Running a thought experiment, I considered someone's weight when they were standing on the Earth with the Moon directly overhead.  My initial thought was that the person weighs less because the Moon's gravity is working "against" the Earth's.  Conversely, someone standing on the Earth with the Moon above the opposite side would conceivably weigh more, since the Moon's and Earth's gravity are working "together".  But, I think this is an incomplete analysis, and here's why:  When considering both the Earth's and Moon's gravity, one must consider the two bodies as an integrated whole, with a mass equal to the sum of each individual body's mass.  Now, here's the kicker:  when performing gravity calculations, the center of this (unified) mass must be used to establish the distance in the formula.  Also known as the .  Thus, in $ F = G \dfrac{m_{1} m_{2}}{r^{2}}$, $m_{1}$ is the combined mass of the Earth and Moon, $m_{2}$ is the mass of the person standing on the Earth (for whom we're measuring the force of gravity), and $r$ is the distance between this person (on the surface of the Earth) and the .  So, someone standing on the Earth directly underneath the Moon is closer to the barycenter than someone standing on the opposite side of the Earth, and my intuitive thinking is that whatever pull the two bodies have working against each other is offset by the reduced $r$.  Conversely for someone standing on the opposite side of the Earth.  Question:  Is the concept I've described above accurate?  Taking this a step further (no pun intended):  Consider someone on the near side vs. far side of the Moon.  I can't conceptualize the above phenomenon being the same since the barycenter is (still) inside Earth, thus a really long distance away from someone on either the near side or far side of our Moon.  Question:  Would someone standing on the far side of the Moon weigh substantially (measurably) different than someone on the near side?    Addendum  No, I don't think  (and its answers) answered mine.  I even linked it below.  But it doesn't mention anything about integrated masses of the two bodies or the concept of center-of-mass-as-barycenter.  I assert that my questions (especially the 2nd one, which is the primary motivation for me asking this to begin with) is not even remotely addressed by the heavier day-or-night question.  Finally, I explicitly did not mention "day" nor "night"; I'm aware of the fact that the Moon can be directly above someone standing on the Earth in broad daylight.    For the record, I have read the below Physics SE questions:    - I'm not convinced that mine is a duplicate of this one!
Using $g = \frac{Gm}{r^2}$, the force on a point mass located at 1 AU from the Sun ($m = 2 \cdot 10^{30} \text{ kg}$) is about ~0.006 N/kg.  Does that mean that, e.g., a 70 kg person is ~42g lighter during the day, and ~42g heavier at night? That seems like it could make a big difference for things like measuring gold bars or other weight-sensitive items. (Gold arbitrage: buy your gold during the day and sell it during the night! Risk-free profit!)  This makes me suspect that I'm overlooking something obvious, because a ~0.05% weight difference seems like something everyone would have noticed long ago. So what am I missing?    Edit: A few answers below indicate that there shouldn't be any weight difference, because the Earth orbits the Sun in free-fall. But if that's the reason, does that mean that a 1:1 tidally-locked Earth-Sun system wouldn't experience differential gravity from the Sun on opposite sides? That doesn't seem right.
Help resizing posted images I often add pictures to my questions. Most of the time they come out the right size. Well, that used to be OK always, up until I started posting screenshots taken by my mobile. For example, this one:  This way, I'm sure you'll agree it is huge, decidedly too big, it takes up the whole body of the question, I mean. At present, the only way I have to resize it is the following code: &lt;sub&gt;(Click image to enlarge)&lt;/sub&gt;  [&lt;img src=&quot;http://i.stack.imgur.com/hKmRp.jpg&quot; width=&quot;210&quot; height=&quot;350&quot; alt=&quot;&quot;&gt;][1]  which I do not remember, and always have to recover from . Is there a simpler, less HTML-hacky (and, I'd expect, more Markdown-based) way of resizing it? PS The output of the above code is: (Click image to enlarge) Disable Tabbing in Firefox 3, Chrome 4 and IE8? Tabbed Browsing may be nice for reading news sites in one sitting, but I still think they are a usability nightmare if you need to keep the page open for a long time, e.g., for testing it while developing or as a reference for something.  