if we do the double slit experiment but we shoot metal balls in a closed box so we don't observe. what pattern comes then on the board? if i understand correctly there will be no interference pattern because for bigger objects the wavelengths are too small. but do we see as a result then evenly divided points all over the wall or do we see a particle pattern in two places?
I read an interesting question here in the forum () and came up with the following doubt: even though its diffraction angle is too small to be detected, if we had the possibility to detect it, what would the diffraction of a soccer ball through a pair of posts look like? I mean, in the case of electrons, there is a screen in which the diffraction pattern can be appreciated. But what about this case? Would we see lots of soccer balls in different places or what?
Unexpected behavior of rule matching a pattern Rectangle with cross on title bar I have installed Ubuntu 14.04 and I think that I have found a bug. When I open terminal and  go to edit profile preferences I get a rectangle with a cross on the title bar. It happened with other applications too. Here are the screen shots: How to reconcile the cashless human/Federation society with human traders and merchants I am getting an 'access is denied' error when I attempt to delete a folder that is not empty. I used the following command in my attempt: os.remove("/folder_name").   What is the most effective way of removing/deleting a folder/directory that is not empty? Multiplicative inverses for $Z_n$ Does untapping an attacker stop its attack? Use Yaml for MongoMapper Config How do I erase an iPhone? I had an iPhone on loan from a friend to try it out and now I need to give it back. How do I make sure that everything is erased from it? SMS, Calendar, Phone Logs, contacts? The binomial sum   $$s_n=\binom{n+1}{0}+\binom{n}{1}+\binom{n-1}{2}+\cdots$$   satisfies the Fibonacci relation.   I failed to prove that $\binom{n-k+1}{k}=\binom{n-k}{k}+\binom{n-k-1}{k}$... Any hints or suggestions? class and private variables “between” vs. “among” I was wondering what is the purpose of piping the downloaded file by curl into sudo -E bash - in the following shell:   curl -sL https://deb.nodesource.com/setup_5.x | sudo -E bash - Note: TL;DR, scroll to bottom for summary. I recently set up my computer for a dual boot between Ubuntu 9.10 and Windows 7. My current drive setup is as follows.  | A1 |     A2       | | B1    |  B2  | B3 |        A1: 100 mb, windows 7 "System   Reserved" boot partition      A2: 230 gb, data section, this one   needs to be shared between the   operating systems      B1: 125 gb, windows 7 OS      B2: 123 gb, ubuntu OS      B3: 2gb, linux swap space     Pretty much I want to have my documents, music, pictures, videos, etc accessible from both operating systems. My first attempt involved making the data (a2) partition NTFS, and moving my home folder from ubuntu to the data partition. However, as I read NTFS does not work nice with permission, and it messed up my home folder. My next idea is one of the following: 1) format the data partition to ext2/3/4 and move my home folder from linux there, and get a driver to read ext partitions in windows 7. The problem with that is that most of the ext drivers/software are not compatible with windows 7 or do not integrate with windows explorer (I really don't want to open a separate software window just to access my data, plus it's probably not compatible with other software.)  looks promising, but I'm not sure how it works with ext4 and windows 7 (not officially supported, when trying in Vista Compatibility Mode, it tells me I need to format the ext drive to use it). My next idea, 2) keep the home folder in ubuntu where it is, but create symlinks for the Documents, Music, etc folders to an NTFS formatted Data (A2) partition, and add those locations to the windows 7 libraries. I'm not totally sure how the permissions would work out, but it should be fine since it's only the documents, music, etc and not the important config files in the rest of /home/user/. Correct me if I'm wrong. Currently, symlinks is my best idea, although i'm not sure how it will work.  Any suggestions, additions to my ideas, links, pointers, whatever would be greatly appreciated. Even if it means I should reformat both my drives and repartition (2 250gb drives if you want to suggest a setup for that), I won't be too opposed if that's the best suggestion (I've gone through the format/install/format/reinstall process 5 times over the past 3 days, once more won't hurt me).    TL;DR, summary:  I have two hard drives. One is partitioned for Ubuntu and Windows 7, the second one I want accessible to both operating systems to store documents, music, pictures, videos, etc. Suggestions on how to set up the data drive.    P.S. bonus if I can get an apache server document root folder working between the two OS's as well (permissions could become very complicated, so don't worry too much about that)  P.P.S. Related question, but data viewing is one way: In  (the original movie), Marty comes back from 1955 10 mins early to prevent Doc from being shot (which he fails at). He then rolls up to the Lone Pine Mall and sees himself hit 88 mph and shoot back to 1955 (again). Marty then carries on with this life as usual. What happens to the Marty that goes back to 1955 at the end of the film?  