Fermi Questions C

[whythelongface]

If I am interpreting this properly, you want the distance light travels in Googolplex^Googolplex lunar months, and how many times that would wrap around an arbitrary equator of the visible universe.

If that's the case, I really can't see how you'd express that as a Fermi answer, unless you accept 10^(100+10^100)-12.

More likely than not though I'm just reading this problem wrong.

[NeilMehta]

How many times would a violet light (assume perfect light) wrap around the universe if it traveled for Googleplex to the Googleplex lunar months?

https://www.wolframalpha.com/input/?i=g ... googolplex
raleway don't do this to us

[Alex-RCHS]

How many times would a violet light (assume perfect light) wrap around the universe if it traveled for Googleplex to the Googleplex lunar months?

I'll admit, I had to look up the radius of the observable universe. I got:

this (Click to Show Hidden Text)

googleplex to the googleplex, as a fermi number, would be 1000googleplex. That's how many months light would be traveling. Divide by 12 to get the distance in light years and you get 999googleplex as a fermi number. Divide by the radius of the observable universe (46 billion light years) and you get 989googleplex as your fermi answer.

Not 100% sure I did that right.

A horse walks into a bar. How fast would he have to be walking into the bar in order to have the equivalent amount of (kinetic) energy as 10 million chickens crossing the road at normal chicken speed?

[whythelongface]

How many times would a violet light (assume perfect light) wrap around the universe if it traveled for Googleplex to the Googleplex lunar months?

I'll admit, I had to look up the radius of the observable universe. I got:

this (Click to Show Hidden Text)

googleplex to the googleplex, as a fermi number, would be 1000googleplex. That's how many months light would be traveling. Divide by 12 to get the distance in light years and you get 999googleplex as a fermi number. Divide by the radius of the observable universe (46 billion light years) and you get 989googleplex as your fermi answer.

Not 100% sure I did that right.

A horse walks into a bar. How fast would he have to be walking into the bar in order to have the equivalent amount of (kinetic) energy as 10 million chickens crossing the road at normal chicken speed?

I don't think that's right, but whatever.

Answer (Click to Show Hidden Text)

Let's assume a horse masses about E3 kg, and that a chicken massess around E1 kg. Ten million chickens would therefore mass E8 kg. KE = 0.5mv^2, Assume chickens move at E-1 m/s. The kinetic energy of ten million chickens would therefore be about 5E5. This means the the horse's velocity squared would be about E3 (m/s)^2, and so the Fermi answer would be [b]2[/b], if we are measuring in meters per second. If we're lucky, maybe we can have supersonic horses by the end of this century.

Supposing pencils can be made out of any wood (not just cedar), if I cut down the entire Amazon Rain Forest and used the wood to solely manufacture pencils, and used them to draw one, straight, unbroken line, how long would the line be, in Angstroms?

[NeilMehta]

I doubt I could ever solve this without info, so I'll post the solution for anyone who wants to check their work:

BREAK GLASS IN CASE OF EMERGENCY (Click to Show Hidden Text)

Solution: 31

[Torterra]

If she sells all the seashells on all the seashores on earth, how much money will she make?

[whythelongface]

If she sells all the seashells on all the seashores on earth, how much money will she make?

Depends on what you define as the seashore, as well as what you define as a "shell". Do fragments count? What about the shells of sea urchins, crabs, horseshoe crabs, etc? And how about the shore? Does the intertidal zone count? What about the littoral zone? Also, many shores have layers of sediment stacked on top of each other. How deep can you go and still call it the shore? Depending on your answers, you could be many magnitudes, perhaps even as far off as five magnitudes, off.

[Torterra]

Sorry, should have clarified. Only shells that are intact and on the surface to an inch deep. Any kind of shell though

[whythelongface]

Business calculations (Click to Show Hidden Text)

Assume E4 intact shells of any kind along any stretch of coastline 1m long and 1 inch deep. That makes E7 shells along a 1 km coast. Assume E6 km of oceanic coastline, to get E13 shells.

Shells vary in rarity, going from bare cents to thousands of dollars. Estimate average cost of shell to be around ten cents, or E-1 dollars. Multiplying, we get E12 dollars. She is the world's first trillionaire.

Absolutely sure I got some approximation completely wrong, if not all of them.

Have you, perchance, heard of Sanibel Island, Florida?

If a 6-pack of Monster got converted into pure energy based on its caloric content, how long could it keep a 60-watt lightbulb on for?

[jkang]

If a 6-pack of Monster got converted into pure energy based on its caloric content, how long could it keep a 60-watt lightbulb on for?

Answer (Click to Show Hidden Text)

5
~200 Calories in monster (16 oz). 1 Calorie is ~4200 joules, 840000 joules per can, 60 W/6 pack -> ~ 84000 seconds. Putting it in fermi, 5.

If we took all the energy used by phones in the world in a day, how long could we produce the same power as the Sun (seconds)?