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Anonymous Wed 28 Oct 2015 08:51:56 No.7621064 View Reply Report Delete

Quoted By: >>7621614

New stupid questions general thread here.

Old thread here:
>>7612327

Anonymous

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Anonymous Wed 28 Oct 2015 08:55:18 No.7621066 Report Delete

Quoted By: >>7621079 >>7621087

I'm in my first Organic chemistry lab this semester and I've been looking over the msds safety data and I have a question

Is there a tangible and/or meaningful difference between a liquid compound being
"Flammable"
"Highly flammable"
"Combustible"
Having "Flammable liquid and vapor"?

Anonymous

Anonymous Wed 28 Oct 2015 09:01:22 No.7621079 Report Delete

Quoted By: >>7621087

>>7621066
take this with a grain of salt as I'm not entirely sure:

>"Flammable"
An open flame will cause it to ignite
>"Highly flammable"
An open flame will cause it to ignite, and it will ignite very easily
>"Combustible"
Enough heat can cause it to catch fire.
>Having "Flammable liquid and vapor"?
The liquid and the vapour can be ignited by an open flame. (think gasoline)

Anonymous

Anonymous Wed 28 Oct 2015 09:05:53 No.7621087 Report Delete

Quoted By:

>>7621066
>>7621079 here

try this:
http://www.differencebetween.com/difference-between-flammable-and-vs-combustible/
http://www.differencebetween.com/difference-between-flammable-and-highly-flammable/

Anonymous

Anonymous Wed 28 Oct 2015 09:23:41 No.7621107 Report Delete

Quoted By: >>7621466

Is there any similarity in the principles of electron clouds of atoms and photon spheres of black holes?
Can a black hole have other orbiting surfaces around it, like a neutrino sphere?

Anonymous

Anonymous Wed 28 Oct 2015 09:36:37 No.7621121 Report Delete

Quoted By: >>7621134 >>7621150

if you were to put a giant copper wire around the earth and put a magnet on orbit around said wire , would that generate "infinite" electricity until the magnet fall back on earth ?

Anonymous

Anonymous Wed 28 Oct 2015 09:45:07 No.7621134 Report Delete

Quoted By:

>>7621121
No

Anonymous

Anonymous Wed 28 Oct 2015 09:46:46 No.7621137 Report Delete

Quoted By: >>7621276 >>7621396 >>7621646

What is the proof that 1+1=2?

Google had nothing.

Anonymous

Anonymous Wed 28 Oct 2015 09:56:50 No.7621150 Report Delete

Quoted By:

>>7621121
The generation of electricity would reduce the kinetic energy of the magnet, causing it to fall faster, so it could not stay in orbit. No perpetual motion machines for you fggt.

Anonymous

Anonymous Wed 28 Oct 2015 10:02:52 No.7621164 Report Delete

Quoted By: >>7621355

Reposting
Is biotechnology and biomedical engineering stuff like nanomachines to improve your body?

Anonymous

Anonymous Wed 28 Oct 2015 10:10:26 No.7621179 Report Delete

Quoted By:

Reposting
Is biotechnology and biomedical engineering stuff like nanomachines to improve your body?

Anonymous

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Anonymous Wed 28 Oct 2015 11:12:26 No.7621254 Report Delete

Quoted By: >>7621365

I have a simple circuit where
$\mathbf{V} = 200\measuredangle{-180^\circ}$
L = 2H
$\mathbf{I} = 100\measuredangle{90^\circ}$

I need to find the voltage across the current source, but the current source itself is throwing me off. What are the step I need to take?

Anonymous

Anonymous Wed 28 Oct 2015 11:37:56 No.7621276 Report Delete

Quoted By:

>>7621137
https://en.m.wikipedia.org/wiki/Peano_axioms

Anonymous

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Anonymous Wed 28 Oct 2015 12:17:23 No.7621327 Report Delete

Quoted By: >>7621360

A quantum mechanical particle in a magnetic field.

If the Coulomb gauge gives that the divergence of A is 0, where does the 2 in the denominator go? Multiple sources say that this is the right answer, but I just can't wrap my head around how it works.

Source of the pic: http://www.tcm.phy.cam.ac.uk/~bds10/aqp/lec5_compressed.pdf

Anonymous

Anonymous Wed 28 Oct 2015 12:33:45 No.7621355 Report Delete

Quoted By: >>7621393

>>7621164
No. It's physics mixed with chemestry, programming and electrical engineering. You won't build nanobots.
You will programm stuff.

