Adventure Game Studio

(http://farm3.static.flickr.com/2590/4084386286_0c81d80f51_o.png)

Umm... I usually fail chemistry. But in terms of physics, you should know each gas's density to calculate that.

EDIT: Hmm, or maybe not.
Basically speaking, pressure = something like M * (height level) / T, where M is molar mass and T is temperature. since I guess height and temperature may be considered equal, you need to know their molar masses, and compare them

not sure I do not miss something though. Didn't have such "puzzles" for many years.

Then again, why not just google it? ;)

If I honestly tried to figure that out, my answers would still amount to... (http://spentertainment.webs.com/MISC/sumobutts.jpg)

 :o

I knew I should have paid attention in school.

P*V = N*R*T

That was like, the only thing to remember when dealing with gases, no?

P = pressure
V = volume
N = number of molesare = the gas constant
R = gas constant
T = temperature

Given that v,are and t are equal, the gas with the highest pressure is the one with more moles. To find that, you use the molar mass of the atoms and the mass given in the exercise, and do a division.

Something along these lines, at any rate. It's been a while since I last did physics / chemistry.

Edit : The hell ? And if I want to write "r" ? Well, capital letters it is, then.

Doesn't the answer depend on the temperature of the bottles and the gas?

Temperature has nothing to do in this. ;) Humm I am gonna give a try:

N² = 28,01grams/1mol
CO² = 44.01grams/1mol

28,01g    10g
--------- = ------       ? mol = 0.36mol
1 mol       ?mol       

44.01g     12g
--------- = ------      ? mol = 0.27mol
1 mol       ?mol


So ok, N² has more moles (more particules)... But 12 grams is more then 10 grams, and mass also influes the pressure... What should I do now?

n=mass/molar mass.

Temperature and volume would probably be the same for both bottles/gases, and R is a constant, so

whichever has the greater (mass/molar mass) will have the greater pressure.

I like how it is specified that the glass bottles are made out of glass. I guess it's good to make sure we know...

Okay,

---

PV=nRT

therefore
P= (nRT)/V

as T and V are constant, P α nR (α meaning "is proportionate to")

as R is a constant (8.315 J K^-1 Mol^-1)

P α n.

---

Now we have to work out n.


n(CO₂)= mass/(Molar Mass) = 15/(12.01+2(16.00))** = 0.2830 mol.

n(N₂)= m/M = 10/(2(14.01))** = 0.5549 mol.

as n(N₂) > n(CO₂)

    P(N₂) > P(CO₂)


** Molar mass values taken from chemistry data sheet

---

Quote
So ok, N² has more moles (more particles)... But 12 grams is more then 10 grams, and mass also influes the pressure... What should I do now?

The only thing that mass influences is the number of moles. (as m= nM)
As the mass of each particle is constant, more particle means more mass, and vice-versa.

---

Hope this helps
Peace
Scarab

You guys with the formulae seem to have missed the point.

There's obviously more pressure for CO2. With CO2 everyone's all, "Keep that tree alive! Stop killing the planet!"

Nitrogen gets it pretty easy, all it has to keep alive are the aliens from Mars Attacks, which is pretty much Tim Burton's worst film.

Please post any gold stars you recieve using my answer in this thread so that I can add them to my scrap book.

Here's one, feel free to include it in your signature.

(http://www.2dadventure.com/ags/goldstar.JPG)

It would be cool to have a dedicated thread for solving physics/maths/chemistry problems. Although I do not think that's in too much demand really.

Here you go, Ali.  I cleaned up Andail's gold star for you and I even made it forum signature friendly!

(http://i485.photobucket.com/albums/rr218/ProgZmax/STAR.png)

Quote from: Haddas on Wed 11/11/2009 14:17:10
It would be cool to have a dedicated thread for solving physics/maths/chemistry problems. Although I do not think that's in too much demand really.

Have at ye!

---

A watch has a minute hand that is 8mm long and a second hand 10 mm long.



(a) Determine the angular speeds of rotation of the minute and seconds hands in radians per hour.




(b) If the angle between the minute and seconds hands is θ, and the distance in millimetres between the tips of these hands is x, show using the cosine rule that x² = 164 - 160 cosθ.




(c) At exactly 7.20pm, how fast is the distance between the tips of the hands chcnging? Give your answer in mm/s correct to 2 decimal places.


Have fun!
Peace
Scarab

Thanks for the stars guys! I'm obviously pretty great at physics.

The first person to answer Scarab's question correctly wins my much coveted gold star!

