I am creating a drain that lets water in but keeps pieces of plastic out. I need to know the size of a standard water droplet.
30611
Assume I have a inverse cone which holds 200ml water. I am going to cut the tip of the cone to create a small hole. How to calculate the maximum radius of the hole that the water will still stay in the container?
32078
This adapter is attached to an outdoor spigot:    It is near the location of a former hot tub so I believe it has something to do with pressure control. When I attach a hose to it and turn it on if the hose is not flowing (e.g. a closed nozzle at the end) the water leaks (more like squirts, lots of it) from this adapter, it appears to be from the holes and not the connection to the spigot or the hose.  What is this thing? Is there a way to utilize it? I'd like to remove it so I can use this spigot but when I try to unscrew it with a wrench it does not budge - in fact the spigot itself starts to rotate. I wasn't really able to tighten a pipe wrench to the spigot due to the available clearance &amp; pipe width.  There appears to be a rivot on the under side of the adapter but it is difficult to tell, it is somewhat rusted over.  Follow Up  My original concern was water leaking from it when it was used. While a plumber was on site doing a different job I asked him to look at this to see if something was wrong. He said the vacuum breaker wasn't working properly and ended up replacing the entire hose bib (with new vacuum breaker), I presume because he was not able to remove the vacuum breaker as well or it wasn't salvageable. Now no water spews out when it is used.
62368
A friend of mine has a homework question and we're having some trouble figuring out what physical mechanisms come into play for this.     An underwater swimmer sees a spherical air bubble that appears to have a diameter $d=$ 1.5 cm. What is its actual diameter?   We are having trouble, specifically, thinking of physical mechanisms that would change the apparent sizes of objects when seen underwater. Is it caused by refraction at the curved surface of the bubble?
103909
I am a blender beginner and after watching some tutorials online I figured I'd give modeling and animation a try. I am trying to make an oil hourglass animation that will move the liquid downwards., I have a physical reference of it. The only issue is that my liquid doesn't seem to want to pass through 2 holes I made on the container mesh, I suspect it's because of the wrong facing normals right outside of it:  I managed to get the liquid to flow inside of the top sphere.  but It won't flow into the rest of the mesh.  Will I have to create a new mesh? Help is appreciated, thank you!
88147
Problem: I have a cylindrical vessel of height $H$ and radius $R$. There are two liquid layers in the vessel. The first has density $D_1$ and height $h_1$, the second has density $D_2$ and height $h_2$. The second liquid is floating on the first liquid (thus $D_2 &lt; D_1$) and they are both entirely within the vessel ($h_1+h_2 \le H$). Now I have a cube of density $D$ and edge length $n$. The cube fits into the vessel ($n \le \sqrt2R$) but it might not be completely inside it ($n&gt;H$ is possible). I place it into the vessel such that one side is horizontal (the cube is not lopsided). Some liquid might overflow as a result. How do I use Archimedes' principle to calculate the topmost liquid level after placing the cube? I can ignore damping and other physical effects. EDIT: Note that the only values I know are $H, R, D_1, h_1, D_2, h_2, n$ and their abovementioned constraints.  My workings: I tried using the formula directly, where $\text{weight of cube}=\text{weight of displaced liquid}$. I first get $Dn^3=D_2n^2s_2$ where $s_2$ is the height submerged in the second liquid. If $s_2 \le h_2$ then the cube is only submerged in the second liquid. So the answer is $\min(H, h_1+h_2+{n^2s_2\over\pi R^2})$, accounting for overflows. This case is simple. But if $s_2&gt;h_2$ then the cube can be submerged in the first liquid too. I thought of subtracting the weight of the second liquid displaced to obtain the height submerged in first liquid, but realize that I don't know that. The weight displaced depends on the final liquid height, which can be affected by overflows, whether the cube fits totally within the two liquids, etc. In fact, it can be the case that after the liquid level rises, the cube now displaces more of the second liquid, causing it to not be submerged in the first (is that even possible?) It seems very messy and I have no idea how to start. :( Does anyone have a nice solution to this problem?
56821
A conical vessel with semi-vertical angle 30 degrees and height 10.5 cm has a thin lid. A sphere kept in the vessel touches the lid. The radius of the sphere in cm is ...
