An example of an internal force is the tension in the spring of a harmonic oscillator. An example of an external force is the gravity on a pendulum.  Are there any forces that are both internal and external? Are the fields generated by a force field internal, external, or (if they exist) internal-external?  EDIT: My question is different from the "suggested" post because I'm asking for whether there exists a third category, while the suggested post is asking for an explanation of the difference between internal and external forces.
When a person lifts a book while standing on the earth's surface. Why is his force considered external to the book-earth system? Isn't his force internal like the case of the forces that cause the exploson of a bomb. I think the forces the person exerts on both the book and the earth are internal since they are equal but opposite therefore resulting in the position of the centre of mass of the system being maintained and if i'm right is the mechanical energy of the system constant?  Am i wrong? Please correct me.
Let's say a 10N cube (in air, on Earth) rests flat on a scale at the bottom of a tank of water, and the scale reads 8N, so there is 2N of buoyant force on the cube.  How does the buoyant force manifest itself on the cube?  I assume the force is normal to the surface of the scale, but if there isn't any water under the cube, how does the water push up on the cube? How do I backup data (SMS/contacts) from a device with a broken screen? I just broke the screen of my (unrooted) Galaxy S.  How do I get back my SMS and contacts?   PS: I tried Kies but it won't detect my phone for some reason. (Kies works fine with another Galaxy I possess but not this one) Is there canon info on shelf life of various potions in Harry Potter? Do we know how long potions (in general, or at least specific ones) are potent (no pun intended) in Harry Potter canon?  E.g. do they last a couple of hours? Minutes? Years?   I seem to vaguely remember some comments of Snape's about Wolfesbane potion therfore I think at least some information exists. Issue in evaluating the performance of my "clustering algorithm" using NMI, ARI when the "ground truth" is available? Term for a person who can read but cannot write I'm looking for a term to accurately describe a person who can only read but cannot write. While I'm primarily concerned with people who have never learned to write, I would also be interested in any additional terms used to represent people who have lost the ability to write as a result of disuse.  Both illiterate and unlettered imply an inability to read as well as to write. What would be the value of $\sum\limits_{n=0}^\infty \frac{1}{an^2+bn+c}$ I would like to evaluate the sum  $$\sum_{n=0}^\infty \frac{1}{an^2+bn+c}$$  Here is my attempt:    Letting    $$f(z)=\frac{1}{az^2+bz+c}$$  The poles of $f(z)$ are located at   $$z_0 = \frac{-b+\sqrt{b^2-4ac}}{2a}$$  and  $$z_1 = \frac{-b-\sqrt{b^2-4ac}}{2a}$$  Then  $$ b_0=\operatorname*{Res}_{z=z_0}\,\pi \cot (\pi z)f(z)= \lim_{z \to z_0} \frac{(z-z_0)\pi\cot (\pi z)}{az^2+bz+c}= \lim_{z \to z_0} \frac{\pi\cot (\pi z)+(z_0-z)\pi^2\csc^2 (\pi z)}{2az+b} $$  Using L'Hopital's rule. Continuing, we have the limit is  $$ \lim_{z \to z_0} \frac{\pi\cot (\pi z)+(z_0-z)\pi^2\csc^2 (\pi z)}{2az+b}= \frac{\pi\cot (\pi z_0)}{2az_0+b} $$  For $z_0 \ne 0$  Similarly, we find  $$b_1=\operatorname*{Res}_{z=z_1}\,\pi \cot (\pi z)f(z)=\frac{\pi\cot (\pi z_1)}{2az_1+b}$$  Then   $$\sum_{n=-\infty}^\infty \frac{1}{an^2+bn+c} = -(b_0+b_1)=\\ -\pi\left( \frac{\cot (\pi z_0)}{2az_0+b} + \frac{\cot (\pi z_1)}{2az_1+b}\right)=  -\pi\left( \frac{\cot (\pi z_0)}{\sqrt{b^2-4ac}} + \frac{\cot (\pi z_1)}{-\sqrt{b^2-4ac}}\right)= \frac{-\pi(\cot (\pi z_0)-\cot (\pi z_1))}{\sqrt{b^2-4ac}}= \frac{\pi(\cot (\pi z_1)-\cot (\pi z_0))}{\sqrt{b^2-4ac}} $$  Then we have  $$\sum_{n=0}^\infty \frac{1}{an^2+bn+c} = \frac{\pi(\cot (\pi z_1)-\cot (\pi z_0))}{2\sqrt{b^2-4ac}}$$    Is this correct?  I feel like I made a mistake somewhere.  Could someone correct me?  Is there an easier way to evaluate this sum? Emails Per Invocation Still Capped at 100? What is the rule of thumb for mixing doughs? Without delving too deeply into specialty doughs, but addressing more than just  and , what concerns can I evaluate when calculating the technique for approaching a dough?  For instance, some recipes say not to mix too much, others would imply that beating every bit of air out of the dough is the best thing for it. What I am looking for is a set of criteria, maybe a flowchart, that outlines how to deal with having a well-reasoned technique (as opposed to just following a recipe).  Crumble, cream, or melt the fat first? Paddle or hook? Beat, fold or whisk? Knead once and rest or knead three times and throw? I would assume that to some extent getting air in or out, activating and mixing ingredients, distributing fats more or less evenly, and so forth play a factor in any decision. However, I have no idea what those different things do, or the food science behind them. What general principles would help me understand what each factor does, so that I can have more confidence when staring down the ingredients, whether for some cookies or pizza crusts or muffins? Let $a_n$ be a sequence of positive reals, such that the partial sums $S_n = \sum_{i = 1} ^ n a_i$ diverge to $\infty$. For given $\epsilon &gt; 0$ do we have $$\sum_{n = 1} ^ \infty \frac{a_n}{S_n^{1 + \epsilon}} &lt; \infty?