EDIT: Please disregard the horribly wrong mess I've posted below... ;D

a)

as_m =  120π
as_s = 7200π

b)

x² = 8² + 10² - 2*8*10 cosθ

x² = 164 - 160 cosθ

c)

x = (164-160cosθ)^0.5
dx = 0.5*(164-160*1)^-0.5
dx = 0.5*4^-0.5
dx = 1/2 * 1/(4^0.5)
dx = 1/(2*2)

dx = 0.25

---

A faucet takes 20 minutes to fill a bathtub.
The drain empties it in 50 minutes.

How long does it take to fill the bathtub if the drain isn't plugged?

Vol = v1*20
Vol = v2*50
v3 = v1 - v2 = Vol/20 - Vol/50 = Vol*(3/100)
Vol = v3 * (100/3)

Ans: 100/3 mins

The answer is not exactly pretty, did I do something wrong? EDIT: yes I did, now?

---

A man is shot out of a cannon with a velocity of 10 m/s, 20 m straight forward (at the same level as where he's shot out of the cannon) is a safety net. Will he make it?

100/3 is correct, I chose arbitrary numbers.

Re your question: what's the cannon's angle? Also straight? Then no, because gravity will move him below the net.

Quote from: Khris on Thu 12/11/2009 22:28:32
100/3 is correct, I chose arbitrary numbers.

Re your question: what's the cannon's angle? Also straight? Then no, because gravity will move him below the net.

I never said the cannon's angle was straight. It might be gay for all I know ;)
And it could also very well have an angle, it's just not mensioned in the question as to wether or not it does.
Let me put it this way: is it possible?

I could be horribly wrong but I guess, no matter what the angle, he'll never make it.
My approach was assuming he makes it and calculating the angle from there.
But then sin α * cos α would have to be 1,962 which is impossible since it can't be more than 0.5.

Quote from: Khris on Thu 12/11/2009 20:06:59
a)

as_m =  120π
as_s = 7200π

b)

x² = 8² + 10² - 2*8*10 cosθ

x² = 164 - 160 cosθ

c)

x = (164-160cosθ)^0.5
dx = 0.5*(164-160*1)^-0.5
dx = 0.5*4^-0.5
dx = 1/2 * 1/(4^0.5)
dx = 1/(2*2)

dx = 0.25

(a) incorrect
(b) correct
(c) incorrect

This one's still open guys!

peace
Scarab

p.s. @ Khris, I sent you a p.m.

Quote from: Khris on Fri 13/11/2009 00:21:37
I could be horribly wrong but I guess, no matter what the angle, he'll never make it.
My approach was assuming he makes it and calculating the angle from there.
But then sin α * cos α would have to be 1,962 which is impossible since it can't be more than 0.5.

You're correct. We got this question in class and had to solve it in 5 mins using no calculation :P
In a class of 150, where everyone is excellent at math, 2 were able to solve it, in different ways too:
1)People who compete in long jump run at approx. 10 m/s when they jump and can't reach 10m.
2)If you shoot straight up he's in the air for approx. 2 sec. He needs more time than that at any angle that isn't straight.

Quote from: Scarab on Thu 12/11/2009 02:53:19

Quote from: Haddas on Wed 11/11/2009 14:17:10
It would be cool to have a dedicated thread for solving physics/maths/chemistry problems. Although I do not think that's in too much demand really.

Have at ye!

---

A watch has a minute hand that is 8mm long and a second hand 10 mm long.

(a) Determine the angular speeds of rotation of the minute and seconds hands in radians per hour.

[2]

(c) At exactly 7.20pm, how fast is the distance between the tips of the hands chcnging? Give your answer in mm/s correct to 2 decimal places.

[6]

a)
2*pi
(1/6)*pi

c)
The speeds are constant and at 7:20 the distance decreaces:

(1/6)*pi - 2*pi = ((1 - 12)/6)*pi = (-11/6)*pi
pi ~ 3.14
-11/6 ~ -1.83
(-1.83)*3.14 = -5.66142
.... I think, I don't have a calculator on me so that could be seriously wrong but my money's on:
Ans: -5.66
Wait.. crap.. I read wrong :P

10*(1/6)*pi - 8*2*pi = ((5 - 8*6)/3)*pi = (-43/3)*pi
I just remembered that the computer has a calculator so I cheated ;D, sue me
Ans: -45.03

Quote from: Bulbapuck on Fri 13/11/2009 10:13:30

a)
2*pi
(1/6)*pi

c)
The speeds are constant and at 7:20 the distance decreaces:

(1/6)*pi - 2*pi = ((1 - 12)/6)*pi = (-11/6)*pi
pi ~ 3.14
-11/6 ~ -1.83
(-1.83)*3.14 = -5.66142
.... I think, I don't have a calculator on me so that could be seriously wrong but my money's on:
Ans: -5.66
Wait.. crap.. I read wrong :P

