From what I can tell, this is definitely from the more difficult end. Easier ones involve much simpler premises and conditions, i.e. what is the distance, in megaparsecs, that a snail travels in the amount of time it takes Pluto to make a full revolution around the sun or something like that.People are actually expected to figure this stuff out....
And we return!
How fast, in decimeters per hour squared, would a 2017 Volkswagen Beetle, which is currently orbiting the Earth at 700 km altitude, have to be accelerating in order to use as much power as the sun emits?
20? We can say P*t = 1/2m*v^2. Solving for v, v = sqrt(2P*t/m) Differentiating wrt t, a = sqrt(P/2mt) The luminosity of the sun is ~10^26 Watts, and I'd imagine a smaller car like the beetle would weigh a ton, or about 900 kg. Assuming we're trying to get the power output of the beetle to match that of the Sun, we set t = 1 second and find: sqrt((10^26 kg*m^2/s^3)/(2*900 kg*1 s) ~ sqrt((10^26 kg*m^2/s^3)/(2*10^3 kg*s) ~ sqrt(10^22) ~ 11 m/s^2. There's 10 meters in a decimeter and 3600 seconds in an hour, so 1 m/s^2 ~ 1e8 dm/s^2. 11 + 8 = 19, and counting all the rounding and numbers I've dropped, the answer might be closer to 20 so I'll go with that. That said, I'd imagine the car would explode long before it got anywhere near this level of acceleration. Also in this analysis I'm basically ignoring the altitude since that's basically the exosphere and unless you wanted me to talk about the car gaining even more potential energy from it falling back down or possibly decreased efficiency in the engine, but that's wayyy too much work beyond the simple algebra I'm willing to do
Unome has high standardsFrom what I can tell, this is definitely from the more difficult end. Easier ones involve much simpler premises and conditions, i.e. what is the distance, in megaparsecs, that a snail travels in the amount of time it takes Pluto to make a full revolution around the sun or something like that.People are actually expected to figure this stuff out....
Let me take a stab:Unome has high standardsFrom what I can tell, this is definitely from the more difficult end. Easier ones involve much simpler premises and conditions, i.e. what is the distance, in megaparsecs, that a snail travels in the amount of time it takes Pluto to make a full revolution around the sun or something like that.People are actually expected to figure this stuff out....
Also, new question:
How many copies of Campbell's Biology would it take to cover the state of Texas?
Campbell's Biology is slightly larger than an 8.5" x 11" sheet of paper, yielding an area of maybe about 5.5E2cm^2. Assuming the state of Texas is the area of half a square 1000km wide, that's 5E5km^2 or 5E15cm^2. Dividing, we get a Fermi answer of [b]13[/b]. Google gives me 1.024E13, so it looks like I was correct.
Let me take a stab:Unome has high standards
From what I can tell, this is definitely from the more difficult end. Easier ones involve much simpler premises and conditions, i.e. what is the distance, in megaparsecs, that a snail travels in the amount of time it takes Pluto to make a full revolution around the sun or something like that.
Also, new question:
How many copies of Campbell's Biology would it take to cover the state of Texas?New question:Campbell's Biology is slightly larger than an 8.5" x 11" sheet of paper, yielding an area of maybe about 5.5E2cm^2. Assuming the state of Texas is the area of half a square 1000km wide, that's 5E5km^2 or 5E15cm^2. Dividing, we get a Fermi answer of [b]13[/b]. Google gives me 1.024E13, so it looks like I was correct.
What is the mass, in kilograms, of a stack of quarters stretching from the sun to the center of the Andromeda galaxy?
I thankfully remember that Andromeda is about 10 million light years away (E7) I also remember that a light year is E13 km Quarter in width is just about 1mm, so E6 quarters to make a km and each quarter is E1 grams Altogether that makes E26
Apparently Andromeda is actually E6 away but quarters are actually E2 grams so they cancel out
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