The nice thing about the Windows Taskbar (before Windows 7 ) is that it's easy and natural to remember where things are. I can have 30 tabs in it, but because I know what each one is, I know where I have to click to open something - it's the most efficient way of working I've encountered so far.  Sadly, Tabbed Browsing completely breaks this efficiency because it is not possible to specifically click on a tab in the taskbar, I have to click on the browser, search the tab and then click it, which means at least 2 clicks.  Long story, short question: Is there an extension or setting to completely disable tabbing on Firefox 3, Chrome 4 and Internet Explorer 8 and have one Window/Taskbar entry for each open page? I know that in Chrome I can just drag a tab out of a Window to create a new one, but if there is a general way, that would be easier.  Edit: IE8 conveniently offers an option for that: Tools - Options - Tabs - Enable Tabbed Browsing. The algebraic closure of a finite field and its Galois group I am trying to evaluate the following:  $\displaystyle \int_{0}^{\infty} \frac{1}{1+x^a}dx$, where $a&gt;1$ and $a \in \mathbb{R}$  Any help will be much appreciated. Eccentricity of an ellipse How is $\frac{PF}{PD} = e = \frac{C}{A}$ ? where e is eccentricity, P stands for any point on the ellipse. $F$ stands for one of the foci. $e$ stands for eccentricity. $D$ is a point on the directrix of the ellipse. 'C' is the distance from the center to the focus of the ellipse 'A' is the distance from the center to a vertex.   This is referring to an ellipse/hyperbola/parabola and their conic sections. The problem is not the proof for how $PF/PD = e$, or how $C/A = e$, but how the two equate to each other. (The letters stem from the points/foci of an ellipse of a cone and its directrix).  What is the answer that does NOT use analytic geometry? (only trigonometry) Do you have to want to hurt someone for the cruciatus curse to work? OK so the scene where Harry sees Voldemort making Draco tutur Rowel brings a few questions to my mind.      he felt the rage that did not belong to him posses his soul, saw a long room, only light by fire light, and the grate blond Death Eater on the floor,screaming and writhing, and a slighter figure standing over him, wand outstretched, while Harry spoke in a high, cold, merciless voice.   "More, Rowel, or shall we end it and feed you to Nagini? Lord Voldemort is not sure he will forgive you this time...You called me back for this, to tell me Harry Potter has escaped again? Draco, give Rowel another taste of our displeasure... do it, or feel, my wrath yourself!"   A log fell in the fire: flames reared, their light darting across a terrified, pointed white face-   I remember in the Order of the Phoenix Harry tries to do the cruciatus curse on Bellatrix, it doesn't work. She say this is because Harry doesn't mean it, he doesn't want to hurt he enough. If she's right about that (I'm sure she is) I don't see how Draco is managing to make to make Rowel writhe and scream, because it is clear he doesn't want to be doing it.   Also Harry was actually angry at Bellatrix and if his anger at her killing his godfather won't work... well you get my drift.   I know Draco is probably better at magic, but I feel Bellatrix is right, you really have to want them to hurt for it to work.  Next thing, Voldemort's having Draco do it why? Is he doing to upset Lucius, because that and lot of effort to go to, seeing as Voldemort is angry and he'd probably do a better job himself. Or is he trying to train himself a new soldier, I mean the other ones are getting old. Object selection box in addon In my addon panel, how can I have an object selection box?  And what data type would I use to store the the selected object?  It should look like this: If I inherit from a base class and want to pass something from the constructor of the inherited class to the constructor of the base class, how do I do that?  