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December 24[edit]

The definition of matter - space and mass[edit]

I've read in a book and also here on the article (matter) that the definition of matter is "something that has mass and takes up space". My question is why do we need both definitions / conditions: mass and space, while apparently one of them is enough for the definition (since mass cannot be without occupying space and vise versa) 93.126.88.30 (talk) 06:02, 24 December 2016 (UTC)[reply]

There are objects which are considered to have mass, but no volume, such as point particles and the gravitational singularity in the center of each black hole. Whether they really have no volume, or are just modeled that way, isn't always clear. Massless particles may not individually have volume, but structures composed of them, such as gravitational waves, do. StuRat (talk) 15:15, 24 December 2016 (UTC)[reply]

It's very difficult up to incomprehensible to understand how could it be a matter that has mass but not volume. Moreover, if it's indeed exists (again I don't understand it logically) then it contradicts the definition of matter (space and mass) because they (both of the examples that you brought) considered as matter, and here is a matter that doesn't occupy a space and the other one doesn't have mass despite it occupy space.

I've read the first paragraph of the point article (particle that has zero dimensional and it's still named particle) and it's neither incomprehensible nor makes sense for me now:

"A point particle is an idealization of particles heavily used in physics. Its defining feature is that it lacks spatial extension: being zero-dimensional, it does not take up space". I'd like to understand the explanation for this issue. 93.126.88.30 (talk) 15:47, 24 December 2016 (UTC)[reply]

Well, when you get into small enough particles, strings, etc., they don't behave according to the laws of physics we are familiar with at human scales. There doesn't seem to be any way to understand these things intuitively. So why do the physicists think they are correct ? Because these models match the math and experiments they have performed. StuRat (talk) 16:14, 24 December 2016 (UTC)[reply]

There has never been a precise, official definition of the word "matter" in physics. The closest term with a precise definition is baryonic matter, which is stuff made of atoms (or their constituents protons, neutrons and electrons). Baryonic matter is pretty close to what people thought of as "matter" before the 20th century. Dark matter is also called matter by physicists, and it does have mass (by definition) but it doesn't necessarily "occupy space": WIMPs and many other dark matter candidates pass through ordinary (baryonic) matter, and each other, with almost no effect. -- BenRG (talk) 21:05, 24 December 2016 (UTC)[reply]

After second reading here, I realized that the all of the examples are for mass without volume. Are there an example for volume without mass? 93.126.88.30 (talk) 16:10, 25 December 2016 (UTC)[reply]

I don't think so. Everything has mass (= energy). -- BenRG (talk) 17:40, 25 December 2016 (UTC)[reply]

are electrons made of by fermions?[edit]

Are electrons made of fermions (quarks and leptons) the same as protons and neutrons, or the electron doesn't subdivided as protons and neutrons do? (meaning to say that that the electrons don't have sub-particles)93.126.88.30 (talk) 06:20, 24 December 2016 (UTC)[reply]

The electron is itself defined as a fermion, and is an elementary particle, so can't be subdivided (at least as far as we currently know. Rojomoke (talk) 06:28, 24 December 2016 (UTC)[reply]

thank you! 93.126.88.30 (talk) 15:48, 24 December 2016 (UTC)[reply]

Question about the redox potential[edit]

Hello everyone. Sorry for my bad and unconventional English, but, I'm Italian...

The redox potential is the tendency of a solution to give electrons, in fact a solution of KMnO4 is a strong oxidizing agent, because the anion MnO4- take electrons easily and his standard potential (25°C, 1 M) is about 1,51 V. If I put a voltmeter or a redox indicator in a solution of permanganate, they tell me "is a oxidized environment".

But if I have a solution of KCl, or MgCl2? They are stable salts. The standard redox potential of Mg⁺⁺ is about -2,37 V, the standard potential of Cl is about 1,36 v. In the solution there is the water (solvent), that has a potential (about -0,83 V) of reduction. The answer is: if I put a Voltmeter in a solution of MgCl2, the value of Volts (and of the redox potential) is positive or negative? If in the same solution I put a redox indicator, it tell me "oxidized environment" or "redocted environment"?

Thank you for the attention and the eventually answers. I take this opportunity for wish merry Christmas and a happy 2017 at everyone. Ciao --87.5.140.83 (talk) 08:16, 24 December 2016 (UTC)[reply]

Sizes[edit]

Are there any scientific sources comparing relative the sizes of anuses, hands, and mouths of humans and other primates? Benjamin (talk) 09:19, 24 December 2016 (UTC)[reply]

The evolution of human and ape hand proportions (((The Quixotic Potato))) (talk) 23:10, 24 December 2016 (UTC)[reply]

Evolution of the human hand: the role of throwing and clubbing (((The Quixotic Potato))) (talk) 23:13, 24 December 2016 (UTC)[reply]

Are there places which don't have lightning & thunder despite they have rain and snow?[edit]

I've been in Ukraine (Kiev) a full winter, and while it was snowy and rainy still I've never seen or listened there neither lightning nor thunders. What is the explanation for that? (I had big windows and the sky was in front of me. In addition the sound of thunders is something that is very difficult to miss)93.126.88.30 (talk) 15:33, 24 December 2016 (UTC)[reply]

