Role of rubber elasticity in the pressure inside a balloon I'm just wondering if there's a pressure difference between the atmosphere and the air inside the balloon due to the rubber elasticity. A kind of analogous to the Laplace pressure in soap bubbles.
Relationship between balloon size and pressure I see numerous websites talking about how a small balloon is at higher pressure than a big balloon. This is a fun counter-intuitive factoid... unless it isn't a fact at all.  Young-Laplace gives  $$T = \Delta p \frac R2 $$  Which means that for a given tension, increasing the radius does decrease the pressure. But why would the tension be constant? A latex balloon is like a 2D spring, right? So the tension in a inflated spherical balloon should be proportional to the surface area, correct?  $$T \propto R^2 $$  But this gives  $$\Delta p \propto R$$  Which means the bigger the balloon, the higher the internal pressure. Which, again, would be the normal intuition - but it would burst the bubble of some simplified explanations. (har har)
Why can't variable names start with numbers? I was working with a new C++ developer a while back when he asked the question: "Why can't variable names start with numbers?"  I couldn't come up with an answer except that some numbers can have text in them (123456L, 123456U) and that wouldn't be possible if the compilers were thinking everything with some amount of alpha characters was a variable name.  Was that the right answer?  Are there any more reasons?  string 2BeOrNot2Be = "that is the question"; // Why won't this compile? Vieta Jumping: Related to IMO problem 6, 1988: If $ab + 1$ divides $a^2 + b^2$ then $ab + 1$ cannot be a perfect square. The famous IMO problem 6 states that if $a,b$ are positive integers, such that $ab + 1$ divides $a^2 + b^2$, then $\frac{ a^2 + b^2}{ab + 1 }$ is a perfect square, namely, $gcd(a,b)^2$.  How about a modification of this problem:    If $a,b$ are (strictly) positive integers, such that $ab + 1$ divides $a^2 + b^2$, then $ab + 1$ cannot be a perfect square.    I am looking for a proof of the claim above, or a counter-example.  One possible approach towards a proof is the following:   Suppose there is such a pair $(a,b)$ as above, then by the famous IMO problem 6 from 1988, $\frac{ a^2 + b^2}{ab + 1 } = g^2$ where $g = gcd(a,b)$. Since $ab + 1$ is a perfect square, $a^2 + b^2 = c^2$ for some integer $c$. So that $(a,b,c)$ is a Pythagorean tripple, therefore there exists positive integers $n,m,l$ with $n$ coprime to $m$, such that $a = l(n^2 - m^2)$ and $b = 2lnm$ - by Euclids formula.  Then by plugging in $a$ and $b$ in terms of $n,m,l$ into the original equation, and solving for $l$, it is possible to obtain the following:  There exists positive coprime integers $n,m$ such that   $\frac{(n^2 + m^2 + 1)(n^2 + m^2 - 1)}{2mn(n+m)(n-m)}$ is a perfect square.  If we put the quotient above into a program, as in this python code snipet:  N = 1000  for n in range(1,N):     for m in range(n+1, N):         A = (n*n + m*m + 1)*(n*n + m*m - 1)         B = 2*m*n*(n+m)*(m-n)         if A % B == 0:             print(A/B)   The quotient always is 2, regardless of whether or not $n$ and $m$ are co prime! So if the very strong implication   $2mn(n+m)(n-m) \mid (n^2 + m^2 + 1)(n^2 + m^2 - 1) \implies \frac{(n^2 + m^2 + 1)(n^2 + m^2 - 1)}{2mn(n+m)(n-m)} = 2$    holds, then the original problem will be solved. What is the error "Every derived table must have its own alias" in MySQL? rstrip not removing newline char what am I doing wrong? Is there a catch-all place on the Stack Exchange network where one can post any question that does not obviously belong to the Stack Exchange sites? Is there a catch-all place somewhere on the Stack Exchange sites where one can post a (any) question that does not obviously belong to any of the Stack Exchange sites?  Sometimes I have questions but I don't want to ask them on other sites (no need for names). I prefer to post it on the Stack Exchange network for various reasons.  If there isn't, are there plans for one? I imagine that if such a site exists, questions on that site can also be moved to other more specific sites when they exists, and it's not to be used to post questions which obviously belong to a specific Stack Exchange site. Assume $\bar A\cap\bar B=\emptyset$. Is $\partial (A \cup B)=\partial A\cup\partial B$, where $\partial A$ and $\bar A$ mean the boundary set and closure of set $A$?   