For example, if I inherit from the Exception class I want to do something like this:  class MyExceptionClass : Exception {      public MyExceptionClass(string message, string extraInfo)      {          //This is where it's all falling apart          base(message);      } }   Basically what I want is to be able to pass the string message to the base Exception class. Why doesn't matter pass through other matter if atoms are 99.999% empty space? The ghostly passage of one body through another is obviously out of the question if the continuum assumption were valid, but we know that at the micro, nano, pico levels (and beyond) this is not even remotely the case. My understanding is that the volume of the average atom actually occupied by matter is a vanishingly small fraction of the atom's volume as a whole. If this is the case, why can't matter simply pass through other matter? Are the atom's electrons so nearly omnipresent that they can simultaneously prevent collisions/intersections from all possible directions? Determining a patients biological age? I am looking for an algorithm that shows the exact coefficients some variables that have on a hospital patient's chronological age versus their biological age.   I understand a lot of the variables that go into determining ones biological age (i.e. Blood pressure, diet, family history of disease, obesity etc.) but not the amount of impact that it has on the actual scoring of their age.  I have found many sites that will calculate it, however I need the actual variable coefficients and influence in order to produce a proper score for thousands of patient records.   Thanks in advance for your input! What it says on the tin, a foreign friend of mine has asked and I can't tell him; apart from it sounding horrible.  For example:  "Why is it raining today?"   Instead of:  "Why it is raining today?" Is it wasteful to use a heating element, instead of doing useful work? Consider a computer CPU consuming electrical energy to perform calculations and consequently emitting heat.  Assumption: That a CPU consuming x Watts of power, emits the same amount of heat as an incandescent heating element which consumes the same amount of power.  The rest of the question assumes that this is true: the energy has to go somewhere, and I can't see that the outcome of the calculations counts as "somewhere".  I've long repeated the claim that "wasteful" appliances such as incandescent lightbulbs, over-full kettles, computers left running, etc. are not wasting power if we make the assumptions:   the appliance is in a room which we aim to maintain at above a certain temperature, warmer than the external temperature (cf a home in winter) the room leaks heat to the outside faster than the "wasteful" appliances emit heat. the room contains an electric heater the electric heater is regulated by a thermostat   In these circumstances, I think (?), if a light bulb produces excess heat, its only effect is to cause the thermostat to switch off sooner, so what energy you've "wasted" in the light bulb, is "saved" at the heater.  I don't think that's controversial.  But my question is:   If a CPU does useful work, and emits all the input energy as heat as a side effect But a heating element emits the same heat, without doing useful work Is it therefore wasteful to use heating elements? Is it more efficient to (for example) run  on a CPU, to get "free" computation as a side-effect of heating?   One reason to heat a room with a heating element, rather than with CPUs, is that the heating element is much cheaper to produce. But ignoring the initial cost of the hardware, would a "heater" made out of 20 CPUs, each running at full capacity and consuming 50W each, output less thermal heat than a 1KW element that does nothing else useful?  To be useful, of course, the CPUs would need to output the result of their computations, which would account for a tiny amount of energy. Let's assume that's negligible.  I'm sure there's a flaw in there somewhere; probably that the assumption above is false. What have I missed? A locally constant sheaf on a locally connected space is a covering space; Proof? As part of my hobby i'm learning about sheaves from Mac Lane and Moerdijk. I have a problem with Ch 2 Q 5, to the extent that i don't believe the claim to be proven is actually true, currently. Here is the question repeated:     A sheaf on a locally connected space $X$ is locally constant if each point $x \in X$ has a basis of open neighborhoods  $\mathcal{N}_x$ such that whenever $U,V \in \mathcal{N}_x$ with $U \subset V$, the restriction $\mathcal{F}V \rightarrow \mathcal{F}U$ is a bijection. Prove that $\mathcal{F}$ is locally constant iff the associated etale space over $X$ is a covering.   