Is there a word for someone born in the UK?

Question: While watching the World Cup game a few minutes ago, I was wondering what if the United Kingdom - and not England - had a National Team, with English, Scottish, Welsh, and Northern Ireland players included. Is there a word we could use to refer to their nationality ? I'm not looking for the word "British". Instead I'm trying to find out whether there is a word derived from "United Kingdom".
Comment: Related (possible dupe):http://english.stackexchange.com/q/117993/8019
Answer: British :

British citizens usually hold this status through a connection with the United Kingdom, Channel Islands and Isle of Man ("United Kingdom and Islands"). Citizens of the United Kingdom and Colonies (CUKCs) who possessed right of abode under the Immigration Act 1971 through a connection with the UK and Islands generally became British citizens on 1 January 1983.
British citizenship is the most common type of British nationality, and the only one that automatically carries a right of abode in the UK.
Comment: You are answering the question with the question itself?
Comment: @BlessedGeek The adjective **is** British, despite it being rejected in the question.
Answer:
I'm trying to find out whether there is a word [for someone born in the UK] derived from "United Kingdom".

There is no single word for a native of the United Kingdom.
There is no word in common use that is derived linguistically from the words united and kingdom and which indicates a person or persons born within the corresponding political territory.
The most common word is certainly Briton. As you note it is derived from what is now the name of a geographic territory not from the name of a current political territory.
Briton means an inhabitant of the island named Great Britain. At least some people born in Northern Ireland might not consider themselves Britons. They'd be right geographically.
You can be an inhabitant (and therefore a Briton), or even a full citizen, without being born there.
Some might argue that "United Kingdom" is a contraction of "The United Kingdom of Great Britain and Northern Ireland" and that "Briton" is therefore derived from one part of that name. That would be incorrect I think.
If this imprecision makes you unhappy, You have to use a longer phrase to express what you want.
As for national team. I'm not sure that the people concerned would all consider the UK to be a nation anyway - the word is mostly used for nations such as England, Wales, Scotland, Ireland and perhaps for subsets of at least the latter.
Comment: "You can be an inhabitant (and therefore a Briton) ... without being born there."

I'm not sure that's quite right: people from other countries living in Britain wouldn't normally be referred to as _Britons_, I don't think. To me it implies birth, citizenship, or both.
Comment: Sorry to be so disagreeable :-), but you say: "As for national team. I'm not sure that the people concerned would all consider the UK to be a nation anyway - the word is mostly used for nations such as England, Wales, Scotland, Ireland". The UK is the _nation_. The others are _countries_ within it. That's my understanding of the legal status.

Football is different, of course.
Comment: @Martin: Don't worry about disagreeing with me :-). The terms are somewhat poorly defined. Many would say the UK is a state, not a nation. Regarding legal status of UK as nation, is that Scottish law or the Law of England and Wales, or elsewhere? ;) Which acts of law define nation?
Comment: "and that "Briton" is therefore derived from one part of that name." Or, it could be derived from "Britain", which is a correct short form of the United Kingdom.
Answer: The OED1 entry for Briton seems apposite:

A. sb. 1, A native of Britain : ... c.  Since the union of
  England and Scotland: A native of Great Britain, or of the
  British Empire; much used in the 18th c.; now chiefly in poetic, rhetorical, or melodramatic use, and in phrases dating to the 'Rule Britannia' period...


Pytest on Python Tools for visual studio

Question: Can debug python tests which are using pytest library on visual studio 2010 ? I added the -m pytest on the Interpreter arguments but the breakpoints are not hit, I can only run the test script without debugging it.
Answer: Pytest recommends creating a standalone startup script.
<code>if __name__ == '__main__':
    import pytest
    pytest.main()
</code>
This is what I do and PTVS debugging works for me.
Answer: It would be great if you can be more specific. When I used pytools I found it very useful and found great help at http://pytools.codeplex.com. You can check http://pytools.codeplex.com/discussions.
Also check Can I debug with python debugger when using py.test somehow? or Python Unit Testing: Automatically Running the Debugger when a test fails for help as they have asked something similar.
Comment: Thanks for the answer fscore.I need to debug pytest tests using python tools for visual studio and not python debugger(pdb). On visual studio, on project properties->debug->interpreter arguments I added "-m pytest" so the test script will be run as "python.exe -m pytest test_scripts.py". I added the breakpoints on the test script, I run it using "start with debugging" but the breakpoints are not hit, the test is just run without debug.
Comment: Re: The codeplex links. "This site, including the discussions and issues, are no longer actively monitored.". They've gone to Github, sadly.


Orchestrating multiple databases

Question: We have a project where our hosting clients are offered a database (MySql) as part of their plan and I was curious how would one orchestrate this. We want to run the databases on different machines to where the web servers run.
I was thinking on using Kubernetes with the new StatefulSet feature but I'm really intrigued by the way compose.io is doing it's job. They seem to guarantee iops and CPU with the limitations being applied on RAM and Disk, as with all the databases they offer. One way would be in Kubernetes to allocate a limited persistent volume and running the container with a limited RAM without anything applied on CPU but that will create a highly unstable cluster as containers/pods start eating CPU and with Kubernetes constantly shifting them on nodes where the required resources are found. Couple that with persistent volumes which slower things down since the pod must be killed before attaching the disk elsewhere and you might get a lot of downtime for customers.
I'm just curious, what is your guess on how they are orchestrating their databases.
Answer: WARNING: State is very hard inside Kubernetes, do more homework than this answer before putting anything even near production!
Each scenario is different, but some best practices are:

Mutli-instance cluster using StatefulSets
Each pod in the set gets its own volume for data
Inter-pod Anti-affinity

Orchestrating multiple databases


Silverlight on Linux

Question: did anyone try to view a Silverlight site on a Linux machine? afaik, there is something called mono on Linux which is a try to port .NET on Linux machines, but did one really try to use it? did it work?
Answer: Moonlight is the mono implementation of SilverLight.  http://www.mono-project.com/Moonlight
In an interview, they said Microsoft gave them the same test suite they use on Silverlight, so the implementation is pretty good I would suppose for being a version behind http://www.hanselminutes.com/default.aspx?showID=157
Answer: Currently Moonlight, the implementation of Silverlight on Linux, is in Beta. According to their roadmap, the final version of Moonlight 1.0 should be out on January 20, 2009. As stated by MrChrister, they're a "release behind", because we already have Silverlight 2. However, according to the same roadmap, they plan to catch up sometime next year. Until then, I don't think Silverlight on Linux is an option to consider.

Silverlight on Linux


How come islanders don't have knowledge to create titan serum even though Reiss family have them?

Question: How come islanders don't know to create titan serum even though the Reiss family have them?
We know that Rod had titan serum to convert someone into a pure titan in the episode where he himself is converted to a giant caterpillar-like titan. But in chapter 107 visitor, it is implied that Islanders don't have the knowledge to make titan serum and they only have those which were gifted to them by visitors. Clearly, someone on the island knows how to make them. Then why can't they find that someone probably in the government who had ties to Rod? Since they overthrow his government, they should know about all his secret technology.
Answer: This might be the knowledge that only the Fritz family knew about, which Grisha Jager killed in the first place. or maybe, only the wielder of the founding titan from a royal bloodline knows how to make it. As you might know, every royal who inherits the founding titan also inherits the previous wielder's memories and king Fritz's will of peace.


How to rotate North oriented Map

Question: I want to rotate my North oriented point shapefile to 45%, is there any way to rotate it QGIS or MapInfo?
Comment: Do you wish to print a rotated Map, or do you wish to rotate the actual data itself?
Comment: i want to rotate the actual data itself.
Comment: Hm, well the question yo claim this is a duplicate of, only asks about QGIS. This question asks about QGIS or MapInfo, so I guess it's not a complete duplicate
Answer: In QGis you need to use the Affine Plugin (plugins->Fetch Python Plugin) or you could do it using PostGIS.
Have a look at this thread for further description on correct usage.
Comment: This doesn't rotate the data, only the Map in the print.
Comment: Editted answer to reflect that the OP wants to rotate the data and not the map.
Answer: In MapInfo Professional you can rotate selected objects by any angel around a specific coordinate.

Make the alyer editable
Select the objects you want to rotate
Select Objects > Rotate and enter the angle and a coordinate you want to rotate the data around
Hit OK

If you data currently is stored in a shape file, you need to save it into a tab file to be able to edit it.

How to rotate North oriented Map


Can you miss a Job?

Question: More than a few jobs are part of sub-scenarios. If I finish a Chapter without doing any sub scenarios do I miss those jobs forever (well until NG+)?
Answer: Some of the Side-quests will expire eventually and you will miss out on those jobs.
Comment: Which ones can be missed?
Comment: @Sterno Sorry, I dont feel like doing a run through the game without doing any sidequests just to find out (that sounds awful being stuck with so few jobs), and its not available online yet.
Comment: How do you know they expire then? Is it kind of on a per-chapter basis or something, like when you move to the next chapter, they're gone? The key information I think this question is really after is how to be sure you won't miss any jobs.
Comment: Yes they will only disappear when you move between chapters. As long as you complete them before the end of the chapter its fine.
Comment: The quest might disappear but Jobs won't. There are quests that change in nature over time...  the most important rewards (i.e. Jobs) are obtainable from other quests later on in the game. You might miss out on some item rewards like equipment though.
Answer: No Jobs are truly miss-able, but they get much harder to get later on as you progress towards the end of the game. I.e., Some item steals and drops can get quite limited and you might find yourself unable to physically win the battles towards the end of the second to last Chapter (unless you Summon a high/max level wrecking ball of a player to your aid).
Be on the look out for new quest later on in the Chapter. E.g., There are two Jobs that you can't get before the end of Chapter Four. You will need to beat the Fire Temple Before you can unlock the Ninja; The Pirate jobs involves talking to Datz (in Eisenburg) about a ghost ship and going out to sea to look for a foggy area.
It's worth noting that the Performer Job in Chapter Three can be miss-able early on, since it has a time limit, but it becomes available again in Chapter Five, if you missed it. There's also the Vampire Genome abilities, some of which are only accessible right after unlocking the Vampire Job in Chapter Four and don't become available until the end of the final final chapter. No content or items in the game re truly miss-able, since you can go back via New Game+.
The game is somewhat passive-aggressive about telling you to do all the quests. A general rule to help you not miss a majority of the content is to:

Do all available side quests (blue marker) before going about doing the main quest
Steal from all unique enemies you encounter

As a side note, you will need to do at least 3 play through in order to get all Event Viewer entries. Getting all the pages for the Encyclopedia might take more though.

