How much actual liquid can a cylinder hold measuring 30" diameter x 60" height?
493858
this will be  169646.00329384883487698274269709 cubic inches
354060
0.2 gal. or 25.6 fluid oz.
732310
5.00/0.5 = 1 to 10 dilution\n\n600/10= 60 mL of stock needed\n\nthe rest (540mL) is water
418026
Volume of Cylinder is pi(r-squared)times height.\nDiameter is 10 inches, so raius is 5 inches\n\npi(25)(12)=942 cubic inches\n\nUsing the volume conversion in the link below, we find this to be roughly 4.08 gallons
239849
.7M x 20mL = .25M x (Volume of new IA solution)\nVol = (.7 x 20) / .25\nVol= 56mL
28543
volume of a cylinder is pi*r^2*height:\npi * (1.5 ft) ^ 2 * (6 ft)\n42.4115008 ft^3\n\nso the volume of the cylinder is about 42.4115 cubic feet. There are about 7.48 gallons in a cubic foot, so: \n\n42.4115008 ft^3 * 7.48051945 gal/ft^3 = 317.260057 gal.\n\nSo a cylinder 6 feet tall with diameter of 3 feet can hold about 317.26 gallons of water.
1111363
graduated cylinder or callibrated pipette are the answers to the test questions in my classroom.  And in scientific situations, volume is measured in liters/litres- except when measured in cc's (cubic centimenters)
804606
For cubic measure:  L x W x H = CI / 231 = G\nL = Length\nW = Width\nH = Height\nCI = Cubic Inches (Volume)\nG = Gallons\nMeasure the Length, Width, and Height in Inches.\n231 Cubic Inches to one Gallon.\nConvert Volume to Cubic Inches\nLength x Width x Height  = Cubic Inches  / 231 = Gallons\n\nFor Right cylinder measure: Pi x r'2 x H = CI / 231 = G\nPi = 3.14159\nr = radius (half of diameter) '2 = squared\nH = Height (or length)\nCI = Cubic Inches (Volume)\nConvert Volume to Cubic Inches. \nPi x r squared x H = CI / 231 = gallons
619237
1st, the volume of the tank is (4/3)π(4.5)^3 = 381.7035 cubic meters.  The part left unfilled is a spherical cap whose volume formula is (π/6)(3r1² + h²)h, where r1 is the radius of the base of the cap, and h is its height.  The height here is 9 - 6.5 = 2.5, and r1² = 4.5² - (6.5-4.5)² = 16.25.  \n\nSo the volume of the unfilled cap is (π/6)(3*16.25 - 2.5²)*2.5 = 55.6324 cubic meters.  We subtract that from the volume of the tank to get 326.0711 cubic meters of water when the tank is filled 6.5 meters deep.  \n\nNow one liter fills a cube 1/10 of a meter on a side (10 cm), so a bucket is 1/100 of a cubic meter.  Multiplying the required volume gives 32,607.11 buckets, which rounds up to 32,608.\n\nMathgirl mistakenly assumed the volume of the tank was proportional to the depth of water, which would be true for a cylinder, but not for a sphere.
246106
You need the exact geometry of the hemispheric ends if you want the exact volume. Or fill the tank with water and measure how much liters you put inside.\nLPG tanks are filled up to an 80% only, they have to leave space for evaporation so I don't think there would be an important error if you take the volume of a cylinder plus the volumes of two spherical caps (the hemispheric ends)\n\nSee link for formula http://mathforum.org/dr.math/faq/formulas/faq.sphere.html
384987
Too few details... you need to disclose base area of tank, shape of it (is your tank cylindrical, spherical, cuboid, pyramid, etc).\n\nBUT, assuming it's a cuboid OR cylindrical (meaning has a round base and sides aren't tapered),\n\nif 210 barrels is 6 feet tall,\nthen 210 barrels => (6 x 12) inches\n     210 => 72 inches\nso   210/2.1 => 72/2.1\n     100 => 720/21 inches\n          = 34 6/21 inches\n          = 2 feet 10 6/21 inches\n          approx 2 feet 10.3 inches
601391
Depends on the height.\n\nV=pi*r^2*h\n\n1 ft^3 = 7.48 gallons (U.S.)
761667
cylinder\n1.volume=pir^2h\n2.LSA=2pirh\n3.TSA=2pir(h+r)\n\nsphere\n1.volume=(4/3)pir^3\n2.area=4pir^2\n\nhemisphere\n1.volume=(2/3)pir^3\n2.basearea=pir^2\n3.LSA=2pir^2\n4.TSA=3pir^2\n\ncone\n1.volume=(1/3)pir^2h\n2.LSA=pirl\n3.TSA=pir(l+r)\n\ncube\n1.volume=a^3\n2.surface area=6a^2\n\ncuboid\n1.volume=lbh\n2.surface area=2(lb+lh+bh)
342773
Want to see something neat??\n\nType "19 gallons in cubic feet." into google.\n\nWorks for all kinds of conversions.
