Can water oxidize zinc?

Question:
According to this textbook, H₂O has lower reduction potential (-0.83 V) than Zn (-0.76 V). This mean zinc is unable to reduce water to hydroxide. If so, why this video demonstrate that by heating zinc in water, a reaction occurs producing hydrogen gas from the bubbles while also turning wet red litmus paper blue?
Comment: The given value of −0.83 V corresponds to pH = 14. You should also be able to find the value for pH = 0 in your list, and then estimate what happens with water at a roughly neutral pH
Comment: Ok so i found out the answer, zinc cannot directly reduce water to form zinc hydroxide because it has lower reduction potential (-0.76V) than water (-0.83V). Zinc does not react with cold water, however it reacts with hot steam or boiling hot water. This is because high temperature increases water autoionization, thereby increasing the presence of H⁺ and OH⁻. So, zinc is actually reducing H⁺ ion to H₂ because H⁺ ion has higher reduction potential. The Zn²⁺ then reacts with OH⁻ to form Zn(OH)₂
Answer: I've seen similar demonstrations with magnesium, and in both cases the explanation is the same.  The metal reacts to produce the hydroxide plus hydrogen; with zinc this would be
$\ce{Zn + 2 H2O -> Zn(OH)2 + H2}$
But, the zinc hydroxide is essentially insoluble in cold water and seals up the metal before the reaction gets very far.
Boiling the water speeds things up in two ways.  First, the metal hydroxide, with either zinc or magnesium, is inherently more soluble.  Second, the turbulent action created by the boiling (the bubbles you see are actually steam, not hydrogen) acts as a stirring mechanism, speeding up the mass transfer so the slight concentration of dissolved hydroxide gets to the litmus paper faster.
Magnesium differs from zinc because the former hydroxide is somewhat more soluble.  Therefore, the basic nature of the product with magnesium can be demonstrated with phenolphthalein, which requires a slightly higher pH to react than litmus.  Some demonstrations with magnesium show a rose coloring of the magnesium metal in cold water with the phenolphthalein showing that the magnesium hydroxide remains essentially on the surface.

Can water oxidize zinc?


How to open QGIS using a customisation file?

Question: I have a customisation file which can provide a much simplified QGIS interface for beginner users. We have QGIS available on a shared machine. I'd like to provide a shortcut which would open QGIS with the simplified interface already active (and another which opens it without customisation). It seems that on Linux this can be done with command line options. Is there some equivalent SIMPLE arrangement for Windows (I'm not a programmer, and I don't have admin permissions on any machines).
Comment: If you installed QGIS using the [OSGeo4W Network Installer](https://www.qgis.org/en/site/forusers/download.html), you could use the **OSGeo4W Shell** command prompt program. Simply entering `qgis` would load up QGIS (without customisation). But I never used customisation files before so can't help you with this part. Hopefully others can advise :)
Comment: Joseph - Thanks. Installation at this stage is on a computer by computer basis as I try to introduce QGIS as a useful tool. In effect I'm a fellow worker encouraging uptake of QGIS for simple map/data viewing by non-GIS colleagues, not a administrator of anything at all. Consequently any solutions need to be simple. I'll be able to have support from our (remote) IT people for something which I can explain clearly to them - but work from the assumption that QGIS is a completely alien thing to them at the moment.
Answer: This question already has an answer. Save a ini file with custom ui setting and add that ini file with the z option to the qgis command when starting QGIS
How to setup default toolbars and menu item?
Comment: Sorry - I don't understand the answer you've linked to. QGIS on Windows is started up using a shortcut which does (for Windows) relatively complex things. I'm unaware of how to start up from the Windows Command line.
Comment: If you right click the QGIS icon, you can see the path and file that QGIS gets run from. If it is a batch (bat), edit it and add the z-option to command starting QGIS. You can also try to add the option to the target field of the desktop icon properties.
Comment: I'm understanding you to mean to edit the batch file itself? Am I right? Again - no expert on batch files,  but presumably the following line in the file is the key one: [ start "QGIS" /B "%OSGEO4W_ROOT%"\bin\qgis-bin.exe %* ]. If so you're meaning to add the option here? Before or after the '%*' ?
Comment: start "QGIS" /B "%OSGEO4W_ROOT%"\bin\qgis-bin.exe -z e:\test.ini%*
Comment: Hmmm... I can see that this might work, but at the moment it's failing. I can't save a new .bat file into my (or anyone else's) Program Files folder (see comment on other answer about why this is). I can set up a shortcut to a new .bat file (stored in the user's folder), but then it complains about not finding qgis-bin.exe (Thanks for your help on this - I'll write it up into a proper answer if it works). (As you'll gather I'm struggling to understand the operation of the .bat file)
Comment: Please note that it is not helpful for the discoverability of good answers to duplicate questions. Please post your answer to the related question if you work it out. Thank you.
Comment: Made this work at long last and have added a comment on the other question's answer. Still can only make this work if the bat file is in the exact same place as the original qgis.bat / qgis-ltr.bat (C:/OSGEO4SW64/bin)- and can only now do this because we're now using the OSGEO4W installer - meaning that as a non-admin user I can save files here.


Equivalences for mCi doses and Rads?

Question: What would be the magnitude of radiation received for a 50mCi dose of radioiodine I-31 as compared to say RADS?  I read that it might be on the order of about 35 RADS.  Does this make sense?
Answer: There is no direct conversion for what you are asking.  Curies are a measure of activity equal to $3.7 \cdot 10^7$ decays per second.  Rad are a measure of absorbed dose equal to 1 Joule per kilogram.  In order to determine how much radiation is received, you need to know things like the amount of time of exposure, distance between source and target, sheilding if applicable (and intervening material regardless), solid angle of target, and mass of the target.
I assume that you meant to write $^{131}I$ and I expect that you are implying that the source was ingested and so all of the radiation is absorbed.  $^{131}I$ always decays by $\beta^-$ emission and about 90% of these decays have an average energy of 191 keV, 7% have an average energy of 96 keV, and 2% have an average energy of 69 keV.  Thus, on average, the $\beta^-$ have an energy of about 180 keV which is about $2.9 \cdot 10^{-14} J$.
A 50 mCi source gives off $1.85\cdot 10^9 \beta^-$ particles per second so, in this case, we are looking at about 0.003 Joules per minute of exposure.  If you divide this by the weight of the patient, it will tell you the absorbed dose in rad.
Now, $^{131}I$ has a biological half-life of about 100 days.  Biological half-life works about the same as radioactive half-life, there is an exponential decay of the effects.  In other words, about 50 days after ingestion, you will be recieving only half of the exposure you would receive if the sample were, say, sitting on a table.  Keep in mind, however, that the radioactive half-life of $^{131}I$ is only 8 days.  You can combine radioactive half-life and biological half-life to determine the expected absorbed dose over time but I will not do it here.
One other thing to consider, however, is that the exposure to radiation in this case is not uniform because Iodine migrates to specific places in the body so that the residence time depends on which organ you are considering.  Additionally, for health purposes we typically use dose equivalent rather than absorbed dose which is measured in Sieverts rather than rads.  Dose equivalent is determined by multiplying the absorbed dose by a quality factor Q as recommended by the International Commission on Radiological Protection (ICRP).  This accounts for the difference resulting from the energy of the incident particles.  To make matters even more confusing, there is often also an "equivalent dose" (not the same as dose equivalent) that includes an additional weighting factor to account for the different nature of radiation from varous kinds of particles.  On top of that, there is an "effective dose" that is different still and includes an additional weighting factor to account for the fact that a dose to the gonads is worse than one to bone.
In conclussion, yes, the suggested result you heard of 35 rad as an absorbed dose received from $^{131}I$ sounds reasonable.  However, simplifying radiation exposure is frought with difficulty, every exposure situation is unique and needs individual and competant medical attention.  Many of us in the industry wish the situation were simpler because the current situation is a public relations nightmare when something like Fukushima happens.  Unfortunately, when you are dealing with the effect a stochastic process on biological systems, it just isn't that simple.
Comment: Are you sure, for example, that the activity was milicuries and not microcuries?
Comment: @S.Danielson Hi, and welcome to Physics Stack Exchange! As dmckee told you earlier, please do not respond to other people's answers by posting another answer of your own. Use the "add comment" link which appears just below this message at the left.
Comment: Thank you so much. I am asking as a patient (thyroid) and am trying to determine the magnitude of the ingested dose I received. One of the personnel told me that I probably absorbed the same amount of radiation as you would normally receive in about a year. This seems like it must be incorrect, and that it must be much higher.
Comment: I'm not sure I would say it is "much" higher. 35 rad would contribute on the order of magnitude of 300 mSv which is certainly quite high but that takes a lot of assumptions into account. If you are very concerned, do a bit of research; I work in the nuke power industry and am not a health physicist at that so my answer is more hueristic than quantitative result.
Comment: That's interesting. I have tried to look up the information but as there is no direct conversion as you said, its difficult for a layman to get an idea of magnitude compared to things we have a sense of already, like X rays. I'm guessing you answered as you did because the dose to the thyroid in RAI therapy is much higher than to the rest of the body, even though that would be still higher than what would be normally absorbed in a year? I wondered about this too because there are so many precautions advised for the people around you when you have the dose, yet they insist that the treatment is
Comment: Yes, it's 50 milicuries, a fairly standard dose for low risk thyroid cancer and maybe even in some other therapeutic applications.
Comment: S. Danielson,  Yes, my answer is based solely on the physics of the question as that is what I have been educated in.  Please note that I edited my answer because I originally neglected to compound the biological half-life with the radioactive half-life.  The half-life of 131I is only 8 days so it actually decays fairly rapidly.  Additionally, I expect there is something like a saturation effect with the amount of iodine the thyroid can take up.
Comment: Nonetheless, a year's worth of dose is a fair amount and it is nothing to sniff at.  One factor I did not mention above is that when the iodine atoms decay, they don't just disappear, they turn into something else (131Xe) which has it's own decay characteristics and so on.  The bottom line is, calculating dose is very complex.  If you are really really interested, you can browse the web for health physics textbooks like Atoms, Radiation, and Radiation Protection by Turner... that's what I used in grad school.
Comment: @S. Danielson, One possible explaination for the very high seeming value of 35 rads is because that dose might be calculated only for the thyroid.  When you are going from Joules per minute exposure to rads, if you divide by the weight of the thyroid rather than the weight of the patient, you could easily gain a couple orders of magnitude in dose.
Answer: AdamRedwine's answer also doesn't take into account the fact that the radioactive iodine will be processed by the body and expelled in the urine over the course of a few days. This is what the comment on the increased exposure on the bladder wall is from, referenced by another comment. So I believe (but am not an expert in this area) that the exposure will actually be much less than just taking the (already short) half-life into account.
[Apologies, I want to provide a comment on the previous answer, but it seems like the well-meaning but misguided moderation system won't let me.]

