Answering physics questions!

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het !6.hhI.UC6Y Sun Aug 31 21:59:27 2014 No.6729178 [Reply] [Original]

Answering physics questions!

>>	Anonymous Sun Aug 31 22:20:50 2014 No.6729208 >>6729178 why are physics students so bad at getting laid they didn't even show up on the chart?

>>	Anonymous Sun Aug 31 22:20:57 2014 No.6729209 Dear Mr Science- Say I have an object free falling in a gravitational field. Does the path of the center of mass follow the same path that a "point mass" of equal mass would follow?

>>	Anonymous Sun Aug 31 22:23:05 2014 No.6729212 >>6729208 >he thinks real life is like his cartoons This place is 18+.

Anonymous Sun Aug 31 22:28:28 2014 No.6729218
File: 62 KB, 1094x386, time average power.jpg [View same] [iqdb] [saucenao] [google]

how the fuck do I handle the big N in the time average power equation? I understand that for energy you see if it converges, but i dont understand what that other variable means in the power equation or how to handle it

see pic

>>	Anonymous Sun Aug 31 22:30:49 2014 No.6729222 >>6729178 I've heard that there have been attempts at axiomatizing physics. Could you provide more information on this?

>>	Anonymous Sun Aug 31 22:34:37 2014 No.6729226 Why is my physics grad friend so seemingly miserable?

het !6.hhI.UC6Y Sun Aug 31 22:40:06 2014 No.6729235

>>6729209
Yes. The force on a body in a gravitational field is proportional to its mass, and so is the acceleration, so the mass of the body itself cancels.

>>6729218
I'm not entirely sure what you are asking, or what the math on your scanned sheet means. Why did you set x = 1? Why did you write E = infinity?

>>6729222
This is a really broad question. The fundamental underlying axiom of physics is the scientific method, but I am assuming this is not what you are asking. Are you asking about which axioms of mathematics best represent the mathematics that the natural universe obeys? Or are you asking what the "furthest back" we can take physics is, before we have to just accept certain facts as axioms?

het !6.hhI.UC6Y Sun Aug 31 22:43:38 2014 No.6729237

>>6729226
>>6729226
I don't know, people are miserable for all kinds of reasons. Maybe their advisor is a bastard, maybe they don't really enjoy physics.

But probably it has nothing to do with physics, and there's something else at play. Maybe they are ill, they are having family trouble, or they feel lonely, inadequate, or bored. Why don't you ask your friend and see what they say?

>>	Anonymous Sun Aug 31 22:45:42 2014 No.6729239 >>6729235 the series is divergent for the given signal meaning that it has infinite energy disregard the physics then. if someone wrote a summation to N starting with n=-N how do you process that shit? Plug in N or something?

het !6.hhI.UC6Y Sun Aug 31 22:46:28 2014 No.6729240

>>6729235
Sorry, my response to >>6729209 didn't really make sense the way I meant it to. What I meant to say is that the force on a body in a gravitational field is proportional to its mass, and force is mass times acceleration. Setting the force F = ma[/spoiler] equal to F = mg[/spoiler], where g[/spoiler] is the gravitational field strength, the mass terms cancel.

>>	Anonymous Sun Aug 31 22:51:30 2014 No.6729244 >>6729237 He's always complaining about his department. Maybe being a physics grad student is a terrible gig until you get some kind of big break for your career?

het !6.hhI.UC6Y Sun Aug 31 22:51:34 2014 No.6729245

>>6729239
Divergence is not the same as infinity. For no n[/spoiler] is \sin (n \pi / 4) [/spoiler] going to be infinity.

What you want to do regarding the second equation is observe that sin(-x) = -sin(x)[/spoiler], so each negative term will cancel with each positive term in that summation. The only term left will be the n = 0[/spoiler] term, which will be divided by 2N+1[/spoiler], so the limit is 0.

het !6.hhI.UC6Y Sun Aug 31 22:54:04 2014 No.6729250

>>6729244
No, lots of physics grad students are happy. I had a great time in grad school, and so did lots of my cohort, whether they ultimately went into industry, academia, finance, whatever. There were a few people who were totally miserable, and a couple people who didn't hate or love it but just were there to get through it, but that's the case anywhere you go.

