Just a simple question but trickier than I expected.  Let us suppose that there is a fly hovering in the moving train. The train moves at a speed much higher than the hovering of the fly. However, the fly moves with the train and does not hit its walls.  Could anyone explain the exact reason for this? Is it because of the air inside the train and the fly is very light? Then if the inside the train is a vacuum, devoid of air, would the fly hit the wall?
A couple of weeks ago I was travelling in a car (120 km/h approximately) and I saw a  flying in front of me (inside the car, near my nose, windows closed). I wonder how was that possible.   Does it mean is really flying at 120 km/h or the fly is being affected by some kind of gravity/force?
What space does the spell Cloud of Daggers fill? I am trying to workout what template  (PHB, p. 222) would be on a grid. The spell states:     You fill the air with spinning daggers in a cube 5 feet on each side, centered on a point you choose within range.   So does that mean it would look like this:     Or like this:    From what I've read online, there seems to be some suggestion that it is a single square (tile) though this does stack with the spell description of "a cube 5 feet on each side".  Clearing this up for my group would be great, and I am sure our DM will be pleased as his monster were nicely diced by this spell. Thanks. Why doesn't connecting a positive terminal of a battery to ground cause current? How to check if list item already exist when adding it with ViewParam vs @ManagedProperty(value = "#{param.id}") How do I create variable variables? My interest in combinatorics was recently sparked when I read about the many things that the Catalan numbers count, as found by Richard Stanley. I picked up a copy of Brualdi's Combinatorics, and while browsing the section on counting sequences I found a nice little puzzle that has definitely puzzled me.     Let $m$ and $n$ be nonnegative integers with $n\geq m$. There are $m+n$ people in line to get into a theater for which admission if $50$ cents. Of the $m+n$ people, $n$ have a $50$-cent piece and $m$ have a $\$ 1$ dollar bill. The box offices opens with an empty cash register. Show that the number of ways the people can line up so that change is available when needed is   $$ \frac{n-m+1}{n+1}\binom{m+n}{m}. $$     I first noted that the first person to enter must be one of the $n$ with a half-dollar. Now the register has a half-dollar change. The second person can be either a person with a half-dollar or a dollar. In the first case, the register will now have two half-dollars, in the second case, the register will now have one dollar bill. So it seems to me that when one of the $n$ people with a half-dollar enters, the number of half-dollars in the register increases by $1$, and when one of the $m$ people with a bill enters, the number of half-dollars decreases by $1$ but the number of bills increases by $1$.   I tried to model this by looking at paths in $\mathbb{Z}^2$. The $x$-axis is like the number of half-dollars, and the $y$-axis is the number of bills. You start at $(0,0)$, and you can take steps forward $(1,0)$ or backwards diagonally $(-1,1)$ corresponding to who enters, but you must always stay in the first quadrant of the plane without crossing over the axes. The goal is to make $m+n$ moves, and I figured maybe the number of such paths is counted by $\frac{n-m+1}{n+1}\binom{m+n}{m}$, but I'm not sure how to show this. I don't know if this observation simplifies the problem at all, as I don't know how to finish up. I'd be happy to see how this problem is done, thank you. Adverb between conjugated verb and infinitive? I'm using Terminator version 0.97. Is there a way that I can scroll unlimited? In the current version I only can scroll back limited history Proving $A+A'B=A+B$ without truth tables How can I prove the Boolean algebraic rule $$A+A'B=A+B$$ without using a truth table?  With the truth table, it is easy to see that the two are equal, but how can I prove it using lesser Boolean identities? How does Java's PreparedStatement work? I am planning to replace repeatedly executed Statement objects with PreparedStatement objects to improve performance. I am using arguments like the MySQL function now(), and string variables.   Most of the PreparedStatement queries I have seen contained constant values (like 10, and strings like "New York") as arguments used for the ? in the queries. How would I go about using functions like now(), and variables as arguments? Is it necessary to use the ?s in the queries instead of actual values? I am quite confounded. Is there a free program to (batch) change photo file's date to match EXIF? I want to specify a directory and have the software find all the photos in the directory and its sub-directories, and if they contain EXIF date/time, it sets their filesystem timestamp to match the EXIF. Regionalism or just bad English? I've encountered a particular type of writing occasionally and it being, derp, in writing, it's hard to tell whether there's an accent behind it. The English used seems to me to be simply incorrect, but I can't help but feel that the following features are part of a dialect or a regional style of English. One of the most glaring (and grating) features is the substitution of &quot;saw&quot; with &quot;seen&quot;.  