Equivalences for mCi doses and Rads?


Why are derived categories natural places to do deformation theory?

Question: It seems to me that a lot of people do deformation theory (of schemes, sheaves, maps etc) in derived category (of an appropriate abelian category). For example, the cotangent complex of a morphism $f:X\rightarrow Y$ of schemes is defined as an object in the derived category of coherent sheaves $X$. I would like to understand why derived category is an appropriate category to do deformation theory. I would appreciate it if someone could give me a good example or motivation.
Comment: Fernando: That seems to be rather a hard line to take. Clearly, many people ask questions like that when they feel like something *may* be useful to them, but they're not sure enough to devote a lot of time to studying it. In such a situation, isn't it quite helpful to hear friendly explanations of why other people find it useful?
Comment: And then other times, some popular thing looks vaguely interesting, even if you're not motivated by too specific a need. Then you'd like to hear what all the fuss is about in a relatively painless way.
Comment: @Fernando Muro I hope you don't mind if I revise my statement into a slightly more presumptuous form. I know a number of great mathematicians who share your view as a matter of personal *conviction*. My impression is that they find it rather hard to follow consistently as a matter of actual practice.
Comment: @Fernando Muro: That's fine of course. I know a number of great mathematicians who essentially share your view as a matter of personal practice. I was just puzzled that you were puzzled by some *other* people having the view I outlined! Clearly these different temperaments can coexist and contribute to mathematics.
Comment: I feel puzzled about all "why do we need" questions. I think all these questions can be answered in the same way: if you don't know why, then you don't need it.
Comment: As I understand it, in the 1950s people are happy with cohomology groups, they are where the obstructions live, or they classifies isomorphism class (as a torsor under some cohomology group), or they control automorphisms. [The magic word here is "gerbe".] For smooth object sheaf cohomology just works fine. However, for singular things, it is more complicated, namely they are not just controlled by cohomologies of some single sheaf, but somehow mixed up. It turns out here is a complex that controls everything, which is the cotangent complex. (to be continued)
Comment: @Daniel, I think that my answer would be more or less the same with the new title. "Being natural" is a very personal viewpoint, unless we're talking about natural transformations. You probably think that's natural because you've attended talks or read papers by people who promote that point of view. Read the papers by those people, if you find it natural afterwards, then it's natural for you, if not, it's simply not natural, at least from your point of view. Maths is not what other people tell you it is, you must form your own viewpoint.
Comment: I have never used deformation theory with derived categories, and I think your question makes full sense and is interesting. I think I can offer a first approximation  answer, as the combination of two
well-known observations: (1) Deformations notions are parametrize by cohomology spaces (like $H^1$ for the deformation tangent space, $H^2$ for obstructions) and (2) derived category is a natural place to do cohomology.
Comment: I am puzzled by the combination of "It seems to me that a natural category to do deformation theory is derived category" and "I would like to know why derived category is an appropriate category to do deformation theory"...
Comment: @Femando OK. Maybe the title is not appropriate; I changed the title.
Comment: @Yemon You are right. I edited the first line.
Comment: The cotangent complex is a more fundamental object. Derived category is where the cotangent complex currently lives. As people work out the derived algebraic geometry, it may live in some more interesting category.
Comment: @36min I would appreciate it if you could explain a bit more detail. You can post your answer :)
Comment: I don't feel like I understand it well enough. Ask Jacob Lurie or any derived AG people.
Comment: @Minhyong Kim, I honestly have a completely different view.
Answer: I'll post an answer to what I think is a reasonable question, hoping that someone more expert will improve on this answer. My apologies if this answer is too chatty.
First, a very simple reason you might hope that there is something like a cotangent complex and that it should be an object in the derived category of quasi-coherent sheaves. Given a morphism $f: X \rightarrow Y$ of schemes (over some base, which I leave implicit), you get a right exact sequence of quasi-coherent sheaves on $X$:
$$f^{*}\Omega^{1}_{Y} \rightarrow \Omega^{1}_{X} \rightarrow \Omega^{1}_{X/Y} \rightarrow 0.$$
Experience has taught us that when we have a functorial half-exact sequence, it can often be completed functorially to a long exact sequence involving 'derived functors', and in abelian contexts such a long exact sequence is usually associated to a short exact sequence (or exact triangle) of 'total derived functor' complexes, the complexes being well-defined up to quasi-isomorphism and hence objects in a derived category. This is the semi-modern point of view on derived functors, which one can learn for instance from Gelfand-Manin's Methods of Homological Algebra. Experience has also shown that not only is the cohomology of the total derived functor complex important in computations, but that the complex itself, up to quasi-isomorphism, contains strictly more information and is often easier to work with, until the very last moment when you want to compute some cohomology.
Once you've seen this work a number of times (say for global sections, for $Hom$, and for $\otimes$), one might ask if there is a total derived functor of $\Omega^{1}$, call it $\mathbb{L}$, which among other things produces an exact triangle
$$f^{*}\mathbb{L}_{Y} \rightarrow \mathbb{L}_{X} \rightarrow \mathbb{L}_{X/Y}$$
such that the long exact sequence of cohomology sheaves begins with the original right exact sequence.
If you believe that such a thing should be useful, then you might go about trying to construct it. One way to do this involves interpreting $\Omega^{1}$ as representing derivations which in turn correspond to square-zero extensions, and this is where deformation theory comes in. So one might begin to ask what 'derived square-zero extensions' should be, and you might guess that you should try to extend not just by modules but by bounded above complexes of modules. When you do this, such a square-zero extension becomes not just a commutative algebra but some kind of derived version thereof, such as a simplicial commutative algebra. In these terms, the cotangent complex $\mathbb{L}$ of a commutative algebra turns out to be nothing but K\"ahler differentials of an appropriate resolution of our commutative algebra in the world of simplicial commutative algebras. Following this idea through and figuring out how descent should work leads, after a long song and dance, to the desired theory of the cotangent complex.
Once you set this all up, it becomes clear that there was no reason to restrict oneself to classical commutative algebras in setting up algebraic geometry, but that one could have worked with simplicial commutative algebras to begin with, and this leads to `derived algebraic geometry'. In some sense, this is the natural place in which to understand the cotangent complex, and here the higher cohomology of the cotangent complex has a natural geometric interpretation.
One should also point out that Quillen's point of view on the cotangent complex was as a homology theory for commutative algebras (search for Andre-Quillen homology), which in a precise sense is an analogue of the usual homology of a topological space. This is described in the last chapter of Quillen's Homotopical Algebra and is also discussed in Goerss-Schemmerhorn's Model Categories and Simplicial Methods.
To summarise. Deformation theory is about square-zero extensions (as well as about other more general infinitesimal extensions). K\"ahler differentials corepresent derivations which in turn correspond to square-zero extensions. For each morphism of schemes $X \rightarrow Y$, there is a natural right exact sequence involving K\"ahler differentials, which it would be useful to complete to a long exact sequence. Even better,
we'd like this long exact sequence to come from an exact triangle of objects in the derived category of quasi-coherent sheaves. Realising this goal naturally leads to derived or homotopical algebraic geometry.
Comment: @temp Typo corrected. Thank you. To expand on your pickiness, for people who haven't encountered this before, there are a priori two different categories that you might consider: localisation at quasi-isomorphisms of the category of complex of quasi-coherent sheaves or the subcategory with quasi-coherent cohomology of the localisation at quasi-isomorphisms of complexes of `$O_X$`-modules. If a scheme is not quasi-compact, these can be different, and it is usually the latter that one is interested in. Jacob Bell's comment is that there are other ways to realise the derived category...
Comment: A typo in the triangle of cotangent complex...
Comment: And the cotangent complex is not constructed as an object in the derived category of quasi-coherent sheaves, it is an object in the derived category of O_X modules whose cohomologies are quasi-coherent. Sorry for being picky.
Comment: @temp: are you really being picky, or just thinking of a particular realisation of the cotangent complex? it is my impression that it can be defined entirely internally to QC, at least if you go full derived.
Comment: ..., for example by basically defining it via descent on affines. Full derived here perhaps means that in order to have descent for complexes up to quasi-isomorphism, you have to do descent with homotopical wisdom.
Answer: Perhaps one way of reading this question is "Why is it important to think about complexes when doing deformation theory?"  One must certainly accept that deformation theory is cohomological in nature:  consider the standard Čech construction of classes in $H^1(X, T_X)$ and $H^2(X, T_X)$ corresponding respectively to deformations and obstructions to deformations of a smooth scheme $X$; or consider the deformations of a regular embedding $X \rightarrow Y$, which are parameterized by $H^0(X, N_{X/Y})$ and obstructed by $H^1(X, N_{X/Y})$.
Let me recall the arguments that yield these classes.  Locally a smooth scheme has only one deformation up to isomorphism, so global deformations arise from gluing the local ones; the gluing data come from the tangent bundle, so deformations are classified by $H^1(X, T_X)$.  Similarly, solutions to an obstructed deformation problem exist locally and are locally unique up to isomorphism, so the obstruction comes from gluing.  One writes down the gluing condition and observes that it is a Čech $2$-cocycle (or observes directly that deformations form a gerbe...).
The argument for a regular embedding is similar.  One first gives a direct construction for the identification of $H^0(X,N_{X/Y})$ and the deformations of the embedding; then one shows that there are no local obstructions, so global obstructions arise entirely from gluing, and this gives a Čech 1-cocycle with values in $N_{X/Y}$.
What about a more general deformation problem?  For example, suppose $X$ is a local complete intersection.  A great deal of the arguments above goes through without modification:  locally, deformations of a local complete intersection are unobstructed, so obstructions should come from gluing.  One should be able to produce a Čech cocycle obstructing the existence of a global deformation.  But a cocycle valued in what?
The answer is the (shifted) tangent complex (dual to the cotangent complex mentioned in another answer).  This is a complex in in degrees $[0,1]$ that coincides with $T_X$ if $X$.  The relative tangent complex of a regular embedding $X \rightarrow Y$ is $N_{X/Y}[-1]$.  Thus the tangent complex recovers the two special cases discussed above.
For a complete intersection $X$ in a smooth scheme $Y$, the tangent complex of $X$ may be constructed as $[T_Y \vert_X \rightarrow N_{X/Y}]$.  If $X$ is only a local complete intersection then this construction works locally in $X$, but it's not going to glue to something global.  However, the local construction is quasi-isomorphic to the restriction of something global.  At last, the derived category appears.
All of this is vastly generalized by the cotangent complex, which works the same way as above for arbitrary schemes, without the lci restriction.  You can read more about that in Chris Brav's answer.
Answer: I do not know about the general deformation theory (if there is such a thing), so I will talk about the special case I am familiar with, namely, representation varieties $R=Hom(\pi,G)$ of representations of finitely-generated groups $\pi$ to a Lie group $G$. In this case, as in some  other cases, each analytical germ $(R,r)$ of $R$, regarded as the "deformation space" of $r$, is completely determined by an appropriate controlling differential graded Lie algebra (dgla) $A^\bullet$. A dgla $A^\bullet$ is a certain chain complex equipped with a binary operation called Lie bracket, satisfying certain axioms. The reason for appearance of dgla's in the context of representation varieties $R$ is the basic fact that representations of finitely generated groups could be identified with holonomies of flat connections on appropriate bundles, which, in turn, are described by differential forms with appropriate coefficients. For the differential forms one has both the exterior differential and a bracket, coming from the bracket of the Lie algebra of $G$, hence, one gets a dgla.
Then, there is a basic theorem observed by Deligne, but going back to the earlier work of Schlessinger and Stasheff:
Equivalence Theorem. A weak isomorphism of dglas induces an isomorphism of analytical germs of deformation spaces.
A proof of this theorem (in the general context of dglas, not only for the ones corresponding to representation varieties) could be found in this paper "The Deformation Theory of Representations of Fundamental Groups of Compact Kaehler
Manifolds", Publ. Math. I.H.E.S., 67(1988)
by Goldman and Millson. The original letter from Deligne could be found here.
Definition. Two dglas $A^\bullet, B^\bullet$ are weakly isomorphic if there is a morphism $f: A^\bullet \to B^\bullet$ which induces an isomorphism on the level of $H^0, H^1$ and an epimorphism on the level of $H^2$.
This, of course, holds if $f$ induces an isomorphism of all cohomology groups, i.e., is a quasi-isomorphism of dglas. This means that the derived category of dglas is the natural framework for the deformation theory whenever it is determined by dglas, since one can expoit sequences
$$
A^\bullet \rightarrow B^\bullet \leftarrow C^\bullet \rightarrow D^\bullet \ldots
$$
of quasi-isomorphisms (or weak equivalences) of dglas in order to "compute" germs of deformation spaces. This is exactly what Goldman and Millson did in their paper  (with a follow-up by Simpson) in order to prove that singularities of representation varieties of Kaehler groups are quadratic.
The intuitive reason for the Equivalence Theorem is that $H^1$ describes infinitesimal deformations while $H^2$ describes obstructions to integrability of all orders for the infinitesimal deformations (both observations go back, I think, to the work of Kodaira and Spencer). In particular, if infinitesimal deformations and all obstructions match then one gets an analytical isomorphism of germs. The isomorphism is merely analytical since one has to deal with infinite power series, as one has to kill obstructions of infinitely many orders. Occasionally, after an analytical reparameterization of the germ, there is only one obstruction (the first one), as in the theorem of Goldman and Millson (in this case, the deformation problem is formal), but this is a bit of a miracle, coming from Kaehler geometry.
Addendum: After looking at this paper by Lurie linked in David's answer to another question (see David's comment), I see that all deformation problems, at least on the level of formal neighborhoods, are determined by some dglas. This is something I did not realize before; I guess, I never think about mathematics in this degree of abstraction. I did find it surprising though, that Lurie did not mention Deligne's letter and the paper by Goldman and Millson, maybe representation varieties do not appear among the deformation problems he was thinking about.
Comment: This is a very nice answer. An over concise summary (as in this question http://mathoverflow.net/questions/385/deformation-theory-and-differential-graded-lie-algebras) is that in (the appropriate refinement of) the derived category any deformation theory becomes simply the study of the inverse processes of taking tangents (leading to dgla's as above) and exponentiating (passing to formal deformation spaces).
Comment: I agree that Goldman-Millson's theory and Deligne's letter, as discussed in Millson's ICM address (together with important insights of Drinfeld, Feigin and Kontsevich) are certainly among the crucial historical origins of the current understanding of deformation theory. In some sense representation varieties are behind what I might consider the origin of this entire line of reasoning, in Quillen's amazing work on rational homotopy theory and the cotangent complex..