50835
Question:  Consider the following square sheet of metal (grey) with the side s = 10 cm. From it, we want to cut the parts of a cylindrical can with a lid and a bottom (bblack). Determine the radius and height of the can when the volume is maximal.   The rectangle and circles have to be cut in entire, singular pieces, so no trickery to minimize waste and no diagonal placement of the rectangle.  Solution has to use optimization / differentiation.  Attempted solution:  In order for this to work out the circumference of the circles and one side of the rectangle has to be identical. Otherwise they will not fit together in a cylindrical can.  This leads us to two cases that needs to be treated separately.  Case 1: The rectangle is standing upright.  Case 2: The rectangle is lying on its longest side.  There are two cases that needs to be considered separately.  Case 1:  The right-hand side of the above image must be the circumference of a circle, so  $$s_1 = 2r\pi$$  The bottom-side must be the height of the cylinder plus two radii of the circles:  $$s_1 = h + 2r$$  Putting these together becomes:  $$2r\pi = h + 2r = 10$$  Solving for h:  $$h = 10 - 2r$$  The volume of the can:  $$V = r^2\pi h$$  Adding in the earlier formula to get rid of h:  $$V = r^2\pi h = r^2\pi (10 - 2r) = 10r^2\pi - 2r^3\pi$$  Taking the derivative:  $$V' = 20\pi r - 6\pi r^2$$  Setting derivative equal to zero:  $$V' = 0 \Rightarrow r = \frac{10}{3}$$  (neglecting trivial r = 0 solution)  Height is then:  $$h = 10 - 2r = 10 - \frac{20}{3} = \frac{10}{3}$$  But this is not really possible given the side of the metal sheet cannot be more than 10 cm and $$2r\pi &gt; 10$$.  Case 2:  The bottom becomes:  $$2r + 2r\pi = 10 \Rightarrow r= \frac{5}{1+\pi}$$  The right side becomes:  $$h = 10$$  However, none of these answers match the solution:  $$r = \frac{5}{\pi}$$  $$h = 10(1-\frac{1}{\pi})$$  Did I get the starting point for the possible cases right? Where did I go wrong? How come it appears that no calculus is needed in the second step?
34884
I got the large round container without a lid, used for washing clothes in the below picture. But I don't know how to call it.     What do you name that plastic washing container?    A plastic washing tub, a plastic basin or something?
109335
After collecting GPS location for road culverts I would like to calculate the $ values for each culvert based on their diameter and length in ArcMap. Existing number fields in the attribute table are Culvert Diameter, Length and the Value field I would like to populate. Unfortunately I have no programming experience so I'm hoping for your help. I adjusted a basic if/then formula I found in this forum and added a multiplication function in the return line. Unfortunately this doesn't work...      def CulvertValue(Diameter):       if Diameter &gt; 0 and Diameter &lt; 14:         return (Length*50)       elif Diameter &gt; 14 and Diameter &lt; 20:         return (Length*100)       else:         return "N/A"   And the result      CulvertValue(!Diameter!)   So let's say the culvert is under 14 inches in diameter and 20 feet long, knowing that a culvert with this diameter costs $50 per foot, the result would be 1,000. I would obviously have to add more elif lines to cover the different price categories. Any thoughts how this could be realized with a Python script?
40942
The  comes in a 6-pin package, 4 of which are connected to the drain, and only 1 to the source. Why this difference? The source sees the same current as the drain, doesn't it?
54567
I am reading a computer science puzzles book.   And I get the following question -  "You have a five quart jug, a three quart jug and unlimited supply of water. How would you come up with exactly four quarts of water ?"  The question is pretty simple and I figured the sequence of "water transfers" to solve it.  But there was an interesting side-note that I did not know.  The side-note said that as long as the two jugs size relatively prime, it is possible to find a pour sequence for any value between one and the sum of the jug sizes.  It is true for 5 and 3.   But I want to know if it is really generally true or not ? If it is, it is really cool! What is the theorem called ? Who discovered it ?
49378
I just bought a house built in 1984 it has PEX plumbing. I need to remove a toilet to install a new floor and I noticed the water supply line to the toilet connects to a water shut off valve that I’ve never seen before. It looks like a PVC bonnet or nut cap thing With a half crescent ring that you turn left or right to turn the water on or off. Are these reliable or should I replace with a quarter turn ball valve. I really don’t want to do anything unnecessary but a PVC shut off valve gives me anxiety as an inexperienced DIY homeowner
123958
I am trying to find a small enough microcontroller for a project, but I am not familiar with the notation used to denote the sizes in this image:  What does that mean when it says 12,20/11,80 SQ and 7,50 TYP.  The image is from .  OR is this not where I find the physical dimensions? If these do not denote the physical dimensions of the part, then where in the datasheet is that written?
39168
When a drop of water falls into a reservoir of water from a high enough altitude, water droplets will splash ():    My question: Does the water in those droplets come from the original drop or from the reservoir?
51007
How can I make water follow a path?
54415
A vessel is filled with liquid,3 parts of which are water and 5 parts syrup.How much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup?  What is the best possible way to solve this question?
121508
I followed @Micha's instruction and successfully used the modules in GRASS. I use r.watershed to get the basin raster and the drainage line：  and I use r.water.outlet to get the watershed of one outlet：  I got the whole basin area of the green outlet point, but all I want is the little part of the original basin area, like this：part B is what I want  This schematic photo shows the idea of what I want：  Can I achieve this goal using r.water.outlet or other modules in GRASS？ Or all I can do is using the intersect tool in Geoprocessing tools in QGIS？  I want to find the watershed area, but using QGIS. I have following files to work with：  stream network shapefile(polyline) point shapefile clipped from the polyline shapefile a whole island(the stream is inside the island) DEM(I got 2 types:.lan &amp; .tiff)  I want to find the total watershed area located upstream of every point in the network, here's the schematic diagram：  the green, red, and yellow line enclose the watershed of each point(here I only take 3 points as the example), and I want to calculate their area. I already got the DEM file, and I must cut the region that contains the stream off, should I polygonize it first？ Can I directly use this form(.lan or .tiff) to work with GRASS？