Fermi Questions C

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Fermi Questions C

Postby Entomology » September 11th, 2016, 10:06 pm

Seeing as SoCal has replaced Game On with Fermi Questions, I thought I would open a thread to further study for this event. Can a mod make it a topic in the study event forum too or...?

Anyone with any advice? (I'm looking at you, samlan) I looked at the wiki page and this seems like a...mathematical event.
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Re: Fermi Questions C

Postby [NerdyTotoro] » September 12th, 2016, 4:46 pm

Seeing as SoCal has replaced Game On with Fermi Questions, I thought I would open a thread to further study for this event. Can a mod make it a topic in the study event forum too or...?

Anyone with any advice? (I'm looking at you, samlan) I looked at the wiki page and this seems like a...mathematical event.
samlan16 is gone though?
:idea:

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Re: Fermi Questions C

Postby appleshake123 » November 11th, 2016, 9:28 am

For help, being knowledgeable on unit conversion and measurements of many things. Also, I was told when estimating, .5-5=10^0, 5-50=10^1 50-500 = 10^2 etc. "Fermi estimates generally work because the estimations of the individual terms are often close to correct, and overestimates and underestimates help cancel each other out."

From what I know of this event, it is quick estimation knowing many conversions and answering using the power of 10. An example from a test this year is, "How many unsharpened pencils would fit between the centre of the milky way and the Sun?" (Example answer: 10^1 = 1, 10^30= 30)

Here's the answer. The Fermi Answer[FA]= 21.
The milky way is about 30kpc wide, Sun is not on the centre, so the distance is probably around 12kpc so 1E4 pc.
A parsec is 3.26 light years(time light travels in a year) 3E0
A day is 86400, or abour 9e5 seconds, so a year is about 9e5*300 or 3E7 s
Light is 3*10^8 m/s or 3E8 m
So a light year is 9E15 meters, about E16
Let's say a pencil is 1/5 a metre, so 5 is E1
E4 pc* E0 ly/pc * E16 m/ly * e1 pencils/ m = E(4+0+16+1)= FA= 21
Actual answer from internet: Distance from milky way to sun is 8kpc which is 9.719e+21 inches. an unsharpened pencil is 7.5 inches, so there are about 1.3 e21 pencils from milky way to sun, so FA= 21 is correct.
Another question! What is the surface area of water on earth in yellow school busses?

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Re: Fermi Questions C

Postby jonboyage » December 13th, 2016, 8:39 am

For help, being knowledgeable on unit conversion and measurements of many things. Also, I was told when estimating, .5-5=10^0, 5-50=10^1 50-500 = 10^2 etc. "Fermi estimates generally work because the estimations of the individual terms are often close to correct, and overestimates and underestimates help cancel each other out."

From what I know of this event, it is quick estimation knowing many conversions and answering using the power of 10. An example from a test this year is, "How many unsharpened pencils would fit between the centre of the milky way and the Sun?" (Example answer: 10^1 = 1, 10^30= 30)

Here's the answer. The Fermi Answer[FA]= 21.
The milky way is about 30kpc wide, Sun is not on the centre, so the distance is probably around 12kpc so 1E4 pc.
A parsec is 3.26 light years(time light travels in a year) 3E0
A day is 86400, or abour 9e5 seconds, so a year is about 9e5*300 or 3E7 s
Light is 3*10^8 m/s or 3E8 m
So a light year is 9E15 meters, about E16
Let's say a pencil is 1/5 a metre, so 5 is E1
E4 pc* E0 ly/pc * E16 m/ly * e1 pencils/ m = E(4+0+16+1)= FA= 21
Actual answer from internet: Distance from milky way to sun is 8kpc which is 9.719e+21 inches. an unsharpened pencil is 7.5 inches, so there are about 1.3 e21 pencils from milky way to sun, so FA= 21 is correct.
Another question! What is the surface area of water on earth in yellow school busses?
When you say yellow school buses, are you referring to one side of a school bus, or the total surface area of a bus if you were to "unfold" it?

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Re: Fermi Questions C

Postby Unome » December 13th, 2016, 9:45 am

For help, being knowledgeable on unit conversion and measurements of many things. Also, I was told when estimating, .5-5=10^0, 5-50=10^1 50-500 = 10^2 etc. "Fermi estimates generally work because the estimations of the individual terms are often close to correct, and overestimates and underestimates help cancel each other out."

