1
Crap! / Re: post music that kwiftee likes
« on: April 22, 2012, 05:42:33 pm »
i am mildly perplexed by this thread
but you guys seem to be doing alright so far.
but you guys seem to be doing alright so far.
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2
Crap! / Re: Silencing Gun
« on: March 16, 2012, 06:51:30 am »
if only it could be used over the internet.
...Oh wait that'd just be the ignore button. FORGET I SAID ANYTHING
Also weird I was just reading about this in a cracked article.
Anyway, it's a neat idea; I can't wrap my head around how it works, cause the article gave me the impression that it's some kind of psychological manipulation thing.
...Oh wait that'd just be the ignore button. FORGET I SAID ANYTHING
Also weird I was just reading about this in a cracked article.
Anyway, it's a neat idea; I can't wrap my head around how it works, cause the article gave me the impression that it's some kind of psychological manipulation thing.
3
General Discussion! / Re: mari0 - mario plus portal
« on: March 06, 2012, 07:33:22 pm »
Okay fine.
5
General Discussion! / Re: Oxidation Numbers
« on: February 25, 2012, 10:26:42 pm »
I still lurk! I do only really post to do Mark's homework for him though.
6
General Discussion! / Re: Oxidation Numbers
« on: February 24, 2012, 07:52:00 pm »
hey fuck that, i was really looking forward to a thread about oxidation numbers
7
Crap! / Re: ITT: Ray diagrams (Physics)
« on: February 15, 2012, 10:29:41 pm »
don't draw them because the ray model is stupid and i hate it. wave particle duality all the way.
Nah, seriously, can't help with this one. Haven't touched ray diagrams since two years ago. I remember I hated them though, if that helps!
Nah, seriously, can't help with this one. Haven't touched ray diagrams since two years ago. I remember I hated them though, if that helps!
8
Crap! / Re: I have returned
« on: February 13, 2012, 09:43:32 am »
seriously don't make it as obvious as 'reverse the username'
Also, I know I recognise that name but I have literally no idea who you are.
Also, I know I recognise that name but I have literally no idea who you are.
9
General Discussion! / Re: Modulus Inequalities Question
« on: February 07, 2012, 06:24:45 am »
The issue is that your final solutions don't actually contradict each other; you're just getting mixed up between AND and OR statements. You're assuming that all the statements are 'and's, but they aren't necessarily.
Firstly, assume (as you did) that both sides are positive (or zero). Then, you have:
3x-2 >= 0 AND 2x+3 >= 0 AND 3x - 2 >= 2x + 3
simplifying to x>=2/3 AND x>=-3/2 AND x>=5.
which simplifies to x >= 5.
But, as you know, that assumption we made isn't the only case. Another case is that left hand side is positive, RHS negative.
3x-2 >= 0 AND 2x+3 <= 0 AND 3x - 2 >= -2x - 3
x >= 2/3 AND x <= -3/2 AND x >= -1/5
Note that this leads to a contradiction; there's no values of x which satisfy all three of those statements. So, the second assumption was just never true.
as for the other two assumptions:
LHS<=0, RHS>=0
gives 3x-2 <= 0 AND 2x+3 >= 0 AND 2 - 3x >= 2x + 3
x <= 2/3 AND x >= -3/2 AND x <= -1/5
hence, -3/2 <= x <= -1/5
LHS<=0, RHS<=0
gives 3x-2 <= 0 AND 2x+3 <= 0 AND 2 - 3x >= -2x - 3
x <= 2/3 AND x <= -3/2 AND x <= 5.
hence, x <= -3/2
Now, since all of these were assumptions, any one of them could lead to a solution. So now, we use OR statements, giving us our answer of x >= 5 or -3/2 <= x <= -1/5 or x <= -3/2
...simplifying to x >= 5 or x <= -1/5
When to use AND and when to use OR requires a bit of intuition sometimes; I'm not quite sure how to explain it. Plus I remember there was a problem I once did where this way of thinking didn't work out; something involving the range of arccos(x2) from memory, but I don't remember.
-
At any rate this isn't the simplest way of doing the above problem; it'd be much easier to just graph both of them and do it visually. Or failing that, you can recognise that both y = |3x - 2| and y = |2x + 3| are continuous functions, solve them equal to each other which tells you at what points they cross and just use intuition once again to work out which one is on top at a given value of x.
Firstly, assume (as you did) that both sides are positive (or zero). Then, you have:
3x-2 >= 0 AND 2x+3 >= 0 AND 3x - 2 >= 2x + 3
simplifying to x>=2/3 AND x>=-3/2 AND x>=5.
which simplifies to x >= 5.
But, as you know, that assumption we made isn't the only case. Another case is that left hand side is positive, RHS negative.