Can you miss a Job?


What is the flow of electrons, cations, and anions in a voltaic cell (please see image attached)?

Question:
This is an IB question. The mark scheme says that the correct answer is A and only A. But I believe B is also correct. The following is my thought process. Since A is more reactive, it undergoes oxidation to form A2+ ions, while B2+ ions undergo reduction and form B. From this we can deduce that A is the anode and B is the cathode. According to my notes and many sources on the internet, electrons and cations both travel from the anode (A in the image) to the cathode (B in the image). Therefore, A and B would both be correct. Please can someone confirm my thinking or point out any mistakes. Thank you
Answer:
According to my notes and many sources on the internet, electrons and cations both travel from the anode (A in the image) to the cathode (B in the image).

The idea of the salt bridge is to prevent electrolytes mixing while providing ion flow. When you have a high concentration of inert ions in the salt bridge, cations in the salt bridge will flow into B, and anions in the salt bridge will flow into A. The solution in the salt bridge is not mixed, so it would take a very long time for the electrolyte in A to diffuse all the way to B.
To conclude, ions flow out of the salt bridge into the solutions A and B, but virtually no ion flows all the way through the salt bridge. This tells you that the salt bridge will be depleted after a certain amount of current flows through the wire. Accordingly, electrochemical cells used in batteries usually don't have a salt bridge but a semipermeable membrane instead, or a clever choice of half reactions where mixing of the electrolytes is not a problem (lead acid battery for example, where the only soluble species, sulfuric acid, is common to both half reactions).
Answer: It is a bad question because "Metal A is more reactive than Metal B" has no real meaning. It is a vague choice of word, A is more reactive, but with respect to what?
Your logic is correct

Since A is more reactive, it undergoes oxidation to form A2+ ions,
  while B2+ ions undergo reduction and form B. From this we can deduce
  that A is the anode and B is the cathode.

All you have to do now is to deduce the sign of A and B. Since A is losing electrons, where do these electrons go? They impart a negative charge to the electrode A. Electricity flows from the negative terminal to the positive terminal, if we take electrons as the charge carriers.*

In older books or electrical engineering you might see something called as the conventional current. The direction of the conventional current is from the positive pole to the negative pole and the charge carrier is considered to be a positive entity.
Comment: The conventional current is still from positive to negative potential, regardless of awareness of the opposite electron current in metallic conductors. And, in case of LHC, the current is based on positive entities.
Comment: The word "reactive" to me is opposite of "noble". In my experience with analyzing corrosion issues, zinc is more reactive than carbon steel which is more reactive than copper which is more reactive than stainless steel which is one of the most "noble" (i.e. "unreactive") metals. Zinc, carbon steel, copper, and stainless steel are all members of the "[Galvanic Series](https://en.wikipedia.org/wiki/Galvanic_series)"; this is an ordered list of metals sorted by the equilibrium cell potentials (a.k.a. "electromotive force" (deprecated); units are volts) of their half reactions.
Answer: It is a bit of a tricky question. The trick is in the wording "flow through". The positive ions on the A side do not actually "flow through" the salt bridge from A to B. Rather positive ions exit the salt bridge on the B side to compensate for the cations depositing on the cathode (B). Likewise negative ions exit the salt bridge on the A side to balance the charge of the cations being liberated from the anode (A).
Some explanations also show ions moving into the salt bridge:
Anion moving in.
Cation and anion moving in.


What are the limits of allocating my CPU cores to VMs?

Question: At least in VMware when creating or tweaking the settings of a VM I choose how many CPU cores I let the VM use. This question came to mind when thinking about at some point potentially running multiple VMs at once. Is there a limit to how many VMs I can run based off of how many CPU cores are allocated to other VMs or my main system? For example I have 6 cores, would I only be able to run 5 VMs; each one using 5 cores leaving one for my system? Or can I run more than 5 VMs and they'll just share cores at an obvious cost to performance?
Comment: You can overprovision your CPU allowing you to assign more vCPUs than you have.  Too much though and you'll see a noticeable performance hit on the host and VMs.  However, you'll generally run out of RAM or have disk IO issues well before your CPU is overtaxed.
Answer: If you're using vSphere (which I assume you are as you tagged vmware, when you allocate 1 CPU core you're allocating 1 vCPU core. The CPU does not bind to a physical CPU core (pCPU).

For vSphere 6.0, there is a maximum of 32 vCPUs per physical core, and vSphere administrators can allocate up to 4,096 vCPUs to virtual machines on a single host, although the actual achievable number of vCPUs per core depends on the workload and specifics of the hardware.

For every workload beyond a 1:1 vCPU to pCPU ratio to get processor time, the vSphere hypervisor must invoke processor scheduling to distribute processor time to virtual machines that need it. Therefore, if the vSphere administrator has created a 5:1 vCPU to pCPU ratio, each processor is supporting five vCPUs. The higher the ratio becomes, the higher the performance impact will be, because you have to account for the length of time that a virtual machine has to wait for physical processors to become available. The metric that is by far the most useful when looking at CPU oversubscription, and when determining how long virtual machines have to wait for processor time, is CPU Ready.

The vCPU-to-pCPU ratio to aim to achieve in your design depends upon the application you are virtualizing. In the absence of any empirical data, which is generally the case on a heterogeneous cloud platform, it is a good practice, through the use of templates and blueprints, to encourage your service consumers to start with a single vCPU and scale out when it is necessary. While multiple vCPUs are great for workloads that support parallelization, this is counterproductive in the case for applications that do not have built in multi-threaded structures. Therefore, while a virtual machine with 4 vCPUs will require the hypervisor to wait for 4 pCPUs to become available, on a particularly busy ESXi host with other virtual machines, this could take significantly longer than if the VM in question only had a single vCPU. This performance impact is further extended as the vSphere ESXi scheduling mechanism prefers to use the same vCPU-to-pCPU mapping to boost performance through CPU caching on the socket.

https://www.vmware.com/pdf/Perf_Best_Practices_vSphere5.5.pdf
You can also search for 'virtual machine cpu overcommit' for more results.
Answer: I find a good rule of thumb if using VMware Workstation is 1 CPU with two cores. This works well for Windows Guests and Linux Guests.
You are not limited by the number of cores you have as to the number of machines. My laptop has 4 cores and I run a Windows 10 Host with 3 guests: Windows 7, Kali, and Ubuntu.
I assign 3 GB of memory to each machine and have 16GB of memory total.
The biggest performance boost I get is from a very fast NVMe SSD.


How can I slim down all my bloated photoshop files at one time?

Question: I have a massive 8 TB worth of 15 years of Photoshop Work. In recent years I have done better at organizing my photoshop layers and deleting unused layers, but is there a solution that would allow me to go back and mass delete empty layers in photoshop of every file on my server. Perhaps a script.
Comment: an empty layer isn't likely adding much data to your files. In other words, I don't think empty layers are your issue.
Comment: Is it just PURE Photoshop files?  Are you including source files and fonts?  I wouldn't trust a script enough to go in and modify layers.  I would see if you can consolidate the work in regards to the typography, origs, supplied, etc. etc.  If you have a finished file that is several years old I see no logical purpose in keeping the client's originals.  I would just keep the source files and design files.
Comment: What @DA01 said. I'd assume an empty layer probably doesn't use more than a few KBs. You'd probably gain a lot of space by deleting hidden layers, but that's incredibly dangerous and probably not something you can do in a batch.
Comment: Also, calculate the cost in time you'd spend doing this vs. the cost of a few hard drives.
Comment: Yes your right @DA01 a 8 TB drive costs $260 so given that you may want to have backup also that would be still ~600  (to a thousand dollars). On the other hand how much would you free up things? 10% so not much worth doing anything for a 60$ retuyrn on investment.
Answer: Just make a batch action from a default PS script that deletes all the empty layers.
The script can be accessed here:
File › Scripts › Delete all empty layers


To display dashboard for sales users

Question: I have created a dashboard with 6 component, now i want this dashboard as the default dashboard in dashboard tab for the sales people.  How can i make this changes in the users production?
Answer: There is no way to assign a single dashboard as a default dashboard to a specific profile ( assuming the sales users are all on one profile).
The first dashboard on the dashboard list will be selected as default for the user. You can login as the user ( if you are a sys admin and have login as (ie) modify all data access) and customize the list to point to the default dashboard you want the user to see (or) send a document out to your sales users on how to choose a default dashboard if there are too many users in your org.
Up vote this idea if you want to support this feature :
https://success.salesforce.com/ideaview?id=08730000000BrcUAAS
Update: How to login as other user:
1) Go to your profile. Click on profile name and search for "modify all data", see if the flag is enabled for your profile, if it is you can login as other user (or) you need to set up delegate user.

2) How to customize the dashboard view:
Once you login as the user, go to the homepage and click on customize page.

