Question: How many helium balloons would it take to lift every human in the United States?

Re: Fermi Questions C

Posted: September 6th, 2018, 10:01 am

by NeilMehta

Let's start this up again!

Question: How many helium balloons would it take to lift every human in the United States?

There are 3E8 people in the USA and one helium balloon can lift E1 g of weight so it takes E2 balloons to carry 1 kg. If the average adult is 50 kg, then it would take about E4 balloons to carry a person or E12 balloons for everyone

One ballon can support 14 g of weight, average weight is 80.7 kg and there are 325.7 million people in the USA.
Solution: 1.877E12

Next question: What is the combined weight of every nickel in circulation in the USA?

Re: Fermi Questions C

Posted: September 7th, 2018, 6:39 am

by TheChiScientist

Let see. There should be about 1E9 nickles at least in circulation. A Nickle should be about 12g? So I'd say 2E9 grams

There about Half a billion nickels in total circulation so then they weigh 5 grams thus that equals 2E9. :)

How many gigaparsecs would it take to travel from Bueno Aires to Easter Island?

Re: Fermi Questions C

Posted: September 7th, 2018, 9:42 am

by Name

How many gigaparsecs would it take to travel from Bueno Aires to Easter Island?

1 Gigaparsecs is 3e9 ly or 3e22 km. Distance maybe around 3e3km /3e22 is -19

distance is just under 5e3km/3e22 is still -19

How many times greater is the sun's yearly energy production then the amount of energy humans use in a year

Re: Fermi Questions C

Posted: October 6th, 2018, 10:19 pm

by PM2017

How many gigaparsecs would it take to travel from Bueno Aires to Easter Island?

1 Gigaparsecs is 3e9 ly or 3e22 km. Distance maybe around 3e3km /3e22 is -19

distance is just under 5e3km/3e22 is still -19

How many times greater is the sun's yearly energy production then the amount of energy humans use in a year

The basis for my approach will be the Kardashev scale. (I'm sorry, I'm sleepy, I'm not going to explain it rn). Basically, Carl Sagan calulated that we are a Kardashev 0.72. The equation, IIRC is score = (log(power consumption) - 6)/10. If we are a kardashev 0.72, our power consumption is 10^13.2 W. The sun outputs e26 watts. so Fermi answer[b]13[/b]

According to wikipedia, the 2014 world power consumption was 155,481 Terrwatt hours, so, over the course of 365.24 days per year, I get 1.77e13 watts. The sun outputs 3.8e26 watts. This means that my fermi answer is indeed [b]13[/b]

How many human hairs were inside the auditorium at CSU during the nationals 2018 awards ceremony?

Re: Fermi Questions C

Posted: October 12th, 2018, 6:39 pm

by Jjshan26

There are 60 teams that attended 2018 Nats. Assume that each team had 15 members, a coach, some supervisors, and a few alternates... Whatever these numbers do not matter as it will not change the order of magnitude (we know there won't be like 80 people per team). As each human has 100,000 hair follicles, your answer will be 60*20*100,000 = 1.2e8 --> 8Next Question: A cool question that I saw on a practice test: how many cubic meters of water are evaporated from Earth's oceans in a year?

Re: Fermi Questions C

Posted: October 12th, 2018, 6:45 pm

by PM2017

There are 60 teams that attended 2018 Nats. Assume that each team had 15 members, a coach, some supervisors, and a few alternates... Whatever these numbers do not matter as it will not change the order of magnitude (we know there won't be like 80 people per team). As each human has 100,000 hair follicles, your answer will be 60*20*100,000 = 1.2e8 --> 8Next Question: A cool question that I saw on a practice test: how many cubic meters of water are evaporated from Earth's oceans in a year?

Actually, there were 120 teams (div b and div c), but it shouldn't change the answer.

Re: Fermi Questions C

Posted: October 12th, 2018, 8:59 pm

by Name

There are 60 teams that attended 2018 Nats. Assume that each team had 15 members, a coach, some supervisors, and a few alternates... Whatever these numbers do not matter as it will not change the order of magnitude (we know there won't be like 80 people per team). As each human has 100,000 hair follicles, your answer will be 60*20*100,000 = 1.2e8 --> 8Next Question: A cool question that I saw on a practice test: how many cubic meters of water are evaporated from Earth's oceans in a year?

Actually, there were 120 teams (div b and div c), but it shouldn't change the answer.

The way the question was phased implies that it's not just hair on the head but on the body. According to google theres about 5 million hair follicles on the entire human body suggesting a fermi answer of about 10

well the surface area of earth is 5e8 km. Oceans are about 70 percent of earth SA so 3.5e8 km, 3.5e14 m. If you leave water out maybe a cm would evaporate in a day so 3.5e12 cubic meter/day or about E15 cubic meters per year. Also the volume of the oceons is about e18 cubic meters which implies a average residence time of about 1000 years which seems somewhat accurate from what little I remember from water quality so final answer 15

according to [url=https://www.windows2universe.org/earth/Water/water_cycle_climate_change.html]this[/url] there's about 4.3e5 cubic km of water that evaporates per year or 4.3e14 cubic m which would round to 14. Also the average residence time of ocean water is 3000 years or so which also implies around 4.3e14 m so final answers 14

By volume, how many moons would fit in the earth?

Re: Fermi Questions C

Posted: October 13th, 2018, 3:33 pm

by PM2017

There are 60 teams that attended 2018 Nats. Assume that each team had 15 members, a coach, some supervisors, and a few alternates... Whatever these numbers do not matter as it will not change the order of magnitude (we know there won't be like 80 people per team). As each human has 100,000 hair follicles, your answer will be 60*20*100,000 = 1.2e8 --> 8Next Question: A cool question that I saw on a practice test: how many cubic meters of water are evaporated from Earth's oceans in a year?