>>	Anonymous Sun Aug 31 23:05:59 2014 No.6729258 >>6729245 but thats wrong. if the energy for a signal is infinity the power will be a real positive number and vice versa. your thread a shit

het !6.hhI.UC6Y Sun Aug 31 23:08:50 2014 No.6729263

>>6729258
I don't know what the physics problem is that you are talking about. I am just looking at the math on that page, and it is wrong. The limit as n goes to infinity of sin(n \pi / 4) [/spoiler] is not infinity. Can you explain what the problem you are working on is?

>>	Anonymous Sun Aug 31 23:15:43 2014 No.6729274 >>6729263 there should be an absolute value sign around the expression hence no negative values to cancel out

>>	Anonymous Sun Aug 31 23:17:38 2014 No.6729276 >>6729274 and a summation. who ever wrote the answer key i am referencing is asleep on the job

Anonymous Sun Aug 31 23:18:43 2014 No.6729277

I'm utterly confused about what this question means, even after seeing the answers others have given. I'm not even sure what exactly is being asked here. This showed up on my first exam on Friday.

au.answers.yahoo.com/question/index?qid=20121203225405AAlUeXO

het !6.hhI.UC6Y Sun Aug 31 23:24:53 2014 No.6729286

>>6729274
I still can't really provide input unless you explain the problem. Why are you summing values of the sines of multiples of \pi/4[/spoiler]?

I did overlook the absolute value sign, which changes my result slightly for the limit on the bottom. For every 8 values of n you are picking up 2 + 2 \sqrt{2}[/spoiler] in the sum. So you can rewrite it as \lim_{\rightarrow \infty} \frac{(2N+1)(2 + 2\sqrt{2})}{8(2N+1)} = (1 + \sqrt{2})/4 = 0.604[/spoiler]

>>	het !6.hhI.UC6Y Sun Aug 31 23:26:53 2014 No.6729288 >>6729277 >au.answers.yahoo.com/question/index?qid=20121203225405AAlUeXO Can you post the figures referenced in the problem? Also, I'd rather not do peoples' homework for them.

>>	Anonymous Sun Aug 31 23:36:03 2014 No.6729295 >>6729288 Nvm, I found an answer I get while looking for a copy of the figures. Since my non-homework is out of the way, here's a different question: How do dimensions beyond the third relate (if at all) to imaginary numbers?

Anonymous Sun Aug 31 23:38:18 2014 No.6729300
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>>6729286
thanks for offering input, all i really need to know is how to handle the "N" variable in the summation sign. that is what is tripping me up.

The equation for energy is shown in the pic. It runs from negative inf to inf. Cool I get that. The power equation shown involves summing from n = -N to n = N. I do not know how to move forward because I don't understand plugging in a variable into another variable when trying to find out how it converges. I've googled all kinds of shit to find out how to do it. If you could point me in the right direction with a link or explain it im set for this assignment.

>>	Anonymous Sun Aug 31 23:39:59 2014 No.6729303 what do you think would be a better plasma-facing material for a tokamak type fusion reactor, beryllium, tungsten, carbon

het !6.hhI.UC6Y Sun Aug 31 23:42:58 2014 No.6729307

>>6729300
I see. Well it's pretty much what I said. The first infinite sum gives you infinity, as you found. The second one can be found by observing that every 8 values of n contribute 2 + 2 \sqrt{2}[/spoiler] to the sum, so (not in general, but within the limit as N goes to infinity) you can rewrite \sum_{n=-N}{N} | \sin (n \pi / 4) | = (2 + 2\sqrt{2}) (2N + 1) / 8 [/spoiler].

>>	Anonymous Sun Aug 31 23:43:34 2014 No.6729309 What kind of a role do you see physics playing in finance?

>>	Anonymous Sun Aug 31 23:47:02 2014 No.6729316 >>6729309 bout 3fiddy k

het !6.hhI.UC6Y Sun Aug 31 23:50:30 2014 No.6729321

>>6729307
That sum should be typeset as \sum_{n = -N}^{N}[/spoiler].

>>6729295
They don't. There are two ways that imaginary numbers come into geometry in physics. One is the use of geometry to describe complex numbers, using the complex plane. This is a way of representing an imaginary number z = a + bi[/spoiler] as an x-y pair (a, b) [/spoiler] of the real and imaginary parts. Representing complex numbers in this way yields many beautiful and non trivial results that are invaluable for certain computations.