3x-2 >= 0 AND 2x+3 <= 0 AND 3x - 2 >= -2x - 3
x >= 2/3 AND x <= -3/2 AND x >= -1/5
Note that this leads to a contradiction; there's no values of x which satisfy all three of those statements. So, the second assumption was just never true.
as for the other two assumptions:
LHS<=0, RHS>=0
gives 3x-2 <= 0 AND 2x+3 >= 0 AND 2 - 3x >= 2x + 3
x <= 2/3 AND x >= -3/2 AND x <= -1/5
hence, -3/2 <= x <= -1/5
LHS<=0, RHS<=0
gives 3x-2 <= 0 AND 2x+3 <= 0 AND 2 - 3x >= -2x - 3
x <= 2/3 AND x <= -3/2 AND x <= 5.
hence, x <= -3/2
Now, since all of these were assumptions, any one of them could lead to a solution. So now, we use OR statements, giving us our answer of x >= 5 or -3/2 <= x <= -1/5 or x <= -3/2
...simplifying to x >= 5 or x <= -1/5
When to use AND and when to use OR requires a bit of intuition sometimes; I'm not quite sure how to explain it. Plus I remember there was a problem I once did where this way of thinking didn't work out; something involving the range of arccos(x2) from memory, but I don't remember.
-
At any rate this isn't the simplest way of doing the above problem; it'd be much easier to just graph both of them and do it visually. Or failing that, you can recognise that both y = |3x - 2| and y = |2x + 3| are continuous functions, solve them equal to each other which tells you at what points they cross and just use intuition once again to work out which one is on top at a given value of x.
10
Crap! / Re: math(s) thread
« on: February 04, 2012, 07:50:42 pm »
for x > 0
x = 1 + 1 + 1 + ... + 1; } x times
x2 = x + x + x + ... + x; } x times
d/dx(LHS) = d/dx(x2) = 2x.
d/dx(RHS) = d/dx(x + x + x + ... + x); } x times
= d/dx(x) + d/dx(x) + d/dx(x) + ... d/dx(x); } x times
= 1 + 1 + 1 + ... + 1; } x times
= x.
but 2x ≠ x.
what
x = 1 + 1 + 1 + ... + 1; } x times
x2 = x + x + x + ... + x; } x times
d/dx(LHS) = d/dx(x2) = 2x.
d/dx(RHS) = d/dx(x + x + x + ... + x); } x times
= d/dx(x) + d/dx(x) + d/dx(x) + ... d/dx(x); } x times
= 1 + 1 + 1 + ... + 1; } x times
= x.
but 2x ≠ x.
what
11
Crap! / Re: math(s) thread
« on: January 31, 2012, 01:49:54 am »By the way, I've been stuck on this physics question for a while now.
When light of wavelength 2.08 * 10^-7 m falls on a photosurface, a voltage of 1.40 V is required to stop the emitted electrons from reaching the anode. What is the largest wavelength of light that will result in the emission of electrons from this photosurface?
I was thinking since V = E/Q then E = 1.40 * 1.6 * 10-19 which gives E = 2.24 * 10^-19 J, then sub in this value for E = hfc and since h is just planck's constant then you just solve for the critical frequency. Then you use λ = c / f where 'c' is the speed of light and 'f' is the critical frequency, resulting in a wavelength of 8.9 * 10^-7 m, but the answer in my textbook says it's 2.7 * 10^-7 m
halp
Well, we know that the stopping voltage is 1.40V. This corresponds to an energy of 2.24*10^-19 J, as you said. That energy represents the maximum possible kinetic energy that an emitted photoelectron can have, when light of the given wavelength is used.
Now, we use EK max = hf - W, in order to get the work function. Or, in this case, we use E = hc/λ -W
2.24*10^-19 = (6.63*10^-34) * (3.0*10^8) / (2.08*10^-7) - W.
solving for W gives W = 7.3225*10^-19 J.
Now, we want the cutoff wavelength; the point at which EK max = 0. (ie. electrons aren't really emitted at all.) Hence, hc/λ - W = 0
W = hc/λ
λ = hc/W
so, λ = (6.63*10^-34) * (3.0*10^8) / (7.3225*10^-19)
which is 2.7*10^-7 m.
The mistake you were making is that the energy you had found was NOT the work function. Rather, it was the amount of energy that was left over once you had subtracted the work function from the energy of the given photons.
If that doesn't make sense I'll try and explain it again; these types of problems were always the most difficult we got in our physics exams. You had to make sure you were visualising what type of energy you were referring to correctly. Talking of which; if you know how to use electron volts as your units rather than joules, it makes it a fair bit simpler.
12
General Games! / Re: Super fast fights
« on: January 26, 2012, 12:04:45 pm »
[p1][/p1] vs [p2]
: show
13
General Games! / Re: Super fast fights
« on: January 26, 2012, 05:57:07 am »
I wonder if you can trick this thing into dividing by zero.
[p1]1/0[/p1] vs [p2]meh[/p2]
[p1]1/0[/p1] vs [p2]meh[/p2]
14
General Discussion! / Re: It's been a while, IWBTF...
« on: January 15, 2012, 08:34:56 am »
who is "sl"
15
Serious Discussion! / Re: Sopa/PROTECT IP Act/Internet Blacklist Legislation
« on: December 30, 2011, 08:29:34 pm »
I have no idea if this got brought up on the forums or not, but a while back Universal claimed copyright on a piece of media they didn't actually own, and youtube took it down.
Apparently, SOPA would allow this. It's ridiculous.
(Although I am getting my info for this from reddit, and they're not exactly perfectly unbiased about SOPA)
Apparently, SOPA would allow this. It's ridiculous.
(Although I am getting my info for this from reddit, and they're not exactly perfectly unbiased about SOPA)