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Mr. Pedantic

Diamond Member
Feb 14, 2010
5,027
0
76
Interesting answers.




What is a notochord?

What is an orchidometer and what does it do?
It actually measures testicle size. No kidding. Look it up.

What was the original usage of the term "Planet X" in reference to, why was its existence theorized, and what happened to it?


What are the names of Mars' moons?


What is phrenology?


Chemically, what is the difference between starch and cellulose?
Both are made of long chains of glucose molecules. There are several differences. Starch is actually made of two different polysaccharides, amylose and amylopectin. The polysaccharides making up cellulose are much longer, up to several thousand residues long. But the one I was looking for, and the one that is relevant to our metabolism of cellulose, is that in cellulose the glucose residues are joined by beta-1-4 glycosidic bonds, which means that each glucose molecule is upside down, so to speak, relative to the glucose molecules it is bonded to.

What are Tanner Stages?
Tanner stages are a measure of development of secondary sexual characteristics - pubic hair, breast, and penile development.

What drug does Sherlock Holmes use most often?


two men are walking towards each other alongside a railway. A freight train overtakes one of them in 20 seconds and exactly 10 minutes later reaches the other man coming in the opposite direction. The train passes this man in 18s. How long after the train has passed the second man will the two men meet? (Constant speeds are assumed throughout)
1 hour, 32 minutes, 42 seconds; or 5562s. If anyone wants the working I'll write it up.
Answers for the rest.
 

angminas

Diamond Member
Dec 17, 2006
3,331
26
91
I name some of the comic character's traditional rivals, you name the character.

1. The Mandarin, Whiplash, The Crimson Dynamo

2. Thanos, Mephisto, Galactus

3. Thor, The Thing, The Abomination

4. Darkseid, Doomsday, Bizarro

5. Rayek, Winnowill, King Guttlekraw

6. Carnage, The Lizard, The Vulture
 

angminas

Diamond Member
Dec 17, 2006
3,331
26
91
I'd be interested in the writeup on the train problem if it's not too much trouble.
 

Locut0s

Lifer
Nov 28, 2001
22,205
44
91
I name some of the comic character's traditional rivals, you name the character.

1. The Mandarin, Whiplash, The Crimson Dynamo

Captain marvel?

2. Thanos, Mephisto, Galactus

The Silver Surfer

3. Thor, The Thing, The Abomination

Hulk

4. Darkseid, Doomsday, Bizarro

Superman

5. Rayek, Winnowill, King Guttlekraw

No idea

6. Carnage, The Lizard, The Vulture

Spiderman

^^
 

Mr. Pedantic

Diamond Member
Feb 14, 2010
5,027
0
76
I'd be interested in the writeup on the train problem if it's not too much trouble.
Dammit. I was hoping to get away without having to do this.

So first, some definitions:

Train speed = A
First person speed = x
Second person speed = y

The speed of the train relative to the first person is (A-x). This speed takes 20s to travel the length of the train. The speed of the train relative to the second person is (A+y). This speed takes 18s to travel the length of the train. So we have:

20(A-x) = 18(A+y)
20A - 20x = 18A + 18y
2A = 20x + 18y
A = 10x + 9y

Next. The distance between the two people, when the train has just finished passing person 1, is the length of the train, plus 10 minutes of travel.

Length of the train = 18(A+y)
Length of the distance between the train and person 2 = 600(A+y)

So at this point the distance between the two people = (600+18)(A+y)
Since A = 10x + 9y
The distance = 618(10x + 9y + y) = 618(10x+10y) = 6180(x+y)

Since (x+y) is the speed that the two people are walking towards each other, we're almost there. 6180 seconds, and this includes the 10 minutes (600 seconds) the train spent traveling between them, as well as the 18 seconds it took for the train to pass person 2.

6180 - 600 - 18 = 5562 seconds. Or 1 hour, 32 minutes, and 42 seconds.

There are lots of ways of doing this. For using the length of the train, 20(A-x) works out as well as 18(A+y), it's just that using y is a bit easier. You can also use the distance by each of the objects in various time periods instead of just using speeds, though I find speeds more intuitive.
 
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zokudu

Diamond Member
Nov 11, 2009
4,364
1
81
Well, since someone answered the last math problem, here's a geometry one:

You're a ball-bearing manufacturer. You've just gotten a good deal on 1cmx1cmx1cm inflexible cubic boxes to ship your ball bearings in. What's the maximum number of 1mm-diameter spherical ball bearings can you fit inside each box?

Going to guess 10000
 

Ken g6

Programming Moderator, Elite Member
Moderator
Dec 11, 1999
16,695
4,657
75
Well, since someone answered the last math problem, here's a geometry one:

You're a ball-bearing manufacturer. You've just gotten a good deal on 1cmx1cmx1cm inflexible cubic boxes to ship your ball bearings in. What's the maximum number of 1mm-diameter spherical ball bearings can you fit inside each box?
Going to guess 10000

This isn't a guessing game. :mad:

Since I'm expecting a proof, let me prove you're wrong.
Volume of a ball bearing = 4/3*Pi*r^3 ~= .52 mm^3. 10,000 ball bearings ~= 5200 mm^3.
Volume of the box = 10^3 mm^3 = 1000 mm^3.
Even melted down, 10000 ball bearings wouldn't fit in the box. (And I want the answer when they're not melted down.)

Edit: I should be clear: I have a proof that my configuration fits. I don't have a proof that it's optimal; but that seems likely.
 
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PlasmaBomb

Lifer
Nov 19, 2004
11,636
2
81
You have units mixed up. Remember that Planck's constant is in units of J.s, c is in units of ms^-1, and lambda is in units of m.

Put another way: the Planck constant describes the minimum total energy of a wave. It does not describe the minimum rate of transfer of that energy.


You can fit at least 1024 in each box.

I suppose that I should have checked the units balanced... (kind of knew when I checked Plancks constant... missremembered it a being in J not J.s)
 

hypn0tik

Diamond Member
Jul 5, 2005
5,866
2
0
Well, since someone answered the last math problem, here's a geometry one:

You're a ball-bearing manufacturer. You've just gotten a good deal on 1cmx1cmx1cm inflexible cubic boxes to ship your ball bearings in. What's the maximum number of 1mm-diameter spherical ball bearings can you fit inside each box?

Close Cubic Packing (CCP) is the way to go. Will work it out later when I have time.
 

Ken g6

Programming Moderator, Elite Member
Moderator
Dec 11, 1999
16,695
4,657
75
Pardon the late bump; I forgot about this thread. My best estimate was 1254 balls. As I showed here 6 years ago, they fit. I wonder if your cannonball stacks have overlapping balls anywhere?