The other relationship is when complex numbers are used to represent geometry. The distance between two points in Euclidean space is the sum of squares of the differences in their cartesian coordinates, or for a 4D space, d^2 = x_1^2 + x_2^2 + x_3^2 + x_4^2[/spoiler]. But we live in what is called a Minkowski space, and the fourth coordinate, time, actually contributes negatively to distance in spacetime: d^2 = x^2 + y^2 + z^2 - t^2[/spoiler]. One thing that is entirely non physical and unrelated to math is to do a coordinate substitution \tau = it [/spoiler], which makes the distance equation look more like Euclidean space: d^2 = x^2 + y^2 + z^2 + \tau^2[/spoiler]. This can occasionally simplify calculations, but before any usable results are obtained you have to substitute back in the real t[/spoiler] coordinate.

>>	Anonymous Sun Aug 31 23:53:14 2014 No.6729325 >>6729307 t-thanks... but I am retarded and do not understand what the big N means. is big N the upper and lower limits of time as specified in the problem? pls anon

Anonymous Sun Aug 31 23:53:39 2014 No.6729327

>>6729240
Thanks, but I'm not sure if you are answering what I intended to ask. Suppose you have an irregularly shaped object, falling in a gravitational field. Perhaps it is turning over and over as it falls. The center of mass of the object traces out some path. Is this the same path as a very small spherical object of equal mass would take given the same initial condition?

het !6.hhI.UC6Y Sun Aug 31 23:55:37 2014 No.6729331

>>6729303
I'm not a fusion engineer, so I can't really claim to know about what material properties are required in tokamaks. If the plasma-facing material is meant to absorb neutrons and radiate heat into a steam generator, I guess it would need to have a high neutron interaction cross section. If cooling is adequate I doubt that the temperature would be that much of a problem, so I'd go with beryllium?

>>6729309
Frankly there's not a lot of physics in finance, but research in physics has yielded and continues to yield results in nonlinear dynamics, chaos, probability, computer science, and other applied-math topics that are highly applicable to the study of finance.

Anonymous Mon Sep 1 00:00:50 2014 No.6729335

Thought of this problem a couple days ago, but couldn't figure it out:

Suppose you have a thin, stiff, horizontal metal rod of length L[/spoiler] and cross-sectional density \delta(x)[/spoiler], so the total mass of the rod is m= \int_0^L \delta(x) \,dx[/spoiler]. Suppose the rod is resting on n[/spoiler] pins at locations
0\le p_1< p_2<\dots<p_n\le L[/spoiler]. Let g[/spoiler] be the acceleration of gravity. What is the downward force F_n[/spoiler] that the rod exerts on the n[/spoiler]-th pin?

het !6.hhI.UC6Y Mon Sep 1 00:01:43 2014 No.6729336

>>6729325
This is the definition of an infinite sum. Any time you see sum to infinity written, you aren't actually plugging in infinity to n, you are taking the limit of value of the nth partial sum as n goes to infinity. Big N is just the outer limit of your sum, and is being used to explicitly write the infinite sum as the limit by which it is actually calculated.

>>6729327
Oh, I see. For a body under no external forces, any rotation is about the center of mass, so you actually won't see any wobbling about happening at all. However, this is thrown out the window if your gravitational field varies on the size scale of the object- if the field is stronger at the left end than the right end, you have a torque and essentially a different system of forces compared to if you had a point mass.

Ultimately this boils down to the fact that a point mass is a completely adequate approximation for a free falling body, if and only if the gravitation field varies slowly compared to the dimensions of the body.

>>	Anonymous Mon Sep 1 00:05:02 2014 No.6729339 >>6729178 I have a question, if you sent a rocket to Europa, wouldn't it make the atmosphere explode since it's mostly oxygen?

Anonymous Mon Sep 1 00:08:43 2014 No.6729343

>>6729339
the atmosphere would have to have something for the oxygen to react to release energy resulting in an explosion. oxygen doesnt just release energy by it's self. its got to have something to reach a lower energy state through an exothermic process

het !6.hhI.UC6Y Mon Sep 1 00:09:20 2014 No.6729344

>>6729335
This is a tough question. The answer is going to be a big system of equations with terms like \int_{p_i}^{p_{i+1}} x \delta(x) dx [/spoiler], constrained by the balancing of torques. I could try and work it out, but it might take a little to get everything right. I'll get back to you.