Comment: I don't see the login as option in production, there is not much users, so thought of going and changing them individually, but only see the login as option in sandbox.  Also how to make this dashboard as the first dashboard, so that it will be default for the user.
Comment: @user8132 look for updated answer
Comment: thanks for the info, was very helpful.  Modify all data is checked still i don;t see login near the names, came across this https://help.salesforce.com/HTViewSolution?id=000089838&language=en_US.
Comment: created a ticket with sf for permission
Comment: @user8132 if its not too much trouble can you mark this solved, helps others and answer ratio too thanks
Answer: @Rao
Its an old post, but still, if anyone is here to get some knowledge, "login as user" facility cant be given only by having Modify all access (as told by Rao). YOu need to go to: Settings|Security Controls|Login Access Policies... There u will have to check the flag: Administrators Can Log in as Any User

To display dashboard for sales users


Solving forced harmonic oscillator differential equation using fourier transform

Question: I am trying to solve the equation of a forced harmonic oscillator using Fourier Transform. I know that if a function $f(t)$ is such that $\lim_{x->\pm \infty} f(t) = 0$, then
$$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \frac{df}{dt}\exp(i\omega t)dt = -i\omega \tilde{f}(\omega)$$
Now if we apply this to the equation of motion of forced harmonic oscillator
$$\ddot{x} + \omega_0^2x = F(x).$$
Then applying Fourier transform to the above equation and assuming that $\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} x(t) \exp(i \omega_0 t) dt = g(\omega)$, then in Fourier transform of $\ddot{x}(t)$ I am stuck here
$$\mathscr{F}[\ddot{x}(t)] = \frac{1}{\sqrt{2\pi}} \left[ (\dot{x}(t) \exp(i \omega t))\vert^{\infty}_{0}+ i \omega (x(t) \exp(i \omega t))\vert^{\infty}_{0} - \omega^2g(\omega)\right
]$$
Now we can say that $x(t = 0) = 0$ and $\dot{x}(t = 0) = 0$. But how can I argue the same at $t=\infty$?
Comment: This question has a close vote for violating the [homework policy](https://physics.meta.stackexchange.com/questions/714/how-do-i-ask-homework-questions-on-physics-stack-exchange). While this is true at the surface, the question has a clearly recognizable conceptual core (and the concrete computation can be seen as just an illustration of this) – @ShubhamDas feel free to edit the question to highlight the conceptual problem to prevent the question being closed.
Comment: Furthermore, while the conceptual problem being a maths problem, I don't feel Math.SE would be a good home for this question as it is about the pragmatic "physicst's way" of doing Fourier transformation in cases where the integrals don't converge.
Answer: It is often helpful to add a regularization factor $e^{-\epsilon \left| t \right|}$ to the integrands in the Fourier transformation to ensure convergence. You then take the limit $\epsilon \to 0$ after computing the integral. This can be understood as approximating the problem by a similar one, that differs from the real one only far away from $t = 0$.
This procedure prevents problems with oscillating terms at infinity. Rigorously the integrals don't exist otherwise (but physicists typically don't care as the regularization procedure gives consistent results).
There are several ways you can understand this procedure in a rigorous way (and why our "arbitrary" regularization gives the correct result). If you have some background about distributions this Math.SE post offers an explanation.
Answer: Without any claiming to be a complete answer with all the details you need, you could start using the (unilateral) Laplace transform
\begin{equation}
  F(s) := \int_{t=0}^{\infty} f(t) e^{-st} dt \ .
\end{equation}
If the system is stable, the solution of the homogeneous equation containing the effects of the initial condition is a decaying transient solution, so it goes to zero. Then, you can get the forced solution setting $s = j \omega$ in the particular solution of the forced system.
Laplace transform of the equation
\begin{equation}
 m \ddot{x} + c \dot{x} + k x = f(t) \ ,
\end{equation}
reads
\begin{equation}
  [ s^2 m + s c + k ]X(s) = F(s) + s m x_0 + c x_0 + m v_0
\end{equation}
with initial conditions $x(0) = x_0$, $\dot{x}(0) = v_0$, and the solution in the equation reads
\begin{equation}
  X(s) = \dfrac{F(s)}{s^2 m + s c + k } + \dfrac{s m x_0 + c x_0 + m v_0}{s^2 m + s c + k } \ ,
\end{equation}
being the first the solution of the forced equation, and the second one the (decaying) solution of the homogeneous equation with the prescribed initial conditions.
So, you get the harmonic solution after the transient is over as
\begin{equation}
  X(j \omega) =  \dfrac{1}{-\omega^2 m + j\omega c + k } F(j \omega) \ .
\end{equation}
Some details about Fourier transform of non-decaying signals. You could either use a decaying regularization factor on the forcing, or cut the forcing after the time interval of your interest, to avoid troubles with the infinity. Namely, you can set your forcing as
\begin{equation}
 \widetilde{f}(t) = \begin{cases}
f(t) \quad & , \quad t < T \\
0    \quad & , \quad t \ge T
\end{cases}
\end{equation}


Tiny unreadable text on everything after install of Debian 9 to mini pc

Question: I have a special pc, a thin client that had a small hard drive! It is a HP-G56N. I love mini pcs. It successfully installed, but upon boot the text is infinitesamally small and unreadable. Looks like little ant tracks. :) why are the default settings like this? Maybe a pc hiccup or had a bad install image, or maybe the hardware being niche caused it?
Answer: Adding the answer i found to help someone else on the Internet with a similar issue.
I found weird bug with xfce4: menu panel font very small which mentioned a custom dpi value. I didn't realize right off that this was related to font size but i unchecked the "use custom DPI setting" checkbox which was checked by default for some reason and text all went back to normal and i sized my text for my 32" tv monitor.
I bet this problem was related to the resolution of a tv as a monitor so use custom dpi was auto calculated based on the tv possibly?
Comment: Note! I also had xfce4 desktop and tools installed and i found this option in an xfce tool


Unzip a zipped file using php

Question: Lets start by looking at what I am trying to accomplish.
I work for a web bureau, and we use almost the same CMS for every website. The CMS contains over 6000 files, so it takes a long time to upload. It is much faster to upload a zipped file, and then unzip it on the FTP server.
I browsed stackoverflow, and found this thread: Unzip a file with php
This is fine. However, when i use this code:
<?php
// assuming file.zip is in the same directory as the executing script.
$file = 'concrete184.108.40.206-Dansk.zip';

// get the absolute path to $file
$path = pathinfo(realpath($file), PATHINFO_DIRNAME);

$zip = new ZipArchive;
$res = $zip->open($file);
if ($res === TRUE) {
  // extract it to the path we determined above
  $zip->extractTo($path);
  $zip->close();
  echo "WOOT! $file extracted to $path";
} else {
  echo "Doh! I couldn't open $file";
}
?>

I get the following problem:
It unzips the entire .zip file. This is including the folder that was originally compressed. Which will create my index.php file, one directory too deep.
So instead of this:
    www.hello.com/index.php
It will come out as:
    www.hello.com/folder/index.php
That was problem #1.
Here is problem #2:
I work on mac, and when I unzip my zipped file, I get the annoying "_macosx" folder too.
Is there a way to fix both of these issues? :)
Answer: Very simple method i am using
system('unzip assets_04_02_2015.zip');

Unzip a zipped file using php


What word describes writing a word someone else said while trying to write something else?

Question: I was trying to write "diagram" and someone said "poop" and I ended up trying to write "diapoop."  I quickly caught myself, but I've always wanted to know what word is used for this phenomenon, if there is a word for it at all.
Comment: I would call it a false portmanteau writing lapsus based on an aural stimulus.
Comment: An aural mis-portmanteau?
Comment: Ah laddie, a piggy-backer, eh? :)
Comment: Oooh, good one.
Comment: I don't think there is a single word for it, but someone should invent one! Portapsralus comes to mind... not sure about the sonic resonances though. Perhaps there's some Latin or Greek work (if we're being all boring and formal about this -- something I sometimes do) meaning something similar which could be adapted??
Comment: I was thinking about using Japanese, because hey, [we got a lot of words from them](https://en.wikipedia.org/wiki/List_of_English_words_of_Japanese_origin), and it wouldn't hurt to have another one.  Don't see how we would go about figuring that out, or if Japanese is a reasonable choice.
Answer: I believe what you are thinking of is commonly referred to as a Freudian Slip, also known as Parapraxis. Now, Freudian Slips are normally considered to encompass repressed or subconscious thoughts. Freudian Slips may also be due to the existence of some locally appropriate response pattern. In your example, someone said poop, and you started to write what you heard. Your response of including it in your writing could be due to a few things.
1) You could have been easily distracted at the time.
2) Hearing the word "poop" aloud may have been funny to you and in mentally processing the word, you physically wrote it.
3) You may have also been harboring a repressed detachment or disliking regarding the diagram and thus, appropriately wrote what you thought of it.

Freudian Slip - IntroPsych.com
Freud believed errors of all types
  were revealing. Like defense mechanisms, errors come in many
  varieties. Freudian slips are errors of language such as word
  substitutions and mispronunciations. Sometimes they involve sex, but
  that is not part of the definition. A Freudian slip is defined as any
  language error that is unintentionally revealing. Source Link:
  http://www.intropsych.com/ch13_therapies/freudian_slips_and_other_errors.html

Freudian Slip - Wikipedia.com
A Freudian slip, also called parapraxis, is an error in speech, memory, or physical action that is interpreted as occurring
  due to the interference of an unconscious ("dynamically repressed")
  subdued wish or internal train of thought.
Classical examples of parapraxes involve slips of the tongue and of
  the pen, but psychoanalytic theory also embraces misreadings,
  mishearings, temporary forgettings, and the mislaying and losing of
  objects.
In contrast to psychoanalytic theorists, cognitive psychologists say
  that linguistic slips can represent a sequencing conflict in grammar
  production. From this perspective, slips may be due to cognitive
  underspecification that can take a variety of forms – inattention,
  incomplete sense data or insufficient knowledge. Secondly, they may be
  due to the existence of some locally appropriate response pattern that
  is strongly primed by its prior usage, recent activation or emotional
  change or by the situation calling conditions.
Answer: This might be a bit more general than you want, but:
slip of the pen:

An inadvertent mistake made while writing, usually the substitution of one letter, word, or phrase for the correct one. Also called a... lapsus calami.
A slip of the pen is regarded as a type of heterophemy, the unintentional use of an incorrect or inappropriate word or phrase in speech or writing.
The expression slip of the pen dates from the mid-17th century.
(here)

Heterophemy is defined as "the (unintentional) use of some other word or phrase in place of the one that was meant" (here).
A heterophemism is presumably an instance of the general tendency.
Answer: I wouldn't call it dyatypesia as a similar "phenomenon" often occurs in speech... similar to the combination of two words which you are trying to say that come out at the same time. I feel like they would be related.
Answer: This is sometimes called a malapropism.

A malapropism (also called a malaprop or Dogberryism) is the use of an
  incorrect word in place of a word with a similar sound, resulting in a
  nonsensical, often humorous utterance. An example is the statement by
  baseball player Yogi Berra, "Texas has a lot of electrical votes,"
  rather than "electoral votes". Malapropisms also occur as errors
  in natural speech and are often the subject of media attention,
  especially when made by politicians or other prominent individuals.

The newspaper comic script Crankshaft, for example, has a running gag of its title character's malapropisms.