He seen him at the shop the other day. I seen that you were looking for... etc.  The use of &quot;away (to)&quot; is also peculiar, if not incorrect.  He was away to open the door when... He was away to get what he had asked for. etc.  I haven't come across these forms very often, and even then it was accompanied by otherwise mostly correct English, which is what makes me wonder what the hell. This is more intellectual curiosity than anything else. less with "update file" like functionality I want to watch a file that gets overwritten every 5 minutes with less. How can I make less follow the new file descriptor instead of keeping the old one displayed? watch "cat file" won't do it because the file is too long to fit into one terminal window. VBA Hide Rows that contain N/A For any prime $p > 3$, why is $p^2-1$ always divisible by 24? When an account is deleted, the username in the user's posts and comments is changed to a generic user&lt;userID&gt;. But the username stays the same on close and delete notifications on posts the user had voted to close / delete1. I think it shouldn't, especially for users who used their real name as their username, if they chose to distance themselves from the site and remove their accounts, it makes sense that they wouldn't really want their name plastered on close and delete notifications.   A perhaps easy way to avoid the actual username remaining visible would be to change it to the generic user&lt;userID&gt; (or anything generic enough, like "anonymous") prior to deletion, and wait a bit for the generic username to propagate all over the site. If I'm not horribly misreading , this was part of the process in the old days.   This could be automated or added as a suggestion to the .  Glaring example can be seen , where the deleted user also posted a comment where their name is properly anonymized:        1 I've seen it on close notifications and self deleted answers. I'm not sure it happens on deleted questions, or deleted answers that aren't self deletions, and I don't have a way to verify it - other than stumbling on them accidentally Identity for convolution of central binomial coefficients: $\sum\limits_{k=0}^n \binom{2k}{k}\binom{2(n-k)}{n-k}=2^{2n}$ It's not difficult to show that  $$(1-z^2)^{-1/2}=\sum_{n=0}^\infty \binom{2n}{n}2^{-2n}z^{2n}$$  On the other hand, we have $(1-z^2)^{-1}=\sum z^{2n}$.  Squaring the first power series and comparing terms gives us  $$\sum_{k=0}^n \binom{2k}{k}\binom{2(n-k)}{n-k}2^{-2n}=1$$  that is,  $$\sum_{k=0}^n \binom{2k}{k}\binom{2(n-k)}{n-k}=2^{2n}$$  My question: is there a more direct, combinatorial proof of this identity?  I've been racking my brains trying to come up with one but I'm not having much success. Will the below code cause memory leak in c++ class someclass {};  class base {     int a;     int *pint;     someclass objsomeclass;     someclass* psomeclass; public:     base()     {         objsomeclass = someclass();         psomeclass = new someclass();         pint = new int();          throw "constructor failed";         a = 43;     } }  int main() {     base temp(); }   In the above code, the constructor throws. Which objects will be leaked, and how can the memory leaks be avoided?  int main() {     base *temp = new base(); }   How about in the above code? How can the memory leaks be avoided after the constructor throws? Checking if file is checked out through CSOM Is there a way to check if a file is checked out by another user? I am going trhough a list file by file and would like to avoid the "file is opened by another user" error. Something like this would be nice:  Guid id = new Guid(slideLibrary); // ID of Slide-Library. List list = ctx.Web.Lists.GetById(id); ListItemCollection items = list.GetItems(CamlQuery.CreateAllItemsQuery()); ctx.Load(items); // Loading all the fields System.Security.SecureString pwd = new System.Security.SecureString(); foreach (char c in "pwd".ToCharArray()) pwd.AppendChar(c);  ctx.Credentials = new SharePointOnlineCredentials(login, pwd); ctx.ExecuteQuery();  // Iterating through all the entries. foreach (var item in items) {     String currentID = item["ID"].ToString(); // Getting the ID of the current item.      if (!item.IsCheckedOut()) // This is the part I am looking for.     {        // Do stuff     }     else     {        Console.WriteLine("Item is blocked, skipping it.")     } }   I have yet not been able to find something fitting. Group theory proof of Euler's theorem ($a^{\phi(m)} \equiv 1\mbox{ }(\mbox{mod }m)$ if $\gcd(a,m)=1$) From A Classical Introduction to Modern Number Theory by Ireland and Rosen, page 33:     Corollary 1 (Euler's Theorem). If $(a,m) = 1$, then $a^{\phi(m)} \equiv 1\,(m)$.      Proof. The units in $\mathbb{Z}/m\mathbb{Z}$ form a group of order $\phi(m)$. If $(a,m) = 1$, $\bar{a}$ is a unit. Thus $\bar{a}^{\phi(m)} = \bar{1}$ or $a^{\phi(m)} \equiv 1\,(m)$.   If I'm interpreting this correctly, this proof implicitly uses the fact(?) that if $G$ is a group and $x\in G$, then $x^{|G|}=1$, where $1$ is the identity element of $G$.  If this is indeed true, can someone explain why? (I have had no prior exposure to group theory.)