As the strong nuclear force is a short range force, and as it is way stronger than Coulomb force, Lumen Learning answered my question as follows: “for low-mass nuclei, the nuclear attraction dominates and each added nucleon forms bonds with all others, causing progressively heavier nuclei to have progressively greater values of BE/A.” But according to wikipedia:  “a neutron has a root mean square radius of about 0.8 fm,” and “the root mean square radius of a proton is about 0.84–0.87 fm.” which means that the diameter of a nucleon is definitely less than 2 fm.  “the  is powerfully attractive between nucleons at distances of about 1 fm but it rapidly decreases to insignificance at distances beyond about 2.5 fm.” which means that the strong nuclear force of a nucleon -surrounded by nucleons- has an effect only on nucleons next to it, and it doesn’t really have any significant effect on nucleons beyond those, i.e. a nucleon doesn’t really form bonds with all other nucleons in medium-mass nuclei. So, why does the BE/A rise with the rise of A, then start dropping away.
Lighter nuclei liberate energy when undergoing fusion, heavier nuclei when undergoing fission.  What is it about the nucleus of an Iron atom that makes it so stable?  Alternatively: Iron has the greatest nuclear binding energy - but why?
If $I$ and $J$ are ideals in a ring $R$ with $1$ such that $I+J = R$, show that $I^m$ and $J^n$ are co-maximal for all $m,n \in \mathbb{N}$ If $I$ and $J$ are ideals in a ring $R$ with 1 which are  co-maximal, i.e $I+J = R$, show that $I^m$ and $J^n$ are co-maximal for all $m,n$ in $\mathbb{N}$  Work done:     Should I proceed using    By taking $$K=I^m+J^m \neq R$$  Let $$S=\{I,J \in R \mid I^m+J^n \neq R \}$$ be the required partially order set ordered by inclusion then  $$I+J \subseteq I^n+J \subseteq I^n+J^m \subseteq \ldots $$  Then, I think this chain doesn't have an upper bound...as $I^m+J^n \neq R$ for all $ m,n \in \mathbb N$. Am I right??  Thanks for any help!! What's the function of the ~ bitwise operator (Tilde) Availability of MXD file for OpenStreetMap data in Esri Geodatabase? From what I understand, an ArcGIS mxd file contains the "styling" information on how to show data.  I downloaded and imported OpenStreetMap data into my geodatabase. Now I need to show them.  Is there any available mxd file match the OpenStreetMap fields/values that I can use?  I check this question(), but the links are all dead.  ps: right now the all the highway/roads/trail display as a single line in my map. I want to use mxd to show them in a better way, like Mapnik or google. Media keys behaving like function keys I've been programming in Java for a while and just got thrown onto a project that's written entirely in C#.  I'm trying to come up to speed in C#, and noticed enums used in several places in my new project, but at first glance, C#'s enums seem to be more simplistic than the Java 1.5+ implementation.  Can anyone enumerate the differences between C# and Java enums, and how to overcome the differences?  (I don't want to start a language flame war, I just want to know how to do some things in C# that I used to do in Java).  For example, could someone post a C# counterpart to Sun's famous Planet enum example?  public enum Planet {   MERCURY (3.303e+23, 2.4397e6),   VENUS   (4.869e+24, 6.0518e6),   EARTH   (5.976e+24, 6.37814e6),   MARS    (6.421e+23, 3.3972e6),   JUPITER (1.9e+27,   7.1492e7),   SATURN  (5.688e+26, 6.0268e7),   URANUS  (8.686e+25, 2.5559e7),   NEPTUNE (1.024e+26, 2.4746e7),   PLUTO   (1.27e+22,  1.137e6);    private final double mass;   // in kilograms   private final double radius; // in meters   Planet(double mass, double radius) {       this.mass = mass;       this.radius = radius;   }   public double mass()   { return mass; }   public double radius() { return radius; }    // universal gravitational constant  (m3 kg-1 s-2)   public static final double G = 6.67300E-11;    public double surfaceGravity() {       return G * mass / (radius * radius);   }   public double surfaceWeight(double otherMass) {       return otherMass * surfaceGravity();   } }  // Example usage (slight modification of Sun's example): public static void main(String[] args) {     Planet pEarth = Planet.EARTH;     double earthRadius = pEarth.radius(); // Just threw it in to show usage      // Argument passed in is earth Weight.  