>>	Anonymous Mon Sep 1 00:09:51 2014 No.6729345 >>6729336 >Oh, I see. For a body under no external forces,... Thank you! This is exactly what I wanted to know. I was interested in the case where the object is small relative to the varying force of gravity.

>>	Anonymous Mon Sep 1 00:09:59 2014 No.6729346 >>6729335 I think the system is unsolveable for more than two pins Try some statics methods <span class="math"> \Sum F = 0 [/spoiler] <span class="math"> \Sum M = 0 [/spoiler] Each give 1 equation, so two equations and n unknowns

>>	het !6.hhI.UC6Y Mon Sep 1 00:11:44 2014 No.6729348 >>6729339 Oxygen itself doesn't just flame or explode. It is highly reactive with other chemicals in the environment. Oxygen is fairly stable on Europa due to the lack of anything for it to react with, whether it is exposed to heat or not.

Anonymous Mon Sep 1 00:17:22 2014 No.6729355

>>6729344
Thanks! I asked the falling body question too. I figured the sum of the forces should equal g*M, where M=\int_0^L \delta(x) dx. I thought I must consider the torques as well, but couldn't decide what fulcrum(s) to use.

I was putting up shelves in my garage and wondered how much weight each bracket would bear given some load. They were rated at 40 pounds. Turned out to be more difficult than I thought it would be.

het !6.hhI.UC6Y Mon Sep 1 00:23:28 2014 No.6729365

>>6729346
>>6729344
>>6729335

OK, here is my analysis. You get the following equations:

 \sum_{i = 1}^N F_i = mg [/spoiler]

 g \int_0^{p_1} x \delta(x) dx = g \int_0^{p_1} x \delta(x) dx - \sum_{i = 2}^N F_i [/spoiler]

 g \int_{p_N}^L x \delta(x) dx = g \int_0^{p_N} x \delta(x) dx - \sum_{i = 1}^{N-1} F_i [/spoiler]

and then for j = 2, 3, 4 ... N-1:
 g \int_0^{p_j} x \delta(x) dx - \sum{i = 1}{j-1} F_i = g \int_{p_j}^{p_N} x \delta(x) dx - \sum_{i = j+1}^{N} F_i [/spoiler]

From here it's just computation.

>>	het !6.hhI.UC6Y Mon Sep 1 00:25:04 2014 No.6729366 >>6729365 Fucked up that last typesetting: <span class="math"> g \int_0^{p_j} x \delta(x) dx - \sum_{i = 1}^{j-1} F_i = g \int_{p_j}^L x \delta(x) dx - \sum_{i = j+1}^N F_i [/spoiler]

het !6.hhI.UC6Y Mon Sep 1 00:29:45 2014 No.6729376

>>6729366
Oh balls, the sums should all be sums of F_i (p_j - p_i) [/spoiler], not just F_i[/spoiler].

So you understand what I did, I just balanced the torques about each pin. So the torque from gravity acting on the beam, which is the integral term on each side of each equation, minus the torques from each other pin, which is the sum of forces-times-distances-from-central-pin that I wrote above, should be equal on either side. The first equation comes from the fact that they have to all add up to the total mass.

>>	Quasi-Shadow Mon Sep 1 00:31:40 2014 No.6729378 Question: what is the interaction between two strong magnetic beams coming into contact? magnetic fields projected as beams intersecting.

het !6.hhI.UC6Y Mon Sep 1 00:35:47 2014 No.6729380

>>6729376
Goddammit, I keep seeing small errors, that's what I get for copying and pasting.