Multiple Feedback Band Pass Filter Noise

Question: I designed a multiple feedback filter using the analogue devices filter wizard. I have never designed this type of filter therefore I followed up by reading about it and manually calculating the components. I used arbitrary values to learn the process.
My first filter has Q of 5 and Fc at 50KHz and about 6dB of gain. I also used Ideal op-amp for simplicity. I measured the noise over a wide spectrum of 1K to 1Meg ignoring 0.1Hz to 1KHz as 1/f noise is heavily suppressed as well as I am using an ideal op-amp.

I have decided to redesign the filter with a higher value for the Q. I designed for Q of 50 and fc at 50KHz and about 6dB of gain. Same as before I used ideal op-amp and the same bandwidth for noise. To my surprise, the filter with higher Q is noisier. What would be the reason for this? I measured the noise produced by R1 as it is 10s higher and the RMS value calculated by LTSpice is about the same. Therefore, the only other major difference is positive feedback.

Could someone explain how this positive feedback loop works? Also why this positive feedback is not necessary for lower values of Q?
Comment: Can you explain where your 2nd circuit (with positive feedback) came from please?
Comment: @Andyaka I used Analog Filter Wizard and reduced the passband from 10KHz (1st circuit) to 1KHz.
Answer:
To my surprise, the filter with higher Q is noisier. What would be the
reason for this?

The noise gain in the 2nd circuit is about 20 dB higher than the first circuit. How do you measure noise-gain: -

I discarded the input source and left it open circuit because the circa 100 Ω resistor was very dominant here. Strictly speaking I should have shorted the input source resistor but, on this occasion it will make very little difference to the result.

The difference in noise gain is about 20 dB hence, the 2nd circuit will handle op-amp noise quite differently compared to the first circuit. And, it is almost entirely down to the use of positive feedback. Here's what the noise gains look like when the feedback is removed in the 2nd circuit: -

Could someone explain how this positive feedback loop works? Also why
this positive feedback is not necessary for lower values of Q?

I might be able to but, you need to link to a document that describes that circuit. It's not one I've come across before so I'm a tad out-of-my-depth here but, I'm willing to run through any derivation about this if you can provide a link to a web source that describes it.
Suffice to say (for now) that it is the positive feedback that creates a significantly higher op-amp noise gain.
Comment: Thank you for your answer regarding noise.
Regarding Q factor I noticed this post:
https://electronics.stackexchange.com/questions/380518/active-band-pass-filter-that-looks-like-a-sallen-key-topology-filter
and reference to this book:
https://archive.org/details/fe_Analog_Electronic_Filters_Theory_Design_and_Synthesis_H._Dimopoulos_Springer_201/page/n481/mode/2up?view=theater
Comment: Page 482 has all the formulas which I used to calculate the Beta (100) of my design. And verify the component values to my satisfaction. Still struggling to see a direct correlation to high Q. Additionally, I noticed a vague explanation of how Beta is selected "The value of Beta could be any positive number but this particular value has been calculated so as to minimize the Q -sensitivity." [Page 482]
I must admit out-of-my-depth here!
Comment: Also, why do I connect the noise source to non-inverting input? Considering that I feed signal to inverting input when the circuit works as a filter? The non-inverting input is my reference voltage to shift the signal.


What's the relationship between "free theorems" and "free objects"

Question: What's the relationship between free theorems and free objects from algebra. They seem quite similar. I'm wondering if there's an underlying principle here.
Answer: There is no relationship.  They both use the word "free", but with different meanings of the word "free".  It's just an accidental collision, which will happen when you have a language like English with a fixed number of words and the number of concepts we want to talk about exceeds the number of words in the language.
A free group is, roughly, a group that is about as generic as possible (it has no special structure).  Here, the word free means "having no non-trivial strucutre", i.e., free of special equalities.
In contrast, a free theorem about some code (e.g., a function) is one that follows automatically from the type signature of that code.  Here, the word free means "at no cost" or "with no extra reasoning about the code required"; it's an automatic consequence of the type signature, regardless of the implementation/code of the function.
Comment: and yet free objects are, in a sense, generated from the axioms. Free theorems are generated (as you say "follow automatically") from the type signature.


Was this a common practice for parents to leave their children in the temple as Hannah did when she dedicated Samuel by bringing him to the temple?

Question: I Samuel 1:27,28 27 For this child I prayed; and the Lord hath given me my petition which I asked of him:
28 Therefore also I have lent him to the Lord; as long as he liveth he shall be lent to the Lord. And he worshipped the Lord there.
Answer: Was this a common practice for parents to leave their children in the temple as Hannah did when she dedicated Samuel by bringing him to the temple?
No. In fact, Israelites were commanded to redeem their firtborns by paying five shekels of silver.

Numbers 18:15
The first offspring of every womb, both human and animal, that is offered to the LORD is yours. But you must redeem every firstborn son and every firstborn male of unclean animals. 16When they are a month old, you must redeem them at the redemption price set at five shekels of silver, according to the sanctuary shekel, which weighs twenty gerahs.