Probably because it was winter. Thunder and lightning are rare in winter, in temperate zones. As for why, it has something to do with the local energy of the storm, which depends on temperature and humidity differentials, which are minimal in winter. This, in turn, is due to less sunlight, as the uneven heating of the Earth due to sunlight (day versus night, cloudy versus sunny, dark versus light albedo, etc.) drives these temperature and humidity differentials. Also, frozen-over lakes and oceans limit the air moisture content (humidity) contributed by evaporation. StuRat (talk) 15:48, 24 December 2016 (UTC)[reply]

In the middle east they appears in the winter only. 93.126.88.30 (talk) 15:50, 24 December 2016 (UTC)[reply]

(edit conflict) It's because for lightning and thunder to occur, you need a thunderstorm to form. Thunderstorms almost always bring rain or snow, but they are much more complicated than that; thunderstorms have convection currents inside of them, which is what creates the necessary difference in potential required to create lightning and thunder. In short, a thunderstorm guarantees precipitation but it doesn't work the other way around! Because of this, they are actually more common in the summer, especially in warm areas. I hope this helps! --T H F S W (T · C · E) 15:51, 24 December 2016 (UTC)[reply]

The frequency of lightning varies dramatically around the world. Lightning requires large convective storms, which can form nearly anywhere, but are much more common in some areas than others. Here is a map of the frequency of lighting around the world compiled by satellites. [1] It is a little hard to tell, but it looks like Ukraine would typically see 2-5 lightning strikes per km^2 per year. This rate is typical of Europe but much lower than places like the US, Africa, India, China, etc. Dragons flight (talk) 16:35, 24 December 2016 (UTC)[reply]

There is a phenomenon called Thundersnow, which as StuRat said is kind of rare. ←Baseball Bugs ^{What's up, Doc?} carrots→ 18:20, 24 December 2016 (UTC)[reply]

Here's The Weather Channel's Jim Cantore reacting to thundersnow in Chicago a few years ago.[2] ←Baseball Bugs ^{What's up, Doc?} carrots→ 18:55, 24 December 2016 (UTC)[reply]

No thunderstorms reported here recently. Count Iblis (talk) 23:49, 24 December 2016 (UTC)[reply]

BTW, it's spelled "lightning". ("Lightening" is also a word, but it means to become lighter, as in "lightening the load".) StuRat (talk) 03:54, 25 December 2016 (UTC)[reply]

Thank you. I corrected it. 93.126.88.30 (talk) 04:43, 25 December 2016 (UTC)[reply]

I don't think the Middle East is an exception to the rule. Their winters are possibly as warm as temperate summers and they may not have summer rain at all. 86.185.150.23 (talk) 11:56, 25 December 2016 (UTC)[reply]

I remember Vancouver never having lightening, or at least none that I can remember. Is that something to do with mountains? Anna Frodesiak (talk) 06:06, 28 December 2016 (UTC)[reply]

The Willamette Valley of Oregon also has very little lightning. I think it's due to most of the rain coming from foggy drizzle with storms being very rare. Causes for this weather pattern would be fun to speculate on, (e.g. most rain occurring in cooler months with less energy for storms, the effect of a large, cool ocean nearby), but those would just be my guesses. Mountains tend to cause extreme weather, and when hiking in the Rockies I was advised to get off the mountains before the afternoon lightning, which was quite common.--Wikimedes (talk) 19:38, 28 December 2016 (UTC)[reply]

Feynman Lectures. Lecture 41 [3][edit]

...
We want the time average of x(dx/dt), so let us take the average of the whole equation, and study the three terms. Now what about x times the force? If the particle happens to have gone a certain distance x, then, since the irregular force is completely irregular and does not know where the particle started from, the next impulse can be in any direction relative to x. If x is positive, there is no reason why the average force should also be in that direction. It is just as likely to be one way as the other. The bombardment forces are not driving it in a definite direction. So the average value of x times F is zero. On the other hand, for the term mx(d2x/dt2) we will have to be a little fancy, and write this as
$mx\,{\frac {d^{2}x}{dt^{2}}}=m\,{\frac {d[x(dx/dt)]}{dt}}-m{\biggl (}{\frac {dx}{dt}}{\biggr )}^{2}.$
Thus we put in these two terms and take the average of both. So let us see how much the first term should be. Now x times the velocity has a mean that does not change with time, because when it gets to some position it has no remembrance of where it was before, so things are no longer changing with time. So this quantity, on the average, is zero.

— Feynman • Leighton • Sands, The Feynman Lectures on Physics, Volume I

Explain please this part in simple words. xF_x= 0 because x can be positive +√(number of steps) and negative -√(number of steps). Arithmetic mean then must be zero. Did Feynman use this argument?

Why $m\,{\frac {d[x(dx/dt)]}{dt}}=0$ ? How "remembrance" is used ?

Username160611000000 (talk) 19:11, 24 December 2016 (UTC)[reply]

x and Fx are uncorrelated so the average of their product is the product of averages. But the average of Fx is zero. In the second case the two values are correlated but we know that the average should not depend on time as the system is homogeneous and stationary. Ruslik_Zero 20:04, 24 December 2016 (UTC)[reply]

But the average of F_x is zero. - In this question we had the same picture with adding many vectors of random direction, and resulting vector = √(Number of vectors). In case of F_x we also must add many random forces, must not we? Username160611000000 (talk) 06:30, 25 December 2016 (UTC)[reply]

the average should not depend on time as the system is homogeneous and stationary. - The system flies away due to diffusion. E.g. if we spill the acetone in one part of room, then in some time we feel the odour. So there are some nonzero speed in all directions and radius.Username160611000000 (talk) 06:47, 25 December 2016 (UTC)[reply]

You can solve a more general equation assuming that