Final results:

Force balancing:
\sum_{i = 1}{N} F_i = mg[/spoiler]

Torque balancing, end cases:
 g \int_0^{p_1} x \delta(x) dx = g \int_{p_1}^L x \delta(x) dx - \sum{i = 2}{N} F_i |p_1 - p_i| [/spoiler]

 g \int_0^{p_N} x \delta(x) dx - \sum{i = 1}{N-1} F_i |p_N - p_i| = g \int_{p_N}^L x \delta(x) dx [/spoiler]

Torque balancing for pins j = 2, 3, 4, ... N-1:
g \int_0^{p_j} x \delta(x) dx - \sum{i = 1}{j-1} F_i |p_j - p_i| = g \int_{p_j}^L x \delta(x) dx - \sum{i = j+1}{N}F_i |p_j - p_i|}

>>	Anonymous Mon Sep 1 00:36:41 2014 No.6729381 >>6729346 Oh, surely it must be solvable. Don't engineers do this sort of thing all the time? Bridge and supports...

>>	het !6.hhI.UC6Y Mon Sep 1 00:37:33 2014 No.6729385 >>6729380 Seriously, fuck this TeX in the tiny reply box. You get what I'm saying, yes? >>6729378 A magnetic beam isn't really a thing. Generally opposing fields cause repulsion, aligned fields cause attraction.

>>	het !6.hhI.UC6Y Mon Sep 1 00:38:42 2014 No.6729386 >>6729381 Of course it's solvable, but whether we can write an analytically exact expression for it entirely depends on what <span class="math">\delta(x)[/spoiler] is.

>>	Anonymous Mon Sep 1 00:39:28 2014 No.6729388 >>6729380 Thanks again! I'll try to follow what you've done. Then I'll be able to finish putting up those shelves. :)

>>	het !6.hhI.UC6Y Mon Sep 1 00:51:15 2014 No.6729403 File: 1.22 MB, 1830x1799, IMG_0047.jpg [View same] [iqdb] [saucenao] [google] >>6729388 I just wrote it out by hand.

Anonymous Mon Sep 1 00:55:54 2014 No.6729411 [DELETED]

>>6729380
g\int_0^{p_1}x\delta(x)dx=g\int_{p_1}^L x\delta(x)dx-\sum_{i=2}^N (p_i-p_1)F_i

g\int_{p_N}^L x\delta(x)dx=g\int_{0}^{p_N} x\delta(x)dx-\sum_{i=1}^{N-1} (p_N-p_i)F_i

g \int_0^{p_j} x \delta(x) dx - \sum_{i = 1}^{j-1}(p_j - p_i)F_i = g \int_{p_j}^L x \delta(x) dx - \sum_{i = j+1}^{N}(p_i - p_j)F_i

Anonymous Mon Sep 1 00:57:41 2014 No.6729414 [DELETED]

>>6729403
Tried tex...


g\int_0^{p_1}x\delta(x)dx=g\int_{p_1}^L x\delta(x)dx-\sum_{i=2}^N (p_i-p_1)F_i

g\int_{p_N}^L x\delta(x)dx=g\int_{0}^{p_N} x\delta(x)dx-\sum_{i=1}^{N-1} (p_N-p_i)F_i

g \int_0^{p_j} x \delta(x) dx - \sum_{i = 1}^{j-1}(p_j - p_i)F_i = g \int_{p_j}^L x \delta(x) dx - \sum_{i = j+1}^{N}(p_i - p_j)F_i
[/spoiler]

Anonymous Mon Sep 1 00:59:52 2014 No.6729422

>>6729403

g\int_0^{p_1}x\delta(x)dx=g\int_{p_1}^L x\delta(x)dx-\sum_{i=2}^N (p_i-p_1)F_i
[/spoiler]


g\int_{p_N}^L x\delta(x)dx=g\int_{0}^{p_N} x\delta(x)dx-\sum_{i=1}^{N-1} (p_N-p_i)F_i
[/spoiler]


g \int_0^{p_j} x \delta(x) dx - \sum_{i = 1}^{j-1}(p_j - p_i)F_i = g \int_{p_j}^L x \delta(x) dx - \sum_{i = j+1}^{N}(p_i - p_j)F_i[/spoiler]

Anonymous Mon Sep 1 01:01:29 2014 No.6729426

>>6729381
I'm a MechE is why I mention it.
It's not solvable by statics 1 methods. (at least not to my memory)
We had systems that had more pins in statics 2 (3- and 4- point bending), but then the beam would be allowed to move
>>6729386
I like your solution and it seems promising, but I think you should do some simple solutions.
 \delta(x) = \lambda [/spoiler]
For 3 evenly distributed pins and then for 4.

