Fermi Questions C

[UTF-8 U+6211 U+662F]

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.

Actual answer (Click to Show Hidden Text)

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.

[sciencegirl03]

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

Still figuring out the Attempt/Hide...

[UTF-8 U+6211 U+662F]

Still figuring out the Attempt/Hide...

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[hide]Attempt|adfjsakdjfksajdfkj[/hide]

for the attempt and either

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[hide]Answer|asdkfjakdsjfksj[/hide]

or

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[answer]aksdjfkasjdfkajf[/answer]

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

[sciencegirl03]

Still figuring out the Attempt/Hide...

Use

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[hide]Attempt|adfjsakdjfksajdfkj[/hide]

for the attempt and either

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[hide]Answer|asdkfjakdsjfksj[/hide]

or

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[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

[NeilMehta]

New question:
On average, how many text messages are sent every nanosecond?

[Unome]

Attempt (Click to Show Hidden Text)

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

Actual (Click to Show Hidden Text)

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?

[Name]

Attempt (Click to Show Hidden Text)

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

Answer (Click to Show Hidden Text)

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

[PM2017]

Attempt (Click to Show Hidden Text)

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

Answer (Click to Show Hidden Text)

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

attempt (Click to Show Hidden Text)

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.

[Name]

Attempt (Click to Show Hidden Text)

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

Answer (Click to Show Hidden Text)

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

attempt (Click to Show Hidden Text)

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

[Tom_MS]

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.