Further, only sons of Levi were allowed to approach the tent of meeting.

21"I give to the Levites all the tithes in Israel as their inheritance in return for the work they do while serving at the tent of meeting. 22From now on the Israelites must not go near the tent of meeting, or they will bear the consequences of their sin and will die. 23It is the Levites who are to do the work at the tent of meeting and bear the responsibility for any offenses they commit against it. This is a lasting ordinance for the generations to come.

The case of Samuel was special. He was appointed as a prophet of the Lord. Generally speaking, Israelites were discouraged to minister in the holy services.
Answer: No, I think the account tells of Hannah's actions because they are radical, along with the fact that Samuel played a huge role in the history of Israel.


Any way to automatically set my sound output to HDMI when the cable is plugged in and off when it's plugged out?

Question: It'd be great if I had a software do this for me, but I guess a bash script is inevitable. However as someone that has never wrote a single line of bash code I'd prolly need a lot of help. Where to start?
Comment: You saw that: [Set HDMI sound output automatically on connect/disconnect](https://askubuntu.com/questions/263248/set-hdmi-sound-output-automatically-on-connect-disconnect)?
Answer: Sorry, I don't have the right answer for you but I can offer you the next best thing:
Create a toggle to the next audio output sink with click of the mouse.
Motivation to the solution:
My system has an HDMI port+audio, a USB audio bar and the green audio socket.
When USB audio is selected, my system does not detect when the audio jack of my earphones is (un)plugged. So I am unable to use hardware events and react on them.
Therefor I decided to use pacmd; a tool to manipulate the audio system.
It can be used interactively (pacmd) or with arguments (pacmd --help).
Short description:
A script and a launcher on my Xubuntu panel. A click on the launcher will toggle to the next audio output sink. If any program is playing audio, its audio input sink is forced to the output sink and notify-send is used to inform which audio output sink is active.
Actions:
The number of audio devices and their names are likely different on your system so some (minor) changes to the script are required.
The comments I put in the script (further down) will help to customize it.
Copy & paste the script into a file named ToggleAudioOutput.
Make it executable. Filemanager -> Rightclik -> Permissions -> Tick Allow...
In Terminal -> Change Directory to the location of the file.

$ chmod +x ToggleAudioOutput

Launcher on Panel:
Right click the panel -> Panel > Add New Items -> Launcher.
Click Add & Close.
On the right the empty launcher appears; right click it.
Properties -> click small plus
Name: ToggleAudioOutput
Command: browse to where you stored the file and click it
Icon: select a nice one
Click Create & Close
Right-click the launcher again -> Move it to a nice spot on the panel.
Link on Desktop:
Open File Manager and browse to the location of ToggleAudioOutput.
Right-click it -> Send to Desktop (create link)
Let me know if its usefull to you !
<code>#!/bin/bash
# -- ToggleAudioOutput -- Script to change audio output sink.
#
# The script is based on the output of this command line.
# Of course the output varies per system. Copy & paste this command line
# in a terminal to find the number of devices and their names.
#
# $ pacmd list-sinks | grep -e 'index:' -e 'alsa.name' | awk 'NR%2{printf "%s",$0;next;}1'
#   index: 0                alsa.name = "HDMI 0"
# * index: 1                alsa.name = "USB Audio"
#   index: 2                alsa.name = "ALC662 rev3 Analog"
#
# Output sequence doesn't change except for the asterix (active device)
# and after (un)plugging usb audio devices. If your system has more or
# less devices, change the case statement accordingly.
##############################################################################
# Determine the current audio output sink, 0, 1 or 2.
OutputIndex=$(pacmd list-sinks | grep '\* index:' | awk '{ print $3 }')

# To force output to USB Audio add the command:  ToggleAudioOutput 1
# in an entry in Session and Startup -> Application Autostart.
if [ "$1" == "1" ]
then OutputIndex=1; Name="USB Audio"
else
   case $OutputIndex in
   0) OutputIndex=1; Name="USB Audio"
      ;;
   1) OutputIndex=2; Name="Speaker/Headphones"
      ;;
   2) OutputIndex=0; Name="HDMI 0"
# I don't use HDMI audio, so delete this and the following line if needed.
      OutputIndex=1; Name="USB Audio"
      ;;
   esac
fi
##############################################################################
# Do the work...
pacmd set-default-sink $OutputIndex
notify-send "Audio output device set to:   >$Name<" -t 5000

# Any programs playing audio? Force them to the current audio output sink.
for InputIndex in $(pacmd list-sink-inputs | grep 'index:' | awk '{print $2}')
do
   pacmd move-sink-input $InputIndex $OutputIndex
done
##############################################################################
#EOF
</code>


User's Login date and login time

Question: I want to fetch user's login time and login date, is there any command in Unix providing User's login date and User's login time ? this problem i want to perform in Shell-script where username is accepting from the end-user and after checking the availability of user, i would like to fetch that user's login time and login date in different variable and then display using 'echo' command.
Answer: For past logins:
last "$USER_NAME"

Also, the command who lists current logins.
If you're looking for the date of the user's last login, some systems provide it directly, for example lastlog -u "$USER_NAME" on Linux or lastlogin "$USER_NAME" on FreeBSD. It's also available in the output of finger, but not in an easy-to-parse form. In any case, it's available in the output of last (on many unix variants, last -n 1 "$USER_NAME" shows the last login; otherwise you can do last "$USER_NAME" | head -n 1). Note that last login may not correspond to the last logout (e.g. a user remained connected from one origin for a long time and made a quick network login recently).
Comment: i know 'who' command is providing the login information of all users , but i wants to have "Login-date" and "Login-time", and i dont wants the user information for past login but for a username accepted from a user and want to fetch the information about that user
Comment: ? last "$username" provides "Login-date" and "Login-time". The second part of your sentence is not clear for me, maybe its my English, maybe its yours... "Username accepted from a user[...]" ?
Comment: @Paresh: I've run out of ways to interpret your question. If my answer still doesn't help, try making your question clearer.
Comment: @Paresh : `last $USER_NAME` will show current users as well. It will show something like `username    ttyp0    192.168.1.100 Tue Sep 13 13:09   still logged in`
Comment: @Gilies let me clear what i want to perform actually, i want to make a shell-script in which i accept user-name as "read $username" and then i want to check whether the user is log-on or not , if yes then i want to fetch their login-time and login-date, thanx for the support
Comment: linux. you beautiful.
Answer: On linux, last -R $username | awk '/still logged in/ {print $3,$4,$5,$6}' will return nothing if the user is not logged in, otherwise a date/time list for each active session.
Other unixes that don't know the -R option to last (which supresses the hostname) will need some modification.

User's Login date and login time


iPhone - In App purchase, no valid products

Question: I continually received the message "No valid Products" when attempting to test the in-app purchase feature in my iPhone/iPad app.  I've checked many lists on the web and was positive that I covered everything.
Answer: After many hours of working on this bug I finally discovered what the actual problem was and therefore, the solution.
Yes, you do NOT need to submit a binary in order to test the in-app purchase.  However, since I could not figure out what the problem was, I decided to submit my binary anyway.  I figured, I've done everything else what could it hurt.
So, Here is what I did:

Archived my app
Clicked the "Distribute" button
Selected "Submit to the iOS App Store" and clicked Next
Click Next again to log into iTunes Connect

BAMM!
I received the message:   "No identities are available for signing"
What?!!! What does that mean?
This was a very good message in order for me to narrow down the problem.  This told me that I couldn't submit my app to the app store, hmmm, so I thought I should double-check the bundle id.  Looks good in the app, what about in iTunes connect?  I logged in to iTunes Connect and went to Manage apps and selected my app.
uh,oh!  what?  I had the wrong bundle id?  how come?
Well, it was because when I created this new version of my app in the latest Xcode I decided to create a brand new project and I changed the name of the project and of course it had a different bundle id.  ahha!
Next step was to change the bundle id to the old one (This has no bearing on the name of the Xcode project)  Problem was that when I selected my Target, the bundle id had in grey the name of my new Xcode project at the end and it wouldn't allow me to delete the name.  I googled the issue and here is what you need to do:
Open your -info.plist and edit the Bundle identifier.  Delete the suffix ${PRODUCT_NAME}.  Now you can set your bundle id correctly.
That was it!  once I set the bundle id my in-app purchase was working!  YAY!
TIP: I got fooled into thinking that since the Product ID for the In-App Purchase Item (com.name.newprojectname) matched my new bundle id (com.name.newprojectname), that all was ok.  nope.  you can put anything you want in the product id.  When you run your application in Xcode and you are trying to make an In-app purchase, it should check that the bundle id of your application that you are running in Xcode exists in iTunes connect and display a message when they don't match.  That would have saved me SO much time.
Finally I can submit my app to the app store.
I hope that this solution has saved some other fellow developers some time and hair pulling.


Is the cosmological Doppler effect more effective at long wavelengths?

