a car weighs 100kg and has Velocity 0. it accelerates and its speed become 10m/s. its kinetic energy would be 100*10*10/2=5000j.  from another view car is at rest and the world accelerates and has now 10m/s speed.  as energy must be constant it should be:5000=10*10/2*(world mass except car) world mass except car=100kg?  am i using conservation of energy in wrong way, or i cant have that point of view?
If the velocity is a relative quantity, will it make inconsistent equations when applying it to the conservation of energy equations?  For example:  In the train moving at $V$ relative to ground, there is an object moving at $v$ relative to the frame in the same direction the frame moves. Observer on the ground calculates the object kinetic energy as $\frac{1}{2}m(v+V)^2$. However, another observer on the frame calculates the energy as $\frac{1}{2}mv^2$. When each of these equations is plugged into the conservation of energy, they will result in 2 distinct results (I think).
How to recursively download a folder via FTP on Linux I'm trying to ftp a folder using the command line ftp client, but so far I've only been able to use 'get' to get individual files. How do you prevent an ENTER key press from submitting a form in a web-based application? Does the  tag position in HTML affects performance of the webpage? A friend of mine asked me to calculate the value of a perticular determinant, but after spending quite some time on it I'm unable to find a solution to it. The given determinant is:  $$         \begin{vmatrix}         0 &amp; 1 &amp; \cdots &amp;n-2&amp;n-1 \\         n-1 &amp; 0 &amp; \cdots &amp;n-3&amp;n-2 \\         \vdots &amp; \vdots &amp; \ddots &amp; \vdots &amp;\vdots\\         1 &amp; 2 &amp; \cdots &amp; n-1 &amp;0 \\         \end{vmatrix} $$  Also my friend is supposed to solve this determinant by more or less elementary methods, like elementary row operations, transpose... and no eigenvalues, nor adjoints.    First of all it seems like the value of the determinant is:  $$         \frac{1}{2n}\begin{vmatrix}         0 &amp; n &amp; \cdots &amp;n&amp;n \\         n &amp; 0 &amp; \cdots &amp;n&amp;n \\         \vdots &amp; \vdots &amp; \ddots &amp; \vdots &amp;\vdots\\         n &amp; n &amp; \cdots &amp; n &amp;0 \\         \end{vmatrix} $$  Anyway I tried adding all the rows to the last one and by taking $\frac{n(n-1)}{2}$ in front to have a row of 1s, but it didn't helped me much.  Then I tried to use the fact that:  $$         \begin{vmatrix}         A &amp; C \\         0 &amp; B \\         \end{vmatrix} = \det(A) \cdot \det(B) $$, where $A,B$ are square matrices. By subtracting the $(\frac{n}{2} + i)$-th column from the $i$-th ($i=0,1,2...$) and then adding the $i$-th row to the $(\frac{n}{2} + i)$-th row, one can obtain the wanted form, but again nothing more. Also this method pretty much fails when $n$ is odd.  My best attepmt was to notice that $A_{ij} + A_{ji} = n$ and using $\det(A^{T}) = \det(A)$, but unfortunatelly  $$\begin{vmatrix}         0 &amp; n &amp; \cdots &amp;n&amp;n \\         n &amp; 0 &amp; \cdots &amp;n&amp;n \\         \vdots &amp; \vdots &amp; \ddots &amp; \vdots &amp;\vdots\\         n &amp; n &amp; \cdots &amp; n &amp;0 \\         \end{vmatrix} \not = \begin{vmatrix}         0 &amp; 1 &amp; \cdots &amp;n-2&amp;n-1 \\         n-1 &amp; 0 &amp; \cdots &amp;n-3&amp;n-2 \\         \vdots &amp; \vdots &amp; \ddots &amp; \vdots &amp;\vdots\\         1 &amp; 2 &amp; \cdots &amp; n-1 &amp;0 \\         \end{vmatrix} + \left( \begin{vmatrix}         0 &amp; 1 &amp; \cdots &amp;n-2&amp;n-1 \\         n-1 &amp; 0 &amp; \cdots &amp;n-3&amp;n-2 \\         \vdots &amp; \vdots &amp; \ddots &amp; \vdots &amp;\vdots\\         1 &amp; 2 &amp; \cdots &amp; n-1 &amp;0 \\         \end{vmatrix}\right)^T$$ I tried to decompose the LHS by using the $n-$linearity of the determinants, by then one gets $2^n$ determinants on the RHS similar to the first one, but things are messy. In the sentence: "Drinking and driving IS dangerous", why does the verb "to be" need to be singular? In the sentence: "Drinking and driving is dangerous". Drinking and driving are both subjects and the word dangerous is the object (I think) but why do I have to