>>	Anonymous Mon Sep 1 01:02:32 2014 No.6729429 >>6729426 *would be allowed to bend problems were still static is statics ii

het !6.hhI.UC6Y Mon Sep 1 01:09:18 2014 No.6729440

>>6729426
Well, you can see what it will be. Considering that we are punching in p_i[/spoiler] and \delta(x)[/spoiler], the integrals are just constants, as are the terms p_j - p_i[/spoiler]. So each equation is just a linear equation in F_i [/spoiler]. We have N linear equations for the N variables F_i[/spoiler]. Solving is simple.

>>	Anonymous Mon Sep 1 01:13:38 2014 No.6729444 >>6729440 Looks like N+1 equations and N unknowns. One may be redundant.

>>	het !6.hhI.UC6Y Mon Sep 1 01:15:24 2014 No.6729446 >>6729444 Yeah, the force balancing equation is redundant and may be omitted.

>>	Anonymous Mon Sep 1 01:35:36 2014 No.6729464 If my dick is 15cm long, and I charge into a woman with 3atm of dick pressurization, how hard will her orgasm be?

>>	het !6.hhI.UC6Y Mon Sep 1 01:36:12 2014 No.6729466 >>6729464 8

>>	Anonymous Mon Sep 1 04:14:50 2014 No.6729619 Can you explain to me how to solve this by separation? Dx + e^(3x) dy = 0

>>	Anonymous Mon Sep 1 04:28:38 2014 No.6729628 >>6729619 Dx = e^3x dy x' = e^3x ?

>>	Anonymous Mon Sep 1 04:29:18 2014 No.6729629 >>6729619 Well, you don't need to use separation to solve that equation, just multiply by e^(-3x), write dy = dy/dx dx, and integrate with respect to dx.

>>	Anonymous Mon Sep 1 04:31:56 2014 No.6729630 >>6729212 I take it that you haven't seen the charts.

Anonymous Mon Sep 1 19:17:42 2014 No.6730467 [DELETED]

>>6729403
Hey Mr Science... are you still around? Looking at terms such as

g \int_0^{p_j} x \delta(x)\,dx

I'm now wondering if these should be

g \int_0^{p_j} (p_j-x) \delta(x)\,dx

because it's supposed to be the torque around the j-th pin due to the mass of the rod, i.e. the pin is the fulcrum.[/spoiler][/spoiler]

Anonymous Mon Sep 1 19:19:55 2014 No.6730468

>>6729403
Hey Mr. Science-
Looking at terms you have like
g \int_0^{p_j} x \delta(x)\,dx [/spoiler] I'm now wondering if these should be g \int_0^{p_j} (p_j-x) \delta(x)\,dx[/spoiler] because it's supposed to be the torque around the j-th pin due to the mass of the rod, i.e. the j-th pin is the fulcrum.

>>	Anonymous Mon Sep 1 19:35:29 2014 No.6730479 >>6729178 How does gravity work? Is it related to the attraction between oppposite charges? The Earth orbits the Sun because of the velocity perpendicular to the pull of the Sun. Where did that velocity come from?

>>	Anonymous Mon Sep 1 21:56:39 2014 No.6730642 >>6730479 Not related to opposite charges just the presence of mass. The velocity relative to the sun originated during the formation of our solar system when you first have gas collapse into a star and then a protoplanetary disk.

>>	Anonymous Tue Sep 2 14:03:39 2014 No.6731527 >>6729178 please answer >>6723022

>>	Anonymous Tue Sep 2 14:29:27 2014 No.6731547 File: 4 KB, 120x120, avatar5302_28.jpg [View same] [iqdb] [saucenao] [google] how does the linear congruential method work as a random number generator? I'm trying to understand this modulo operation but either i'm retarded or i'm retarded

>>	Anonymous Tue Sep 2 14:39:10 2014 No.6731552 >>6731547 >I'm trying to understand this modulo operation but either i'm retarded or i'm retarded it divides a number by another number then gets the remainder

>>	Anonymous Tue Sep 2 14:43:12 2014 No.6731556 >>6731552 ok, i have this example in the book: 3 mod 7 = 3 I really don't see how to get that result

>>	Anonymous Tue Sep 2 14:48:35 2014 No.6731561 is it correct, that the earth had to make a full revolution in about 10 minutes in order to compensate all of it's gravitational force at the equator?