Question: When observing the Doppler effect of a particular galaxy, is the wavelength change greater for long wavelengths?
According to the simple Doppler effect formula, the wavelength change is proportional to the wavelength at rest and the observed speed.
Answer: Yes, cosmological redshift, and gravitational redshift in general, is generally proportional to wavelength. Within general relativity, light rays in vacuum are modeled as propagating along null geodesics, and the only difference in this description as applied to one wavelength of light (as seen by some observer) versus another is a rescaling of the geodesic's affine parameter, which uniformly rescales its associated frequency everywhere along the light ray's trajectory.
To be more explicit, if a light ray propagates along a null geodesic $\gamma(s)$ (here $\gamma$ is a map describing the null geodesic curve; it maps some real interval into the spacetime manifold) between observer $A$ at $\gamma(s_A)$ and observer $B$ at $\gamma(s_B)$ whose four-velocity vectors are $u_A$ and $u_B$, then the frequencies of the light ray as seen by these observers are
$$\omega_A = \langle \gamma'(s_A), u_A\rangle \quad \quad \omega_B = \langle \gamma'(s_B), u_B \rangle,$$
where the inner product is evaluated according to the spacetime metric.
To describe the same procedure for a light ray with initial frequency $\tilde \omega_A$ as measured by $A$, all that changes is that we describe the null geodesic via a reparameterization $\tilde \gamma(r)$ of $\gamma$ given by $\tilde \gamma(r) := \gamma(\frac{\tilde \omega_A}{\omega_A} r)$. Setting $r_A := \frac{\omega_A}{\tilde \omega_A} s_A$ and $r_B = \frac{\omega_B}{\tilde \omega_B} s_B$, this yields
$$\tilde \gamma'(r_A) = \frac{\tilde \omega_A}{\omega_A} \gamma'(s_A) \quad \quad \tilde \gamma'(r_B) = \frac{\tilde \omega_A}{\omega_A} \gamma'(s_B),$$
so that
$$\tilde \omega_B = \langle \tilde \gamma'(r_B), u_B \rangle = \frac{\tilde \omega_A}{\omega_A} \langle \gamma'(s_B), u_B \rangle = \tilde \omega_A \frac{\omega_B}{\omega_A},$$
and hence $\frac{\tilde \omega_B}{\tilde \omega_A} = \frac{\omega_B}{\omega_A}$. That is, the ratio between the the frequencies (and therefore wavelengths) observed by $A$ and $B$ is the same, irrespective of the initial frequency observed by $A$. Notice that precisely the same formalism and argument applies no matter the four-velocity of $A$ (so cosmologically with $A$ a galaxy, whether it is comoving or not), meaning the "Doppler" component of redshift thought of as due to relative motion is indeed captured here.
A caveat to the conclusion is that, in reality, light rays don't quite travel on null geodesics when they pass through matter due to nontrivial indices of refraction, and dispersion within media can mean that different frequencies would travel slightly different paths (not just differently parameterized paths), which can result in differing relative redshifts for different initial frequencies. I rather expect this is utterly negligible for cosmological observations, however.
Comment: I want to fully understand what you wrote, but first, I want to understand the relationship between the wavelength change caused by the Doppler effect of the short wavelength and the wavelength change of the long wavelength. The more blue the light, the smaller the wavelength change?
Comment: If the observed speed is constant, is it right that the size of the red-light red-shift should be greater than the size of the blue-light red-shift?
Comment: @빛나는밤 Indeed, the bluer the light, the smaller the wavelength change. The conclusion of my discussion was that the ratio $\frac{\lambda_B}{\lambda_A}$ is constant, no matter the initial wavelength $\lambda_A$. This implies what you say. That ratio depends only on the path the light travels through spacetime and the four-velocities of the observers $A$ and $B$.


How big of a neutron star would be needed to form a quark star inside of it?

Question:
A quark star is a hypothetical type of compact exotic star composed of quark matter. These are ultra-dense phases of degenerate matter theorized to form inside particularly massive[citation needed] neutron stars. -Wikipedia

If you add enough mass to a neutron star it forms a black hole but how much mass is needed to form a quark star? When do neutrons and protons start to break down into a mush of quarks and gluons?
Comment: Nobody yet knows whether this is possible at all.
Answer: A quark star may or may not be possible - as the wikipedia page you refer to says - they are "hypothetical".
As the mass of a neutron star increases, generally speaking, it become smaller and the central density increases.
Quarks are ``confined'' within nucleons by a strong force that increases with quark separation. However, at very small separations it may be that quarks attain an asymptotic freedom where they can to first order be treated as non-interacting fermions.
At densities of $\sim 5-10\rho_{0}$ (where $\rho_0 = 2.7\times 10^{17}$ kg/m$^3$ is the density of normal nuclear matter) it is possible that the nucleons dissolve into a quark-gluon plasma, containing the lightest three quarks, u,d,s. There will then be a weak-force equilibrium between the species, which results in equal populations of each. Since the rest mass energies of the quarks are thought to be $2-150$ MeV/, they will be extremely relativistic at these densities, with a pressure that approaches $\rho c^{2}/3$. This is a soft equation of state that could not support the overlying neutron star (or indeed a star made entirely of quark matter).  However, it is inconceivable that there are no strong-force interactions between the quarks (or why would nuclei have the size they do?) and these may harden the equation of state sufficiently to form a stable object with a quark matter core.
As there equal amounts of u,d,s quarks these are sometimes called Strange stars or described as being made of Strange Matter. However, these terms are more usually reserved for stars that are entirely made of such material, which is something more exotic entirely.
It is unknown whether such objects exist and therefore it is very difficult to answer your question. It may be that quark matter is just an ephemeral state on the way to the collapse to a black hole and that no quark stars in equilibrium exist. A possible avenue of exploration is that neutron stars with a quark matter core should cool more rapidly than "normal" neutron matter, because the direct URCA neutrino cooling process is feasible. It would be interesting to see whether neutron stars of different masses follow radically different cooling tracks, but those measurements are not available yet.
Comment: I completely agree with your answer and have upvoted it.  My answer  is intended to augment yours.
Answer: As @RobJeffries suggests, "It may be that quark matter is just an ephemeral state on the way to the collapse to a black hole and that no quark stars in equilibrium exist." But if there are stable quark stars, the stability mechanism that produces neutron stars may offer some theoretical clues to look for.  Neutron stars are stable because of a phenomenon called the saturation of nuclear matter.  That is, as more and more nucleons are added to a large nucleus the central density and average binding energy per nucleon approach constant values (discounting for the moment the repulsive coulomb forces between protons which actually leads to  maximum binding in the iron region of the periodic table).
Prior to the development of QCD, an understanding of the saturation of nuclear matter was achieved via a relativistic mean field model of nuclei (see  1972 Phys. Rev. C 5 241)  that is sometimes called the $\sigma$/$\omega$ model.  This model resulted in a self-consistent mean field (or Hartree) single nucleon potential that was a near balance between an attractive scalar potential and a repulsive vector (time-like component) potential for a Dirac single nucleon equation.  For normal nuclei the attraction is slightly greater than the repulsion leading to bound states. Both of these potentials were obtained by folding the respective nucleon densities (both scalar and vector) with the Yukawa interactions for the scalar $\sigma$ and vector $\omega$ meson exchanges.
Now the vector density is the probability density so it remains normalized to 1 for a single nucleon. The scalar density on the other hand has no set normalization, and its integral tends to decrease as the vector density becomes confined to a smaller and smaller region.  That means that an external pressure (such as gravity would exert on neutron star matter as it crushes the star) would be met by an increased repulsion due to the vector meson exchange (beginning to dominate over the attractive scalar meson exchange).  This would cause a supernova collapse to halt if the resulting neutron star was not massive enough to allow gravity to win and produce a black hole.
If stable quark matter stars exist, it is likely that an analogous mechanism must hold.  I am not an expert on QCD, but qualitatively it would mean that after nucleon boundaries begin to break down and quarks begin to intermix in quantities larger than 3 (under the pressure of gravity), the initial increase in quark density would have to halt as even more repulsive QCD forces came into play.  Quarks are already highly relativistic in this regime (much more so than the nucleons in the $\sigma$/$\omega$ model, so this is quite possible but whether it actually occurs is anyone's guess at this point.
Answer: This article might be of interest to you! :)
Extract from Quark Stars

Vuorinen joined forces with Aleksi Kurkela at ETH Zurich in Switzerland and Paul Romatschke at the University of Washington in Seattle to examine how the pressure of strange quark matter depends on its density – a relationship described by the star's "equation of state". Physicists have looked at this before, but only using highly simplified models of quark interactions. Instead, Vuorinen's group has employed perturbation theory – a technique that approximates mathematical solutions in stepwise fashion, which on the whole is far more accurate.
The result may surprise other physicists. Current thinking has it that quark stars should be smaller than neutron stars, and indeed that compact stars above a certain size – typically about twice the mass of our Sun, or two solar masses – must be pure neutron stars with no quark core. However, Vuorinen's group conclude almost the opposite: that the biggest quark stars can be larger than neutron stars, perhaps up to 2.5 solar masses. In other words, as Vuorinen points out, the detection of a compact star with a mass near that limit would be a "strong indication" of a quark star

Image Source: Quark Stars


In an automatic transmission, what decides to change gears?

Question: So say there is an automatic transmission in a car. Is there a computer inside the transmission that decides to change gears, or is there a computer in the car that tells the transmission to change gears? If the computer is in the transmission, how does it decide to shift? What about it's in the car?
Comment: I had checked the wiki, and it didn't help me. That's why I asked the question...
Comment: http://en.wikipedia.org/wiki/Automatic_transmission
Comment: Without reading the wiki, is it vacuum and engine speed?
Answer: It depends on the transmission manufacturer and the model of the transmission. In the '80s and before, an automatic transmission was controlled by a vacuum actuator and a cable which ran to the throttle (called a TV cable). Later in the '90s, when electronic engine management came into play, transmissions started being controlled by a main computer in the car (or Powertrain Control Module - PCM). This controlled a lot of aspects of the tranny, to include shifting, line pressures, torque converter lockup and other operations. Sometime later, some manufacturers started dividing the Transmission Control Module (TCM) and Engine Control Module (ECM) back out so there are actually two controllers doing the work, while talking to each other.
Shifting the gears is dependent on several different factors, such as engine load, throttle position, speed of the vehicle, engine speed, etc. Every manufacturer does it a little different, so I won't try to go into it.
Answer: Engine speed : Only if engine spped is high enough it can shift to next gear
Speed of car:It will not downshift unless spped of car is lesser, else the engine will over rev
Load on Engine:If load is too high, then it must use a lower gear to obtain the required pull. Like on slopes, or a heavier car.
Pedal position:If your pedal is floored, you directly order the engine to downshift for quick power.
Apart from this there are some automatics there which learn your driving patterns.
Then there are AMTs..
Comment: Good point about the AMT.


Why are there no gamma-ray bursts detected in our galaxy?

