## Search found 10 matches

- March 20th, 2012, 7:30 pm
- Forum: 2012 Study Events
- Topic: Fermi Questions Marathon
- Replies:
**166** - Views:
**26949**

### Re: Fermi Questions Marathon

In response to the 1,000,000th prime number, the mathematical answer uses the prime number theorem: The number of primes less than n asymptotically approaches n/ln(n), which can be manipulated to show that the nth prime number is approximately n*ln(n) or 1E6*ln(1E6) = 6E6*ln(10)~ 3E7, so 7.

- March 14th, 2012, 5:13 pm
- Forum: 2012 Study Events
- Topic: Fermi Questions Marathon
- Replies:
**166** - Views:
**26949**

### Re: Fermi Questions Marathon

If cars were to run on the ink out of printer cartridges with the same "ink efficiency" as gas efficiency, how much would you spend on ink to drive from Dallas, TX to Chicago, IL? Assumptions: Car gets 20mpg, Dallas to Chicago is 2E3 miles, Ink cartrige is 20 ml and costs 25$, 2L is 1 gallon Then i...

- March 7th, 2012, 8:53 pm
- Forum: 2012 Study Events
- Topic: Fermi Questions C
- Replies:
**82** - Views:
**20384**

### Re: Fermi Questions C

I haven't faced any factorials in any of the tournaments I've been to. They must not be that popular in the Kansas region... But just in case, how would one ideally (quickly, fairly accurately) find a factorial during a Fermi Test? Or do you just have to memorize general ones? All you really have t...

- March 7th, 2012, 5:33 pm
- Forum: 2012 Study Events
- Topic: Fermi Questions Marathon
- Replies:
**166** - Views:
**26949**

### Re: Fermi Questions Marathon

If we build a data storage device that was the size of the moon, with our present maximum storage capacity density, how many seconds of 1080p video could it hold?

- March 7th, 2012, 5:19 pm
- Forum: 2012 Study Events
- Topic: Fermi Questions Marathon
- Replies:
**166** - Views:
**26949**

### Re: Fermi Questions Marathon

(e^(pi^(gamma^(phi)))) I assume my gamma you mean the E-M one, We fill in approximate values, 2.7^3.1^0.58^1.6 estimate 0.58^1.6 = 0.45 because 0.6^2 0.36 and 0.6^1 = 0.6 3.1^0.45 ~ sqrt(pi) = 1.77 (memorized for probability theory), rounding down I'll take 1.6 2.7^1.6 is going to be slightly large...

- March 7th, 2012, 5:00 pm
- Forum: 2012 Study Events
- Topic: Fermi Questions Marathon
- Replies:
**166** - Views:
**26949**

### Re: Fermi Questions Marathon

We assume an average height of 1750, 1500 of which are fillable feet, dimensions 15*1*1, and that Orangutans achieve maximum compression at the volume of a 3 ft by 2 ft by 2 ft cube. That gives us total volume of 5*1.2E2^2*1500 ft = 6E4*1.5E3 = 9E7 ft^3, and 12 ft ^3 for orangutans gives 9E7/12 = 7....

- March 7th, 2012, 3:14 am
- Forum: 2012 Study Events
- Topic: Astronomy C
- Replies:
**176** - Views:
**32626**

### Re: Astronomy C

Are there any highly recommended websites on general astronomy info (so far I am using like the first two pages of that from google, 3 glossaries, apod/NASA, bunches of other things)? Thank you very much to anyone who responds. I like to look at college astronomy classes that have some or all of th...

- March 6th, 2012, 12:52 pm
- Forum: 2012 Study Events
- Topic: Astronomy C
- Replies:
**176** - Views:
**32626**

### Re: Astronomy C

Yes, those are the things you should review.JumpUp12 wrote: My competition is tomorrow! Any tips for the most important things to review? Theory? Math? DSOs? Type Ia Supernovae?

- March 6th, 2012, 11:39 am
- Forum: 2012 Study Events
- Topic: Fermi Questions Marathon
- Replies:
**166** - Views:
**26949**

### Re: Fermi Questions Marathon

100!: n! ~ sqrt(2pi*n)*n^n*e^-n) By Stirling logn! = log(sqrt(6.5*100)+log(100^100) + log(e^-100) = 1 + 1/2log(6.5) +100*log(100) -100loge = 1 +0.4 +200 - 43.4 = 200 - 42 = 158 Number of e- in atlantic. Assume average depth is 4000m, atlantic covers 20% of earth, R_(+) = 6E6m, ocean is 100% water, t...

- March 5th, 2012, 8:47 pm
- Forum: 2012 Study Events
- Topic: Fermi Questions Marathon
- Replies:
**166** - Views:
**26949**

### Re: Fermi Questions Marathon

Number of redwoods to absorb all CO2 emissions: It will be difficult to estimate the amount of CO2 absorbed per tree, so I am going to try a more inventive method. Plants are responsible for 0.25 the total C02 absorption. From a memorized list of prominent state park sizes, I know that the major red...