$$  For $\epsilon \ge 1$ we can resolve this quickly by noting $$\frac{a_n}{S_n ^ 2} \le \frac 1 {S_{n - 1}} - \frac 1 {S_n}$$ so for sufficiently large $n$ we can bound $\frac{a_n}{S_n^{1 + \epsilon}}$ by $\frac 1 {S_{n - 1}} - \frac 1 {S_n}$ as well. I'm wondering if this is true for arbitrary $\epsilon &gt; 0$. I know that the series in question diverges for $ \epsilon = 0$, so all that is missing is what happens in $(0, 1)$. Since Ubuntu 11.10 use a new kernel, it's very difficult to have a decent support for virtualization.   VirtualBox doesn't support guest additions for ubuntu 11.10, so I can't copy to and from my ubuntu desktop and windows, which I absolutely require, plus FreeBSD seems not to be able to use DHCP without guest additions.  Virt-manager instead gives an error on launch:  Unable to open a connection to the libvirt management daemon.  Libvirt URI is: qemu:///system  Verify that:  - The 'libvirt-bin' package is installed  - The 'libvirtd' daemon has been started  - You are member of the 'libvirtd' group   unable to connect to '/var/run/libvirt/libvirt-sock', libvirtd may need to be started: Permission denied  Traceback (most recent call last):   File "/usr/share/virt-manager/virtManager/connection.py", line 1146, in _open_thread     self.vmm = self._try_open()   File "/usr/share/virt-manager/virtManager/connection.py", line 1130, in _try_open     flags)   File "/usr/lib/python2.7/dist-packages/libvirt.py", line 102, in openAuth     if ret is None:raise libvirtError('virConnectOpenAuth() failed') libvirtError: unable to connect to '/var/run/libvirt/libvirt-sock', libvirtd may need to be started: Permission denied   The problem is solved by running virt-manager as root, but I don't like that. How do I change permissions to run Virt-Manager as user?  Is there a way to install guest additions on Ubuntu 11.10? Is it OK to call a professor by his first name when he/she signs emails by only first name? If a professor in a North American country presents his- or herself by his/her first name in email messages, does this mean that students can refer to him/her by his/her first name? Or is this generally not a good idea, unless the professor has explicitly mentioned that he/she can be referred to by first name? I've noticed that most professors, who prefer to be referred to more formally, do not sign their emails with just their first names, but usually initials or first and last name. How can I make sharp geometric 3d models look more realistic? I'm making a scene in Blender with manufactured mechanical and electrical objects - cylinders, metal objects and so forth. I want to make these look real, be fully present with a sense of mass and substance.  3DCG has a tendency to make everything very sharply geometric, often looking mathematical and artificial. In other scenes, some organic curves, normal maps or particle systems would help make object look real. Like for example, making a simple plastic dog model look more believable, one can add more vertices for better definition, add texture layer, add fur etc.  My scene shows clean resistors, wires, metal strips and such, all sharply geometric. It's tricky to find ways to make the result less mathematical and more substantial, realistic and interesting to the eye.  What are some effective ways to add realism to these objects? Can moving tables to another MySQL database help performance? I want to know if I using different databases can help performance.  I have one big table support_chat with more than 5 million rows. I need to scan this all the time (3s with ajax) to check new messages on support.  If I move this table to another database in the same computer, can it help anything? (let users, sells, comments table faster)  For example:  db1  users table sells table comments table support_chat table // move this table to db2   db2  support_chat table   They want to keep old things due legal reasons, to have a user support history. What's the difference between real and effective aperture? I understand that the simple idea that f-stop is focal length divided by aperture really applies to a simple single-glass lens, not the compound lens systems used to provide good image quality in modern lenses.  I've heard that in this case, what matters is the ratio of the aperture as it appears through the front element. Why the front element, and not the back (which would be the "point of view" of the sensor or film)?  What about for effects such as diffraction? Does the literal physical aperture size matter, or is it the effective size? Given $$1\cdot 2^1 + 2\cdot 2^2 + 3\cdot 2^3 + 4\cdot 2^4 + \cdots + d \cdot 2^d = \sum_{r=1}^d r \cdot 2^r,$$ how can we infer to the following solution? $$2 (d-1) \cdot 2^d + 2. $$   Thank you What is the opposite of "preaching to the choir"? A hard problem on exponential integration Suppose $a : [0 , 1] \to \Bbb R$ is an infinitely smooth function. For $\lambda\ge1$, define $$F(\lambda) := \lambda \int_0^1 e^{\lambda t} a(t) \, dt.