475972
The volume of a cylinder is given by\n\nV = pi*r^2*h\n\nAnd the surface area of a cylinder is given by\n\nSA = 2*pi*r^2 + 2*pi*r*h\n= 2*pi*r(r + h)\n\nNow, we don't have enough information to specifically find either the radii or heights - we only have a volume.  But we can still work with that based on the fact that the cylinders are similar.\n\nIf the cylinders are similar, their measurements are proportional.  In other words, the radii are proportional and the heights are proportional (r1/r2 = h1/h2 and r1/h1 = r2/h2).  So let's just assume that the larger cylinder has a radius of 3 and a height of 3 (plug in to the volume formula and it works out to 27*pi cm^3, but for the sake of this argument, we'll ignore the pi for now).  For the smaller cylinder to be similar, it would have to have a radius of 2 and a height of 2 (again ignoring pi).\n\nThe surface area of the larger cylinder is then\n\nSA = 2*pi*r(r + h)\n= 2*pi*3(3 + 3)\n= 36*pi\n\nAnd the smaller one would have a surface area of\nSA = 2*pi*r(r + h)\n= 2*pi*2(2 + 2)\n= 16*pi\n\nSo your ratio would then be 36*pi / 16*pi = 9/4.  (If we went back and corrected out radii and heights for pi, they would have all cancelled out by this step.)\n\nNow, what if my initial assumption was wrong and the radii and heights weren't 2 and 3?  Well, we can still work with that, but we'd end up with the same ratio of 9/4.  If the smaller cylinder had a height of 4, then the radius would have to be sqrt(2) for the volume to work out to be 8.  So the ratio of radius to height would be sqrt(2)/4.  Because the two cylinders are similar, this ratio must hold for the larger one.\n\n(Crazy math warning!)\nSo r/h = sqrt(2)/4, so r = h*sqrt(2)/4.  We can find the radius and height of the larger cylinder now:\n\nV = pi*r^2*h\n27 = pi*(h*sqrt(2)/4)^2*h\n27 = pi*(h^2/8)*h\n27 = (pi*h^3)/8\n216 = pi*h^3   <-  we'll drop the pi at this point - same reason as before\n216 = h^3\nh = 6\n\nr = h*sqrt(2)/4\n= 6*sqrt(2)/4\n= 3*sqrt(2)/2\n\nNow let's compare surface areas of the two:\n\nSA(big) = 2*pi*r(r + h)\n= 2*pi*(3*sqrt(2)/2)(3*sqrt(2)/2 + 6)\n= (3*pi*sqrt(2))(3*sqrt(2)/2 + 6)\n= 9*pi + 18*pi*sqrt(2)\n= 9*pi(1 + 2sqrt(2))\n\nSA(small) = 2*pi*r(r + h)\n= 2*pi*(sqrt(2))(sqrt(2) + 4)\n= 4*pi + 8*pi*sqrt(2)\n= 4*pi(1 + 2sqrt(2))\n\nAnd if you took this ratio of big to small, you have 9*pi(1+2sqrt(2)) / 4*pi(1+2sqrt(2)) = 9/4!\n\nSo my original assumption is right, and the ratio turns out to be 9/4 for these two cylinders.\n\nNow I used cylinders for this (I have no idea why), but it turns out that it works for any solid.  There's a relationship among these measurements - if we compare cubes:\n\nRatio of a side = s1/s2\nRatio of area = (s1)^2/(s2)^2\nRatio of volume = (s1)^3/(s2)^3\n\nSo for our case, the ratio of the area can be found by taking the cube root of the ratio of volumes, and then squaring it - 9/4.
1162775
if u evr seen a water tank they r usually round\nhttp://www2.warnerbros.com/main/homepage/homepage.html\n\nvolume = sqr ft x ht\nsqr ft  (of a circle) = pi x half the diameter(radius) squared\n\nvol = pi x (1/2x38) ^2x 100\nvol=3.1415926535897932384626433832795 x (1/2 x 38)^2 x 100\nvol=113,411.49479459153590850142613639 cubic feet
936045
Boy, you probably need to consider the liquid containing bottle's dimensions too. \nThe total average height of liquid will be 0.8 m + half the height in the bottle (say.\n\nAssuming the average height of liquid in the bottle = 100 mm = 0.1 m,\ntotal average height = 0.8 + 0.1 = 0.9 m\n\npressure exerted = potential energy= densityx acceleration due to gravityx height\n\nLength of needle can be neglected because they are usually installed horizontally and hardly add to the height. Also, it can be presumed that the flexible tubing and the needle were fiiled with the liquid at the comencement of the I.V drip. and the pressure difference if any due to this may be neglected.\n\nKinetic energy imparted = potential energy\n\n1/2 m v^2 = mgh\nv^2 = 2 gh\nv^2 = 2 x 9.8 x 0,9\nv= (2 x 9.8 x 0,9)^1/2\nv = 4.2 m/s= 420 cm/sec\n\nvolume to be injected/time in seconds\n= 490/ (35x60) = 1.4 cm3/ sec\n\nvolume flow (V) = speed (S)x cross section area\narea = V/S= 1.4/420 = 1/300 cm^2 \narea = 1/2 pi r^2\n1/300 = 1/2 x 22/7 x r^2\n\nr^2 = 1/300 x 2 x 7/22 cm2\n      = 7/ 3300 cm2\n\n  r    = .046cm = 0.46 mm\ndiameter (d) = 2r= 2 x 0.46= 0.92 mm................. Answer