Consider a container filled with two liquids, of densities $\rho_1$ and $\rho_2$, such that $\rho_2 &lt; \rho_1$. Now imagine placing a cylinder with the top base removed, of density $\rho,\space \rho_2&lt;\rho&lt;\rho_1$, such that the cylinder floats between the liquids. The second liquid is barely filling the cylinder, whose interior diameter is $l_1$ and whose exterior diameter is $l_2$. An illustration can be found below:    Now I am a little bit confused about the buoyancy force in liquid $2$: I thought I could express it as (assuming the height of the cylinder in liquid $2$ is $h$):  $$F_{B2} = (S_2-S_1)h\rho_2g=(\pi\frac{l_2^2}{4}-\pi\frac{l_1^2}{4})h\rho_2g=\\=\frac{\pi h\rho_2g}{4}(l_2^2-l_1^2)$$  Because the liquid actually enters the cylinder and only the walls displace a volume of liquid. However, the authors of the problem and my teacher claim that:  $$F_{B2}=\frac{\pi l_2^2h\rho_2g}{4}$$  Which means that the entire part of the cylinder that lies in liquid $2$ actually displaces a certain part of its volume, rather than the walls only. I have strong reasons to believe that I am wrong, because this is a tiny sub-part of a problem from a national contest, and the official solutions claim that the latter is correct. Which version is correct and why? What part of the cylinder displaces some of the second liquid, the walls or the cylinder itself as a whole (well, the part of them that lie in the second liquid) and why?
A thin-walled container of mass $m$ floats vertically at the separation surface of the two liquids of density $ρ_1$ and $ρ_2$  . The whole mass of the container is concentrated in the part of height $h$.  The question is to determine the immersion depth $h'$of the container in the lower liquid if the bottom of the container has a thickness $h$ and an area $S$ and if the container itself is filled with the liquid of density $ρ_1$.    I applied the fundamental principle of the statics on the container  $Balance$ $Sheet$:  The weight of the container ($-mg$)  The buoyancy force applied by the fluid with density $ρ_2$ ($ρ_2gSh$)  The weight of the fluid with density  $ρ_1$  (-$ρ_2gSh'$)  I summed up the forces and set them to zero and find my $h'$  but then  realised that the fluid with density  $ρ_1$  applied also a buoyoncy force on the top of the container , and the fluid with density  $ρ_2$  applied a buoyoncy force on the other fluid.  Where's the problem ?
Help with proof that $\mathbb Z[i]/\langle 1 - i \rangle$ is a field. Can you read my proof and tell me if it's correct? Thanks.  Let $V$ be a vector space over a complete topological field say $\mathbb R$ (or $\mathbb C$) with $\dim(V) = n$, base $e_i$ and norm $\|\cdot\|$. Let $v_k$ be a Cauchy sequence w.r.t. $\|\cdot\|$. Since any two norms on a finite dimensional space are equivalent, $\|\cdot\|$ is equivalent to the $l^1$-norm $\|\cdot\|_1$ which means that for some constant $C$, $\varepsilon &gt; 0$, $k,j$ large enough, $$ \varepsilon &gt; \|v_j - v_k\| \geq C \|v_j - v_k\|_1 e_i= C \sum_{i=1}^n |v_{ji} - v_{ki}| \geq |v_{ji} - v_{ki}|$$ for each $1 \leq i \leq n$. Hence $v_{ki}$ is a Cauchy sequence in $\mathbb R$ (or $\mathbb C$) for each $i$. $\mathbb R$ (or $\mathbb C$) is complete hence $v_i = \lim_{k \to \infty} v_{ki} $ is in $\mathbb R$ (or $\Bbb C$) for each $i$. Let $v = (v_1, \dots , v_n) = \sum_i v_i e_i$. Then $v$ is in $V$ and $\|v_k - v\| \to 0$:  Let $\varepsilon &gt; 0$. Then $$ \|v_k - v\| \leq C \|v_k - v\|_1 = C \sum_{i=1}^n |v_{ki} - v_i| \leq C^{'}n \varepsilon$$  for $k$ large enough. Commas for multiple adjectives I've been taking photographs for a while, and have started to think about moving to full-frame. I'm on a very limited budget, but I currently have a good macro lens and am quite willing to buy something like a EF 50mm f/1.8 to start off with. I could possibly stretch my budget as far as a 70-200 f/4 L or 24-105 f/4 L.  