Silverblade5

Math and science

287 posts in this topic

@Caesura Ah! The "Good Stuff". I would consider myself lucky if I ever perform the Barking Dog experiment. It is amazing. 

Edited by Hood
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@Hood I've never seen the Barking Dog experiment before! That's super cool, I'm gonna show my professor :D

Have you ever seen aluminium foil burned in bromine? One of my professors likes demonstrating it in his labs - the first time I saw it, it melted the bottom of the beaker it was in.

Luminol + hydrogen peroxide is really cool too; a friend of mine has been working out how to take it on a science roadshow thingy we do.

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@Caesura I might have seen the aluminium one (ha! Aluminium is so much better than the heretic Aluminum). But I haven't seen the Luminol+hydrogen peroxide. Tbh, my chemistry lab was as impoverished as it can be. And apart from synthesising "fruity" smelling esters and "patience building" titrations, we never did anything else. Except maybe generate H2S to take petty revenge on some seniors. 

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@Hood we didn't do much in high school - titrations and flame tests was about as good as it got. University chemistry is a whole different ballpark though :D

I sent a video of the barking dog experiment to the group that do crazy demonstrations for the science roadshow and they're going to look into it.

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Going in and finishing my degree in programming and web development classes, I will come bug you all when I have to get through my Calc classes. Your future help will be appreciated! Haha.

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Hey y'all! I've got a BS in Biotech, so if you need to bend life to your wi-I mean if you want to know about genetic engineering stuff, I'm around!

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I've got a physics question for y'all. So I've taken high school physics, and it got me thinking:

So when you stretch a spring, you are storing energy in it, because Elastic Potential Energy = 0.5*k*x^2. Because 0.5 and k are constants assuming your spring does not warp, change in elastic potential energy is solely dependent on the change in length of the spring.  (We'll revisit this later.) Therefore, if a spring isn't moving, it is not storing or releasing energy, right? No work is done because no distance is traveled, so there's the energy conservation. 0 change in energy stored in the spring, 0 work done.

So if we have a block sitting on a table at rest, and we attach 2 identical stretched springs to opposite sides of the block, and attach the other side the springs to immovable anchor points equidistant from the block, no motion occurs. The energy remains stored in the springs, but the forces exerted by the springs balance out. No work is done, because there is no distance for a force to be exerted over.

Question 1: Assuming all of my reasoning is correct so far, that means that these springs could exert a constant force forever assuming no changes to the system. This doesn't make intuitive sense to me: I've stored a finite amount of energy in each spring, but each can now exert a finite amount of force for an infinite amount of time? How did this infinity find its way into the system? (If it involves calculus I've got through Calc 1 down.) Is it because the time has nothing to do with the amount of energy in the spring, because time is not in the formula for work?

Now let's consider a similar example. A person attempts to lift a small but super-dense cube straight up, but their lifting force is exactly equal to the force of gravity and so they can't lift the cube at all. Because the person is lifting perpendicular to the surface, there is no friction, and because this cube is so small air resistance can be disregarded. The only forces that we need to consider are the cube's weight and the force applied upwards by the person. Just like the springs, we have two equal forces canceling each other out, but something's different here.

Question 2: The springs could oppose each other with matched forces without spending any energy, so why would the person's muscles grow tired after fighting gravity? One of two things must be the case: either the person has done no work on the cube, or the person has done work on the cube but gravity has also done work on the cube, and the work has canceled out. If it's that the person has done no work on the cube, why then do their muscles grow tired after outputting no energy? If it's that the person has done work on the cube, how is it that the springs did not lose energy? (Because therefore they would have done work on their block, and when a spring loses energy the only thing that can change about it is the extension length, per paragraph 2.)

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On 27.5.2017 at 2:17 PM, Overstorm said:

And now for a riddle:

main-qimg-5b0690e302a38cf2a8068158199e7a

Without looking further into the thread, this does not happen to be a comic version of Fermat's problem?