Question: I found from Wikipedia and other sites that there are no GRBs detected in the Milky Way. Can someone give a feasible reason for that? Why are there no GRBs detected in the Milky Way galaxy?
Comment: - We are still alive. - The aliens who use them as close to lightspeed "starship" launchers have decided that we should be left alone yet a while.
Comment: This question seems to be based on a false premise: that there _should_ be frequent GRBs in our galaxy. Could you explain why you think this?
Comment: Seems to be a pretty reasonable question to me. Voting to leave open.
Comment: Pretty certain this is the "Law of Big Numbers" - GRBs are rare, we live in 1 galaxy, and there are near infinite number of galaxies we don't live in.  Odds are very strong then that we won't see them in our Galaxy.
Comment: @OscarBravo, yes I believe that behind this should be a reason but the source of GRBs still a mystery, if we know that then we confidently enough answered that,
Comment: @sundar3492 Ok - you think we should have frequent GRBs in our galaxy. Can you explain why? It seems to me that conventional wisdom is that GRBs are very rare and that we might never observe one in our galaxy.
Comment: @OscarBravo, right now I have not exact answer that's why I am not posting here. But my current ongoing research on this topic. If I will succeed in my research on this topic, obviously I will post it here.
Answer: All models of gamma-ray bursts involve extremely energetic phenomena: particular types of supernovae, the coalescence of binary compact objects, strong magnetar flares, or tidal disruption events. It turns out that these events are quite rare - so rare, in fact, that GRBs would be expected to occur in a low-redshift Milky Way-like galaxy at a rate of only one every few million or tens of millions of years (Zhang & Meszaros 2003). If the Milky Way was undergoing a period of intense star formation, yielding more massive stars and therefore more supernovae, then this rate would go up, but still not significantly. Bear in mind that the Galactic supernova rate, for example, is believed to be only several supernovae per century, and the vast majority of supernovae don't lead to gamma-ray bursts.
We know of many extragalactic GRBs for several reasons. It helps that we can observe large numbers of galaxies (thanks to how bright the bursts are), and if we could look at millions of Milky Way-like galaxies, it wouldn't be surprising if we could detect $\sim$1 event per year. (There's also the advantage that star formation peaked at a redshift $z\sim2$, and so high-redshift objects would be more likely to produce more GRBs!)
Comment: The Soft Gamma-ray Repeater bursts are one class of gamma-ray bursts that are local. They  originate from extreme magnetic field neutron stars aka magnetars. These bursts have different characteristics than the classical bursts : shorter duration, repeating and spectra that have proportionally more low energy photons in there x-ray to gamma-ray spectra.
Comment: In particular, it's not odd that the Milky Way has provided no GRB observations, because only a tiny minority of the galaxies we've monitored have done so.
Comment: The related aspect is that we can detect gamma ray burst over a large part of the universe. The largest distance measured is at z=8.2. [link](https://www.nasa.gov/mission_pages/swift/bursts/cosmic_record.html). Aside from measuring individual burst distance the uniform sky distribution tells us that they must be mostly very far away not to show any structure such as voids/super-clusters.


Omitting を in 十数えました

Question: I came across the sentence その子は指で十数えました in writing, and I was under the impression that, in written Japanese particles are not omitted. I can see that that sentence means, "That child counted to ten with her fingers". However, I'm having difficulty understanding if there should be an を (or another particle) after 十. I feel like 十 should be the direct object of 数えました, but maybe my understanding of the verb 数えました is incorrect.
Comment: Numerical words are basically an adverb.
Comment: I'd have thought the missing particle was まで rather than を if the meaning is count **to** ten. Can't comment on why/when it would be omitted though.
Comment: That may also be the case, I *feel* like there is a particle missing, and my guess that it is を may very possibly be incorrect.
Answer: No particle is omitted in this sentence. We rarely say 十を数える regardless of it's written or spoken, in the first place. This 十 is like an adverb that directly modify a verb without any particle, and this happens very often.
By the way, this 十 can be read as じゅう and とお, both of which are fine but the former is common.
Here are some formal ways to use 数える:

五【ご】数える count five (aloud / with a finger)
五【いつ】つ数える count five  (aloud / with a finger)
ボールの数を数える count the number of balls
ボールを数える count the balls

You can omit を in the last two examples in casual conversations.
Answer: "十数えました" or "十数える" is correct. We native Japanese speakers feel comfortable using this expression, so it is probably idiomatic.
"十を数えました" or "十を数える" is correct and almost natural. (Some native Japanese speakers might feel these expressions are unnatural.)
If you count "人数" or "車の台数", you have to use を resulting in: "人数を数えました" or "人数を数える", "車の台数を数えました" or "車の台数を数える."
For native speakers of Japanese, the expressions "人数数えました", "人数数える", "車の台数数える" and "車の台数数えました" would be unnatural, and we would need the を.

Omitting を in 十数えました


How to access sharepoint intranet from internet?

Question: We have a static internet website and like to have SharePoint intranet link for employees only which should connect intranet from outside of the network. How to acheive this? In other words, Is there any way to access SharePoint intranet from outside of the network? Thanks in advance
Answer: it's called "Extranet";

Create A-record in your DNS for your domain name for what you want that points to your public IP address. We generally use team.companyname.com
Set your firewall to port forward Port 80 to your internal server running SharePoint
In Alternate Access Mappings under the SharePoint Central admin control panel change the site to the match the A-record you created.

Its under Application Management
Alernate Access Mappings, you can leave the type at Default.

If there is anything you want secured I'd look at SSL and moving to port 443.


React Loader Issue depends on directory?

Question: I've been playing with https://www.npmjs.com/package/react-calendar-timeline and I have an issue.
If I want to overload the timeline class, I simply do
import React from 'react'
import Timeline from 'react-calendar-timeline';
export class MyTimeline extends Timeline {
     //Here I can override methods
}
export default MyTimeline;

This compiles, and works correctly.
However, if I want to make the import statement:
import Timeline from 'react-calendar-timeline/src/lib/Timeline';

(I will eventually also have to update the .../lib/Item/item.js file and therefore need to inherit timeline.js, and /lib/items/items.js to make sure that we load my overridden item.js file instead of that branch's file).
I get loader issues, and don't really understand why.
Module parse failed: PATH\node_modules\react-calendar-timeline\src\lib\Timeline.js Unexpected token (36:19)
You may need an appropriate loader to handle this file type.
|
| export default class ReactCalendarTimeline extends Component {
|   static propTypes = {

I don't understand this error because if I take everything from the react-calendar-timeline/src/lib directory inside of node_modules, and put it directly into my project, I can now inherit from the Timeline.js file (I know physically moving the directory out of node modules and into my project is bad practice).
So why is webpack able to compile the same code when it is in my actual project directory instead of node_modules?
My current webpack.config
loaders: [
    {
        test: /\.(js|jsx)$/,
        loader: 'babel-loader',
        exclude: /node_modules/,
        query: {
            presets: ['es2015', 'react'],
            plugins: ['transform-class-properties']
        }
    },
    { test: /\.css$/, loader: 'style-loader!css-loader' },
    { test: /\.scss$/, loader: 'style-loader!css-loader' },
]

Answer: You need to allow babel to transform class properties using the babel-plugin-transform-class-properties plugin. Link: https://babeljs.io/docs/plugins/transform-class-properties/


Why did Harry fall in the Interrogation Court Room in GoF but not in subsequent memory viewings via the Pensieve?

Question: Why did Harry fall in the Interrogation Court Room in GoF when viewing Karkaroff's trial but not in subsequent memory viewings via the Pensieve, such as Bob Ogden, Tom Riddle at the Orphanage, Morfin Gaunt, Hokey the House-Elf/Hepzibah Smith, wherein he was instead viewing it from a different perspective?
Answer: In all of the cases you mention, Harry did indeed fall:
Bob Ogden

Harry bent forward, took a deep breath, and plunged his face into the silvery substance. He felt his feet leave the office floor; he was falling, falling through whirling darkness and then, quite suddenly, he was blinking in dazzling sunlight. Before his eyes had adjusted, Dumbledore landed beside him.

Tom Riddle at the Orphanage

Harry bent over the Pensieve; his face broke the cool surface of the memory and then he was falling through darkness again.... Seconds later, his feet hit firm ground; he opened his eyes and found that he and Dumbledore were standing in a bustling, old-fashioned London street.

Morfin Gaunt

Harry stepped up to the stone basin and bowed obediently until his face sank through the surface of the memory; he felt the familiar sensation of falling through nothingness and then landed upon a dirty stone floor in almost total darkness.

Hokey the House-Elf/Hepzibah Smith

Harry got to his feet and bent once more over the rippling silver contents of the stone basin until his face touched them. He tumbled through dark nothingness and landed in a sitting room in front of an immensely fat old lady wearing an elaborate ginger wig and a brilliant pink set of robes that flowed all around her, giving her the look of a melting iced cake.
Comment: What about Snape's memories in Deathly Hallows?
Comment: @KyleV There too: *He fell headlong into sunlight*


Outlet drop from 3 way Switch that doesn't have incoming power

Question: I have a light in my kitchen with three-way switches. Power comes into the first switch and leaves with a red traveler wire. Black white and red wires come into the second three-way switch. I want to drop an outlet for the second three-way switch. I do not need to have both switches remain active. I don't need the second switch at all. I have pictures of the first and second switch, but these are after I unwired and rewired them when my first attempt didn't work, I'm not sure I rewired them back correctly, in fact I'm sure I didn't because the second switch is not functioning. [
Comment: With a pair of three way switches, one of the switches has constant power coming to it, and the other switch sends power to the light.  Do you you know which switch you are "operating" on here?  Also, you might not want both switches, but code may require that they both work, if that's something you're worried about.
Comment: Also look at some three way wiring diagrams.  Each switch has a black screw for the "common" wire that will be power coming in or power going to a light, and two brass screws for "travelers" that go directly from switch to switch.  If you mix up the common and travelers, there are a variety of symptoms, and one switch not doing anything is one symptom.
Comment: Thank you. I'm pretty sure I'm working off the one that doesn't have constant power, I sometimes get a reading from it and sometimes not
Answer: Can't get a receptacle off that circuit
Sorry. Kitchen receptacles need to be on 20A circuits, and need to serve only receptacles.  Further, the circuit must serve receptacles only in the kitchen, dining and pantry areas.  (they can also serve the receptacle that powers a gas range, or a wall clock, but no other loads).
Tapping a kitchen receptacle off a light circuit is Right Out.
In some cases, a switch there may be required
Depending on the building codes and the paths into the room, a switch there may be required so the lights can be turned on.  I couldn't speculate further without knowing more about the room layout.
To roll that back to dual 3-ways, note how 3-way circuits work.
Your misconception is that there is only 1 traveler. Actually, there are two.  Here's a wiring diagram for 3-ways that I think is closest to yours (although the wire colors have been colored to depict function, your wires are not!)