Primarily, I'm interested in taking photos of small wildlife like reptiles, amphibians and insects, but I have a healthy interest in landscapes, birds and street photography.   I currently have an EOS 30D, and I love the way the camera handles, especially with the dual control dials. For this reason, I think shooting with a Rebel would be a significant step down.   However, I have seen 5Ds and occasionally 1Ds Mark IIs on ebay for roughly the same price as an EOS 650D. I'm sure these cameras have better IQ than my 30D, since they are pro-level bodies from around the same time, and full-frame besides. The question is, are these cameras now so old that the image quality from them is a substantial step down from current Rebels?   I realise I'll be giving up continuous shooting speed and some amount of low-light performance, and that I'd also need to get an external flash, but besides these, are there any substantial caveats I need to be aware of before I go ahead and take the plunge? What is the world's biggest Schrodinger cat? Apply a modifier as a shapekey into existing shapekey or merge shapekeys? I am making a shapekey of a model shrinkwrapped to another model. The shrinkmap modifier is too inaccurate to wrap to the model alone. So, my first step was to roughly hand position the geometry around of the first model around the second, then use shrinkwrap. The rough positioning was done as a shape key. I can then apply shrinkwrap, but its as a separate shapekey. The two do not work well together. I need them to be the same shapekey. How do I apply a modifier to an existing shapekey, or merge/combine/join the two shapekeys? How to print only defined variables (shell and/or environment variables) in bash The bash builtin command set, if invoked without arguments, will print all shell and environment variables, but also all defined functions. This makes the output unusable for humans and difficult to grep.  How can I make the bash builtin command set print only the variables and not the functions?  Are there other commands which prints only the shell variables, without the functions?  Note: bash differentiates between shell variables and environment variables. see here I'm a Sri Lankan passport holder. I'm planning to travel to China in January and have to transit in Malaysia. I wonder whether I can obtain a Malaysian transit visa in KLIA 2 when I'm flying back to Sri Lanka. Equaling 3 Displays Total (or 2, if the laptop display cannot be used).  I work at home on two large monitors, but at the office on a laptop with a single large monitor. Is it possible to attach two (or more?) external monitors to a laptop without having them clone each-others display? Crick’s Central Dogma — Counter Cases Solving $\sqrt{x+5} = x - 1$ I'm currently learning about radicals and simplifying them, and I came across this problem on the internet and tried to solve it:  $$\sqrt{x+5} = x - 1$$  So I used this logic:  $$ \begin{align} \sqrt{x+5} &amp;= x - 1 \\ x + 5 &amp;= (x-1)^2 \\ x + 5 &amp;= (x-1)(x-1) \\ x + 5 &amp;= x^2 - 2x + 1 \\ 0 &amp;= x^2 - 3x - 4 \\ 0 &amp;= (x-4)(x+1) \\ \end{align} $$  Therefore, $x = -1$ and $x = 4$ satisfy the equation $0 = (x-4)(x+1)$.  But then I tried to plug them in the original problem $\sqrt{x+5}=x-1$:  $$ \begin{align} \sqrt{4 + 5} &amp;= 4 - 1 \\ \sqrt{9} &amp;= 3 \\ 3 &amp;= 3 \end{align} $$  So using 4 works as expected, but when using $-1$:  $$ \begin{align} \sqrt{-1 + 5} &amp;= -1 - 1 \\ \sqrt{4} &amp;= -2 \\ 2 &amp;\ne -2 \end{align} $$  At what stage am I going wrong?  And according the , the solution is $x = 4$. Creating a virtual microphone "In" vs. "after" for future talk The class will be over in 10 minutes. The class will be over after 10 minutes.  I know the first is correct but why? View list of files in ZIP archive on Linux I'm trying to compute the shadow of 3D objects on the ground. To do this I'm assuming parallel rays from Sun origin.   