Ubuntu 11.10 - computer with no HDD

Question: I'm new to Linux and Ubuntu [less than a month using Ubuntu 11.10].
this old machine i have:
<code>amd athlon x2 64, 4gb ram
no hdd
no cd/dvd drive
usb pen 4gb with live cd [Ubuntu 11.10 i386]
usb pen 8gb --- intended as hdd
sd card 2gb --- with Ubuntu 11.10 i386 iso file
fast internet 25 mib/s
</code>
what im trying to do is:

start computer with 4GB pen Ubuntu 11.10
install Ubuntu in 8GB pen [not create a live cd]
work from the 8GB pen with Ubuntu installed in it.

is this possible ?
if so, any tips how i go to do this ?
Answer: As found here in this previous post:
How can I save settings on a Live USB?
You could use the USB stick as the boot device and even save documents and settings on the USB.
The article above references this entry from the Ubuntu wiki that should be helpful:
https://wiki.ubuntu.com/LiveUsbPendrivePersistent
Answer: 2016 Ubuntu 11.10 is no longer supported, but a normal install should work for all versions, as long as you unplug your internal HDD first.
To customize the install to suit pendrives, (optional). When installation gets to Partitioning select "Something else".
Leave the first partition FAT32 so items in that partition can be accessed by Linux or Windows.
Make the second partition ext4 and select "/" for mount point. This is where the system will be installed.
You can optionally make a third partition ext2 or ext4 and set mount point "/home". This can save the data in your home folder during upgrades.
You can make a fourth partition as swap but many people don't use swap on flash drives as they are afraid of making too many writes to the pendrive,
Pendrives are normally good for over 10000 writes, at 50MB/s, this is a long time, years or decades depending on pendrive size. pendrives also use wear leveling to distribute writes over the drive. if it ever does wear out it will turn read only and data will not be lost.
After partitioning continue with the install.
If you are OK with a persistent install try mkusb, it will make a Live USB install with persistent partitions. left over space is turned into a NTFS first partition that Linux and Windows can both use.
See Running an install from a flash drive for comparison between Persistent and Full install.

Ubuntu 11.10 - computer with no HDD


Can you explain 'but that's beside the point'?

Question: Does it mean, but that is not what I am trying to point out? That is only a trivia, what I am trying to say is something else.
Comment: If you type in 'beside the point' as a search in this site, you will find lots of usages of it. Look as some of them, and if you have not got a clear idea of the meaning after that, cut one out and paste it into your question.
Answer: "Beside the point" is an idiom meaning that what the person just said is irrelevant, out of context, or not important to the topic being discussed.
Example:
A: Those tacos have red shells. Yuck!
B: That's beside the point. I ate the tacos earlier and they taste the same, even though their shells are a weird color.
Comment: I don't think your example really fits.  Usually, the expression is used in an argument when one person brings up something that is irrelevant to the discussion.


Can a Rank Three Tensor act as a Trilinear, Bilinear, or Linear Map?

Question: Can a rank three tensor act as a trilinear, bilinear, and linear map? Similarly, a matrix (a representation of a rank two tensor) can be bilinear, taking in two vectors and spitting out a scalar for instance, but it can also be a linear map on just one vector. What's going on here?
Answer: Let $v,w,$ be fixed vectors, and $x$ be arbitrary.
Then $T_1(x)=x\otimes v\otimes w,$ $T_2(x)=v\otimes x\otimes w$ and $T_3=v\otimes w\otimes x$ are all linear maps.
To see why, let $x_1$ and $x_2$ be arbitrary vectors, with $\alpha$ any scalar, then
$$T_3(\alpha x_1 +x_2)=v\otimes w\otimes (\alpha x_1+x_2)=v\otimes w\otimes (\alpha x_1)+v\otimes w\otimes x_2=\underbrace{\alpha(v\otimes w\otimes x_1)+v\otimes w\otimes x_2}_{=\alpha T_3(x_1)+T_3(x_2)}.$$
The other two maps have identical arguments to show linearity.
Similarly, fixing one input, say $v\otimes x_1\otimes x_2$ defines a bilinear map.
All of this is a result of the how tensors are defined. That is, they must satisfy $x\otimes (v+w)=x\otimes v +x\otimes w$, and they're balanced over the scalar field
$$\alpha(v\otimes w)= (\alpha v)\otimes w=v\otimes (\alpha w).$$
Answer: Let's simplify this a bit and take a $(1,1)$ tensor over a real vector space $V$. We might take this, by definition, to be a bilinear map $V\times V^* \to \mathbb R.$ So we can think of it as taking a vector and a covector argument and looking like $T(v,v^*) = a.$
However, we can also think of it as a linear map $V\to V,$ or as a linear map $V^*\to V^*.$ To see how this works, take $v^*\in V^*$ fixed, and consider $T(v,v^*)$ as a function of $v.$ This is a linear map that takes a vector $v$ and produces a real number. In other words, it is a covector. So we can also define the tensor as the map which takes a covector $v^*$ and returns the covector $T(\_,v^*)$ we just just discussed. This is a way of viewing it as a linear map $V^*\to V^*.$ Similarly we can view it as $v\mapsto T(v,\_)$ a map from $V\to V.$ (Recall that a linear map that takes a covector and produces a vector is just a vector).
This reasoning works out on higher ranks as well. An $(n,m)$ tensor can be looked at a multilinear map $V^m\times (V^*)^n \to \mathbb R,$ or we can think of it as a map that takes a vector (or another tensor) and contracts it with the tensor to produce a lower-rank tensor. For instance if you give an $(n,m)$ tensor a vector, it will act linearly on that vector and return an $(n,m-1)$ tensor. In other words, it's equally well viewed as a linear map $$ V\to V^{\otimes (m-1)}\otimes V^{\otimes n}$$
Comment: So just to check, all n-linear maps are also m-linear to m variables, when m is between 0 and n. Multilinear maps are a type of linear map on a variable.
Comment: @BenjaminThoburn Yes. I think the other two answers addressed this explicitly. If you fix some of the arguments in a multilinear map, the map defined this way is multilinear in the remaining arguments
Answer: A multilinear map is linear in all its arguments. If you make some of them constant, you obtain a multilinear map of lower order, which can be described by a tensor of lower order.


Coprimes satisfying a pair

Question: We know that number of coprimes less than a number can be found using Euler's totient function.
But if there are two numbers $p$ and $q$ and we need to find number of numbers less than $q$ and coprime to $p$, is there any efficient method?
Comment: You seem to be suggesting that computing the Euler totient function provides an easy way to find out the number of coprimes less than a number. But that's the _definition_ of the function -- is there any known efficient way of actually computing it?
Comment: It involves computing the prime factorization of the number, which isn't known to be easy. Also, I've just noticed that the article you linked is just a rip-off of Wikipedia, so I've changed that to point to the original. Please don't encourage Wikipedia rip-offs.
Answer: No.  This is as hard as factoring, and there is no known efficient algorithm for factoring.  In particular, there is a standard reduction to show that if you could compute Euler's totient function efficiently, then you could factor efficiently.

Coprimes satisfying a pair


Restrict user with sudo permission to access home folders

Question: I have a linux development servers with some projects in home folders belonging to different users. Due to the nature of the server, some users can do sudo -s for some administrative task. Is there any way I can prevent users from snooping other home folders (or a specific folder)?
Comment: Yes, don't give them root access via sudo or any other method.
Comment: Avoid giving them root access by using a program that runs a command as root *but only whitelisted commands*. No idea if this would work.
Answer: If a user runs a command with sudo then they are running the command as if they were root, because they are root. So you cannot prevent a user that can use sudo from switching to a directory.
If you don't want users to access files in a directory, you can't give them root access on the machine.
Comment: Encrypting wouldn't help. That protects against someone who steals the hard drive, not against someone who can run commands on the machine while the files are accessible.
Comment: You can encrypt a file with gpg for example. You password encrypt it and only you know that password, other users do not.
Comment: But if someone is root then they can read the file as soon as the legitimate user enters the password, and they can read the password too.
Answer: If someone can run arbitrary commands as root, then they can switch to any user account, so they can do whatever that user can do.
In practice you have three options.

The privileged users only need to perform very specific tasks. Then let them only run some specific commands with sudo. This is secure only if those commands don't allow "shell escapes", i.e. if there's no way to leverage those commands to execute other programs. For example, package management (allowing sudo apt-get install …) often allows shell escapes by injecting the commands inside a package (this can be ok if package signatures are checked properly from known sources).
The privileged users need administrative powers, but they don't need it on the "real" computer, all they need is to have it on some computer (e.g. to play with network settings). Then create virtual environments (e.g. a virtual machine, or an lxc container, or an ad hoc Linux namespace) for the privileged users, and don't make other users' data available in the virtual environments.
Use social rather than technical controls: give the users broad permissions but let them know that any violation of privacy will be harshly dealt with. This is often ok if the users are employees, but it depends on how sensitive the data is.
Answer: Depending on what they need to be able to do, you can give them sudo access to only certain commands by dropping a line like this into your sudoers file:
%devgroup ALL=(ALL): /bin/commandname

which would make it so members of the group devgroup can only execute /bin/commandname as root (and of course with this particular example, you'd also want to remove them from the wheel group).


Can you use the default resource pack as a template?

Question: I need to create a resource pack for an adventure map and I don't feel safe downloading a template. Is it possible that I can just copy the default Minecraft resource pack and paste it into the resourcepack folder and edit it to my liking?
Answer: Yes this is possible.

Navigate to your .minecraft folder (%appdata%/.minecraft/ on Windows usually)
Open versions folder
Open the folder for the version you would like to create a resource pack for
Open the .jar file with an archiving software such as WinRAR (right click, Open with WinRAR)
Navigate inside the assets folder and extract the minecraft folder
That folder is now your resource pack template! Rename the folder to whatever you would like to name your resource pack and copy it into the Resource packs folder. You can now edit it as you wish!