The most important note is that the travelers go on the brass colored screws, and travelers are always in the same cable.  Since no traveler is white, I gather white is handling neutral, hence this diagram being a fit.


How to intersect more than two layers? Is there a quick way?

Question: We have three layers, income, obesity and fast food restaurants for England and we wanted to find districts where they all intercepted. So we did a combination of two layers at a time to find where all three crossed over, but is there a quicker way of doing this in one step??
Comment: Does the [Intersect tool](http://resources.arcgis.com/en/help/main/10.1/index.html#//00080000000p000000) not do the trick?
Comment: Could you please update your post to include which licensing you have?  FYI, you may be better off using QGIS if your ArcGIS license is not sufficient.
Answer: With ArcGIS for Desktop Basic and Standard licenses, the number of input feature classes or layers is limited to two. (source: arcgis help)
So if you don't have the advanced license, you need to run the tool twice. If you want to have the intersect tool for three layers because you will often need it, you can use model builder to make your own reusable "three layer intersect" tool.


Is the mass of an individual star almost constant throughout its life?

Question: Could we say that an individual star is almost a closed system?, as many of the stars like the sun don't exchange significant amount of their mass with their environment (except solar flares & other material ejections)
Answer: Mass loss is one of the most important features of stellar evolution! I suppose much depends on what you mean by "until its death".
It is true that most of the mass loss occurs during the last 10-20% of their lives - either in the asymptotic giant branch phase for stars less than about 10 solar masses or the supergiant and Wolf-Rayet phases for massive stars. During these phases the star could easily lose more than half its mass in powerful, radiatively driven winds. It is because of this that stars of initial masses up to 8 solar masses leave behind white dwarfs of a solar mass or less, and why even stars of 20-30 solar masses may end up producing 1-2 solar mass neutron star remnants. Without mass loss there would be many more (and more massive) black holes.
Comment: It's a really great question of scale right? The energy loss by the Sun during normal Hydrogen burning is minimal but actually sustains life on this planet! Closed system indeed!
Comment: It is important to distinguish between the conversion of mass to thermal and neutrino energy in fusion (which is quite a minor matter compared to the stellar mass) and the loss of gas and dust migrating off the surface under the influence of the star's own radiation pressure which is what Rob is referring to here. Also, cue the beautiful picture of the SN1987a shockwaves interacting the the shells of earlier ejected mass.
Answer: It does not exchange mass directly, but it radiates energy.
From equation $E^2 - (pc)^2 = m^2c^4$, you can see that the mass of the star decreases if energy is radiated.
So, the mass of the star does not remain constant.
Comment: The asker said "...exchange _significant_ amount..." - how significant is the mass lost due to radiation?
Comment: "Significant" is open to interpretation. It's significant enough to power all life processes on this entire planet.
Comment: I have sugegsted an edit of Your answer. if you wrap text in (double) dollars it will be rendered using $\TeX$ interpreter.


Given son-in-laws old imac, it is still registered under his iCloud and need to switch it to mine.

Question: How do I switch iCloud accounts on a gifted imac desktop from son-in-laws to mine. It won't update because it says it is registered to another account. Son in law did give me his iCloud signin info. This is the first Mac I have ever owned so a lot out my experience.
Comment: You should really just erase the hard drive before doing anything. Google if you need help, there are many many articles on how.
Answer: You can sign out of his iCloud and then in with yours (or create your free iCloud account if you don't have one). This can all be done in System Preferences.

Open the  (Apple) menu in the top left corn
Open System Preferences
Click on iCloud
If your son-in-law's account is still signed in, click the "Sign Out" button in the bottom left corner

At this point, the iMac should be signed out of his iCloud account. On this same screen, you can sign in with your iCloud account and turn on/off the various services you would like to use.
Like a commenter mentioned above, another option is to start fresh by erasing the hard drive. If that is the route you'd like to take, this Apple reference can help walk you through it:
https://support.apple.com/en-us/HT204904
Comment: This is fine... until any of the apps or even the OS itself needs updating - then you'll discover you need the old ID details before it will update. The apps & OS don't belong to the machine, they belong to the original purchaser's Apple ID. I'd just go with option 2. Erase & start over.


JMX client that can persist gathered information

Question: We've added performance measures to our application and are exposing them using
JMX. Now we would like to gather and store performance data for analysis, in files or in a database.
Does anyone know of a JMX client, open source or commercial, that can persist gathered information?
Answer: JRockit Mission Control can persist JMX data to log-rotated XML-files that you can open up and view in the gui.
Answer: you can attach logging functionality to jconsole itself and log mbean output etc.


Save files with modification time of last change in Notepad++

Question: In notepad++, many times I have hundreds of tabs and leave unsaved text files for weeks before I finally save them. When I do, I would like to save them not with the current timestamp, but with the timestamp of when I actually last modified them (i.e. the timestamp of the appropriate backup file in c:\Documents and Settings\<USER>\Appplication Data\Notepad++\backup\. Is it possible in Notepad++? How?
Comment: By unsaved files do you mean existing files (e.g. `effects.js`) or new files? (e.g. `new44`). Or both?  Also, are you open to any coding in order to achieve requested behavior (e.g. writing your N++ plugin)?
Comment: @miroxlav: actually, both :-) in fact, both variants will be in the backup directory with the right timestamp so it shouldn't be a problem. And I would prefer not to code of course (but writing a script would be much easier than having to compile and build something)
Answer: I think this is exactly what plugins are for. I am not sure about existing plugin for this, but you should be able to achieve your goal with relatively simple plugin. Create a plugin which will create a command "Save with last modified timestamp" (Edit: ...and map it to Ctrl+S originally used for Save).
When invoked:

Asks for save file name. (It is trivial using standard Windows FileSaveDialog. Almost a copy-paste from N++ source for Save As....)
Saves the file into specified location. (Almost a copy-paste as well...)
Retrieves last modification timestamp from file backup of that file. (Location of the backup file is known to N++. Once you have the file path, getting the timestamp should be a single API call.)
Sets last modification timestamp to file saved in step 2. (Again, should be nothing more than a single API call.)

And you can enjoy timestamp values you expect.
Here are Plugin Resources including Plugin Development Quck Start Guide and Technical information on coding a plugin. Based on your StackOverflow profile, this should not be very diffictult thing for you. Checking the resources, plugins can be developed in 4-5 various programming languages. Perhaps C or C# could suit you the best. Also it is possible you can write the entire thing in Python script. Just install N++ Python Script plugin and you can go way of scripting. However, I am not sure whether all necessary N++ internals (e.g. location of backup file) are available for running scripts. If not, you might need to work that around.
Comment: Thanks!! It's great that it can be done using a plugin!!! I thought patch will be necessary. And scripting is great too!! Though I don't know python.... Anyway t would be desirable *not* to create a new command, but to change the behavior of "Save". Because I know myself and I am 100% sure that I will forget not to use save but to use some special new command.... So the best solution is to add a settings option that tells Save to use either current or last modified date as the timestamp. Is that possible in a plugin?
Comment: @Tomas: Your plugin can have options, although not in main Options dialog, but on its own. I believe that modifying Save implemented in core to check value of option set somewhere is not possible without changes in the core. But I think you still *can* create a new command (as described in the answer) and then simply map it to `Ctrl`+`S` which was used for Save. (Using the Keyboard Mapper.) And then hide the toolbar with original Save icon to prevent accidentally clicking it.


What are the differences between "mimic" and "copy"?

Question: I found this thread on wordreference.com in which it is claimed that "mimic" has frivolous connotations and "copy" has derogatory connotations.  Specifically I am wondering how this distinction may relate to interpretations of Wittgenstein.  If one can mimic another's gesture or expression, can one also copy it, and vice-versa?
Comment: I find the difference between the two words quite transparent, myself.  A _mimic_ of something, used as precisely possible, does not require a faithful reproduction of something, whereas a a legitimate _copy_ of it does.
Answer: To me, copying involves a mechanistic approach, such as constructing the copied outcome from the source's details by following a systematic process. Mimicking, however, entails a convergent approach, where the mimicked outcome emerges from the usage of a more experimental or trial-and-error method.
This is compatible with others' claim that you cannot copy somebody else's voice, you can only mimic it; that is, you try to imitate and perfect the result it until you get the desired outcome. But you can copy somebody elses's notes; that is, you get access to the source and mechanically reproduce it piece by piece.
As an aside, mimicry is the technique used by animals (or plants) to "copy" each other's appearance or behaviour.
Answer: The difference lies primarily in their usage in real life.
You can mimic somebody's voice;you can't copy it.
You can copy somebody's notes.
A related term to consider in context would be 'imitate'.
Answer: The OED has that mimic means to imitate or copy (a person, action, etc.) esp. for the purposes of ridicule or satire, or to entertain.  Thus, I would say that mimic has a different connotation than copy, but has no formal difference in meaning when referring to persons, actions, etc.
Of course, as the OED points out, one wouldn't mimic a thing, one would copy a thing, i.e., the difference in meaning and usage rests in the thing being copied or mimicked. One could both copy and mimic a person's speech, but one could only copy a person's notes (as abhiii5459 correctly pointed out).


New HP Omen 32 monitor is capable of 2560x1440 but in Windows 7, 32 bit, it maxes out at 1920x1200

Question: I just bought a new HP Omen 32 monitor and it is capable of 2560 x 1440. But in Windows 7, 32 bit, it maxes out at 1920 x 1200 when I go to the display tab in the Control Center. My NVidia display also supports 2560 x 1440. How can I get it to display maximum resolution? How do I talk to my video driver? Thanks.
Comment: How is it connected?
Answer: I just ran into same problem but think I found a solution.
In  NVIDIA control panel -- display -- change resolution, select omen monitor, then click customize. Then on pop-up window,  choose create custom resolution. You just type in 2560 and 1440 respectively, and let it run a test. Mine shows test successful and allows me to keep this custom resolution. The next time I try to change resolution for the monitor, there is this 2560 x 1440 option showing up.
I am no computer expert but this works so far. Good luck! (probably belated..)