10*(1/6)*pi - 8*2*pi = ((5 - 8*6)/3)*pi = (-43/3)*pi
I just remembered that the computer has a calculator so I cheated ;D, sue me
Ans: -45.03

Not quite,
Still open!
Focus on the units people!

when I did this question, it needed 14 lines of working, so I think it needs about 10 minimum.

peace
scarab

(c)
x^2 = 164 - 160*cosθ
x = sqrt(164 - 160*cosθ)
dx/dt = dx/dθ * dθ/dt
dx/dθ = 40*sinθ/sqrt(41 - 40*cosθ)
dθ/dt = pi/30 - pi/1800 = 59*pi/1800
dx/dt = (59*pi/1800)*40*sinθ/sqrt(41 - 40*cosθ)
dx/dt = (59*pi/45)*sinθ/sqrt(41 - 40*cosθ)
Angle between them is 120. Plug in to get
dx/dt = (59*pi/45)*(sqrt(3)/2)/sqrt(41 + 20)
dx/dt = (59/90)*pi*sqrt(3/61), or approximately 0.46 mm/s

***DING!DING!DING!DING!***
We have a winner. The answer is really -0.457, because dθ/dt = -59/1800 pi, but all the working was there.

Peace
Scarab

I believe this math question has not yet been solved, but since I don't know how to approach it at all I decided to post it for some insight. There are many variations of the question, this is just one.

---

You're the participant at a game show and you've won a prize. But which prize? There are two boxes filled with money, you have no clue as to how much. However, the host tells you that one box contains twice the amount of money then the other one.
You pick one of the boxes. Now the host asks you if you want to change box. Do you?

---

Logically it doesn't matter, right? However, the twist to the problem is this:
You get to open the box.
So now you have a certain amount of money. Say N dollars. Changing the box means a 50% chance that you end up with N/2 dollars, and a 50% chance that you end up with 2N dollars. Meaning you bet N/2 dollars on a 50-50 game and can win 3N/2 dollars.

But this quite obviously means you should change the box.

???

Where in my reasoning did I go wrong?

Some kind of variation on the "three doors" problem? What I understand from such problems is that conditional probabilities are counter-intuitive. :P

(What do you mean "not helping"?)

Quote from: Bulbapuck on Tue 17/11/2009 17:09:10
Logically it doesn't matter, right? However, the twist to the problem is this:
You get to open the box.
So now you have a certain amount of money. Say N dollars. Changing the box means a 50% chance that you end up with N/2 dollars, and a 50% chance that you end up with 2N dollars. Meaning you bet N/2 dollars on a 50-50 game and can win 3N/2 dollars.

But this quite obviously means you should change the box.

Where in my reasoning did I go wrong?

Why are you betting N/2 dollars? As far as I can see you arre betting N dollars with a 50/50 chance of going double or 'half'. (Please correct me if I've misunderstood)

The fact that you get to open the box changes nothing because the amount is arbitrary.

The Monty Hall Problem (or The Three Doors as Luf said it) is counter intuitive because it appears that the third outcome is removed, leaving a 50/50, when actually it is not. This problem just seems perfectly straightforward to me. (however I remember saying something similar when doing the Monty Hall Problem the first time as well  :-\)

Diagram:
(http://scarab117.webs.com/photos/happy/logicdiagram.png)

So I think whether you know the contents or not is irrelevant, and is just thrown in there to throw us off.

Well, I said "bet N/2" because there's no scenario where you end up with less than N/2. Hence you can only lose N/2 dollars. And also, if you don't know the amount of money there is a simple solution a mathematician came up with that states that you can't use probability theorems like that when you don't know the amount. But you're right, I don't really think it matters either.

Your diagram shows my first point perfectly, it doesn't matter.

However, to show my second point: Imagine this fair game:

You bet N dollars on a coin toss to come up heads. If the coin comes up heads, you get 2N dollars. If it comes up tails, you lose the N dollars.

This game is obviously fair. Acording to probability theory your projected winnings are as much as you bet:
(1/2)*0 + (1/2)*2N = N

Now imagine this one:

You bet N dollars on a coin toss to come up heads. If the coin comes up heads, you get 2N dollars. If it comes up tails, you lose N/2 dollars.

Acording to probability theory, this game is in your favour:
(1/2)*(N/2) + (1/2)*2N = 5N/4 > N
Hence you should play the game.

Well, betting N/2 isn't the same as winning at least N/2.

Here's one I particularly like (rather a riddle than a real math problem though):
You work at a place that deals with counterfeit money. Your specialty is coins.
Just before your shift ends, eight big boxes full of coins come in. The coins all look exactly the same, and the real ones are all supposed to have a weight of 10 grams.