>>	Anonymous Tue Sep 2 14:49:55 2014 No.6731565 >>6731556 how many times fits 7 into 3? 0 times! and what is the difference between 0 and 3? 3!

>>	Anonymous Tue Sep 2 14:50:51 2014 No.6731568 When will we have Alcubierre drives?

>>	Anonymous Tue Sep 2 14:51:25 2014 No.6731570 >>6731568 never

>>	Anonymous Tue Sep 2 14:52:43 2014 No.6731571 File: 15 KB, 447x444, 5138266+_cb2d6da46d20d43ae647fb664fef5dcb.jpg [View same] [iqdb] [saucenao] [google] >>6731565 fuck man i've been trying to understand that for the last 30 minutes thanks bro

>>	Anonymous Tue Sep 2 14:57:11 2014 No.6731577 >>6731571 hey, i forgot something: you're not interested in the difference between 0 and 3, but in 70 and 3 so e.g.: 9 mod 7 -> 7 fits 1 time in 9 -> abs(71-9) = 2

>>	Anonymous Tue Sep 2 15:08:46 2014 No.6731584 >>6731577 oh i see, i understood you wrong before then

>>	Anonymous Tue Sep 2 17:31:47 2014 No.6731734 >>6729178 I'm trying to learn thermodynamics. Is the P-V energy of a simple system consisting of a gas in a container part of its internal energy?

>>	Anonymous Tue Sep 2 18:01:11 2014 No.6731793 >>6730468 paging Mr. Science.

>>	Anonymous Tue Sep 2 18:18:50 2014 No.6731839 >>6729178 Who are you?

>>	Anonymous Tue Sep 2 20:28:06 2014 No.6732118 >>6731839 just search image on google. not hard to find out.

>>	Anonymous Tue Sep 2 20:37:46 2014 No.6732144 Monkey hate math. Make monkey button to make machine go? Make monkey button to make machine go!

Anonymous Tue Sep 2 22:01:54 2014 No.6732283

>>6729403
So I'm thinking this isn't sufficient to determine the forces. If what you mean is this: g
\int_0^{p_j} (p_j-x) \delta(x) \,dx + \sum_{i = 1}^{j-1}(p_j - p_i)F_i = g \int_{p_j}^L (x-p_j) \delta(x)\, dx + \sum_{i = j+1}^{N}(p_i - p_j)F_i
[/spoiler], then re-arranging we see this is equivalent to g \int_0^{p_j} (p_j-x) \delta(x)\, dx + g \int_{p_j}^L (p_j-x) \delta(x) dx
= \sum_{i = 1}^{j-1}(p_i - p_j)F_i + \sum_{i = j+1}^{N}(p_i - p_j)F_i [/spoiler] or g \int_0^{L} (p_j-x) \delta(x) dx = \sum_{i =
1}^{N}(p_i - p_j)F_i [/spoiler]. But this equation will be satisfied by any set of F_i[/spoiler] that
satisfy g \int_0^{L} \delta(x) dx = -\sum_{i = 1}^{N} F_i [/spoiler] and g \int_0^{L} x
\delta(x) dx = -\sum_{i = 1}^{N} p_i F_i [/spoiler]. That is, the F_i[/spoiler] need only have the
opposite total force and center of force to the rod. That's only two linear conditions, which will only yield unique F_i[/spoiler] when N=2[/spoiler].

>>	Anonymous Tue Sep 2 22:04:54 2014 No.6732286 File: 20 KB, 660x239, chan29.gif [View same] [iqdb] [saucenao] [google] >>6732283 blah. pic related.

>>	Anonymous Wed Sep 3 15:32:04 2014 No.6733655 >>6729466 :^)

>>	Anonymous Wed Sep 3 16:09:11 2014 No.6733707 What is a time paradox?

>>	Anonymous Wed Sep 3 21:05:12 2014 No.6734208 anyone here? i need help with a problem. you can laugh at how basic it is and how stupid i am

>>	Anonymous Wed Sep 3 21:30:34 2014 No.6734244 File: 84 KB, 313x313, 1353975396071.jpg [View same] [iqdb] [saucenao] [google] >>6729178 Why is my dick so small? Did gravity do this to me?

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