More weight means more pulling force, and thus more tension force.   Like this , however much force you pulling with is equal in magnitude to the tension force.  If that is case, how then can tensions forces in this photo be equal if their sources are of different weight:
Assuming $T_1$ is the force that acts on box $1$ and $T_2$ is the force that acts on box $2$.  Exactly what causes the Tension? Why does $T_1 = T_2$?  The problem is we are told to memorize that $T_1 = T_2$ for mass less ropes, but I do not understand why, especially in an Atwoods machine.  When I think about it, I know it has something to do with the force $m_2g$ which causes the box with mass $m_1$ to rise, and vice versa.  But I cannot apply physics terminology to this, or really understand whats going on.
I have a product key for windows 8.1 (Single language edition or not that i can't remember). I have currently windows 10 pro installed.  I am planning to install windows 10 Home so i activate with windows 8.1 key.  So my question is will i be able to activate it if the Product Key is Windows 8.1 Single Language and i install Windows 10 Home(Multiple language edition) or vice versa. How can I calculate the number of work days between two dates in SQL Server?   Monday to Friday and it must be T-SQL. If $f:E\rightarrow S$ is an elliptic curve over a scheme $S$ (so $f$ is proper and smooth of relative dimension one with geometrically connected fibers of genus one, equipped with a section $0:S\rightarrow E$), then is the sheaf $\underline{\omega}_{E/S}:=f_*\Omega_{E/S}^1$ actually free of rank one? According to the statement of Grothendieck-Serre duality in Hida's book Geometric Modular Forms and Elliptic Curves, there should be a canonical isomorphism $\underline{\omega}_{E/S}\cong\mathcal{Hom}_{\mathscr{O}_S}(R^1f_*\mathscr{O}_E,\mathscr{O}_S)$, and the first result on elliptic curves in this book is that $R^1f_*\mathscr{O}_E\cong\mathscr{O}_S$. It's also not clear to me whether $S$ is being assumed (locally) Noetherian, but if that's necessary for Grothendieck-Serre duality to hold, then I'm fine with assuming it. I can't find another reference with a statement of Grothendieck-Serre duality in this generality which does not use the language of derived categories (which I unfortunately don't understand).  The reason I'm kind of skeptical about this is that in Hida's book, as well as in Katz-Mazur, it is said that $\underline{\omega}_{E/S}$ is invertible, so that an $\mathscr{O}_S$-basis $\omega$ for $\underline{\omega}_{E/S}$ can be found locally on $S$. If the invertible sheaf in question were really trivial then one would be able to choose a global $\mathscr{O}_S$-basis for $\underline{\omega}_{E/S}$, and there would be no reason to talk about doing so locally. Hida goes on to say that, choosing an $\mathscr{O}_S$-basis $\omega$ locally on $S$ allows one to regard $(\Omega_{E/S}^1,\omega)$ as a relative effective Cartier divisor in $E/S$, which also doesn't make complete sense to me because if we can only find $\omega$ locally, how are we getting a global section $\omega\in H^0(E,\Omega_{E/S}^1)=H^0(S,\underline{\omega}_{E/S})$ (unless $\underline{\omega}_{E/S}$ really is trivial)? A better serial voting trap .NET - How can you split a "caps" delimited string into an array? How do I go from this string: "ThisIsMyCapsDelimitedString"  ...to this string: "This Is My Caps Delimited String"  Fewest lines of code in VB.net is preferred but C# is also welcome.  Cheers! When I create a clan in the multiplayer menu, it says clan created but I cannot add or invite people to my clan. I go to create a new one assuming the request was canceled or not entered correctly, it says clan name already exists as if someone else created it.   Is this a bug? Did I do it wrong? Or am I missing something?   I'm on the PS3. Will this also happen on my Xbox 360 Version? Is there a deb package to install Firefox ESR? I would like to have  on my system but I would like to install it via apt-get. Is there a special package to install this?  Download link would be: $|a_{n+1}-a_n| \leq \frac{1}{2} |a_{n}-a_{n-1}|$. I'm trying to show that $(a_n)$ is convergent by showing that $(a_n)$ is Cauchy.  let $|a_2 - a_1 |  =x$.   $|a_3-a_2| \leq \frac{1}{2}x$  $|a_4-a_3| \leq \frac{1}{2}|a_3-a_2|\leq \frac{1}{4}x$  In general, for every $n$:  $|a_{n+1}-a_n| \le \frac {1}{2^{n-1}}x$  Let $\epsilon &gt;0$  for every $m,n &gt; 0 $, say $m&gt;n$ the following holds:  $$|a_m-a_n| = |a_m-a_{m-1} + a_{m-1} - a_{m-2} \dots a_{n+1}- a_n| \le$$  $$|a_m-a_{m-1}| + |a_{m-1} - a_{m-2}| + \dots |a_{n+1}-a_n| \leq \frac{x}{2^{m-2}} + \frac{x}{2^{m-3}} + \dots + \frac{x }{2^{n-1}}$$.  