postby Riptide » February 10th, 2018, 9:22 pm

Looks correct but I asked for moles -10
Sigh, my bad :oops:
Question: If the earth became the density of a marshmallow what would the diameter (In meters) have to be to have the same gravitational attraction
Earth's density is around 5 g/cm^3. Let's assume a marshmallow is 1/5 of that. The force of gravity is equal to G*mass/radius^2. G = 6.67E-11, mass = 6E24 kg. The force of gravity on Earth is about 10 m/s^2. The mass is divided by 5, and we use that to solve the equation 10 = 6.67E-11*1.2E24/Radius^2, we get the radius to be 3E6 meters. The diameter would be 2 times this so the answer is 7.
Here is a list of the precise values i found: 5.972E24 kg for mass of the Earth, 9.807 m/s^2, 5.51 g/cm³ density of earth, 0.5 g/cm³ density of marshmallow. Plugging all this into the equations gives us a more precise radius of 1.92E6 meters, which would keep the diameter at 6.
If quantum mechanics didn't exist, how hot would the sun's core have to be to fuse hydrogen nuclei into helium (kelvin)?[/quote]
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Re: Fermi Questions C

Postby Name » February 10th, 2018, 9:37 pm

Looks correct but I asked for moles -10
Sigh, my bad :oops:
Question: If the earth became the density of a marshmallow what would the diameter (In meters) have to be to have the same gravitational attraction
Earth's density is around 5 g/cm^3. Let's assume a marshmallow is 1/5 of that. The force of gravity is equal to G*mass/radius^2. G = 6.67E-11, mass = 6E24 kg. The force of gravity on Earth is about 10 m/s^2. The mass is divided by 5, and we use that to solve the equation 10 = 6.67E-11*1.2E24/Radius^2, we get the radius to be 3E6 meters. The diameter would be 2 times this so the answer is 7.
Here is a list of the precise values i found: 5.972E24 kg for mass of the Earth, 9.807 m/s^2, 5.51 g/cm³ density of earth, 0.5 g/cm³ density of marshmallow. Plugging all this into the equations gives us a more precise radius of 1.92E6 meters, which would keep the diameter at 6.
If quantum mechanics didn't exist, how hot would the sun's core have to be to fuse hydrogen nuclei into helium (kelvin)?
[/quote]

If u decrease the radius you also decrease the total mass. The volume of the object (which increases the mass) is cubed and the radius in the equation is only squared so if the radius is increased by N times the gravitational pull is increased by N^3/N^2 times or by N times. Because marshmallow is 10x less dense the earth, then the radius would be increased by 10x yielding 6E4 km radius or 8 meter radius

Idk how quantum mechanics works so I'll let someone else do it
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Re: Fermi Questions C

Postby Riptide » February 10th, 2018, 9:40 pm

Looks correct but I asked for moles -10
Sigh, my bad :oops:
Question: If the earth became the density of a marshmallow what would the diameter (In meters) have to be to have the same gravitational attraction
Earth's density is around 5 g/cm^3. Let's assume a marshmallow is 1/5 of that. The force of gravity is equal to G*mass/radius^2. G = 6.67E-11, mass = 6E24 kg. The force of gravity on Earth is about 10 m/s^2. The mass is divided by 5, and we use that to solve the equation 10 = 6.67E-11*1.2E24/Radius^2, we get the radius to be 3E6 meters. The diameter would be 2 times this so the answer is 7.
Here is a list of the precise values i found: 5.972E24 kg for mass of the Earth, 9.807 m/s^2, 5.51 g/cm³ density of earth, 0.5 g/cm³ density of marshmallow. Plugging all this into the equations gives us a more precise radius of 1.92E6 meters, which would keep the diameter at 6.
If quantum mechanics didn't exist, how hot would the sun's core have to be to fuse hydrogen nuclei into helium (kelvin)?
If u decrease the radius you also decrease the total mass. The volume of the object (which increases the mass) is cubed and the radius in the equation is only squared so if the radius is increased by N times the gravitational pull is increased by N^3/N^2 times or by N times. Because marshmallow is 10x less dense the earth, then the radius would be increased by 10x yielding 6E4 km radius or 8 meter radius

Idk how quantum mechanics works so I'll let someone else do it[/quote]

Completely didn't consider the change in mass due to the change in the radius. Nice question!
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