The answer to the question would be a simple "no".

sol.jpg.99a72d87b35261e61dfa86de8b7ad828.jpg

The problem is obviously symmetric in apples, bananas and pineapples, so a coordinate transformation might be in order.

@Elenion

The problem with muscles is that to apply a constant outward force, fibers contract and relax again and again, so your body gets tired although no mechanical work is done. Your body will produce heat though, which might be indicated by you starting to sweat. The human body is energetically speaking a highly inefficient machine, only a small amount of the used energy (provided by metabolism) is in fact converted to mechanical energy.

The problem with your spanned springs is solved quite easily. Force != Energy. Your springs stay deformed in your configuration, so the stored energy stays in the springs as elastic energy (well, some of it will be used to deform the spanned-in body).

With a constant force parallel to a given trajectory along which a body is moved, work=force*pathlength. So without a movement, no mechanical work is done, no energy values are changed.

 

On 2.6.2017 at 9:02 PM, Silverblade5 said:

Ahh. Then perhaps you could explain to me the difference between the strong and weak nuclear forces?

Ok, I've got an answer for you, though I am quite late. I try to keep it short, but no guarantees.

The strong nuclear force is the force holding the atomic nuclei together. It is a short range interaction with a cut-off at around 1.4fm, which coincides with the "radius" of the nucleons. As a result, the strong nuclear force is effective only with adjacent nucleons, leading to saturation effects and the easiest models for nuclei (liquid drop models and Bethe-Weizsäcker mass formula).

The strong nuclear force is in some models described as an interchange of mesons (quark-antiquark pair) between nucleons.
The strong nuclear force is no fundamental force but only a residual force, originating from the strong force between quarks and gluons. In an analogy, the nuclear force relates to the strong force like Van-der-Waals forces relate to the Coulomb force in electromagnetics (force between neutral molecules due to fluctuating dipole moments).

The weak nuclear force plays a role in nuclear decay and the interaction of neutrinos with matter. Neutrinos only interact via the weak nuclear force, which makes them so hard to detect (trillions of neutrinos from the sun travel through our body each second, without any measureable effects). The interchange particles of the weak nuclear force are the Z, W+ and W- bosons with masses around 90 GeV/c^2  (Z) and 80 GeV/c^2 (W). These particles are virtual, that is they have a very low lifetime of 3*10^(-25) seconds. During that time they can travel a distance of c*dt=9*10^(-17) meter (0.09 fm), which is an order of magnitude lower than the range of the strong nuclear force. This leads to a very low cross section for particle interactions.
Electromagnetic and Weak interaction can be brought into a unified theory.

 

Edited by Pattern
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50 minutes ago, Pattern said:

Without looking further into the thread, this does not happen to be a comic version of Fermat's problem?

The answer to the question would be a simple "no".

There are integers that fit this equation. They are not obvious by any means.

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5 hours ago, Mestiv said:

There are integers that fit this equation. They are not obvious by any means.

So then it's at least not a wild goose chase...

But a deep decent into the rabbit hole of number theory and diophantine equations. Solutions and related stuff can be found here:

https://www.quora.com/How-do-you-find-the-positive-integer-solutions-to-frac-x-y+z-+-frac-y-z+x-+-frac-z-x+y-4

http://ami.ektf.hu/uploads/papers/finalpdf/AMI_43_from29to41.pdf

Spoiler

apple = 437361267792869725786125260237139015281653755816161361862143‌7993378423467772036;

banana = 368751317941299998271978115652254748254929799689719709962831‌37471637224634055579‌;

pineapple = 154476802108746166441951315019919837485664325669565431700026‌63489825320203527799‌9;

Of course, due to the symmetry of the equation, numbers and fruits can be permuted ad libidum.

 

Edited by Pattern
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I just stumbled across this, and feel the need to share it somewhere.

22491962_1160393894093321_4915488849209940180_n.jpg.12fac109615ef26e64a8a21de29f04ab.jpg

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Line integrals and inverse trig should never have been brought together. That is all. 

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