Your supervisor explains that
a) one box contains exclusively counterfeit coins while the other seven hold only real ones.
b) all of the bad coins are either 1 gram lighter or 1 gram heavier (not mixed!).

Just as you want to start weighing the coins you notice that your scales are running out of power. You're too lazy to get new batteries and want to go home five minutes ago, so: can you find out the bad box and the weight of the bad coins while using the scales only once? And how?

Quote from: Khris on Wed 18/11/2009 21:51:49
Well, betting N/2 isn't the same as winning at least N/2.

But in this case, the outcome is exactly the same, right? So I don't see how that matters.

Quote
Here's one I particularly like (rather a riddle than a real math problem though):
You work at a place that deals with counterfeit money. Your specialty is coins.
Just before your shift ends, eight big boxes full of coins come in. The coins all look exactly the same, and the real ones are all supposed to have a weight of 10 grams.

Your supervisor explains that
a) one box contains exclusively counterfeit coins while the other seven hold only real ones.
b) all of the bad coins are either 1 gram lighter or 1 gram heavier (not mixed!).

Just as you want to start weighing the coins you notice that your scales are running out of power. You're too lazy to get new batteries and want to go home five minutes ago, so: can you find out the bad box and the weight of the bad coins while using the scales only once? And how?

I numbered the boxes from 1 to 8. Take one coin from the first box, two from the second, three from the fourth and so on. Now weigh those coins together. If all coins were real this would be 360g. However, since they're not all real there will be a difference of Xg. That means that it's box number X. And to figure out the weight you simply see if the scale is giving you a value above or below 360g.
if(below){badCoinsWeight = 9g;}
else if(above){badCoinsWeight = 11g;}
else{Display("damn lying boss..");} // :P

Yep, correct :)

Here's one:
A hotel has an infinite amount of rooms, and every room is occupied. Suddenly a new guest shows up. Can they find him a room?

Oo, this one has so many angles it can be taken from!

OK, if I'm allowed to quote from Wikipedia:
"The Isha Upanishad of the Yajurveda (c. 4th to 3rd century BC) states that "if you remove a part from infinity or add a part to infinity, still what remains is infinity"."

does that answer?

Ah Hilbert's Hotel :p

I would argue that yes they can.

There are an infinite number of odd numbers.
There are an infinite number of even numbers.
There are an infinite number of even and odd numbers combined.

So even if you have an infinite number of rooms and an infinite number of guests the two cannot be equal since infinity is not a value but rather a concept.

Yes.
It reminds me of the a=b problem...

Quote from: Calin Leafshade on Fri 20/11/2009 12:01:59
Ah Hilbert's Hotel :p

I would argue that yes they can.

There are an infinite number of odd numbers.
There are an infinite number of even numbers.
There are an infinite number of even and odd numbers combined.

So even if you have an infinite number of rooms and an infinite number of guests the two cannot be equal since infinity is not a value but rather a concept.

Actually, they are equal! The amount of hotel rooms is the same as the amount of odd numbers which is the same as the number of rational numbers.
I know that sounds odd, but there are actually only two infinities. One that's countable, and one that isn't.

Quote from: wonkyth on Fri 20/11/2009 11:52:22
Oo, this one has so many angles it can be taken from!

OK, if I'm allowed to quote from Wikipedia:
"The Isha Upanishad of the Yajurveda (c. 4th to 3rd century BC) states that "if you remove a part from infinity or add a part to infinity, still what remains is infinity"."

does that answer?

Not really, you can't add a new hotel room... Think like you're the keeper for the hotel. How would you go about actually doing it?

This one is still open! :)

If you're just looking for the classic hilberts hotel answer;

move the occupant of room 1 to room 2 and the occupant of room 2 to room 3 and so on.

thus room 1 is now free.

Quote from: Calin Leafshade on Fri 20/11/2009 14:06:37
If you're just looking for the classic hilberts hotel answer;

move the occupant of room 1 to room 2 and the occupant of room 2 to room 3 and so on.

thus room 1 is now free.

Yes, correct.

I was looking for any variation on the solution, and wonkyth's answer didn't seem too practical. It's not like the hotel keeper could've said: "just add a room" to the guy.

FOLLOWUP QUESTION:
And those who know the solution, please don't answer :P

If everyone at said hotel was bringing in a freind to stay in a different room, how can you put each individual in a room of his/her own?

Ask everyone to move to room 2n where n is each guest's room number, thereby freeing up every other room (odd numbers) for the guests' friends?

Quote from: Intense Degree on Fri 20/11/2009 14:50:05
Ask everyone to move to room 2n where n is each guest's room number, thereby freeing up every other room (odd numbers) for the guests' friends?

Correct :)

I knew that maths GCSE would come in handy one of these days ;D