use the verb to be in singular? centering column headers in siunitx table column R: Trouble using SMOTE package "invalid 'labels'" Using SMOTE package from DMwR library. After loading the data frame, I try to perform sampling as follows :   crime_bal$target &lt;- as.factor(crime_bal$target) crime_bal &lt;- SMOTE(target ~ .,crime_bal,perc.under = 200, perc.over = 100)   But it always ends up to this error :   Error in factor(newCases[, a], levels = 1:nlevels(data[, a]), labels = levels(data[,  :    invalid 'labels'; length 0 should be 1 or 2 In addition: Warning messages: 1: NAs introduced by coercion  2: NAs introduced by coercion    Details of my dataset :   &gt; summary(crime_bal)      text               url            target    Length:6326        Length:6326        0:5994    Class :character   Class :character   1: 332    Mode  :character   Mode  :character   Why do I always end up with the error? I have a strange issue with Windows 10  When I open .nef files they load fine. However after a second or two the contrast changes noticeably. I tried Windows 10 Photos and Picasa Image Viewer.   What am I trying to accomplish? I just want to open files quickly to check for focus and composition before importing into Lightroom  Lightroom does not have this issue   Here is the example opening with Windows 10 Photos    And here is after a second when it 'fully' loads    Picasa Photo Viewer after it loads a bit. It initially opens fine but then it goes all purple What's the Pythonic way to combine two sequences into a dictionary? Is there a more concise way of doing this in Python?:  def toDict(keys, values):   d = dict()   for k,v in zip(keys, values):     d[k] = v    return d Apologies if this has been asked somewhere before, but I didn't see what I was looking for after several pages of Google results.  I was reading Jech's The Axiom of Choice and was introduced to the equivalence of the ultrafilter lemma and the compactness theorem for FOL. I gather that these are also equivalent to the completeness theorem. I am primarily interested in trying to understand the "ultrafilter lemma $\implies$ completeness/compactness" direction. Jech's own proof takes an odd route through "binary messes" and I don't really follow the last couple of steps. (It's on pp.17,18 of the 2008 Dover reprint, if one is interested).  I'm wondering if anyone knows a good reference for, or can provide, a readable proof that the ultrafilter lemma implies completeness or compactness, with a preference for completeness. Alternatively, if you have Jech's book on hand, helping me bridge the concluding paragraph of his $\textrm{(ii)}\implies\textrm{(iii)}$ step with the preceding steps would clear up much of my puzzlement. (Though in this case I can't offer any help in saying what, specifically, is missing in the way of a bridge). Proving $(\forall a\in\mathbb{Z^+})(m\in\mathbb{Z^+}\to a^m\equiv a^{m-\phi(m)}\pmod{m})$ Problem: $(\forall a\in\mathbb{Z^+})(m\in\mathbb{Z^+}\to a^m\equiv a^{m-\phi(m)}\pmod{m})$  My work: Start by letting $m=p_1^{a_1}p_2^{a_2}\cdots p_r^{a_r}$. If $(a,p_i)=1$ for some integer $i$, then we have that  $$ a^{\phi(p_i^{a_i})}\equiv 1\pmod{p_i^{a_i}} $$ by Euler's theorem. Since $\phi(p_i^{a_i}) \mid \phi(m)$, we have that $p_i^{a_i}\mid (a^{\phi(m)}-1)$ if $(a,p_i)=1$.   Beyond this, I am stuck at the moment. Drawing on an image with TikZ Recently I have learned how to draw simple stuff with TikZ. I really like it. Now I wonder if it is possible to draw with TikZ on a image. Instead of pasting a image in Illustrator and adding some vectors to point something in the picture I would like to do it using TikZ. Is it possible?   For example, on a picture of a computer (PNG,etc) I would like to draw some arrows and labels saying "Keyboard", "Monitor", etc. I know that new arXiv papers usually appear in Google Scholar within a few days or weeks. My question is: if someone submits a replacement of his/her arXiv paper, does the new version of the paper also appear in Google Scholar? I have to determine how many six-integer subsets of $\{1, 2, 3, \ldots, 20\}$ are possible if no two consecutive integers are in a set. A movie about artificial intelligence gone rogue Inertial frames of reference I'm struggling with the notion of an inertial frame of reference. I suspect my difficulty lies with the difference between Newtonian and relativistic inertial frames, but I can't see it.   