samlan16 is gone though?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.
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?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.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?
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).
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.
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
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
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=95If 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?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
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
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.
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
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
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.
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
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