Fermi Questions C

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Re: Fermi Questions C

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

Attempt
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

Actual Answer
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

Unome wrote:
Attempt
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

Actual Answer
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?

Attempt
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.

Solution
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

NeilMehta wrote:new question: how many µL of ink would it take to print one copy of Shakespeare's Romeo and Juliet?

Attempt
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.

Solution
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

Attempt
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

Actual
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

Unome wrote:How many pixels are on the screens of all iPhone 5's (standard version only) ever produced?

Attempt
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

Actual
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

Name wrote:
Unome wrote:How many pixels are on the screens of all iPhone 5's (standard version only) ever produced?

Attempt
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

Actual
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?


Attempt
(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.

Actual
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

Riptide wrote:
Attempt
(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.

Actual
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
Attempt
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

Actual
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

Name wrote:Looks correct but I asked for moles -10

Sigh, my bad :oops:

Name wrote: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


Attempt
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.

Actual
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

Riptide wrote:
Name wrote:Looks correct but I asked for moles -10

Sigh, my bad :oops:

Name wrote: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


Attempt
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.

Actual
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

Name wrote:
Riptide wrote:
Name wrote:Looks correct but I asked for moles -10

Sigh, my bad :oops:

Name wrote: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


Attempt
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.

Actual
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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Re: Fermi Questions C

Postby Name » February 10th, 2018, 9:43 pm

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


Lol I didn't make the question. It was on my regionals test and I did the same thing as you did, but afterwards when I thought about it I realized that mass would decrease when radius decreases
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Re: Fermi Questions C

Postby Riptide » February 11th, 2018, 7:33 am

Name wrote:Looks correct but I asked for moles -10

Sigh, my bad :oops:

Name wrote: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


Attempt
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.

Actual
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)?
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Re: Fermi Questions C

Postby PM2017 » February 11th, 2018, 8:47 am

Riptide wrote:
Name wrote:Looks correct but I asked for moles -10

Sigh, my bad :oops:

Name wrote: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


Attempt
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.

Actual
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)?

Check out the discussion on Gamow peaks in the astronomy question marathon for more info on this...
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Re: Fermi Questions C

Postby Name » February 11th, 2018, 12:49 pm

Riptide wrote:If quantum mechanics didn't exist, how hot would the sun's core have to be to fuse hydrogen nuclei into helium (kelvin)?

Probably very wrong answer
I'll assume the core makes all the sun's energy which is 4E26. Mass of all the hydrogen is around 4E9 kg from using e=mc2 or around 2E36 hydrogen atoms. Or around 36 helium is made/second. Using the tip from PM2017 I found the columbs barriers equation (which is the same as coloumbs law). Charge of the hydrogen is 17 in coloumbs squared is 34 plus the 9 from the constant 43/ radius (I'm gonna assume borhs radius) 2E53 joules of energy. Radius of sun's core is .2(sun's radius) or around E5 or around 3E15 kilometer or 3E30 cm or E23 joule per cm. Density of around E2 g/cm specific heat of hydrogen is E1 joule for a gram for kelvin. So final temperature is 20 kelvin. That took way to long.... And it's still probably wrong



Question: how many brain cells did I lose from doing that question? (Jk)
Actual question: Suppose betelguese retained the same mass but decreased it's radius to that of the earth. If you travel at it's escape velocity for a gigamole of planck seconds how many times can that travel across Earth's equater
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Re: Fermi Questions C

Postby Unome » February 12th, 2018, 10:46 am

Attempt
Earth's escape velocity is about E4 m/s. Betelgeuse's mass is about 4E1 solar masses, hence (2E30/5E24)*4E1 Earth masses - which is about 1.5E7. Using Newton's law of universal gravitation, the force should be proportional to this, and the escape velocity proportional to the square root of this - hence, E4*sqrt(1.5E7) = E4*5E3 or so = 5E7 m/s escape velocity. The Planck time is about 5E-44 seconds, therefore a gigamole of them is 6E32*5E-44 = 3E-11 seconds. Multiplying gives a distance of 3E-11*5E7 = 1.5E-3 meters. The Earth's equator is 4E7 meters, hence 1.5E-3 divided by 4E7 gives ~1/3E-10, so Fermi Answer: -11

Actual
Earth's escape velocity is 1.12E4 m/s. Betelgeuse's mass is closer to 10 solar masses, which gives 3.33E6 Earth masses. Square root of this is 1.825E3, multiply by the escape velocity of Earth gives 2.044E7 m/s. A gigamole of Planck seconds is 3.24E-11 seconds, so the distance traveled is 6.634E-4 meters, which divided by the Earth's diameter (1.2742E7 meters) gives 5.2E-11, so Fermi Answer: -10

How much is the mass of all the live rhinoviruses in the world in units of the mass of a standard No. 2 wooden pencil?
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