A block floating on water contained in a vessel is pushed inside until it touches the bottom of the vessel and then released. Will the block come back?   Bouyant force acts because the pressure on the bottom of the block is more than that at the top.   If the block was cubical and its top surface is at a distance $h$ from the surface of water, and had a side length $a$, then the pressure on top will be $P_0+\rho gh$  Pressure on bottom surface will be $P_0+\rho g(h+a)$  Pressure difference = $\rho g a$.   Force acting is = $\rho g a\cdot a^2=a^3\rho g$  If the bottom surface of the cube were to touch the bottom of vessel, then there is no water beneth it to exert  pressure upwards. So won't the block remain in the bottom even if its density is less than that of the liquid?
Consider a cement block of the same density as sea-water resting on the (perfectly polished) seafloor. I also assume the side-faces of the cement block are highly polished, so that no vertical hydrostatic forces can attack on them. A cable is attached to the top of the block.  My question ist: How much force must one apply to the cable before the cement block begins to lift off the seafloor?  The normal response would be to say that only a negligible force is needed - as the block is practically suspended in the water.  However, as the bottom surface of the cement block is not in contact with the water, it is NOT experiencing the hydrostatic pressure from below - only from above.  Before the crane starts lifting, the (zero-sum) forces on the cement block are:  $$F_{Seawater Pressure} +F_{Gravity on Cement Block} - F_{Seafloor Resistance} = 0$$  as the crane starts applying Force, the Seafloor will reduce its Resistance in the same measure, leaving the forces on the cement block zero:  $$F_{Seawater Pressure} +F_{Gravity on Cement Block} - (F_{Seafloor Resistance}+F_{Crane}) = 0$$  Just before the Cement block lifts, the Seafloor will stop applying Force and the Crane will be applying the only upward Force:  $$F_{Crane} = F_{Seawater Pressure} +F_{Gravity on Cement Block}$$  Once the Cement block lifts off the sea-floor, hydrostatic pressure kicks in on the lower cement surface, suspending the block, but before that - boy does it take a lot of force to lift it!  What is wrong with this reasoning?
Given the following data:  972, 975, 985, 993, 993, 995, 998, 1001, 1004, 1007, 1008, 1009, 1011, 1015, 1016, 1020, 1022, 1032  I calculate the lower quartile as follows:  Number of values: 18, Number of 'gaps': 17.  Lower quartile is at 17/4 + 1 = 5.25th value. 5.25th value is 993 + (995 - 993) x 0.25 = 993.5  Upper quartile is at 3 x 17/4 + 1 = 13.75th value. 13.75th value is 1011 + (1015 - 1011) x 0.75 = 1014  Both the Excel QUARTILE function and the R quantile function agree with me. However the book I am using (Understanding Statistics by Graham Upton and Ian Cook) give different answers (993 and 1015 respectively). I'm confused. Which is correct? How can I test whether my camera is working after it was dropped? Issue : How can I evaluate my cameras performance synthetically?  Scenario : Bought a Nikon D3100 an year back. Photos looked good enough. The camera fell, zoom lens effected.  But works. Cleaned lens using soap, water etc.  Carefully. Now photos look much worse.  How do I know if Photos are really bad?  Is there a tool?    Sample Photo  Now (Bad)   Original (Good)   Please help me realize whether my camera has gone bad.  Can you tell your personal opinion also? Javascript && operator as if statement? is there a way to get the option "Organize Desktop by Name" in Ubuntu 19.10 same as in 18.04? the option in Ubuntu 18.04 is "Organize Desktop by name"    But there is no option available in Ubuntu 19.10 Do I have 32-bit or 64-bit? Minecraft 1.7.4 server issues with loading world Class not found loading JDBC org.postgresql.Driver Wording of "possible/exact duplicates" when closing the question When marking question for "duplicate", the reason in dialog window is "exact duplicate", yet when the process is completed there's a comment added that says "possible duplicate". Maybe those two should be evened out to either "possible duplicate" or "exact duplicate"?  Edit: Take  for instance, it says "Possible" on top but "exact" at the bottom    and the top remark was added by the system only after the question was closed, i.e. after it was concluded that it was an exact duplicate rather than a possible one.  So the question is: can the text on top be changed to "Exact Duplicate" please? How do I replace the *first instance* of a string in .NET? single email limit exceeded salesforce very frequently..? Can I add some HTML in my SO Profile 'About me' text box to show some button from linkedin? Bootstrap 3 - Why is row class is wider than its container? 