From what I know of this event, it is quick estimation knowing many conversions and answering using the power of 10. An example from a test this year is, "How many unsharpened pencils would fit between the centre of the milky way and the Sun?" (Example answer: 10^1 = 1, 10^30= 30)

Here's the answer. The Fermi Answer[FA]= 21.
The milky way is about 30kpc wide, Sun is not on the centre, so the distance is probably around 12kpc so 1E4 pc.
A parsec is 3.26 light years(time light travels in a year) 3E0
A day is 86400, or abour 9e5 seconds, so a year is about 9e5*300 or 3E7 s
Light is 3*10^8 m/s or 3E8 m
So a light year is 9E15 meters, about E16
Let's say a pencil is 1/5 a metre, so 5 is E1
E4 pc* E0 ly/pc * E16 m/ly * e1 pencils/ m = E(4+0+16+1)= FA= 21
Actual answer from internet: Distance from milky way to sun is 8kpc which is 9.719e+21 inches. an unsharpened pencil is 7.5 inches, so there are about 1.3 e21 pencils from milky way to sun, so FA= 21 is correct.
Another question! What is the surface area of water on earth in yellow school busses?
Most likely you mean the top; either way, the difference is probably ~E0.5 or so, in which case the lower number is safer, since I don't know for sure.

Earth radius is around 6000 km
Surface area of earth is around 6000^3 times pi times 4/3 which is around E8
Times 70% which should about bring down the extra from 6^3
Convert to meters so E14
Top of a school bus is around 2 meters by 10 meters, so 2E1
E14/2E1
Fermi Answer = 13
Actual Answer from internet
Water on earth's surface = 3.61 E14 square meters Standard school bus = 2.59 meters wide, varies between 31 and 38 feet long; I used 35, so my surface area on the top is 2.769 E1 meters Fermi Answer from internet = 13 (yay!)
Question: If you accelerated a bullet to 0.01% of the speed of light, how far would it penetrate into a solid wall of arbitrary thickness composed of Kevlar? (assuming that the wall is securely anchored).

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Re: Fermi Questions C

Postby appleshake123 » December 22nd, 2016, 5:24 am


Question: If you accelerated a bullet to 0.01% of the speed of light, how far would it penetrate into a solid wall of arbitrary thickness composed of Kevlar? (assuming that the wall is securely anchored).
I'm trying
So I believe I saw somewhere that a bullet travels 1000 ft/s which is about 300 m/s. .01% of the speed of light is about 30000 m/s. I'm just going to assume penetration is linear with velocity, so it will penetrate 100x as far. If 1 cm kevlar can protect against a 300 m/s bullet, then I will say 100 cm, or FA of 2. I probably used the wrong units though.
How many electrons are there in the universe?

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Re: Fermi Questions C

Postby Unome » December 22nd, 2016, 8:31 am

Assuming you mean the observable universe, since the amount of matter in the universe is generally accepted to be infinite
I think I read somewhere that there are E68 kg of matter in the observable universe (hopefully this is close :? ). I'm just going to assume it's all hydrogen, so E71 g of hydrogen is 6E23 * E71 = 6E94, so I'll go with: FA=95
From the internet
It's apparently E53 kg of matter in the observable universe, and the best real answer I can find for the number of electrons is FA=80
If there was a planet with the same mass as Earth, with a diameter such that the Earth could be inscribed in a rectangular prism, which was itself inscribed in this hypothetical planet, what would be the density of this planet in kilograms per cubic meter?

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Re: Fermi Questions C

Postby appleshake123 » December 22nd, 2016, 9:29 am

Assuming you mean the observable universe, since the amount of matter in the universe is generally accepted to be infinite
I think I read somewhere that there are E68 kg of matter in the observable universe (hopefully this is close :? ). I'm just going to assume it's all hydrogen, so E71 g of hydrogen is 6E23 * E71 = 6E94, so I'll go with: FA=95
From the internet
It's apparently E53 kg of matter in the observable universe, and the best real answer I can find for the number of electrons is FA=80
If there was a planet with the same mass as Earth, with a diameter such that the Earth could be inscribed in a rectangular prism, which was itself inscribed in this hypothetical planet, what would be the density of this planet in kilograms per cubic meter?
Assuming the earth is a perfect sphere
Earth has a radius of 6e6m or a diameter about e7m. If it is inscribed in a prism which is inscribed in another sphere. If I can geometry good, then the diameter of the new planet is sqrt(2)(or about 1.4)* D of earth which is still about e7m. The radius is half of that which still rounds to e7m. Then V = 4/3 * r^3= about e21m, maybe round down later. The density is mass of earth: 6e24/V = e4, round down so Fa= 3
From the internet
So I can't geometry, but it shouldn't change the answer too much. A sphere inscribed in a cube has a cube edge length of Diameter(D). A cube inscribed in a circle has a diameter of Dsqrt(3) and a radius of Rsqrt(3). The Earth's radius is 6e6, so the new radius is 1e7. MofEarth/V = Density. (5.97e24)/(4/3*((e7)^3) = 4477.5 or FA = 3.
Within the average high schooler's sleep during the school year, how many periods of radiation of a Cs-133 atom moving between two hyperfine level of the ground state occurs? I probably worded this poorly.