$$ If $\sup_{\lambda\ge1}|F(\lambda)|\lt\infty$, then $a$ is the identically zero function. Below are the some results I have derived:   Derivatives of any order of $a$ vanishes at $t = 1.$ For any $\delta\lt1$, $a(t)$ has a zero in the open interval $(\delta , 1).$  The same is true for all the derivatives of $a(t)$. This tells us that the $n^\text{th}$ derivative of $a$ has infinitely many distinct zeros for all natural $n$. If $a$ is analytic, then I can show that $a(t)\equiv 0.$ If $a(t)\ge0$ on $[0 , 1]$, then it is obvious that $a\equiv 0.$   I would appreciate any hint.  Actually $(1)$ and $(3)$ follows from $(2)$. For $(2)$, suppose that $a(1)\gt0$ (the case that $a(1)\lt0$ can be argued in the same way as the following), then there exists $\delta\lt1$ such that $a(t)$ is strictly positive on $I = [1-\delta , 1]$. Then write $F(\lambda) = \lambda\int_0^{1-\delta}e^{\lambda t}a(t)dt + \lambda\int_{1-\delta}^1e^{\lambda t}a(t)dt$. Now there exists $c\gt0$ such that $a(t)&gt;c$ on $I$ since $I$ is compact, so that the second term is bounded below by $c (e^{\lambda}-e^{\lambda(1-\delta)})$. But the first term is only $O(e^{\lambda(1-\delta)})$, so this contradicts the fact that $F(\lambda)$ is bounded. Morever, using the same idea and integration by parts, one can see that $n^{th}$ derivative of $a$ must vanish at $1$. I have a High poly mesh and I have decided to make the low poly by merging as it is easiest for me. Would it be possible to change the W Hotkey , so that it merges to the center automatically? It is time consuming pressing W > Merge > Center. In  it says that flagged comments can be handled automatically if flagged repeatedly (3 + (score / 3) flags = deletion). Which is great; mods are busy people.  In , the venerable Bill the Lizard says that if all comments on a post are obsolete, you can flag the post (or one of the comments) with a custom flag saying that they're all obsolete, and the mod can remove them with one fell swoop. Problem is, that requires a mod.  What should I do when some of the obsolete comments are mine? I can't flag my own comments, and I don't like raising custom flags if I can avoid it.  You're wondering why I don't just delete them. There are two scenarios where I don't, at least not immediately:   Someone has commented with incorrect information. In that situation, I reply to the comment telling them how they've got it wrong. (Sometimes, of course, they quite correctly reply telling me how I got it wrong, which is neither here nor there&nbsp;&mdash; we still end up with obsolete comments. :-) ) Frequently, they reply saying "you're right" but (near universally) leave the comment lying around. I can (and do) then comment again saying "Let's clean these up" and flag their comments as obsolete. But I can't flag mine. Nor do I want to remove mine while the incorrect comment is still there; it leaves a confusing situation for anyone coming to the post later. Someone has commented with correct information. In that situation, if appropriate (e.g., the post author isn't busily editing at the same time and I know enough to do it properly), I'll edit the comment into the post and reply with "You're right, edited that in." and flag their comment as obsolete. These are rarely removed by the commenter. And again I can't flag mine.   The first scenario is the more problematic, since I can go back 10-20 minutes later and remove my "edited that in" comment and there's no harm if their comment is still there.  I currently bookmark and check back periodically, but that's labor-intensive and I'd like a better way.  Ideally, I'd want to tie my comment to one I've flagged so that it's deleted automatically if the other one is (by flagging, or by the commenter). That's a fair bit of development, though, I'm guessing.  What should I do? Go with a custom flag saying "The conversation between X and me is now obsolete," even though that means dumping it on a mod? If "No One" uses a face from the Hall of Faces to become someone then how did he do this In season five of Game of Thrones, when Arya becomes blind there is a very interesting detail most of us might have missed.   Let's not stretch the question:    Arya returns after assassinating Ser Meryn Trant Jaqen H'ghar "No one" comes in the scene, Waif is present as well Arya and Jaqen talks, he appears to be giving the poison to Arya but drinks it and dies. Arya cries, says he was her friend when Waif asked her something (I don't recall what) Waif removes her face, reveals she is actually Jaqen Arya removes face of Jaqen from the dead body, keeps removing it until her own face appears.   What sorcery is going on in here? How could they use a face of a person who is still alive. They remove the faces of a dead person for a reason? So that the face can be used by the Faceless-men.   Is there something I am missing or can the Faceless-men do these kinds of magic as well?