I can say that for big enough $n,m$, $|a_m-a_n|&lt;\epsilon$ and therefore, this sequence is Cauchy? It sounds not good enough to me yet I don't know what else to do here. I am able to use the split command successfully to split large file into multiple smaller files. This is being achieved by following command  split -b 1G $temp_path $final_filepath   But only caveat is that these files many times contain last line which is split across 2 files. Is there any way to avoid that using split or any other command ? How do I calculate standard errors for sums of OLS coefficients? The Monster Manual describes Lair Actions as occurring "On initiative count 20 (losing all ties)..." (p 11).  What does "initiative count 20" mean? I can't find that phrase anywhere in the PHB or in any other context in the MM... Texture is missing in Cycles? Because I am pretty new to blender I watched the tutorial  from BlenderCookie.com.  I loaded  onto a sphere but then in all views the texture is just some kind of blurry with no details. Why isn't "Column ordering" available in my list settings? I want to change the column order visible when viewing or editing a list item. I have accessed the "Column ordering" interface via the list settings page on other lists in the past, but it isn't visible in this list's settings. I need to display math in my report, but is it possible to remove the vertical space around align*?  A little example is the following (where I just want the align* to have the same vertical space as a \\  \documentclass[10pt,danish,a4paper,oneside,fleqn]{report}  \usepackage[english]{babel} \usepackage{amsmath}  \begin{document}  ssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssss \begin{align*}   A{\cap}B\ &amp; =\ {\left\{{b,d,e}\right\}}{\;}{\cap}{\;}{\left\{{a,b,f,g}\right\}}   \\   \nonumber\ &amp; =\ {\left\{{b}\right\}} \end{align*} ssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssss \\ ssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssss  \end{document} Let $x_0 \in \mathbb{R}$ and $f,g: \mathbb{R} \to \mathbb{R}$  such that $f$ is discontinuous at $x_0$ but $g$ is continuous at $x_0$, then $f+g$ at $x_0$ is...   My approach: After some graphical attempts to construct such a pathological case as given above I claim that the described scenario isn't possible. That is $$f \text{ disc. at } x_0 \text{ and } g \text{ cont. at } x_0 \implies f+g \text{ disc. at } x_0 $$  Proof: I was trying to come up with a contradiction, because if they happen to work out I find these kind of proofs to be most satisfying.   Assume $f+g$ is continuous at $x_0$ that is $$\forall \epsilon &gt; 0 ,\exists \delta_1 &gt; 0 : \forall x \in \mathbb{R}, |x-x_0| &lt; \delta_1 \implies |f(x)+g(x)-f(x_0)-g(x_0)| &lt; \epsilon  $$  Since this statement must be true for all $\epsilon &gt; 0$ I could choose $\epsilon:=1$. Next I know that also $g$ is continuous at said point $x_0$, thus  $$\forall \epsilon &gt; 0, \exists \delta_2 &gt; 0 : \forall x \in \mathbb{R}, |x-x_0| &lt; \delta_2 \implies |g(x)-g(x_0)| &lt; \epsilon $$  and I have that $f$ is NOT continuous at $x_0$ therefore: $$\exists \epsilon^* &gt; 0, \forall \delta&gt;0: \exists x \in  \mathbb{R}, |x-x_0| &lt; \delta \wedge |f(x)-f(x_0)| \geq \epsilon $$  My problem is to bring all these three statements nicely together. Mainly I get confused by the quantors. In order to match all three statements I assume I have to choose $\epsilon &gt; 0$ such that the last statement is met, so say $\epsilon:= \epsilon^*&gt;0$, so statement 1 and 2 are still true for said $\epsilon$  I also need to choose $\delta$. Lets say $\delta = \min{(\delta_1,\delta_2)}$ now I should be able to work with all three statements at the same time right? But then I stumble with my argumentation.  Let $|x-x_0|&lt; \delta$, then we obtain $$ |f(x)+g(x)-f(x_0)-g(x_0)| \leq |f(x)-f(x_0)| + |g(x)-g(x_0)| &lt; - \epsilon + \epsilon =0 $$ which means that $|f(x)+g(x)-f(x_0)-g(x_0)| &lt;0$ and I hope that this is a contradiction because $|.| \geq 0$. Please comment (or answer) if I am right or absolutely on the wrong path. I'd also appreciate it if you criticize my formulation. Misplaced Accents in Cyrillic Text In educational material for the Russian language and linguistic literature the need to put accents on Cyrillic vowels often arises. However when I tried to do this with luaLaTeX the accents where misplaced.  \documentclass[11pt]{article} \usepackage[utf8]{inputenc} \usepackage{fontspec} \setmainfont{Linux Libertine:style=Regular} \begin{document}  Е́сли вы изу́чите  \'{Е}сли вы из\'{у}чите  \'{E}сли вы из\'{y}чите % Here "E" and "y" are Latin glyphs, but this workaround does not                         % work with all Cyrillic letters e.g. Я, Ю \end{document}     The last line would looks nice, but unfortunately these are Latin glyphs.  Is this a problem with the font used (Linux Libertine) or with LaTeX? The essential gist of my question relates to what the real, unvarnished realities are of graduate, post-graduate, research, fellowship, and professorship attainment - when you aren't a single person with no pressing demands other than feeding yourself.   [EDIT: I am humbled by the incredibly helpful and thoughtful responses so far, all of which are +1 from me. I'll leave the question open a few more days before clicking 'accept' on any one, but you all have my gratitude and I hope others will benefit anywhere near as much as I have. ]  What I don't know is what succeeding in this path truly requires in time committment and family support.   Right now I can handle a 12-14 credit course schedule with 3.7-4.0 GPA (in the top 5-10% for college and program, dean's list, etc), work part-time, and still have time to be home and not have a burn-out/breakdown - and the academics are the easy part by far! I could do more, but I need to work more too.   What I really need to know is how sharply different is the time/dedication curve of graduate work and above. I realize each step is a filter that removes most of the entrants (most people don't get a degree, most who do don't attend/finish graduate school, most who do don't seek a PhD,  most who do don't seek/gain professorship...and on and on). But I don't know how severe the stepping is - I'm not talking "tenured professor at MIT or bust" - I realize that bar is many orders of magnitude higher.  Background Details: As an "adult, returning" undergraduate student with a (disabled) wife and kids, I'm halfway through a bachelor degree now. I've located a field of study (in a focus of Industrial Engineering) that has made me very interested in altering my life path to include seeking a Masters and maybe even a PhD/professorship (my father was a professor, so its not a purely novel concept to me). I've met and spoken with my advisor and 2 other professors with experience in related fields, and all is encouraging; I have a graduate school in mind, I've received lots of great advice on preparing for school (on this forum and in person), I test well (no GRE fears remain), good grades/class rank, good prospects for future strong referal letters, etc - a comparitively charmed existence for which I am very grateful! I just have to investigate whether I can make it through with an intact, cared for family!  Any guidance you could provide me, and other aspiring "non-traditional" students alike, would be appreciated immensely - thank you! A finite $p$-group cannot be simple unless it has order $p$ I am to trying to prove this theorem: A finite $p$-group cannot be simple unless it has order $p$.  I have this:   Let $G = P$ and $|G|=p$; then there exists $N$, a normal subgroup of $G$ by Lagrange's theorem, such that $|N| \mid |G|$ so, $|N|= |e| = 1$ or $|N| =  p$  If $|N|=p$ so $N=G$.  But I don't know how I can prove the part in the case of $|N|=|e| = 1$  Is my start correct? Should I defragment my SSD? I applied for a standard visitor visa to the UK for seven days. I earn £278/month and have been working at the same organization for more than a year. I have saved up enough funds for the trip and I attached bank statements to prove it. The ECO refused my visa stating:      I note that you stated that your trip to the UK will cost you £1100, this is approximately 4 times your monthly income. I do not find it credible that you spend this amount on a trip.   I am a single woman, with no financial responsibilities and I saved up for this trip because I really wanted to visit the UK.  How do I prove to the ECO and the authorities that spending £1,100 for my trip wouldn't have caused me any financial harm neither would I have had to access UK government funds?    Another concern of mine is that if I reapply without the partial funding of my father or get sponsored by a different person, does the amount given by my father previously cause hindrance in my visa application?  Won't they still reject it on the basis of my pay being too insignificant?