Own a duplex in Georgia, tenant on one side believes tenant on other dealing drugs due to traffic

Question: I own a duplex in Georgia. The tenant in one side has notified me she believes that the tenant in the other side is dealing drugs based on the amount of foot and car traffic coming to the location. She said she has confronted them and asked them to stop standing in front of and blocking or parking on her parking pad. Previously the accusing tenant only notified me of her parking pad being used. I notified other tenant to stop that behaviour.
Legally, what do I do now that the accusing tenant has stated she believes there is drug use or sales occurring?
Comment: Which Georgia? The ex-USSR republic, or the US state?
Comment: @dollarbill ignoring it is a good way to lose possession of the property; see ohwilleke's answer.
Comment: Ignore it entirely - because you do not know who they are and what information they have on you and the other tenant. It's easily deducible on their side to know "who called the cops" if that's what you was intending to do. In the event that it may cause you harm, it's likely a better option to not get involved. If they're really committing a crime such as drugs, they'll be found out sooner or later, otherwise they may choose a different path in life. Given the economic situation across the US, making any extra $$ is what's on everyone's mind.
Comment: You need to contact a lawyer on this.  You may have a duty to protect the tenant from the one allegedly selling drugs.
Comment: @Greendrake eastern European Georgia still exists today.
Comment: @NeilMeyer I know it does and I didn't imply that it doesn't. Referring to it as the ex-USSR republic just seems much easier way to identify what/where it is.
Answer: You are in a tough spot. The main risk is "civil forfeiture" of the duplex if you have sufficient knowledge that it is being used for purposes of drug dealing and take no action. See, e.g., a case from October 1, 2021 when a house was seized on this basis.
But the tenant has property rights under a lease that you can't abridge unless you can prove that there is a violation of the law. And, it isn't obvious if you know enough to trigger civil forfeiture exposure based merely on uncorroborated statements from another tenant that observe only conduct often associated with drug dealing and not deals themselves. Still, the knowledge you do have might trigger a duty to inquire if you want to escape civil forfeiture liability.
Under the statute, to defeat a civil forfeiture action (assuming it has been used for drug dealing), you must show:

(e)(1) A property interest shall not be subject to forfeiture under
this Code section if the owner of such interest or interest holder
establishes that the owner or interest holder:
(A) Is not legally accountable for the conduct giving rise to its
forfeiture, did not consent to it, and did not know and could not
reasonably have known of the conduct or that it was likely to occur;
(B) Had not acquired and did not stand to acquire substantial proceeds
from the conduct giving rise to its forfeiture other than as an
interest holder in an arm's length commercial transaction;
(C) With respect to conveyances for transportation only, did not hold
the property jointly, in common, or in community with a person whose
conduct gave rise to its forfeiture;
(D) Does not hold the property for the benefit of or as nominee for
any person whose conduct gave rise to its forfeiture, and, if the
owner or interest holder acquired the interest through any such
person, the owner or interest holder acquired it as a bona fide
purchaser for value without knowingly taking part in an illegal
transaction; and
(E) Acquired the interest:
(i) Before the completion of the conduct giving rise to its
forfeiture, and the person whose conduct gave rise to its forfeiture
did not have the authority to convey the interest to a bona fide
purchaser for value at the time of the conduct; or
(ii) After the completion of the conduct giving rise to its
forfeiture:
(I) As a bona fide purchaser for value without knowingly taking part
in an illegal transaction;
(II) Before the filing of a lien on it and before the effective date
of a notice of pending forfeiture relating to it and without notice of
its seizure for forfeiture under this article; and
(III) At the time the interest was acquired, was reasonably without
cause to believe that the property was subject to forfeiture or likely
to become subject to forfeiture under this article.
(2) A property interest shall not be subject to forfeiture under this
Code section for a violation involving only one gram or less of a
mixture containing cocaine or four ounces or less of marijuana unless
said property was used to facilitate a transaction in or a purchase of
or sale of a controlled substance or marijuana.

The language in bold is the most threatening part that effectively establishes a duty of inquiry and a duty to take action if your inquiry establishes that drug dealing is going on. If you can't establish that you were ignorant of wrongdoing, and reasonably so, under the circumstances of being told by one of your tenants that something is going on, you are at great risk of losing the duplex to civil forfeiture if the DA can prove that drug dealing is taking place by far less than a reasonable doubt standard.
You might want to hire a private investigator to look into the situation carefully, and to try to evict the allegedly drug dealing tenant if the private investigator corroborates the fellow tenant's claims.


Is ROM primary memory or secondary memory?

Question:
According to this picture ROM is classified under Main Memory , but isn't ROM a secondary storage because it's external and non volatile ?
Any clarification with reference links will be highly appreciated.
Thank you.
Answer: The distinction between primary and secondary storage relies on several joint characteristics which the prototypical examples – RAM and hard drives – satisfy:

Primary storage can be accessed directly by the CPU. Access is fast. Data is volatile – it is lost when power is turned off.
Secondary storage cannot be accessed directly by the CPU. Access is slow. Data is non-volatile – it is retained even when power is turned off.
Tertiary storage is similar to secondary storage, but can only be accessed via a network device. Access is even slower.

Where does ROM fall into this classification? The typical usage of ROM in computing (according to Wikipedia, now falling out of use), is for booting. When the computer is started, the CPU is directed to start running from some address, which is memory mapped to read-only memory* containing instructions for booting the system. This kind of memory satisfies the first two properties of primary storage listed above, but not the third property.
(In reality, ROM can be writable – say it is really an EPROM – and sometimes it is really battery-backed RAM.)
Complicating the picture, some I/O devices are memory mapped, and so in this sense can be accessed directly by the CPU. This is (or at least, used to be) common for video. Does video memory count as primary storage or as secondary storage?
What these examples show is that the classification has its limits. The purpose of such classifications is to help us conceptualize how computers work. They shouldn't be taken too seriously – they don't apply in a clean way to all possible situations.
Answer: The terms "primary memory" and "secondary memory" are intended to distinguish between memory that is directly accessible from the main processor and memory that isn't. The basic idea is that the processor can read and write directly to primary memory. On the other hand, to access the contents of secondary memory, it needs to copy the content into primary memory and arrange for a data transfer, or to use peripheral input-output instructions rather than memory load-store instructions.
This is not a hard-and-fast distinction and there are a lot of ambiguous cases. Most types of memory (RAM, ROM, Flash, EEPROM, …) can either be primary or secondary, addressable or not, depending on the hardware architecture. Classifying memory like this is pointless, and it's wrong unless you're talking about a specific architecture (e.g. a PC's main processor, or a particular family of microcontrollers, etc.).
ROM and EPROM are non-volatile by definition. They aren't necessarily external: many chips contain a small amount of ROM. (E)EPROM and flash are more often external because the erasure process generates heat which is harder to dissipate if the memory is part of a multi-function chip – an engineering concern, not a scientific distinction. ROM is often primary memory because that's where the boot code is stored, and accessing secondary memory requires some code to run, so that can't be done until the boot code has run.


plug our own consensus in any one of ethereum client?

Question: Actually i try to add new consensus and will through  modifying some consensus to see its effect on transaction throughput.
so what platform should i use that provide me this facility or should i use clique platform to plug new consensus.
Answer:
plug our own consensus in any one of ethereum client?

Take a look at Parity's Pluggable Consensus feature:

https://wiki.parity.io/Pluggable-Consensus

You can either choose to experiment with the consensus algorithms Parity ships with (Ethash, Instant Seal, Aura), or write your own and plug it into the framework. (See the engines part of their Github repository for an idea of how to do this, together with the wiki page above.) Note that everything's written in Rust.
I don't think Geth provides anything similarly pluggable, though there are various tutorials on how to get Clique working (e.g. this one).


How do Message Authentication Codes provide source authentication

Question: I have heard that Message Authentication Codes give source authentication.
But I am a little confused. Message authentication makes a hash code by hashing data or message and when the packet arrives at the destination, it checks the MAC code with the hash code it got from hashing packet message.
At this point, many websites say that by doing this, it can give the source authentication because of a secret key.
But hashing doesn't need a secret key.
So how does it work?
Answer: The client would take a piece of data, typically the body of your message itself and use a hashing algorithm that takes two inputs - your input and a long random secret pre-shared key, normally the key would be concatenated to the message before hashing to make it harder to determine the key from a block of cipher text.
This creates a HMAC - hashed message authentication code, it can be created with many hashing algorithms such as md5 or the SHA family. The HMAC offers a guarantee of data integrity as the hash would would not match if any part of the message has been modified.
The client performs an operation in the form MAC = Hash(key, data) and sends this with the message.
On receipt, the server which knows the key is able to perform the same operation and verify it came from the claimed source. To be more accurate, it only proves it came from a source that knows the secret key.
The HMAC security is dependant on the algorithm chosen and key length, common attacks are brute force attempts on the key.
Further info
RFC 2104 HMAC
Wiki - Message authentication code
Security Stackexchange question


How do I display a VisualForce page on the Reports tab?

Question: I have built a custom report using SOQL and displayed it using the DataTables JQuery plugin.  Now I need to add a link to it on the Reports tab so that users see it alongside the rest of their reports.  I don't want to create a custom tab, if possible, since I only want to "inject" this VF page as a report on the standard Reports tab.
Answer: This is not possible. You can't add visualforce pages to the reports tab and therefore you need to provide s separate tab or a link elsewhere.
Answer: An alternative you could consider is to create a home page component entitled 'Key reports' that you show on the sidebar. The component would contain a set of home page custom links ..

some of which point at regular SFDC reports
and one that would point at your VF page
and perhaps one that simply reads as 'More ...' that links to the Reports tab

This way, users look on the sidebar first for the hot reports, including your cool JQuery report
Comment: I like this idea - thx!
Comment: A bit better for me: I created a VF page with links to my custom reports and one link to the standard Reports tab.  Then I added a VF tab linked to the VF page listing the reports.  Just a slight redirection from the main reports, which is fine since users really want to see their pre-defined reports more than the full-blown report tab.