I can prove that $\partial (A \cup B)\subset \partial A\cup\partial B$ but for proving $\partial A\cup\partial B\subset \partial (A \cup B)$ it seems not trivial. I tried to show that for $x\in \partial A\cup\partial B$ WLOG, $x\in \partial A$ so $B(x)\cap A$ and $B(x)\cap A^c$ not equal to $\emptyset$ but it seems not enough to show the result. Allow embedded HTML5 YouTube video How to show ascii art at the top of the terminal when it's opened? Understanding MongoDB BSON Document size limit The category Set has the property that any morphism f can be decomposed into an epimorphism $e$ and a monomorphism $m$ with $f$ = $m\circ e$. (The intermediary set object is the image of $f$.)  Is there a name for this construction?  Is there a name for this property of categories?  In Set, the intermediary object is unique up to isomorphism. Is this the case for all categories with the property? How can I disable the "Open File Security Warning" when I want to run downloaded executable files? Can I travel to the UK one day before my visa starts? I am planning to travel to the UK with my mother. Her visa is valid from 15th June and my visa's validity starts on July 1st.   I will be leaving from Chennai on June 30th, with a transit in Mumbai, but the arrival in the UK is late on the 30th of June. Will I be permitted to travel or will I be prevented from boarding because of the July 1 validity?   Do I need to apply for a vignette transfer? I'm confused about it, as some sites say it is only for those with a new passport. If I do need a vignette transfer, what is the procedure and what documents are required? Strange arrow mark with TikZ edge and anchors. When I draw an edge between two nodes with anchors and globally set option [-&gt;], I get an additional arrow tip symbol at the beginning of the last edge in the path. For example in  \documentclass{article} \usepackage{tikz}  \begin{document} \begin{tikzpicture}     \node (A) at (0,0) {$A$};     \node (B) at (1,0) {$B$};      \draw[-&gt;] (A.north east) edge (B.north west);      \draw (A.south east) edge[-&gt;] (B.south west); \end{tikzpicture} \end{document}   the first \draw command produces incorrect output, while the second produces correct output:    Am I doing something wrong, or is this a bug in TikZ? Is JavaScript a pass-by-reference or pass-by-value language? The primitive types (number, string, etc.) are passed by value, but objects are unknown, because they can be both passed-by-value (in case we consider that a variable holding an object is in fact a reference to the object) and passed-by-reference (when we consider that the variable to the object holds the object itself).  Although it doesn't really matter at the end, I want to know what is the correct way to present the arguments passing conventions. Is there an excerpt from JavaScript specification, which defines what should be the semantics regarding this? I have the following code and I expect the resulting angle to be 135 degrees but whatevet I try, I get 90.  \documentclass{article} \usepackage{tikz} \usetikzlibrary{calc} \begin{document} \begin{tikzpicture}  \coordinate (A) at (1,0); \coordinate (B) at (-1,1);  \pgfmathanglebetweenpoints{A}{B} \edef\angleAB{\pgfmathresult} \node at (0,0) {\angleAB};  \end{tikzpicture} \end{document}   I also tried the following command without knowing what it means, it also prints 90 except one case. The only working case is when A=(1,0) and B=(-1,0), which it prints 180.  \pgfmathanglebetweenpoints{\pgfpointanchor{A}{center}}{\pgfpointanchor{B}{center}}   Why is that? Couldn't find any reference document for "pgfmathanglebetweenpoints" by the way, that's why I am asking here.  EDIT: B is corrected from (-1,-1) to (-1,1) Ubuntu 18.04 opening system settings logs me out A symmetric operator has only real eigenvalues and different eigenvectors corresponding to different eigenvalues are orthogonal.   These are exactly what we want for a physical observable. I think this is the all required.   So why is quantum mechanics not satisfied with symmetric operators, but wants more? Continuous bijections and their inverses Are there any theorems characterizing the continuity of the inverses of bijections between certain classes of particular spaces. E.g. it seems to me the inverse of any continuous bijection between connected subsets of $\mathbb{R}$ is continouos. It seems that most counterexamples require highly specific spaces. I'm thinking there might be something along the lines of any continuous bijection between homeomorphic spaces has a continuous inverse. Any of this make sense? A problem about Vandemonde matrix No continuous function switches $\mathbb{Q}$ and the irrationals Is there a way to prove the following result using connectedness?  Result:  Let $J=\mathbb{R} \setminus \mathbb{Q}$ denote the set of irrational numbers. There is no continuous map $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $f(\mathbb{Q}) \subseteq J$ and $f(J) \subseteq \mathbb{Q}$.