Calculate weight on each planet:     double earthWeight = Double.parseDouble(args[0]);     double mass = earthWeight/pEarth.surfaceGravity();     for (Planet p : Planet.values())        System.out.printf("Your weight on %s is %f%n",                          p, p.surfaceWeight(mass)); }  // Example output: $ java Planet 175 Your weight on MERCURY is 66.107583 Your weight on VENUS is 158.374842 [etc ...] Can I determine if an argument is string literal? Is the electromagnetic spectrum discrete? Why marinade meat with acid or enzymes? I have a multi-line equation that I want to have matching sized parentheses. The advice I read online was to use \left( equation \right. \\ left. rest-of-equation \right) . However, this syntax leaves first half of the equation unaware of what is in the second half. For example, if you include a square root term after the break, the second parenthesis will be larger than the opening parenthesis. When you are using the aligned environment to split a long equation this can look somewhat offputting.    Compare the top equation to the bottom one. The parentheses match in size for the bottom equation, but the left hand parentheses for the top one is  noticeably smaller.  Top Code:   \begin{aligned}[t]   \gamma_{\text{u}} + \frac{1}{108}   \left(57\gamma_{\text{er}} \right. &amp;+  38\gamma_{\text{se}}\\ &amp; {}\pm  \left.\sqrt{2601 \gamma_{\text{er}}^2 + 3648  \gamma_{\text{er}}\gamma_{\text{se}} + 1444 \gamma_{\text{se}}^2}  \right)   \end{aligned}   Bottom Code:  \begin{aligned}[t] \gamma_{\text{u}} + \frac{1}{16} \left(9\gamma_{\text{er}}  +  6\gamma_{\text{se}} \pm  \sqrt{49 \gamma_{\text{er}}^2 + 76 \gamma_{\text{er}}\gamma_{\text{se}} + 36 \gamma_{\text{se}}^2} \right) \end{aligned} Why does shell Command Substitution gobble up a trailing newline char? As per the following example, and as in  In bash, where has the trailing newline char gone?, I want to know "why" it happens    x="$(echo -ne "a\nb\n")" ; echo -n "$x" | xxd -p  # Output is: 610a62  # The trailing newline from the 'echo' command #   has been "deleted" by Command Substitution   I assume there must be some very significant reason for a shell action, namely Command Substitution, to actually delete some data from the command output it is substituting... but I can't get my head around this one, as it seems to be the antithesis of what it is supposed to do.. ie. to pass the output of a command back into the script process... Holding back one character seems weird to me, but I suppose there is a sensible reason for it...  I'm keen to find out what that reason is... Can I submit an edit for approval after earning 2k reputation? I have now enough reputation points on Stack Overflow to edit every post directly. But sometimes (especially if this is about a field where I'm not savvy, or when I try to edit a question title without having understood the question itself) I might be not totally sure about whether my edit really is an improvement.  For now, I have these options:   Do the edit anyway, hope that someone else will catch it if it is wrong Don't do the edit, and hope that someone else will see this post and edit it in the right way Let a comment and hope that someone will see it (either after editing or instead of it). Flag the post for moderator attention (but will the moderators know it better?) Search the right chat room to find someone who knows more about the topic.   All these are not really optimal, I think.  I would like to have a check box     [_] Submit this edit for approval   which, when checked, would submit the edit into the usual edit approval queue. Then I would get at least one or two other pairs of eyes looking over my edit, and maybe even improving it.  (I don't really care about the +2 reputation for the edit - I think it should be without any reputation difference here.)  There's an  for nearly giving everyone the ability to suggest edits that the post owner can approve, but I'm simply interested in being able to insert them into the general edit queue where anyone can approve them.  Actually, I could do this right now by having a low-reputation sock puppet (or a totally anonymous edit), but I think this is not the intended use for sock puppets, is it? Also, I would like to be able to get feedback for my edit. In the Prisoner of Azkaban, Harry and Lupin could practice the patronus on the boggart, but when Harry was in the final task in the Goblet of Fire, he tried to use the patronus on the boggart, it didn't work and he then moved on to use ridikkilus.  