Anonymous

Anonymous Wed 28 Oct 2015 12:37:15 No.7621360 Report Delete

Quoted By: >>7621368

>>7621327

You have to use the product rule. Perhaps it will make more sense if you think of applying the operator to a wave function:

$-\frac{q}{2m}(\vec{p} \cdot \vec{A} + \vec{A} \cdot \vec{p}) \, \psi = \frac{q i \hbar}{2m}\Big(\nabla \cdot (\vec{A} \psi) + \vec{A} \cdot \nabla \psi) \Big)$

$= \frac{q i \hbar}{2m}\Big((\nabla \cdot \vec{A}) \psi + 2 \vec{A} \cdot \nabla \psi) \Big)$

$= \frac{q i \hbar}{2m}\Big(0 \, \psi + 2 \vec{A} \cdot \nabla \psi) \Big)$

$= \frac{q i \hbar}{m}(\vec{A} \cdot \nabla) \psi$

Anonymous

Anonymous Wed 28 Oct 2015 12:44:07 No.7621365 Report Delete

Quoted By:

>>7621254
Suppose both sources separately, then use superposition of sources.

Anonymous

Anonymous Wed 28 Oct 2015 12:46:15 No.7621368 Report Delete

Quoted By: >>7621426

>>7621360

Ah, right. Using operators which do not always operate on something does confuse me sometimes. Thanks!

Anonymous

Anonymous Wed 28 Oct 2015 13:03:48 No.7621393 Report Delete

Quoted By:

>>7621355
Disappointing...
Thanks anyway!

Anonymous

Anonymous Wed 28 Oct 2015 13:06:38 No.7621396 Report Delete

Quoted By:

>>7621137
http://quod.lib.umich.edu/cgi/t/text/pageviewer-idx?c=umhistmath&cc=umhistmath&idno=aat3201.0001.001&frm=frameset&view=image&seq=401

H Wed 28 Oct 2015 13:29:28 No.7621426 Report Delete

Quoted By:

>>7621368
They will always operate on something, so you can always use that technique. It is probably worth getting used to manipulating operators directly though.

Anonymous

Anonymous Wed 28 Oct 2015 13:48:03 No.7621466 Report Delete

Quoted By:

>>7621107
Can someone please answer this?

Anonymous

Anonymous Wed 28 Oct 2015 13:49:15 No.7621469 Report Delete

Quoted By: >>7621523

Is this a stupid question?

Anonymous

Anonymous Wed 28 Oct 2015 14:18:37 No.7621523 Report Delete

Quoted By:

>>7621469

Yes, because if it were not a stupid question it wouldn't belong here, thus making it quite a stupid question indeed. Its being a stupid question therefore makes it worthy to receive an answer here, which I have just provided.

Anonymous

Anonymous Wed 28 Oct 2015 14:27:53 No.7621551 Report Delete

Quoted By: >>7621612

Are the shapes of atomic orbitals the actual shape of the electron cloud? Can the electron cloud be spherical?

Anonymous

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Anonymous Wed 28 Oct 2015 14:53:24 No.7621612 Report Delete

Quoted By:

>>7621551

No, the shapes illustrated in books are idealizations that convey some spatial characteristics of the electron cloud density given by $|\psi|^2$ or are polar plots of the spherical harmonics like pic related. See here for some other visualizations:

https://en.wikipedia.org/wiki/Atomic_orbital
https://en.wikipedia.org/wiki/Spherical_harmonics

In s-states the electron cloud isn't spherical, just spherically symmetric.

Anonymous

Anonymous Wed 28 Oct 2015 14:53:38 No.7621614 Report Delete

Quoted By: >>7621629

>>7621064
Is there any real chance of us finding a way to extend our life expectancy past 150 years in our lifetime?

Anonymous

Anonymous Wed 28 Oct 2015 14:55:33 No.7621620 Report Delete

Quoted By: >>7621639

If I've shown that not B => not A, is that equivalent to showing that A => B? If so, I can then assume that the latter statement is true, right?

Anonymous

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Anonymous Wed 28 Oct 2015 14:59:37 No.7621629 Report Delete

Quoted By:

>>7621614

This man thinks so, and he looks twice that age right now!

https://en.wikipedia.org/wiki/Aubrey_de_Grey
https://www.youtube.com/watch?v=T0lvxTm2iLg

Anonymous

Anonymous Wed 28 Oct 2015 15:03:17 No.7621639 Report Delete

Quoted By: >>7621640

>>7621620

Yes, that is the contrapositive, which is a logically equivalent statement.

https://en.wikipedia.org/wiki/Contraposition

Anonymous

Anonymous Wed 28 Oct 2015 15:04:59 No.7621640 Report Delete

Quoted By:

>>7621639
Thanks mate

Anonymous

Anonymous Wed 28 Oct 2015 15:08:34 No.7621646 Report Delete

Quoted By:

>>7621137
Because of the way addition is defined, 1 + 1 is the same thing as saying the number directly after, or the successor of, 1. 2 is defined as the successor of 1, hence 1 + 1 = 2. My proof implicitly uses the Peano axioms, so read up on those if you're interested.

tl;dr - that's just the way it's defined.

Anonymous

Anonymous Wed 28 Oct 2015 15:14:14 No.7621659 Report Delete

Quoted By:

Are there any engineers with a B.S degree in engineering that are willing to be interviewed?
I would need your name, university, employment, job title/description, and some answers to basic questions.

If not here, does anyone know a good site/ forum to interview an Engineer?

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