I don't have a problem (i think) showing the "covering" $\rightarrow$ "locally constant" direction of the implication. My problem is in the reverse direction. In fact, in the process of trying to solve this problem i think i have come up with a very simple counterexample.  Here's my attempt at a counterexample:  Let $X = \{a,b,c\}$ be  finite space with open subsets $\{\{\},\{a\},\{a,b\},\{a,b,c\}\}$. I claim this is hyper-connected and locally connected.  Let $\mathcal{F}$ be the sheaf with $\mathcal{F}(\{a,b,c\}) = \{q\}$, $\mathcal{F}(\{a,b\}) = \{r,s\}$, and $\mathcal{F}(\{a\}) = \{t\}$, where $r = q|_{\{a,b\}}$, and $t = q|_{\{a\}} = r|_{\{a\}} = s|_{\{a\}}$. I claim this defines a locally constant sheaf. Indeed we can have $\mathcal{N}_{a} = \{\{a\}\}$, $\mathcal{N}_{b} = \{\{a,b\}\}$, and $\mathcal{N}_c = \{\{a,b,c\}\}$.  Now we can calculate the corresponding etale space $p : E \rightarrow X$ as follows: $E = \{\dot{q}(a), \dot{q}(b), \dot{q}(c), \dot{s}(b)\}$, and has these four distinct elements since $\dot{q}(a) = \dot{r}(a) = \dot{s}(a) = \dot{t}(a)$, and $\dot{q}(b) = \dot{r}(b)$. The action of $p$ follows by construction.  However, $p$ does not appear to me to be a covering space. In particular, the only neighborhoods of $b \in X$ are $\{a,b\}$and $\{a,b,c\}$ and we have $p^{-1}(\{a,b\}) = \{\dot{q}(a),\dot{q}(b),\dot{s}(b)\}$ and this cannot be partitioned into homeomorphic images of $\{a,b\}$ because $2 \nmid 3$. Similarly, $p^{-1}(\{a,b,c\}) = \{\dot{q}(a), \dot{q}(b), \dot{q}(c), \dot{s}(b)\}$ and this cannot be partitioned into homeomorphic images of $\{a,b,c\}$ since $3 \nmid 4$.  So i think i have a locally constant sheaf on a locally connected space whose corresponding etale space is not a covering. What have i done wrong? I have seen the claim mentioned elsewhere so i'm convinced i'm wrong somewhere. Perhaps someone knows where a correct proof is published otherwise? I will give this question a little context. Firstly, as I understand it, as soon as I "close the switch" on a circuit, electric current pretty quickly establishes a steady state where, at any given cross section along a wire, the average kinetic energy of that slice is both constant and equal anywhere along the wire. (Please correct me if I am misunderstanding).   If the average kinetic energy is constant anywhere throughout the wire in the circuit (which confuses me, as I would think that the electrons should all accelerate after exiting the resistor since they are no longer being impeded as much), that means that the average velocity is constant. So, in the example of a simple circuit with a battery of v Volts and a resistor of o Ohms, my question is the following:  Because a 'voltage drop' is known to occur across the resistor, what type energy is being "traded" to generate the heat that radiates from the resistor? I would think that the reflexive answer is "electrical energy"...hence the VOLTAGE drop (the sacrificed energy is clearly not kinetic, as the velocity is constant everywhere). However, I find this confusing. Does this mean that if I had a positive test charge, it would be easier for me to bring it to the beginning of the resistor as compared to bringing it to the end of the resister?     Further, if the correct answer IS "electrical energy", why exactly DOES electrical energy get "dissipated" as the electrons pass through the resistor? I always here the comment Oh! It's because all the electrons are running into densely packed crap which impedes their flow but that to me sounds like a reason for their KINETIC energy to be reduced. However, clearly that's not the case. So, ultimately, I guess the real question is:  **What goes on in the resistor that is literally removing electrical energy **. I am sure that this is a quantum mechanics question, but my quantum mechanics knowledge is not terribly strong. If I could get an answer that is devoid of crazy wave function equations, I would greatly appreciate it. Thanks! Make flash drive bootable with MULTIPLE Windows Installers Error: You are missing the following 32-bit libraries, and