Let's assume the Sun direction is given by azimuth $\varphi$ and elevation $\theta$ angles. So the direction unit vector is retrieved as:  $$ \hat{\mathbf{s}} = \mathbf{R}_y\left(\theta\right)\mathbf{R}_z\left(\varphi\right)\hat{\boldsymbol{\imath}}$$  where $\mathbf{R}_{x/y/z}$ are the rotation matrices along the $x, y, z$ axes given by, for example, Euler-Rodrigues equation, and $\hat{\boldsymbol{\imath}}$ is the unit vector along $x$-axis.  A point on a ray-line has the equation:  \begin{equation} \mathbf{r} = \mathbf{r}_p + \lambda\hat{\mathbf{s}}  \tag{1} \end{equation}  being $\mathbf{r}_p$ a point on that line.  A point on a general plane is given by the equation:  \begin{equation} \left(\mathbf{r} - \mathbf{r_0}\right)\cdot \hat{\mathbf{n}} = 0 \tag{2} \end{equation}  being $\mathbf{r}_0$ a point on the plane and $\hat{\mathbf{n}}$ the unit vector normal to that plane.   Ensuring the validity of both equations (1) and (2) allows to compute the $\lambda$ multiplier and retrieve the position vector of the projected point (writing $\tilde{\mathbf{r}} = \mathbf{r_p} - \mathbf{r_o}$)  \begin{gather} \lambda = - \left(\frac{\tilde{\mathbf{r}}\cdot \hat{\mathbf{n}}} {\hat{\mathbf{s}}\cdot\hat{\mathbf{n}}}\right) \\ \mathbf{r}_\text{proj} = \mathbf{r}_p - \left(\frac{\tilde{\mathbf{r}}\cdot \hat{\mathbf{n}}} {\hat{\mathbf{s}}\cdot\hat{\mathbf{n}}}\right)\hat{\mathbf{s}} \tag{3} \end{gather}  Applying eq. (3) to all the vertices of a mesh $\left\{\mathbf{r}_{p_i}\right\}$ .   A problem can be noted: $\hat{\mathbf{s}}\cdot \hat{\mathbf{n}}$ can be zero (e.g. Sun is horizontal, i.e. the ray is ortogonal to the normal of plane).   Are there some conceptual mistakes in my reasoning? Is there a way to rewrite the operation as a linear operator applied to the set of vertices position vectors?   $$ \left[\mathbf{A}\right]\left\{\mathbf{r}_{p_i}\right\} = \mathbf{r}_\text{proj}$$ How to *quickly* get a list of files that have bad sectors/blocks/clusters/whatever? My wife's Windows 7 (64-bit) box has suddenly developed a SMART "disk is bad" status.  I'm attempting to copy everything off (no admonishments about lacking a backup regimen, please, I know already  :(  ) by creating a System Image across the network to a different machine, but it gets to a certain point and starts taking forever.  Doing a chkdsk reveals that certain files cause this by having many bad blocks (like dozens of thousands in a row, if the event log is any indication) and causing the system to do its standard try-to-recover-and-relocate-upon-access thing.  But this is taking so long, I'm afraid the disk will fail completely before I can get the damned thing copied.  However, several of the files so far have been ones that she has copies of elsewhere, so I am able to just delete them prior to retrying the backup to speed things up considerably.  So:  is there some tool or procedure that will try reading each file, and upon hitting a bad block, just tell me about it and skip to the next file?  So I can see which ones I can just dump and which I need to let it try to recover? What do I need in a camera to take non-blurry photos of a fast-moving child? Proving that expression is equivalent to the definition of derivative Let $f$ be differentiable at $x=a$.   Prove that if $x_n \to a^+$ and $y_n \to a^-$ then:  $$\lim_{n\to \infty} \frac{f(x_n)-f(y_n)}{x_n-y_n}=f'(a).$$  Every option that I think about seems to my very trivial, so I believe that I am doing something wrong.  Both numerator and denominator approach zero as $n\to\infty$ as the case of the formal definition of derivative, but it isn't guaranteed that the limits are equal (“$\frac{0}{0}$”).   Any direction? Value passed with request.setAttribute() is not available by request.getParameter() How to change geometry for one page without using \newgeometry or \restoregeometry