Fermi Questions C

Test your knowledge of various Science Olympiad events.
UTF-8 U+6211 U+662F
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Re: Fermi Questions C

Post by UTF-8 U+6211 U+662F » May 11th, 2018, 8:34 pm

I just realized I meant fall at terminal velocity, but it doesn't really make the question invalid, so I'll stick with free fall.
Paper would travel, in 10 hrs, 4.9 m/s^2 * (10 hrs * 3600 s/hrs)^2, which is 6.35E9 meters. The distance to Alpha Centauri is 4.132E16 meters, giving a Fermi answer of 7.
(I think I did that right: I'm new to this)

Someone else can ask a question.

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Re: Fermi Questions C

Post by sciencegirl03 » May 11th, 2018, 8:39 pm

sciencegirl03 wrote:Distance to Alpha Centauri: 4.5light years; i.e. 4.5 * 10^16m\
Paper falls at 1m/sec so in 10 hrs= 36000meters
Fermi answer 11
Sorry, no drag....
Distance to Alpha Centauri: 4.5light years; i.e. 4.5 * 10^16m
Distance paper falls: .5 * 10 * 36000^2 = 6.5 * 10^9m
Fermi answer: 7

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Re: Fermi Questions C

Post by UTF-8 U+6211 U+662F » May 11th, 2018, 8:41 pm

sciencegirl03 wrote:Still figuring out the Attempt/Hide...
Use

Code: Select all

[hide]Attempt|adfjsakdjfksajdfkj[/hide]
for the attempt and either

Code: Select all

[hide]Answer|asdkfjakdsjfksj[/hide]
or

Code: Select all

[answer]aksdjfkasjdfkajf[/answer]
for the answer.
The vertical bar is right above the enter button on my QWERTY keyboard.

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Re: Fermi Questions C

Post by sciencegirl03 » May 11th, 2018, 9:19 pm

UTF-8 U+6211 U+662F wrote:
sciencegirl03 wrote:Still figuring out the Attempt/Hide...
Use

Code: Select all

[hide]Attempt|adfjsakdjfksajdfkj[/hide]
for the attempt and either

Code: Select all

[hide]Answer|asdkfjakdsjfksj[/hide]
or

Code: Select all

[answer]aksdjfkasjdfkajf[/answer]
for the answer.
The vertical bar is right above the enter button on my QWERTY keyboard.
Thanks UTF-8 U+6211 U+662F
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Re: Fermi Questions C

Post by NeilMehta » May 14th, 2018, 6:18 am

New question:
On average, how many text messages are sent every nanosecond?
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Re: Fermi Questions C

Post by Unome » May 14th, 2018, 7:08 am

Approximately 1E12 text messages per year, divide by 3E7 to get to seconds, divide by 1E9 to get to nanoseconds, yields 3E-5 so Fermi Answer: -5
8.6E12 text messages per year, arund 2.7E-4 per nanosecond, so Fermi Answer: -4
How much energy in ergs is transferred to the Earth by the moon reflecting sunlight over the course of one Svedberg?
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Re: Fermi Questions C

Post by Name » May 14th, 2018, 7:31 am

The surface area of the moon is around 3E7. Distance to sun E8 so surface area is around 2E17. 3E7/2E17 is 1.5E-10 times 4E26 is 6E16. Distance to moon E5 or about 2E11 surface area. Surface area of earth 5E8/2E11 is 2.5E-3 times 6E16 is 1.5E14 Assuming a 1 percent reflection 1.5E12joules/sec. One svedberg is E-13 one erg is E-7. So 1.5E6 or Fermi 6
energy reflection is around 10 percent but the distance to moon is 3E5 not E5 and would cancel out each other so E6
Assuming a uniform density of earth how many times stronger is Earth's gravitational force at it's surface compared to if you went underground 1000 kilometers
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Re: Fermi Questions C

Post by PM2017 » May 14th, 2018, 7:41 am

Name wrote:
The surface area of the moon is around 3E7. Distance to sun E8 so surface area is around 2E17. 3E7/2E17 is 1.5E-10 times 4E26 is 6E16. Distance to moon E5 or about 2E11 surface area. Surface area of earth 5E8/2E11 is 2.5E-3 times 6E16 is 1.5E14 Assuming a 1 percent reflection 1.5E12joules/sec. One svedberg is E-13 one erg is E-7. So 1.5E6 or Fermi 6
energy reflection is around 10 percent but the distance to moon is 3E5 not E5 and would cancel out each other so E6
Assuming a uniform density of earth how many times stronger is Earth's gravitational force at it's surface compared to if you went underground 1000 kilometers
well, the Earth has a radius of 1100 km iirc, so the new distance is 1000, for a ratio of 10000/11000, or 10/11. Squaring this is still very close to one (1/1.21) so Fermi answer 0.
I'll add the actual once I can get to my laptop.

What is the 273rd Fibonacci number?
To clarify, 0 will be considered the first Fibonacci number.
Last edited by PM2017 on May 14th, 2018, 10:25 am, edited 1 time in total.
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Re: Fermi Questions C

Post by Name » May 14th, 2018, 8:28 am

PM2017 wrote:
Name wrote:
The surface area of the moon is around 3E7. Distance to sun E8 so surface area is around 2E17. 3E7/2E17 is 1.5E-10 times 4E26 is 6E16. Distance to moon E5 or about 2E11 surface area. Surface area of earth 5E8/2E11 is 2.5E-3 times 6E16 is 1.5E14 Assuming a 1 percent reflection 1.5E12joules/sec. One svedberg is E-13 one erg is E-7. So 1.5E6 or Fermi 6
energy reflection is around 10 percent but the distance to moon is 3E5 not E5 and would cancel out each other so E6
Assuming a uniform density of earth how many times stronger is Earth's gravitational force at it's surface compared to if you went underground 1000 kilometers
well, the Earth has a radius of 1100 km iirc, so the new distance is 1000, for a ratio of 10000/11000, or 10/11. Squaring this is still very close to one (1/1.21) so Fermi answer 0.
I'll add the actual once I can get to my laptop.

What is the 273rd Fibonacci number?
Wait a screwed up the question. It is E0 but I should've said something like 3000 where the new mass is around E22 and radius is 1000 where the gravitational force is E1 stronger while on the surface
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Re: Fermi Questions C

Post by Tom_MS » May 14th, 2018, 10:55 am

Name wrote: Wait a screwed up the question. It is E0 but I should've said something like 3000 where the new mass is around E22 and radius is 1000 where the gravitational force is E1 stronger while on the surface
I feel like this is being overthought. According to the shell theorem (essentially Gauss's law), the only mass that matters should be within a spherical shell at the radius being considered. If you do the math, this gives a linearly decreasing force for a linearly decreasing radius. Thus, just taking the ratio of 5/6 (5000 km compared to around 6000 km) gives you fermi 0.

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