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 $x = 4$. I have noticed that I hardly ever use the \negthinspace command  \!.  This is because I don't know how to use it properly (I do use other spacing commands quite a bit, because I can better understand the guidelines of when they might be useful). When might it be considered appropriate to use this spacing command? Simple Past vs. Present Perfect: "was" vs. "has been" In a preadditive category, products are coproducts $\renewcommand{\phi}{\varphi}$ $\require{AMScd}$ My question is different from .  I want to see how in a preadditive category $\mathcal{C}$ a product is a coproduct, following . I'll write the proof up to the point where I get stuck.  Let $A=A_1,B=A_2$ be objects and suppose their product $P$ does exist, with projections $\pi_i:P\rightarrow A_i$. By considering a wedge  $$\begin{CD}A @&lt;{id_A}&lt;&lt;  A @&gt;{0_{AB}}&gt;&gt;  B\end{CD}$$  and the obvious analogous one with $A$ and $B$ exchanged, where $0_{A,B}$ is the zero morphism, we see that the product property implies the existence of two unique morphisms $k_i\in\mathcal{C}[A_i,P]$ s.t.  \begin{align*} \pi_j\circ k_i=\begin{cases} \mathrm{id}_i &amp;\text{if }i=j\\ 0_{A_i, A_j} &amp;\text{if }i\neq j \end{cases}. \end{align*} Using these equations and the definition of zero morphism, it's easy to see that the morphism $m\equiv k_1\circ \pi_1+ k_2\circ \pi_2$ satisfies $\pi_i\circ m=\pi_i$.   This is where I get stuck: In the linked proof, they now state that this implies $m=\mathrm{id}_P.$ But I don't see why this should be true.   Specifically, if $\phi\in\mathcal{C}[S,P]$ for some arbitrary object $S$, it is not apparent why $m\circ\phi=\phi$. All that we know is that there are some morphisms $x_i\in\mathcal{C}[S,A_i]$ that factor through $P$ via the universal property, but these are not necessarily unique.  And we learn even less about $m$ by looking at morphisms in $\mathcal{C}[P,S]$.  What am I missing here? Acne underneath skin Introduction  Okay, I've almost won my war. The battle, its almost won... However, I must still claim victory in my final battle. The battle against the little enemies, the ones that have fought with my blade of benzoyl peroxide every single day.  Problem  Through spot treatments I have been able to fell every single one of these spots, but only the ones on my face. I've been using a spot treatment with benzoyl peroxide in, which has worked really well for me. However, my issue comes here:     There are spots underneath my skin, I think. The areas where they come through are a bit red.   So, because there is still slightly red skin on my face, and I think this is causing my acne to constantly return. I need to stop this, and end my war once and for all.  What I've tried   Washing my face with a rag and sponge every day, three times a day Changing my pillow case every day Using a benzoyl peroxide spot treatment Minimizing how much I touch my face Not interfering with the spots with my fingers, not picking at them, not playing with them. Just letting spot creams eradicate them.   Conclusion  Seriously, this driving me up the wall. I spot cream in the evening, by the time tomorrow has ended and its time for a shower, they've all come back. Its like they're coming in waves, damn waves! I must put an end to this.  So, how can I prevent acne from coming back within the space of a single school day? How can I get rid of these once and for all? What types of measures could I try? I have no reason to visit my doctor, its only acne, but I have no idea what I can try now.  Is there even a point in me trying? I acknowledge that I am a teenager, but no others that I know seem to have acne/spots like I do. How can I get distribution name and version number in a simple shell script? Why is `[` a shell builtin and `[[` a shell keyword? How to prove that if $\forall x \in (a,b)$ Lebesgue integral $\int_{(a,x)}fd\lambda=0$, then $f(x)=0$ $\lambda$-almost everywhere? Let $f:(a,b)\rightarrow\mathbb{R}$.  The statement to prove is that if $\forall x \in (a,b)$ Lebesgue integral $\int_{(a,x)}fd\lambda=0$, then $f(x)=0$ $\lambda$-almost everywhere.  So if it wouldn't be true then we would have $\forall x \in (a,b):\int_{(a,x)}f_+d\lambda=\int_{(a,x)}f_-d\lambda\neq0$, so every interval contains positive and negative values of the function. It is  such an $f$ so my guess would be that this function is not Lebesgue measurable and we can't take this integral.  Thanks in advance!