Steam may not run: libXtst.so.6 [...] I'm aware that many people are having problems on launching Steam (downloaded from Steam not the repos) on Ubuntu 16.04 64-bit with this error message:  svetlana@svetlana-desktop:~$ steam Running Steam on ubuntu 16.04 64-bit STEAM_RUNTIME is enabled automatically Error: You are missing the following 32-bit libraries, and Steam may not run: libXtst.so.6 libXrandr.so.2 libgobject-2.0.so.0 libglib-2.0.so.0 libgtk-x11-2.0.so.0 libpulse.so.0 libgdk_pixbuf-2.0.so.0 Installing breakpad exception handler for appid(steam)/version(1463778979) Installing breakpad exception handler for appid(steam)/version(1463778979) Installing breakpad exception handler for appid(steam)/version(1463778979) Installing breakpad exception handler for appid(steam)/version(1463778979) [2016-05-22 16:33:15] Startup - updater built May 20 2016 18:57:58 [2016-05-22 16:33:15] Opted in to client beta 'publicbeta' via beta file You are in the 'publicbeta' client beta. Looks like steam didn't shutdown cleanly, scheduling immediate update check [2016-05-22 16:33:15] Checking for update on startup [2016-05-22 16:33:15] Checking for available updates... [2016-05-22 16:33:15] Download skipped by HTTP 304 Not Modified [2016-05-22 16:33:15] Nothing to do [2016-05-22 16:33:15] Verifying installation... [2016-05-22 16:33:15] Performing checksum verification of executable files [2016-05-22 16:33:15] Verification complete [2016-05-22 16:33:16] Shutdown   My question has anyone found a solution yet? proving that elements of a sequence are pairwise coprime For the sequence $a_n = 6^{2^n} + 1$:   How can I prove that the elements of this sequence are pairwise coprime, i.e. prove that if $m$ is not equal to $n$ then $\gcd(a_m, a_n) = 1$. (begin by proving that $a_n∣ (a_{n+1} - 2)$) iOS 7 Locked Bug via Ubuntu 13.10 I wanted to bring up an issue with the latest Ubuntu release that I hope can be solved. Although I will agree with the community that iOS 7 is not the most open platform release, it still is my desire to be able to grab Photos, Music, and Files off my Phone similar to what I would do to any other device.  Every time I connect my device to Ubuntu, I receive the following notification:  The device "Justin's iPhone" is locked. Enter the passcode on the device and click "Try again".   On my iPhone, I select the following   Unfortunately, this is where problems start to arise. It appears that Ubuntu either does not recognize the permissions from my iPhone or the iPhone is not allowing itself to truly be unlocked. As we can see the permissions are   Browsing askubuntu, I encountered the following   After running the code, I receive the following error:  justin@justin-CT15:~$ idevicepair unpair &amp;&amp; idevicepair pairERROR: Device d1fea2c56e51199f901eb34020a1581ea31a2a50 is not paired with this host   If I call the command again, i receive another error:  justin@justin-CT15:~$ idevicepair pair ERROR: Could not pair with the device because a passcode is set. Please enter the passcode on the device and retry.   Ultimately, I am stuck to experiment with solutions that end up not working. I am troubled by the lack of information and support for the iOS 7 platform on Ubuntu's latest release. If anyone has a workable solution or experiment for a workaround, I am happy to try  Thanks for reading and have a wonderful day! How do I find out what tool a shortcut is triggering? I used to have problem with connecting two objects, because when I smooth the final object, there was a darker line on the connected parts (for example when I was making head, I've had eyes darker this way). Then I found out that I can get rid of it by hitting Ctrl+N, but I don't realy know what it does. I can just see that the lines are gone. So, what does Ctrl+N do? I understand when we say current, we mean charge (protons/electrons) passing past a point per second. And the charges have energy due to the e.m.f. of the power supply.  Now tell me, if a lamp has resistance and you hook it in the circuit, how will the current stay the same? The charges obviously lose energy in the lamp and so become SLOWER, which should mean current decreases, right?  [Edit] All answers explained a bit of everything, so it was hard to choose one. If YOU are looking for an answer, please check the others too, in case the accepted one doesn't answer your question.