Actually, there were 120 teams (div b and div c), but it shouldn't change the answer.

The way the question was phased implies that it's not just hair on the head but on the body. According to google theres about 5 million hair follicles on the entire human body suggesting a fermi answer of about 10

well the surface area of earth is 5e8 km. Oceans are about 70 percent of earth SA so 3.5e8 km, 3.5e14 m. If you leave water out maybe a cm would evaporate in a day so 3.5e12 cubic meter/day or about E15 cubic meters per year. Also the volume of the oceons is about e18 cubic meters which implies a average residence time of about 1000 years which seems somewhat accurate from what little I remember from water quality so final answer 15

according to [url=https://www.windows2universe.org/earth/Water/water_cycle_climate_change.html]this[/url] there's about 4.3e5 cubic km of water that evaporates per year or 4.3e14 cubic m which would round to 14. Also the average residence time of ocean water is 3000 years or so which also implies around 4.3e14 m so final answers 14

By volume, how many moons would fit in the earth?

since I'm silly, I haven't studied for fermi since state 2018, so I don't remember the radius of the moon. I'm going to simply assume it has the same density as the earth, and since the moon has a surface gravity 6 times weaker than the earth, and through some clever manipulation of F=GMm/r^2, and density = m/v, you get F = G*density*r*m, and since m is the same, you can ignore it. so you get 1/6 = G*density*r, and since G and density are assume to be the same, radius of the moon seems to be 1/6 that of the earth, or at least on that order of magnitude. So, the volume would be 1/216 that of the earth, so the fermi answer would be [b]2[/b] moons fit in the earth.

Well, I never expected my equation to be accurate, and it definitely wasn't. The radius of the moon is actually about 1/4 of the earth. But, since 4^3 and 6^3 both are fermi answers of [b]3[/b], I was correct. I probably could have used the fact that the sun and moon seem to have the same angular diameter (thus allowing for near perfect eclipses), and the fact that the sun is 11 (fermi) meters from the earth, and the moon is about 8 (fermi) meters, to find the radius of the moon, and compare with the earth's radius of 6.5ish (fermi) meters, but whatever

In the amount of time it would take a person typing at average typing speed to copy the entirety of the Harry Potter series, how many times would the moon complete an orbit around the earth?

Re: Fermi Questions C

Posted: October 13th, 2018, 4:57 pm

by Name

Actually, there were 120 teams (div b and div c), but it shouldn't change the answer.

The way the question was phased implies that it's not just hair on the head but on the body. According to google theres about 5 million hair follicles on the entire human body suggesting a fermi answer of about 10

well the surface area of earth is 5e8 km. Oceans are about 70 percent of earth SA so 3.5e8 km, 3.5e14 m. If you leave water out maybe a cm would evaporate in a day so 3.5e12 cubic meter/day or about E15 cubic meters per year. Also the volume of the oceons is about e18 cubic meters which implies a average residence time of about 1000 years which seems somewhat accurate from what little I remember from water quality so final answer 15

according to [url=https://www.windows2universe.org/earth/Water/water_cycle_climate_change.html]this[/url] there's about 4.3e5 cubic km of water that evaporates per year or 4.3e14 cubic m which would round to 14. Also the average residence time of ocean water is 3000 years or so which also implies around 4.3e14 m so final answers 14

By volume, how many moons would fit in the earth?

since I'm silly, I haven't studied for fermi since state 2018, so I don't remember the radius of the moon. I'm going to simply assume it has the same density as the earth, and since the moon has a surface gravity 6 times weaker than the earth, and through some clever manipulation of F=GMm/r^2, and density = m/v, you get F = G*density*r*m, and since m is the same, you can ignore it. so you get 1/6 = G*density*r, and since G and density are assume to be the same, radius of the moon seems to be 1/6 that of the earth, or at least on that order of magnitude. So, the volume would be 1/216 that of the earth, so the fermi answer would be [b]2[/b] moons fit in the earth.

Well, I never expected my equation to be accurate, and it definitely wasn't. The radius of the moon is actually about 1/4 of the earth. But, since 4^3 and 6^3 both are fermi answers of [b]3[/b], I was correct. I probably could have used the fact that the sun and moon seem to have the same angular diameter (thus allowing for near perfect eclipses), and the fact that the sun is 11 (fermi) meters from the earth, and the moon is about 8 (fermi) meters, to find the radius of the moon, and compare with the earth's radius of 6.5ish (fermi) meters, but whatever

In the amount of time it would take a person typing at average typing speed to copy the entirety of the Harry Potter series, how many times would the moon complete an orbit around the earth?

The number of moons that fit into earth can be determined by (radius earth/radius moon)^3
(6371/1737)^3 ends up to be 49.3... which would round to 1

iirc HP is about 500 pages average, 7 books, and there's usually about 500 words per page. I think my typing speed is about 60 wpm? Anyways about 1.7e6 words so about 1.7e6 seconds. Moon around earth is about 30 days times 86400 around 2.5e6 seconds. 1.7e6/2.5e6 Fermi answer 0

apparently there's about e6 words and the average (and my) wpm is 40 so 1.5e6 seconds to type out HP. It's about 2.6e6 seconds in 30 days. 1.5e6/2.6e6 is .57 or Fermi 0

Question: according to a interesting video I was watching (by PBS spacetime, it's probably worth checking out if your bored) the maximum amount of information (in bits) that can be stored in a system is = surface area (in plank units)/4. Given that the actual information stored in the universe is closer to around 10^90 bits how many times larger is the radius of the observable universe then the radius of the volume of sphere the total information in the observable universe can be stored in?