Hope I helped. If you need any more details or would like me to create a template for you to use (I guarantee it will be safe!) just comment.
Comment: I tried this but i don't have any applications that can open the .Jar file. If I download a template do i need Archiving software to edit it?
Comment: No you do not. However, I would recommend getting winRAR or 7zip, as they are very useful tools. You do not need archiving software after extraction either, so you can just get the winRAR 30 day trial and then un-install it.
Comment: Ok thanks I probably will get them after I make this adventure map, I want to finish the resource pack before 1.9 comes out.
Comment: How do you open it *without* the archiver? Say your administrator doesn't let you download such?
Comment: @CalebWoodman You can simply rename the file to a .zip (replace the .jar with .zip). You should then be able to open it without any additional software.
Comment: For some reason, this didn't work on my computer, but it works elsewhere :P
Answer: I have gotten stuck in this before. But I have a fix.
Download a texture pack and keep the folders and delete all but one of their contents then run it in minecraft by placing it into  %appdata%/Roaming/.minecraft/recoursepacks on windows, Unfortunately I have not seen the place where to place it.
THEN play minecraft by going into the minecraft Java /options/resource packs.


What to do when I get "mysqld dead but subsys locked" in my amazon ec2 micro instance?

Question: I have an Amazon ec2 micro instance that holds a WordPress site. Since yesterday, my mysql db keep crashing saying:
[ec2-user@ip-xxx-xx-xx-xxx ~]$ sudo service mysqld status
mysqld dead but subsys locked

I tried restarting the mysqld but after every 5 mins it keeps crashing with the same message - "Error establishing a database connection"
This is what my DB log file reads -
160123 05:44:41 mysqld_safe mysqld restarted
160123  5:44:43 [Note] /usr/libexec/mysql55/mysqld (mysqld 5.5.46) starting as process 7928 ...
160123  5:44:44 [Note] Plugin 'FEDERATED' is disabled.
160123  5:44:46 InnoDB: The InnoDB memory heap is disabled
160123  5:44:46 InnoDB: Mutexes and rw_locks use GCC atomic builtins
160123  5:44:46 InnoDB: Compressed tables use zlib 1.2.8
160123  5:44:46 InnoDB: Using Linux native AIO
160123  5:44:46 InnoDB: Initializing buffer pool, size = 128.0M
InnoDB: mmap(137363456 bytes) failed; errno 12
160123  5:44:46 InnoDB: Completed initialization of buffer pool
160123  5:44:46 InnoDB: Fatal error: cannot allocate memory for the buffer pool
160123  5:44:46 [ERROR] Plugin 'InnoDB' init function returned error.
160123  5:44:46 [ERROR] Plugin 'InnoDB' registration as a STORAGE ENGINE failed.
160123  5:44:46 [ERROR] Unknown/unsupported storage engine: InnoDB
160123  5:44:46 [ERROR] Aborting
160123  5:44:46 [Note] /usr/libexec/mysql55/mysqld: Shutdown complete
160123 05:44:46 mysqld_safe mysqld from pid file /var/run/mysqld/mysqld.pid ended

The common solution that found out there when I Googled for a solution for the issue - mysqld dead but subsys locked, is to create a swapfile. I did exactly as most of the solution suggest but since I'm not well versed with UNIX, I don't know what I have done is right or not.

Or, Is there any other solution for this issue. Kindly help.
Answer: Have you tried this mysql is dead but subsys locked
Seems quite straightforward. Also please consider moving to RDS instead of running a mysql server in ec2. Saves you from troubles of this kind.
Comment: Thanks Sougata. I applied the solution suggested in the link in your answer. I'll wait and see if this works. Btw, can you help me understand(or any link to read) if I can use RDS with my WordPress installation?
Comment: Yes you can use rds with wordpress. I'm not very experienced in wordpress but from what i remember, you simply add the hostname,username and password in a configuration file. RDS is nothing but a managed mysql server. https://aws.amazon.com/rds/ . Its quite simple actually.
Comment: Thank you Sougata for the info.
Comment: Crashed again. Let me try RDS.
Comment: Looks like its a pretty costly affair. Looking for another solution.


How do you show the exception handler in a class diagram?

Question: I have one exception handler class. I want to show this class in the uml -- class diagram. But, I do not know how to represent relationship between class <X> and exception handler class. How can I represent relationship between class in class diagram?
Answer: From what I've seen, stereotype is usually used to represent the relationship to an exception.

a better approach would be to use UML's stereotype notation with the keyword exception as the stereotype enclosed in guillemets, as in «exception».

Source: How can I represent an exception in UML?
This answer on Stack Overflow also suggests stereotypes, but uses <<throws>> instead. The author is confident that it's ok to use a stereotype which doesn't exist conventionally, and in this particular case, I agree with him.
As for the proper term, given that a stereotype can be a noun or a verb, both <<exception>> and <<throws>> seem equally correct. The proper terminology also depends on the language you use. For example in Python, you don't throw, but raise an error or an exception, so <<raises>> looks as the most appropriate one; unless your intention is to separate errors and exceptions, in which case <<exception>> and <<error>> may be more expressive.
Finally, Agile Modeling suggests (search for Figure 4) to embed the name of the exception like this:
+ findAllInstances(): Vector {exceptions=NetworkFailure, DatabaseError}

but IMO, while it's well suited for well-known exceptions used by the framework, it is too poor for custom exceptions you have created, because the relation is not visual enough.
Answer: One take on it is that the exception handlers are not a part of the problem domain, but rather a part of the non-functional requirements. That is, the architecture of the system. In that sense, they don't really belong in a class diagram of a domain model any more than a Logger class belongs there. (Or a BeanFactory, or garbage collector, etc.)
Class diagrams display information about classes in the model as well as relationships between them. But an exception handler has a relationship with many or all of the classes -- in that the control flow might suddenly halt and go to the exception handler instead. Worse still, the relationship between the exception handler and the domain classes is managed by the framework of the application, so there isn't a direct relationship to the domain classes (by design).
If your class diagram is for the framework classes, then it's easy. There will most likely be some sort of controller with a "has a" relationship to the exception handler. But if the class diagram is for the domain model, there are lots of bits from the background I wouldn't put in the class diagram.


Parent phone vs. parent's phone

Question: Is it correct to say: parent phone?
I just want to describe that the phone belongs to the parent.
Answer: A "parent's phone" is a phone belonging to a parent.
"parent phone" would indicate some sort of relationship between two or more phones (although this would be very strange wording).
Answer: Since you stated that you just want to describe that the phone belongs to a parent, then "the parent's phone" is correct. Basically, in most cases, adding an apostrophe and an "s" to a noun indicates ownership like "the dog's bowl" or "the man's hat."
Answer: If by "Parent" you mean Mother or father the I believe it has to be "parent's phone". But if you are referring to both like "Mother and Father" then its better to be "Parents' Phone Numbers" or in a rough patch go for "Parents' phones".

Parent phone vs. parent's phone


How do I improve my vocabulary?

Question: Hello guys I wanted some tips on how to improve my vocabulary without sounding too weird. I'd hate to sound like a robot. Do you guys have any suggestions?
Comment: Welcome to English language learners. This question is probably better directed at http://languagelearning.stackexchange.com
Answer: Watch movies. English movies with original soundtrack (as opposed to English movies with Chinese voices).
That way, you are exposed to the correct pronunciation of words.
It may be helpful to also use the English subtitles / captions.
If you practice with Google, you will sound Google-ish - not what you want :)
NOTE: In the context of this answer, you watch the movies concentrated to learn, not to have fun. It means, you listen carefully to the pronunciation, maybe even try to repeat yourself.

How do I improve my vocabulary?


What is the explanation of greater torque having greater "rotatory effect" on a stationary body?

Question: Why does moving a constant force further from center of mass (and thus increasing torque) increase angular acceleration? I know that the explanation for my question is that during the same angular displacement the same amount of force would be exerted over longer (linear) displacement, hence doing more work, and since conservation of energy is a thing the angular acceleration would increase, but it (at least for me) doesn't explain why moving a weight on a balanced beam makes it move (because when it is not yet moving, no work is being done and I can't mathematically prove that it would move).
Answer: I used to ask myself a similar question: How does a load at one end of a lever know how far away along the lever another force is being exerted? Forgive the anthropomorphic phrasing, but I hope you get my drift.
I solved my problem by realising that the lever had to have some internal structure. I considered the case of a lever of length 6$a$, pivoted at a point one third of the way along. I envisaged a lever in the form of a pin-jointed lattice of thin weightless struts and ties. The bottom of the lever is a thin horizontal rod of length 6$a$, resting on the pivot. A distance $a$ above the bottom rod is another thin horizontal rod. Between the two rods are 6 rods of length $\sqrt 2 a$, angled at 45°, each at 90° to its neighbour(s), to form a lattice running from one end of the bottom rod to the other. Two of the angled rods are one side of the pivot; the other four are on the other side; the top horizontal rod needs to be only $4a$ long.
You will find simply by applying force resolution at each joint that if a force $W$ is applied downwards at the end of the long arm of the lever, a force $2W$ is needed at the end of the short arm, in order to have equilibrium.
I then convinced myself that this result is independent of the particular internal structure chosen. This is almost certainly a very eccentric method of establishing the law of the lever (principle of moments) but I found it instructive and convincing!
Comment: Clever derivation! With a little calculus, this can actually demonstrate that you can derive all of the laws of rotational motion from linear motion, and likewise all of the laws of linear motion from rotational motion.  Typically we break motion apart into linear and angular motion simply because the math is easier if we do so, not because there's some fundamental difference between them.
Comment: Thank you for your answer. Since the most force resolution I've ever done is decomposing a force into x, y and z components, I can't wrap my head around it. This is how I got stuck: if I decompose the force to each beam, then the net force on each beam should be the initial force divided by number of beams on each side and the net force on each side should be net force on each beam times number of beams of that side. This is when I get that both forces would be the same.
Comment: @Henry05  I'll append details in an hour or two – technology permitting.
Answer: You can experience this effect if you go to your local playground and use the seesaw. Ask a friend to sit at a fixed position, then sit down at different leverages. The seesaw will rotate.
This effect is not "due to conservation of energy". However, we can use "energy considerations" to calculate the point of equilibrium. This is motivated in the following picture:

The left picture sketches the initial state and the right picture the equilibrium state.
Comment: Thank you for your answer. I apologize, I have not been clear when I asked for proof. I am convinced that it would happen. What I don't understand is why. The only explanation I could find involved work and it was not satisfactory for me, since it can't prove why it starts to move (as far as I can tell).
Comment: @Henry05 Why don't you tell us where you found the explanation being work. Cite a reference
Comment: @BobD It was explained like this at my elementary and high school. Of course, there's also the answer (which is more common) that simply just reiterates the formula - which doesn't explain anything, only describes it.
https://qr.ae/pvdevq

https://physics.stackexchange.com/questions/242713/why-does-torque-exist#:~:text=At%20a%20greater%20distance%20from,you%20will%20need%20less%20force. (which is literally the answer google gives you when you ask this question)
Comment: @Henry05 I thought you were talking about work done moving the weight horizontally when you stated " moving a weight on a balanced beam ", not the work done due to the net torque resulting from moving the weight
Comment: This seems like another way to visualize the center if gravity explanation from one of the other answers, and I like it. At the same time, it immediately leads to a question about why the center of gravity is important.
Answer: Not sure this will address your concern, but here goes:
When the beam is stationary, the center of mass due to the weights on both sides is directly over the fulcrum and there is zero net torque as measured about the fulcrum. If one weight is moved, the c.m. shifts so it is no longer over the fulcrum. There is then a net torque about the fulcrum due to the weights, considered to be acting at the center of mass. The more the weight is moved, the more the c.m. is shifted away from the fulcrum therefore it has more effect so that's how the radius comes into play.
Answer: We can imagine a limit situation, where a mass $m$ is very far from the pivot point, that is when the distance $r \to \infty$. And the rod joining it to that point has a very small mass compared to $m$.
At a first approximation, the movement is linear for an observer close to $m$, and by the second Newton's Law: $\mathbf F = m\mathbf a$.
If we suppose the force orthogonal to the radius $r$:
$$Fr = mar = mr^2\frac{a}{r} \implies \tau = I\alpha$$
Comment: Thank you for your answer. If I understand correctly, you used Fr, or the definition of torque (or rather a special case), in the equation, but that is what I struggle with - why is that equation a thing; meaning - why do we have to multiply force by radius to find it's "rotational accelerative effect"? Knowing this would also explain for me why there's an r^2 in the moment of inertia.
Comment: It is a matter of names. It is possible to say that $Fr$ is proportional to $\frac{a}{r}$, and that the constant of proportionality is $mr^2$. The names: torque, angular acceleration and moment of inertia are only convenient conventions.
Answer: Does it though? Increasing the arm (for a constant force) always increases force, but the angular acceleration doesn't necessarily have to. Consider two point masses connected by a rod of length $2r$ (see figure below). To each point mass a force of strength $F$ is applied always at 90 degrees to the rod. I choose this setup to simplify the situation as much as possible.
In this case the angular acceleration can be found simply by
\begin{align}
\alpha&=\frac{\tau}{I}\\
&=\frac{2rF}{2mr^2}\\
&=\frac{F}{mr}
\end{align}
We find that, for constant force, the angular acceleration actually decreases! This makes sense if you imagine a very long and imagine how fast it rotates.
Comment: Thank you for your answer. I have already thought of this example (or similar), and it was one of the main reasons why this problem confused me, because it proved the opposite (which happens way too often to me with physics).
Answer: Maybe look at it the other way around -- in terms of work done.
The force involved comes from gravity.
When the two weights are balanced... if {the weight that we are going to move outwards (away from the fulcrum)} were to go down... the work done by gravity would be the same on the two weights, except that one would be negative.
If we move the weight away from the fulcrum -- say to 2 * {the original distance} -- then... if we imagine that it is moved down 1 unit of distance... the other weight is moved up 1/2 a unit of distance.
p.s.  It feels tangentially relevant to say that... when doing calculations for, e.g. pushing down on a table, one imagines that the table is pushing back with equal force.  This is strictly a fudge, to get the correct value for the equation, that results in the table not moving (up nor down).
Answer: Its already implied if we accept that a balanced beam wont rotate. Lets look at the case of an essentialy massless beam, and place the two equal weights $w$ a distance $d$ to the right and to the left of the fulcrum, respectively. Of course the beam wont rotate, otherwise it would mean that nature disobeys an almost trivial symmetry. If we then move the weights so they are both directly above the fulcrum, it still wont rotate, because of the same trivial symmetry. We deduce that a force acting directly through the fulcrum, wont cause the beam to rotate. Lets go back to the balanced beam with the weights a distance $d$ from the fulcrum. But now look at the situation with the left weight as the reference point. Seen from there, the normal force from the fulcrum is acting at a distance $d$ and with a force $2w$ upwards, and the rightmost weight is acting at a distance $2d$ with a force $w$ downwards. Now remove the fulcrum, but replace it with an equal force of magnitude $2w$ acting upwards, i.e. the same situation. Put the fulcrum directly under the leftmost weight, i.e. at the current reference point. The leftmost weight cant contribute, since its acting through the fulcrum (if you want, you can elongate the beam leftwards so that the beam is symmetric about the new fulcrumpoint, but that doesnt change anything as we consider the beam essentially massless). It follows that for nature to obey the trivial symmetry, it has to be the case that a force $2f$ at a distance $r$ from a point, must have the same tendency to rotate a beam as a force $f$ at a distance $2r$ from the point.


Wireless icon not showing in the panel

Question: I just installed the Ubuntu 14.04 and I cant see wireless icon in the panel or any settings related to the wireless icon.
Answer: Use the following command to list the interfaces,
sudo iwconfig

Please check whether you have installed wireless adapter
To check what drivers your wireless adapter is currently using, you may run the following command:
sudo lshw -C network

lshw lists information on your hardware
-C network filters the output to only show the network class.
In the output, look for the entry with description: Wireless interface and the driver.
Alternatively, you can use the command:
lspci -nnk | grep -A2 0280

lspci lists information on your PCI connected cards
-nnk instructs lspci to output more information about these cards (including the driver being used)
| pipes the output to the next command
grep 0280 filters the output to show lines containing 0280, which is the PCI class code for wireless PCI controllers in Ubuntu.
-A2 shows two more lines of information.
If no driver is installed please follow the below link install Broadcom 802.11 Linux STA driver
1) Depending on your system download 32 or 64 bit package
from http://www.broadcom.com/support/802.11/linux_sta.php
2) Go to a terminal to the dir with package.
3) tar -xvzf hybrid-portsrc-.tar.gz
Replace version with one you have
4) Install kernel headers (eg. in fedora "yum install kernel-headers")
5) make -C /lib/modules/uname -r/build M=pwd
if There were no errors
6) Unload Modules from memory
rmmod ndiswrapper b43 ssb bcm43xx b43legacy
7) For WPA
modprobe ieee80211_crypt_tkip
8)Check if Compiled Module is working
insmod wl.ko
Now Probably a new interface will be there for wireless (wlan0 or eth-1)
To check it run
iwconfig
iwlist scanning
AND YES !! ITS Working !!!!!
B)Making it load on boot
1)cp wl.ko /lib/modules/uname-r/kernel/net/wireless/
2)depmod -a
3)rmmod wl
4)modprobe wl
5)Add following Lines to /etc/rc.local (in openSuse /etc/rc/rc.d/boot.local)
Once it is done check whether you can able to see wireless icon.
Hope this helps!
Comment: Why do you believe he has a Broadcom device?
Comment: I am not sure but I think you are asking me to install the wireless driver again?
Because right now I just have issue with no wireless icon in the ubuntu panel although i can connect to wireless from settings> Network.

Please let me know if there is any possible way to get the wireless icon in the panel so that i have not to go in the settings always for a connection.

Thanks :)
Answer: You can access the driver manager and check if your wireless driver is enabled or not. This obviously depends on your network card, but that's the way I always set up my wireless connection.
Comment: But internet is working fine right now, the issue is just that I cant see any icon in the ubuntu panel.
For now if I have to connect from a connection then i can do this from the settings > network. And there i can see all the available connections.

Wireless icon not showing in the panel


Why didn't David help the alien in killing Daniels and Tennessee?

Question: In final scenes of Alien: Covenant, the alien enters the ship and kills Upworth and Ricks. Mother alarms all the crewmembers about it.
So Daniels and Tennessee kill the alien with David's help. Why would David help kill the alien when he greatly dislikes the human race?
Comment: If you want to mark an answer as the correct one for you, you can click on the checkmark by the voting buttons as per the [tour].
Answer: I believe that David, in this case, is making a sacrifice of a single Xenomorph in order to better spread the Xenomorph race to the new colony world. As he states to Daniels (albeit in a context where she doesn't understand his meaning):

"But that's what we're doing too, isn't it? Creating a new world on Origae-6 ... Honestly, I could use a new world."

And further direction from the script:

 David strides in.

 The Colonists. Tier after tier after tier of them in hypersleep.

 Also the rows of embryos.

 He walks forward.

 Gazing up at the sleeping Colonists.

 His children. His slaves. His subjects. His.

 The Wagner swells. Grand and triumphant.

 And David smiles.

 Dreaming of the future.
Answer: I ran into a second good theory here. David has been doing his experiments to create the ultimate predator species and the deleted prequel scenes show that he's gone through several generations that were not sufficiently advanced for his purposes. He may have aided Daniels and Tennessee in defeating the creature because he wanted a more challenging test, one which it didn't quite overcome, although now he has more subjects to work with, including Daniels, who he may be targeting to create the first Queen.


Living place in USA

Question: If iam travelling to work in USA, alone without my family, and I dont want to rent an apartment or studio, is there is the option of renting only a room and how much does it cost?
Comment: Yes there's that option. Many people rent a room in a house or apartment. The price varies
Comment: average  Price  ?
Comment: The average price depends on the region you are going to. In San Francisco, count 3000-4000 USD/month. In rural areas in the middle of nowhere, think of 300-500 USD/month. AirBNB has some long-term rentals available and you may want to have a look for options in the region you are going to get an idea.
Answer: Housing cost in the US varies by location. The middle of a big city or developed area (like Silicon Valley) will be very expensive, while smaller cities and towns will be cheaper. Craigslist.org is a good resource for shared housing.
I just looked at CraigsList for San Francisco, California. The cost for renting a room in a shared house ranges from $800/mo to $3000/mo.