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Re: Fermi Questions C

Postby Unome » December 22nd, 2016, 11:01 am

This is from atomic clocks right?
Let's estimate 6 hours per day, so over ~270 days that's ~E3 hours which is ~E6 seconds. If I remember correctly, the period of transition is something like E-8, so E6 / E-8 = E14 FA=14
From the internet
Average hours of sleep is 7-7.25 per day The transition is actually E-10 (I had the right frequency, but I rounded the 9 in the wrong direction to get E-8) So the answer comes out at 6.7E16 or: FA=17
How many blades of grass are there over all land area on the Earth classified as grasslands? (idk I couldn't think of anything good)

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Re: Fermi Questions C

Postby jonboyage » December 22nd, 2016, 5:21 pm

My answer
Lets assume in one square meter of grasslands there's about 10^3-10^4 blades of grass. I'm going to take a wild guess and say that there is 10^6 km^2 of grasslands on the earth. 10^4(blades/m^2)*10^6(m^2/km^2)*10^6(km^2) would give around 10^16 blades of grass on the Earth.
Internet
10 blades of grass per square centimeter, 100.000 blades of grass per square meter, 100.000.000.000 blades of grass per square kilometer. The total surface area of earth is 5.1 * 10^8 km2. The surface of earth is roughly 70.8% water, which means that 29.2% is land. Of these 29.2% approximately 20% is grass. This means the surface area of all grass areas combined is: Area_of_grass = 5.1 * 108 km2 * 29.2 % * 20 % = 29 million km2 This amounts to blades = 1011 blades of grass / km2 * 29 million km2 = 3 * 10^18 blades of grass Keep in mind that this answer does not take into account that the original question asked for just grasslands, so the actual value could be closer to 10^17
What is the length of all of the atoms in a single average person's body lined up in a single atom-wide line in terms of Milky-Way-radii?

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Re: Fermi Questions C

Postby Unome » December 22nd, 2016, 7:13 pm

Nice units
Assumption: the human body is mostly water (will round down the length slightly to account for the non-water parts containing heavier elements that aren't proportionally wider). Approximate human mass of 7E4 grams. If this is all water, that's about 5E3 moles of water, so about E28 atoms. Since the average width is probably about 1 nanometer (estimated from DNA width) or E-9 meters, the total length of those atoms is about E19 meters (or E16 kilometers, since that's better for converting). Convert to light years using 3E8 km per light year, and with the rounding for correction mentioned earlier, E11 light years. The Milky Way is about E5 light years in radius, so E11 / E5 = E6 FA = 6
From the Internet
[url=http://www.madsci.org/posts/archives/2000-06/962225341.Bc.r.html]Elemental composition of the human body[/url] Average human mass is 8E4 grams Running the math, that's 3250 moles of oxygen, 1200 moles of carbon, 8000 moles of hydrogen, and 170 moles of nitrogen (just using those four because the rest are pretty tiny, not much change caused in the length) Molecules: 1.95E27 oxygen, 7.2E26 carbon, 4.8E27 hydrogen, E26 nitrogen Using their actual radii of 60, 70, 25, and 65 picometers respectively, 1.17E17 meters oxygen, 5.04E16 meters carbon, 1.2E17 meters hydrogen, 6.5E15 meters nitrogen. Adding those up, that's a total length of around 5.84E17 meters, or 5.84E14 kilometers (looks like my correction factor wasn't large enough) Convert to 6.2E4 light years The most commonly cited estimate of the Milky Way's radius is 5E4 light years, so 1.24 Milky Way diameters, hence: FA = 1 (wow I was off by a lot)
Over the course of one year, how many skin cells will be produced on a single adult human?

