It is round with a side arm on it and the old water evaporated after 7 yrs of use. There were specific filling instructions originally that we have lost.
202149
Atmospheric pressure will support a column of air about 10 metres high.   The exact height varies from day to day depending on atmospheric pressure; high and low pressure systems.   To start with, partly fill your container with coloured water, with the hole at the bottom submerged in more coloured water.   If there was no air in your transparent ball, it would need to be 10 metres in diameterfor you to see a boundary between the empty space at the top and the water column supported by the atmosphere pushing down on the liquid outside, but because you've trapped some air inside, this air pushes down on the water so the water isn't as high.
274579
i haven't done this in a while.\n\nThe volume of the reservoir is equal to (1/2)*(height)*("base")*length.  the reservoir is a rectangular prism with the "base" being the distance across the surface of the water.  We are given the shape of the full volume so we can scale down to the level of the water to find what the volume in the reservoir at its current level.\n\nif you think about the cross-section of the volume, and use half of that forming a right triangle, we can determine the ratio of the surface (the "base") to the height (water depth).  In this case, it is (40/2)/(20*sqrt(3)) or 1/sqrt(3).  so ("base"/2)/(height) = 1/sqrt(3)  and then "base" = (2*height)/sqrt(3).\n\nsubstitute that into the volume equation so that\nvolume = [(height^2)*length]/sqrt(3) or literally,\nV = [(h^2)*L]/sqrt(3).  differentiate with respect to time,\n(dV/dt) = [2*h*(dh/dt)*L]/sqrt(3) and substitute the numbers in:\ndV/dt = 200 - 50 = 150\nh = 10*sqrt(3)\nL = 450\nand solve for dh/dt which is the rate of change of the water level in yds/hour.  these are odd units, you might want to change to inches/hour.\n\ni did it this way and got 1/60 yds/hr or 0.6 inches/hr.\n\nbut again it has been a long time....
610214
there are a few diff ways for this.  the way i do it is with a rotating lazer level. you can rent them at most rental places, another way is with a transit/builders level but if you dont know how...you will be lost cause you wont know how to read it. or the cheap and easy way.   get some clear plastic hose hold the ends up and fill it with water. create a "benchmark". use a post or somthing else you can tie it to and walk away without spilling. on the other end make a leg for you to rest it on roughley as high as the post. now as you move around you can see if it isflat by marking the level in the tube cause water will seek its own level. i prob left out some things in the explanition but im sure you will pick it up.....it was good enough for the romans and it can work for you.
407814
ist a holeeeee\n\nbut u kinda can take it out...well...cover it with cement anyway..so its technically not there anymore
650484
Able to be dissolved by water.
402593
LOTS OF WATER, FRUIT ECT.
310896
There is a proverbial slip between the cup and the lip. How much a cup can hold is flexible  you can half fill a big cup or overfill a small cup with the same content. aesthetically , one should be sure to leave some space on the top.\n\nActually the organ in question keeps avrying in size so a lil small or big is all adjustable
667545
the water was in the gun
530495
First of all, you need to determine the volumne equation of the tank, which is inverted cone, \n\nV = (1/3)πr^2 h\n\nSince you know the relationship between the radius (r) and the height (h):\n\nr / h = 3 / 11\nr = 3h/11\n\n(note that the unit conversion is not needed above, because both would cancel with each other)\n\nNow, sub the above equation into the volume equation (the reason you do this is to get rid of the variable "r", since when you take the derivative, you can get dh/dt)\n\nV = (π/3) * (3h/11)^2 *h\nV = 3πh^3 / 121\n\nNow, you can take the derivative because on the left side, you can get dV/dt (which is what you want to find) and on the right side, you can get dh/dt (which you are given):\n\ndV/dt = (3π/121) * 3h^2 * dh/dt\n\nSince you know the rate of change of the height (dh/dt) is 21 cm/min at height 250 cm (just to convert back to cm for consistency because the rate is in cm^3/min), sub these information into the above equation yields:\n\ndV/dt = (3π/121) * 3 * (250)^2 * 21\ndV/dt ~ 306694.74 cm^3/min\n\nRemember, this is NET rate. In other words, (flow in - flow out). Otherwise, if flow out > flow in, the water level will not rise!! So,\n\ndV/dt = flow in - flow out = 306694.74\nflow in - 7800 = 306694.74\nflow in = 314494.74 cm^3/min\n\n(sorry, I mis-read the height as cm instead of m before!! The above should be the right answer)
8389
It is a clear liquid.
262716
You have not provide suficiant information. The hieght is of no cosequence to the diameter since you failed to provide a base dimension of lenght or hieght of which could be the same. The water level is not realitive because no parementers have been established. Is the vessel full? Further more: a diameter of a spout is meaningless. Absolute stupidity.
25485
They probably pump it up and fill it. Lke allwater towers there are pipes going both up and down. Once filled though, they can leave it alone for awhile. Usually there are sensers inside that relay to a computer and alerts when the tower is getting low. In small towns I know sometimes they put too much water in and it isn't used. Thats why they sometimes flush it. But I would research the topic in an online enclycopedia.
307906
It's like those "Moen" shower or kitchen sink heads with the flexible "thingy". Check the image below. lol. Oh, it would use a gasket for water use, yes.
525835
Let the height be h (inches), radius r.\n\nNow by proportion, h = 5r at every level.\n\nHence volume is given by\n\nV = (1/3)pi*r^2*h which is\n       (5/3)*pi*r^3\n\n"Rate at which water is drained" means find dV/dt\n\ndV/dt = 5*pi*r^2*(dr/dt),\n\nand dr/dt = (1/5)*(dh/dt)\n               = -2/5\n\nYou can finish from there.
321464
It may be surface tension.
463073
We have had some problems with it in the Bay area. Something to do with leaking cavity fluids, or the like. Legislation was proposed. I do not remember if there was any outcome, or not. It has been a while and I was not paying strict attention, as I am a living organism sort of guy.
662209
No realistic container is perfectly impervious to slow leakage, so the contents will always be lost over time.  However, with industry standard aluminum beverage cans, this should not happen within only a few months, or even a few years.  (Decades, maybe, but even that would be doubtful.)  So there *must* have been some form of leakage, albeit very small.  The fact that they were stored in a cupboard suggests that a tiny imperfection, most likely at a seam, was formed over time due to internal gas pressure.  Such a leak may still be so minute as to be unnoticeable via simple visual inspection, but still significant enough to allow slow evaporation over time, especially with the assistance of internal pressure.\n\nTry this:  Take one of those empty, seemingly-leakless cans and submerge it in water for a few days, or even a few weeks.  The deeper the water, the better, so that there is some pressure outside the can.  Then, take it out of the water and shake it a bit to see if perhaps a few drops have somehow found their way inside.\n\n(This all, of course, assumes that someone did not *intentionally* form a tiny leak and drink away all the contents.  It is possible to create a leak large enough for drinking [albeit very slowly], but small enough to be difficult to detect.  The easiest method is to make the leak at the top, usually where the pull-tab is anchored to the can, and then drink it upside down, shaking the can a bit to build up internal pressure and thus helping push the contents out more quickly.)