Why could he use a patronus against the boggart, but then not in another? Can we really not tell if we are moving? It has been a while since I've thought about physics, however, I remember something about how if you are on a train with no windows that is going perfectly straight and is perfectly smooth, there is no experiment that you could perform to prove that you are in motion or at rest. So I have a couple of questions related to this.   Is this still held to be true? I was just wondering if anything in physics has come to light since I learned about this (about 10 years ago) that would allow an observer on the train to tell if they are moving. If this is true, why couldn't you accelerate a particle in different directions to a certain mass and measure how much energy it took to get it up to that mass. If you were moving in any direction, wouldn't you measure less energy in your direction of travel? Would having an outside observer affect the aforementioned experiment in any way? How do I tail a log file and keep tailing it when the latest one changes date? Specify a field in attribute table with raster calculator Is it possible to prove that $1+1 = 2$? Or rather, how would one prove this algebraically or mathematically? What is Hawking radiation and how does it cause a black hole to evaporate? From the limit definition of differentiability, determine where the function $f(x) = \sqrt{x}$ is differentiable? For the function $f(x) = \sqrt{x}, \hspace{2 mm} f: [0, \infty) \rightarrow \mathbb{R}$, I wish to find all points $a$ at which $f(x)$ is differentiable, from the definition $$ f'(x) = \lim_{x \rightarrow a}\frac{f(x)-f(a)}{x-a} $$  Clearly, for our function $f(x) = \sqrt{x}$, this gives us $$ f'(x) = \lim_{x\to a}\frac{\sqrt{x} - \sqrt{a}}{x-a} $$  So the question boils down to showing for what values of $a$ this limit exists. How would this be done? Intuitively, I know that the answer is that the limit exists for all $a \in [0, \infty)$ and is given by $\frac{1}{2 \sqrt{a}}$, but I'm unsure of how to prove this from the limit definition of differentiability. Search among posts you've voted for Related posts:    These posts are different from mine in that they merely ask for the &quot;votes&quot; tab back or ask if it is at all possible to get such information, while I propose how to implement this as a search feature. Background I recently found myself facing a problem very similar to one that I had solved before. I remembered that I found the solution on SO and I remembered upvoting the post containing the solution. I wasn't able to find that particular post by just searching with relevant keywords. It would have been nice if I could search only among the questions I have voted on. I searched on meta and found that there had been a tab for looking at what you've voted on, but that it was removed due to lack of interest. This is understandable, since just listing the hundreds/thousands of posts you've voted on isn't particularly useful. Proposal What I want is to be able to search for posts I've voted on similarly to the way I'm able to search on my posts within a specific tag. For instance: user:310121 [iphone] apple  This will return all my posts tagged iPhone, where I mention Apple. This is brilliant. This is just how I want search to work. Now what I want to do is to extend this to include votes. The exact syntax isn't relevant, but I had to make one up to show you what I mean: To show all posts you've voted on: user:310121 {votes}  To show all posts you've upvoted: user:310121 {upvotes}  To show all posts you've downvoted: user:310121 {downvotes}  You could probably omit the the user part, since you're only allowed to see your own votes. (Unless it's desirable to be able to look at other peoples votes. Maybe you should be allowed to do this if you have enough rep? Well, that's another discussion.) I guess that even if you're allowed to see other peoples votes it would make sense to display your own votes if the user part is omitted. So in my case I could have done the following search and found the post I was looking for: {upvotes} keyword  Lastly I just want to suggest that clicking on your vote count will work just like clicking on a tag. If I click on a tag in my profile the following search will be done: user:310121 [tag]  So if I click on my upvote count the following search should be made: user:310121 {upvotes}  The same goes for downvotes: user:310121 {downvotes}  This would be a very useful feature and it wouldn't be intrusive in any way. The users that don't care for this feature might never know that it's there. And just for the record the favorite system doesn't replace this feature. How do I load an mp4 video with multiple audio tracks into blender?