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Re: Fermi Questions C

Postby appleshake123 » December 23rd, 2016, 4:19 am

Nice units
Assumption: the human body is mostly water (will round down the length slightly to account for the non-water parts containing heavier elements that aren't proportionally wider). Approximate human mass of 7E4 grams. If this is all water, that's about 5E3 moles of water, so about E28 atoms. Since the average width is probably about 1 nanometer (estimated from DNA width) or E-9 meters, the total length of those atoms is about E19 meters (or E16 kilometers, since that's better for converting). Convert to light years using 3E8 km per light year, and with the rounding for correction mentioned earlier, E11 light years. The Milky Way is about E5 light years in radius, so E11 / E5 = E6 FA = 6
From the Internet
[url=http://www.madsci.org/posts/archives/2000-06/962225341.Bc.r.html]Elemental composition of the human body[/url] Average human mass is 8E4 grams Running the math, that's 3250 moles of oxygen, 1200 moles of carbon, 8000 moles of hydrogen, and 170 moles of nitrogen (just using those four because the rest are pretty tiny, not much change caused in the length) Molecules: 1.95E27 oxygen, 7.2E26 carbon, 4.8E27 hydrogen, E26 nitrogen Using their actual radii of 60, 70, 25, and 65 picometers respectively, 1.17E17 meters oxygen, 5.04E16 meters carbon, 1.2E17 meters hydrogen, 6.5E15 meters nitrogen. Adding those up, that's a total length of around 5.84E17 meters, or 5.84E14 kilometers (looks like my correction factor wasn't large enough) Convert to 6.2E4 light years The most commonly cited estimate of the Milky Way's radius is 5E4 light years, so 1.24 Milky Way diameters, hence: FA = 1 (wow I was off by a lot)
Over the course of one year, how many skin cells will be produced on a single adult human?
Many die
I read somewhere that thousands of skin cells die a second. So assuming that the cells get regenerated at a similar rate and that there are 86400s in a day, 365 days, so about e7s in a year, e7s*e3cell/s = E10 = FA = 10
Actual answer from the internet
So according to[url=http://book.bionumbers.org/how-quickly-do-different-cells-in-the-body-replace-themselves/]this website[/url] skin cells replace themselves 10-30 days, or 12 - 36 times a year. There are 36 billion skin cells. so between 432 - 1296 billion a year or FA = 11 or 12
Since I'm unoriginal, what is 262! [EDIT: Brute force got it even if doubles arent large enough]

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Re: Fermi Questions C

Postby raxu » January 23rd, 2017, 2:30 pm

Many die
I read somewhere that thousands of skin cells die a second. So assuming that the cells get regenerated at a similar rate and that there are 86400s in a day, 365 days, so about e7s in a year, e7s*e3cell/s = E10 = FA = 10
Actual answer from the internet
So according to[url=http://book.bionumbers.org/how-quickly-do-different-cells-in-the-body-replace-themselves/]this website[/url] skin cells replace themselves 10-30 days, or 12 - 36 times a year. There are 36 billion skin cells. so between 432 - 1296 billion a year or FA = 11 or 12
Since I'm unoriginal, what is 262! [EDIT: Brute force got it even if doubles arent large enough]
Let's See how I do
At this size we use stirling approximation. sqrt(2pi*262) is about E1, (262/e)^262 is about E524, so I'm going with 525.
Actual answer from the internet
Wolfram alpha gives E521.4. The problem is that 262<100e...
In chains (the Imperial Unit), how wide is an Iridium atom?
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Re: Fermi Questions C

Postby Unome » May 21st, 2017, 7:31 am

Restarting this in preparation for Fermi I'm return next year.
My Answer
* Iridium being a metal, it's probably a bit large, let's say 100 picometers (estimating from what I know of microbiology) * Equals around E-10 meters * No idea how much a chain is, I'll take it as a bit longer and multiply by E-1 * Final answer: -11
Actual Answer
Wow Turns out a chain is 20.1168 meters (exactly 22 yards), and an Iridium atom has a Van der Waals radius of 200 pm. Although my estimates for the two values were off, the difference s were nearly identical and cancel out perfectly. True Answer: -11
Starting with a semi-introductory level question: How many steps could an average person walk in 100 days, assuming no rest and a constant brisk pace?
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Re: Fermi Questions C

Postby Raleway » May 21st, 2017, 8:04 pm

Restarting this in preparation for Fermi I'm return next year.
My Answer
* Iridium being a metal, it's probably a bit large, let's say 100 picometers (estimating from what I know of microbiology) * Equals around E-10 meters * No idea how much a chain is, I'll take it as a bit longer and multiply by E-1 * Final answer: -11
Actual Answer
Wow Turns out a chain is 20.1168 meters (exactly 22 yards), and an Iridium atom has a Van der Waals radius of 200 pm. Although my estimates for the two values were off, the difference s were nearly identical and cancel out perfectly. True Answer: -11
Starting with a semi-introductory level question: How many steps could an average person walk in 100 days, assuming no rest and a constant brisk pace?
What's hiding lol
My Answer
Assuming the "walking pace" is andante or 60 BPM (Haha @Bernard), then a person walks 60 paces in a minute, 3600 in an hour, and 86400 in a day or approx 10^6-. 100 days is 10^2 which then yields 10^8- but rounds to 8 * Final answer: 8
Actual Answer
Hoooo As the average walking pace is 3.1 MPH, and the average stride length between man and woman averaged is approx 28 inches, that averages to around 7015 steps per hour, which multiplied by 24 (to days), 100 (for 100 days) to finally yield 16835657 which is 1.7*10^7 DANG IT haha maybe my approximation was incorrect so sadly I was quite close :((( True Answer: -7
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