This obviously is a recursive problem where it would create another timeline and Marty would keep coming back 10 mins early to view another Marty zip back to 1955? Observe the following results for searching the text "dictionary" on Google:    Notice that there is a custom search box at the bottom of the listing for dictionary.reference.com. Is there any documentation floating around that explains the criterion by which Google decides to display such a search box? I encountered the integral $$\int_0^1 \frac{\log \left(1-x^2\right) \sin ^{-1}(x)^2}{x^2} \, dx = -0.9393323982...$$ while researching the evaluation of harmonic sums. Mathematica 11 is not able to evaluate this integral, and is not able to evaluate the underlying indefinite integral.  I considered using the Maclaurin series for the expression $\sin ^{-1}(x)^2$ to evaluate this integral. Using this Maclaurin series, it is easily seen that the problem of evaluating the above integral is equivalent to the problem of evaluating the series $$\sum _{n=1}^{\infty } \frac{2^{2 n-1} H_{n-\frac{1}{2}}}{(1-2 n) n^2 \binom{2 n}{n}}=-0.9393323982...$$ which Mathematica 11 is not able to evaluate directly.  I also considered using the Maclaurin series for the expression $\log \left(1-x^2\right)$ to evaluate the above integral. Using this Macluarin series, it is easily seen that this integral is equal to $$\frac{1}{4} \left(16 \pi  C-\pi ^3-\pi ^2 \log (4)\right)+\sum _{n=1}^{\infty } \frac{\, _3F_2\left(\frac{1}{2},\frac{1}{2},1;\frac{3}{2},n+1;1\right)}{n^2-2 n^3}=-0.939332...,$$ but Mathematica is unable to evaluate the above infinite series.  What is a closed-form evaluation of $\int_0^1 \frac{\log \left(1-x^2\right) \sin ^{-1}(x)^2}{x^2} \, dx$? What techniques can be applied to evaluate this definite integral? Proof that every finite dimensional normed vector space is complete Can you read my proof and tell me if it's correct? Thanks.  Let $V$ be a vector space over a complete topological field say $\mathbb R$ (or $\mathbb C$) with $\dim(V) = n$, base $e_i$ and norm $\|\cdot\|$. Let $v_k$ be a Cauchy sequence w.r.t. $\|\cdot\|$. Since any two norms on a finite dimensional space are equivalent, $\|\cdot\|$ is equivalent to the $l^1$-norm $\|\cdot\|_1$ which means that for some constant $C$, $\varepsilon &gt; 0$, $k,j$ large enough, $$ \varepsilon &gt; \|v_j - v_k\| \geq C \|v_j - v_k\|_1 e_i= C \sum_{i=1}^n |v_{ji} - v_{ki}| \geq |v_{ji} - v_{ki}|$$ for each $1 \leq i \leq n$. Hence $v_{ki}$ is a Cauchy sequence in $\mathbb R$ (or $\mathbb C$) for each $i$. $\mathbb R$ (or $\mathbb C$) is complete hence $v_i = \lim_{k \to \infty} v_{ki} $ is in $\mathbb R$ (or $\Bbb C$) for each $i$. Let $v = (v_1, \dots , v_n) = \sum_i v_i e_i$. Then $v$ is in $V$ and $\|v_k - v\| \to 0$:  Let $\varepsilon &gt; 0$. Then $$ \|v_k - v\| \leq C \|v_k - v\|_1 = C \sum_{i=1}^n |v_{ki} - v_i| \leq C^{'}n \varepsilon$$  for $k$ large enough. I just edited an answer on Unix&amp;Linux and when scrolling down, the semi-transparent right sidebar boxes "How to edit" and "How to format" kept their fixed positioning and overlapped with the page footer.     While posting this, I have seen several (not) duplicate questions about this issue, and although from my personal taste, I would prefer the boxes to stop scrolling before the footer is reached, I can accept the development teams decision to let it overlap.   However, on all those other posts, the boxes had an opaque background, so that the footer was hidden and the box content was readable. Here the boxes are transparent though, so the texts get mixed and are no longer well readable. The verb let means  “allow”, “permit”, “not prevent or forbid”, “pass, go or come”  and it's used with an object and the bare infinitive.         Are you going to let me drive or not?     Don't let him off the hook.     Before we let our children surf the Internet, they have to do their homework.     She lets the cats out before leaving.   He let us into the house.      The full form let us can be used similarly        Let us know as soon as possible.      Please, let us help you.   They will never let us forget.      None of the above can be contracted; however, when let us is used for making a suggestion; giving self-encouragement; expressing a consequence or plea,  it is often contracted to let's      Let's go out   Let's have a party   Let's see what happens   Let's stand together in this emergency   Let's not forget those who sacrificed their lives     Questions   I believe that let + us is the only instance where this type of contraction occurs. Is there an explanation as to why verbs such as (i) give + us, (ii) get + us or (iii) take + us were not similarly contracted? When was the apostrophe used to replace the omitted letter "u" in let us?  What is the origin of let's? Viruses: Adaptation to a new host through repeated host jumps