I've read that Newton's laws apply in any non accelerating frame of reference, which are called inertial frames. So, if I play pool on a train moving with uniform velocity, the balls behave in the same way as if I were playing pool in a pool hall. So the train is (to a good approximation, ignoring forces caused by the Earth's rotation and movement around the Sun etc) an inertial frame.   I've also read that in an inertial frame, Newton's first law is satisfied. So, if I slide a rock on a sheet of ice and I could somehow eliminate the frictional forces between the rock and the ice, the rock would carry on sliding for ever, as predicted by Newton's first law.   Question 1 - is the  sheet of ice also therefore a good approximation to an inertial frame?  Question 2 - are these two definitions of inertial frames saying the same thing in different ways?  Both these examples occur in gravitation fields, which doesn't seem to matter as both the train and the ice sheet are presumably good approximations to inertial frames.  Question 3 - is gravity irrelevant when defining these two inertial frames?  In special relativity, I've read (Foster and Nightingale) that an inertial frame is also one where Newton's first law holds. But as it's special relativity there cannot be an inertial frame if there's a gravitational field, so I'm assuming the above two examples are not inertial frames in special relativity.  Question 4 - how come you can use the same definition of an inertial frame (satisfied Newton's first law) but in the case of special relativity, the train and the ice sheet aren't inertial frames. Is this to do with not being able to synchronize clocks in a gravitational field?  In general relativity, I've read (Schutz) that a freely falling frame is (on Earth) the only possible (and local) inertial frame. So again the first two examples would not be inertial frames in general relativity.  Question 5 - am I right to think that my two examples aren't even approximate approximations to an inertial frame in relativity?  Question 6 - have I missed anything else that might be useful?  Many thanks  Edit. On reflection, am I right in thinking that because Newtonian mechanics assumes universal, absolute time we don't need to worry about synchronizing clocks in a Newtonian inertial frame. Therefore we don't need to worry about gravity in a Newtonian inertial frame, because in such a frame gravity does not affect time.  This is not the case in spacetime, as here gravity does affect time and the only way to have synchronized clocks in an inertial frame in spacetime is: 1. have no gravitational field, or 2. use a local, freely falling frame.  Am I on the right track here? Keyboard shortcut/command for opening a new nautilus window and raise it What is the shortcut for opening a new window (opening folders). for that I have to go to folder options and right click and then need to click on open new window.  or I have to make focus on the folder icon or open a folder then press  Ctrl+N  for opening a new window is there any direct command spcific to open a new window in ubuntu. If integral is zero and function is continuous and non negative, then what about the function? In what kind of rings a divisor of a product is a product of divisors? In a , if $a|bc$, then $a$ can be written as $a = a_1a_2$ so that $a_1|b$ and $a_2|c$.  Is this property of a commutative ring strictly weaker than the property of being a UFD?  If so, does it have a name, or is it equivalent to some other well-known property? Why I can't remove linux partitions on macOS Sierra I'm trying to remove linux partition from my iMac (21.5-inch, Mid 2011) on MacOS Sierra 10.12 Beta 2, So I tried existing solutions by using Disk Utility, I tried from Recovery as well, But it says      You can not remove this volume because the previous volume can not be resized         Some  use Terminal but I have no idea about that.  I need a simplest and easiest way to remove these linux partitions and grow my mac partition.