Living place in USA


Relating condition number of hessian to the rate of convergence

Question: While minimizing a Lipschitz continuous strongly convex functions, the rate of convergence of the gradient descent method depends on the condition number of the hessian of the function, where a high condition number leads to slow convergence. Can someone give an intuitive explanation explaining why the condition number of the hessian and the rate of convergence are related?
Answer: The classic zig-zag picture is misleading since it suggests that the slow convergence is due to overshooting. But for ill-conditioned problems, convergence is still slow even on a purely quadratic function with perfect "ridge-line" line search, where you step to the exact ridge-line of the mountain so that there is no overshooting at all.
The correct intuition is that, since the gradient points in the steepest direction, gradient descent cannot effectively explore a direction in parameter space until it has already eliminated all other directions in parameter space that are steeper. So even in the best case scenario of perfect line search, it will take one iteration to eliminate the steepest direction, then the next iteration to eliminate the second steepest direction, then the next iteration to eliminate the third steepest direction, and so forth.
Imagine climbing up a mountain using the method of steepest ascent. You will not walk straight towards the peak. Rather, you
will climb up the steepest face of the mountain until you reach a ridge, then
follow the ridge to the top, making a dog-leg path.

(image credit to Mountains to Sound Greenway Trust)
Were it possible to climb an N -dimensional mountain, the path of steepest ascent would first climb to the top of the (N − 1)-dimensional ridge perpendicular to the steepest direction, then to the top of the (N − 2)-dimensional ridge perpendicular to the first two steepest directions, and so on. The more the eigenvalues of the Hessian are separated, the more the path will resemble the set of $N$ edges on a box, traversing from one corner of the box to the opposite corner. This is illustrated below for ill-conditioned functions $\mathcal{J}(q)$ in increasing numbers of dimensions.
When climbing to a ridge, only minor progress will be made in the direction of future, less steep, ridges. This minor progress is what the classic convergence bounds for gradient descent rely on. The rate of this minor progress will be proportional to the ratio of steepnesses: the greater the difference in steepness, the less progress will be made on the future ridge during the process of climbing the current ridge. The eigenvalues of the Hessian characterize these different steepnesses. The condition number of the Hessian, being the ratio of the largest eigenvalue to the smallest, is the ratio of the steepest ridge's steepness to the shallowest ridge's steepness, which is why it enters into the convergence bounds.
2D

3D

ND

This is discussed in greater detail in Section 3.1 of my Ph.D. thesis here:
https://repositories.lib.utexas.edu/bitstream/handle/2152/75559/ALGER-DISSERTATION-2019.pdf?sequence=1
Comment: amazing intuition, I find difficult to interpret SGD artifacts on high dimensions and this brings some knowledge to the table, thanks!
Answer: The steepest descent method "zigzags" as it approaches a minimum.  See this figure from Wikipedia.  This phenomenon becomes much worse for a badly conditioned problem.
The reason that the method zigzags is that the level curves of the objective function are not perfectly circular.  If they were circular, then the steepest descent direction would point straight to the minimum of the function and the method could converge to the minimum in a single iteration (assuming that the step length was selected properly.)
With elliptical level curves, the steepest descent direction doesn't point straight to the minimum.  Even with an "exact line search" that minimizes along that steepest descent direction, the method ends up zigzagging.
For a very badly conditioned problem, the steepest descent direction can be nearly orthogonal to the direction that would take you to the minimum, so that the method must zigzag many times to get close to the minimum.
Comment: Can you tell intuitively why it zigzags and why the phenomenon becomes worse for a badly conditioned problem?
Comment: @user2684957 See https://math.stackexchange.com/questions/887433/geometric-interpretation-of-an-ill-conditioned-matrix and https://en.wikipedia.org/wiki/Newton%27s_method_in_optimization#/media/File:Newton_optimization_vs_grad_descent.svg
Answer: To give a more visual answer, would suggest you download and play with this fantastic package (I am not the author): https://github.com/lilipads/gradient_descent_viz
In particular, try out gradient descent (both the continuous and stepwise versions). Here is a visualization I generated using this package (also see the full resolution version):

Let the function we are trying to minimize (which generates this convex paraboliod surface) be $f(x)$. Recall that if the surface is an upward-facing paraboloid, it is given by the vector-valued function $f: \mathbb{R}^n \rightarrow \mathbb{R}$, where $f(x) = \frac{x^TAx}{2} - x^Tb$. Things to keep in mind:

$A \in \mathbb{R}^{n \times n}$ is a particular kind of matrix called Symmetric Positive Definite. An SPD matrix is analogous to a positive scalar ($\alpha > 0$) when the number of dimensions is $n \gt 1$. The important property is that $x^TAx > 0, \forall x\neq 0$, which is what makes the parabolid open upwards rather than downwards. The values in $A$ determine how curved the paraboloid is.
$b$ is some vector which we assume to be non-zero. Varying b changes what the minima of this surface is.
The vector which gives the direction of the (negative) gradient at any point $x \in \mathbb{R}^n$ is: $ - \nabla f(x) = b - Ax$.
In our current discussion, $x \in \mathbb{R}^2$.

Notice that in the screencap below, we are currently at $x = (\chi_0, \chi_1) = (-1.68, 0.97)$. Even though $f(x)$ is convex, the gradient (black arrow) does not point towards the minima (red region). You can conjecture that, given a certain $A$ and $b$, there are many values of $x$ which cause this to happen, particularly when $\chi_0$ and $\chi_1$ have relatively large magnitude.

Now, when we update the ball to the second location in the video, we are at $x = (-0.56, 0.04)$. Notice that $\chi_1$ here is much smaller than before (0.97 to 0.04). As a result, when we compute the gradient at our second location, is mostly along the $\chi_0$ (notice the stubby little blue arrow pointing east).
Whether or not the gradient vector points in the direction of the minima is immaterial of where the minima is located; it is simply the tangent to the loss surface at some point $x$.
One case where the gradient vector definitely points to the minima is when $A$ is the identity matrix and $b=0$, then the gradient vector is $-x$, which will point exactly at the minima (which will be the origin).

Now, the previous surface was nicely convex: the width and height of the parabola were roughly equal, and we could converge quickly.
Let's consider an ill-conditioned problem, i.e. due to some skewed values in $A$, the paraboloid $f(x)$ is highly stretched along one axis ($\chi_0$ as per the image below). In this case, the value $f(x)$ will fluctuate rapidly with small changes in $\chi_1$, i.e. the component of the gradient vector $- \nabla f(x)$ corresponding to $\chi_1$ is large. This means the gradient vector will mostly point along $\chi_1$, which may or may not be in the direction of the minima.
On the other hand, assume even that in this stretched parabola, we start due south of the minima. In that case, the gradient will generally point towards of the minima, so gradient descent will converge quickly. The issue occurs when you start somewhere on the far left or the far right. In particular, if your random initialization starts you at some point far from the minima in $\chi_0$ (the stretched axis) then you start getting the zig-zag pattern.
To avoid this, we often precondition the problem, so that the contour plots looks like circles. In that case, no matter where you start, convergence takes roughly the same amount of (short) time.

(image credit: https://mc.ai/my-machine-learning-diary-day-65/)


Wordress admin page is fetching error You do not have sufficient permissions to access this page.

Question: Wordress admin page is fetching error You do not have sufficient permissions to access this page. Mysite url is https://www.artvault.co.in but its admin panel url - https://www.artvault.co.in/wp-admin is rendering an unknown error You do not have sufficient permissions to access this page. Where I thoroughly checked the $table prefix : wp_ is same in databse usermeta tables. I tried to deactivate the all plugins nothing is helpful. Anyone can tell something what happeed.?
Answer: Navigate to your WP database in phpmyadmin. Here in option table locate WordPress Site (URL) and WordPress Address (URL) row. change these two field. https://www.artvault.co.in in WordPress Site (URL) and https://www.artvault.co.in/wp-admin in WordPress Address (URL)


Interpretation of values contained in a confidence interval

Question: Are the values within a confidence interval those that do not significantly differ from the point estimate? Or, put differently, how do we interpret the values contained in a CI given that the CI is just about the coverage rate? EDIT: Notice that this question is not about the formal equivalence of the t-statistic and the CI (i.e., CI's as significance tests) but how to interpret the values contained in an observed interval.
Comment: This is actually not a duplicate because the other source clarifies the equivalence between a t-test and a CI for the mean difference. My question is about the meaning of the values contained in an observed CI.
Comment: If you make that clarification as an edit to the original post, it will enter a reopen queue, so that it can be answered.  Mark the edit prominently, like this       EDIT
Comment: Done, see above.
Answer: This is something that I wrote recently for students that may help: A 95% confidence interval of the effect has a 95% probability (in the sense of long-run frequency) of containing the true effect. This probability is a property of the population of intervals that could be computed using the same sampling and measuring procedure. It is not correct, without further assumptions, to state that there is a 95% probability that the true effect lies within the interval. However, if we have only weak prior beliefs about the possible values of the effect, then it is valid, though possibly misleading, to state that there is an approximately 95% probability that the true effect lies in the interval (Greenland and Poole 2013, Gelman, 2013). Perhaps a more useful interpretation is that the interval contains the range of effects that are consistent with the data, in the sense that a t-test would not reject the null hypothesis of a difference between the estimate and any value within the interval (this interpretation does not imply anything about the true value).
Another paper related to the question is
https://link.springer.com/article/10.3758/s13423-013-0572-3


Proof regarding n-connectedness

Question: If one has to prove that $R^n - \{0\}$ is not $(n-1)$-connected, is it necessary to prove formally that there exists a non contractible $(n-1)$-sphere or can that simply be stated. If one must formally prove this, how may that be done? Can one, for example, consider the intersection of the $(n-1)$-sphere and $R^n$ with an arbitrary 2D plane and then say (for the sake of contradiction by the definition of 1-connectedness) that the given sphere is contractible if and only if the circle that corresponds to the intersection of that sphere is contractible in the subspace topology defined by the intersection of the plane and $R^n$?
Comment: I think you have an index shift in the first line. $R^n \setminus\{0\}$ is $n-1$-connected. $R\setminus \{0\}$ is not connected, $R^2\setminus \{0\}$ is not simply connected (i.e. not $1$-connected).
Answer: This is not immanent in the definitions, so there is need for a proof.
Your approach, however, is not sufficient. In general, there could be a contraction of the $n-1$-sphere outside the plane, without the existence of a contraction inside the plane. I think your approach would say something about relative homotopy groups.
The standard approach (at least I don't know an easier way) to show that $R^n\setminus \{0\}$ is not $n-1$-connected goes by first showing that $R^n \setminus \{0\} \simeq S^{n-1}$ (homotopy equivalence). Therefore, all homotopy groups will be isomorphic.
So it remains to show that $S^{n-1}$ is not $n-1$-connected. It is well known that $S^{n-1}$ is $n-2$-connected, so by the Hurewicz theorem, $\pi_{n-1}(S^{n-1}) \cong H_{n-1}(S^{n-1})$. Homology is not so hard to calculate, in particular $H_{n-1}(S^{n-1})\cong \mathbb{Z}\neq 0$, so $S^{n-1}$ is not ${n-1}$-connected.
See also Wikipedia for this.
Comment: Do you need further help with this topic? Should I extend an explanation? If you are satisfied with the answer, you might consider to accept it.

Proof regarding n-connectedness


Has anyone implemented Google Pagespeed Module into Magento?

Question: Has anyone installed Google Pagespeed Module into Magento? Is the implementation too hard to do it?
Comment: Why don't you use the actual server side Google Pagespeed extension? This isn't something Magento should be doing. Here's some starter configuration for you if using MageStack https://www.sonassihosting.com/help/magestack/optimal-varnishpagespeed-configuration/
Comment: I thought Google was killing it off?
Comment: No, they stopped the PageSpeed hosted product, not the open source module. See https://groups.google.com/forum/#!topic/page-speed-service-discuss/h4jyUpv6Bc8
Answer: This is the one good module which allows you to configure google pagespeed easily.
Play with it
https://github.com/mediarox/pagespeed
I strictly recommend you to use this one in your dev. environment. Once you test all the features, You can move into production.Cheers.
Comment: Tks so much for the tip. However, have you implemented it in a website? Is  it too hard to do it? Does it work well with magento?
Comment: No, Its not that hard. It is very easy. Tanmay Kamath point is good.
Accept mine as answer if it satisfy your quench LOL. :)
Comment: Tks so much! ;) I will try it out.
Answer: The module was easy to implement and worked as expected. However with FPC on it stopped working, so we uninstalled it.
Answer: Nope its not hard. Download the file from the link above. Place the files in appropriate folders. Edit the config file as per your needs. Login to magento admin panel. Configuration path: System > Configuration > ADVANCED > Pagespeed
You will be able to see the pagespeed module there.
Comment: Tks very much! ;)
Answer: I just thought I would share an outline of my recent efforts using the pagespeed_mod on Apache.  It greatly improves page loading speed but is a pest when trying to update the site.  This is the approach I took to make it less of a pain:
1: In the pagespeed module conf I set the cache directory to be inside the Magento var folder to make it easier for web scripts and my remote access user to get to it.
2: Wrote a script to recursively delete the contents of that said folder. See this stackoverflow post on this.
3: Added a new button to the clear cache admin page to call said script.  You can do this by adding admin layout xml similar to this:
<code>     <adminhtml_cache_index>
        <reference name="cache.additional">
            <block type="core/template" name="jh-pagespeedcache" template="pagespeedcache.phtml"/>
        </reference>
     </adminhtml_cache_index>
</code>
You can base your phtml contents on the existing clear cache buttons.
Just saves a bit of time and annoyance when you need to update the live site.


How can I install Ubuntu on a second hard drive with Windows being on the first?

Question: I have recently got an Alienware M17x R4 and since it has two hard drive bays I was thinking of putting my hard drive from my old laptop and installing Ubuntu on it without making any changes to the drive that has Windows on it. Also if possible to add a entry to the Windows boot loader to boot Ubuntu instead of using GRUB.
Answer: This is an updated answer for Ubuntu 18.04 LTS where Windows is installed in UEFI mode
Preamble: Getting ready

Download the official Ubuntu ISO file and create a LiveUSB (Installation USB) following the instructions in the official Ubuntu site.
Make sure Secure boot option is turned off in UEFI. And Fast boot is turned off from within Windows.
Boot from the LiveUSB in UEFI mode. If you boot from the LiveUSB in the UEFI mode you will see this screen:

You are good to go. Select Try Ubuntu without Installing. Try it out and make sure all the hardware works.
Note the Install Ubuntu icon on the desktop.
If you see a different screen like:

immediately followed by:

then you have not booted in the UEFI mode!
Option 1: Only One ESP
This option will add some files to the (U)EFI System Partition (ESP) located in the first drive, where Windows is installed. ESP is part of the UEFI and not Windows. In this sense this option does not change anything in the Windows partitions.
After Ubuntu installation is complete, the default bootloader will be grub and it will give you a choice to boot Ubuntu (the default OS) or Windows every time you turn on or restart the computer. The default OS can be changed later.

Click the Install Ubuntu icon to start in installation. Continue the process until you come to the Installation Type window. To manually select the second drive for Ubuntu, select **Something Else..."

You will see a list of disks and partitions like this:

In this picture the first virtual drive is /dev/vda and the second virtual disk is /dev/vdb.
If you have two SATA physical drive, they may be /dev/sda and /dev/sdb. If you have a NVMe drive it may be /dev/nvme0n1.
Partitions in the SATA drive are /dev/sda1, /dev/sda2, etc. and partitions in NVMe  drives are /dev/nvme0n1p1, /dev/nvme0n1p2 etc.

Make a New GPT Partition Table. To do this highlight the new disk, in my
case /dev/vdb and click on the button New Partition
Table.... You will see a warning:

Click Continue. Now you will see a new row below /dev/vdb called free space.

Highlight the free space below /dev/vdb and click on the + to create a partition. This will be the main partition.

Create root system partition / in ext4 format

Recent versions of Ubuntu do not need a separate swap partition. At this point you may want to create a separate /home partition in the ext4 format, but it is not part of the default Ubuntu installation.
Make sure the bootloader is installed in the first drive, /dev/vda in my case. As long as you have booted the LiveUSB in UEFI mode, the installer will find the EFI System Partition (ESP) in the first drive and install part of the grub bootloader there. Do not select a specific partition (like `/dev/vda1) here.
Continue the rest of the installation as you normally would by pressing  Install Now button.
That's all. Done.
Continue reading if you don't want to modify the ESP in the first drive.
Option 2: Two Drives Two ESPs
For some reason you may not want Ubuntu installation not to write files into the ESP located in the first drive. Then you will need to create a new ESP in the second drive when you manually prepare the second drive for Ubuntu installation.
Special Note: If you use this option grub will be loaded in the second drive, but it will not display when you power on or restart the computer. The computer will boot to Windows as before. To boot Ubuntu you will need to press the UEFI defined function (Fn) key when the computer boots to select the boot drive. The specific Fn key varies by computers.
Bug Alert!
There is a bug (bug number: 1396379) in the Ubuntu installer that installs the grub bootloader on the first ESP it finds. See https://ubuntuforums.org/showthread.php?t=2182302 for discussions on this.
To workaround this bug you will need to take a few extra steps.
Workaround

When you boot from the LiveUSB in UEFI mode, use the Try Ubuntu
without Installing option.
Open the Gparted application when the Ubuntu desktop starts.
Select the Windows (first) drive within Gparted at the top right
corner:

The NVME drive in this picture does not have Windows. However, it has an EFI System Partition (ESP).

Highlight the ESP in the bottom list of partitions as shown above. Right click on the highlighted partition and select Manage Flags from the context menu. You will see:

You should see two flags, boot and esp checked. Uncheck those two flags. Click Close button to exit the Manage Flag window. Click the green "Apply" ✅ button if it is clickable.
You will need to restore those check marks once Ubuntu is installed using the same above process so that Windows can boot from the first drive.
Close Gparted.
This is the end of the workaround.
Install Ubuntu
Follow the instructions to install Ubuntu in the second drive as described in Option 1 above. That is:

Select Something Else... as before.
Create a new gpt partition table for the second disk if not there.
Create a new EFI System Partition (ESP)

Highlight the Free Space under /dev/vdb and click on the + to create the first partition. The first partition will be a 200 MB EFI system partition. Change the size to 200 MB and the Use as:

The Create Partition window should now look like:

Create root system partition / in ext4 format below the ESP:

Once done, your disks and partitions will look like:

Recent versions of Ubuntu do not need a separate swap partition. At this point you may want to create a separate /home partition in the ext4 format, but it is not part of the default Ubuntu installation.
Make sure the bootloader is installed in the second drive, /dev/vdb in my case.  Do not select a specific partition (like `/dev/vdb1) here.
Continue the rest of the installation as you normally would by pressing  Install Now button.
Once the installation is complete, you will get an choice to:

Continue exploring Ubuntu live session
Restart Computer

Do not restart the computer yet!
Select the continue to explore Ubuntu and open Gparted again. Go through the process in the Workaround section above, and reverse the changes made then.
Close Gparted. Restart the computer by clicking on the right top bar and selecting the power button from the drop-down menu.
That's all. Done.
Answer: That's simple:

First download the ISO file then burn it to a CD/DVD or use Unetbootin to get a bootable USB, select the Installation media to boot first from your BIOS.
Once you're in the Live Session, choose Install Ubuntu.
When you get to the Choose an Option screen, select Do Something Else.
Locate the Second drive and choose to Install the System on the second drive.
Choose to also install the bootloader on the 2nd drive, make sure it's the drive and not a partition (i.e. /dev/sda and not /dev/sda1)
Comment: Ok how do I use the windows boot loader to boot ubuntu
Comment: You can download EasyBCD and then add the GRUB entry to the Windows bootloader
Comment: Thanks man ill give it a try hopfully don't mess up my windows bootloader haha


InDesign Data Merge with hyperlinks

Question: I want to run a data merge with text and photos and I want to be able to hyperlink the photo to a website. I have about 400 entries with the data merge that will each be on a separate page but I can't for the life of me figure out how to hyperlink the pictures automatically!
Comment: Hi Nicole. You'd be better off asking this question on InDesign Secrets, or possibly the SuperUser site on StackExchange. On gd.se this falls under the heading of "Technical Support," and is considered off-topic. Sorry about that.
Comment: I'm open to these kinds of questions here (despite the rules) but you'll get better answers elsewhere. I think you'll have the best luck with an XML workflow. Try asking around on the [Adobe InDesign forums](https://forums.adobe.com/community/indesign). If you want something *really* efficient, go to the [InD Scripting forum](https://forums.adobe.com/community/indesign/indesign_scripting).
Answer: InDesign native Datamerge won't allow you to create hyperlinks on the fly. You need at least Scripting to achieve that. A plugin such as EasyCatalog is also a good candidate.

InDesign Data Merge with hyperlinks