what dimensions could i use if i want a cylinder to hold 250,000ml of liquid?
447026
this is 250 liters, that is 250 cubic meters.\n\nVolume of the cylinder is (area of the base)*height.\nAs the base is a circle, its area is (squared radius)*Pi.\n\nIf you assume Pi almost equal to 22/7, you can take its inverse, 7/22, and multiply by 10: you get 70/22. Take this value in meters as the height and 5 meters as the radius of the base.
797268
MM = mass * 22.4 liters/volume\nMM = .125 * 22400/93.3 = 30.01
168817
3L because 1L equals to 1000ml and 3000/1000 equals to my answer
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
696169
Here you are: http://www.bourne.k12.ma.us/Class%20Links/Smith/Smith/images/graduated%20cylinder.jpg\n\n"A graduated cylinder is used for more precise volume measurements."
940101
you go on google, find out the meter length of a beverage can\nmultiply that by 1 billion (1 X 10^9 i believe) and divide by 1000 (giving you km)
720499
7.3m (width) by 2.4m (height)
968095
The density of alcohol (ethyl) is 0.789 g/mL, so it would take up more space (less dense).  Water is 1 g/mL
554878
look at grainger.com for a dual action air cylinder, (probably 1/2" diameter x 6" stroke), also look for a manual control valve.\nadd an air compressor and it's almost built for you.\n\nmake sure to add some guards as the air cylinder will deliver at least 500 lbs force
732197
The density can be assumed to be 1.0 g/ml.  \n\nSo you have 25 ml x 1.0 g/ml = 25 g H2O\n\nThe molar mass of water is 18 g/mol, so you have:\n\n25 g H2O/18 g H2O/mol H2O = 1.4 mol H2O
185076
Get the dimensions of the rectangular tank into the same units, then get the volume. the unit could be in^3, cm^3, m^3, ft^3, mi^3, mm^3, km^3, etc. Have them be converted to cm^3 using the ff. conversion units:\n\n1 mi = 5280 ft\n1 ft = 12 in\n1 in = 2.54 cm\n1 m = 10 dm = 100 cm = 1000mm = .1 dam = .01 hm = .001 km\n\nNow, 1 cm^3 = 1 mL. Have it converted into liters using 1000 mL = 1 L. To get gallons, use the ff. conversion unit:\n3.8 L = 1 gal\n\nHope this helps ^_^
70150
so you have to find the surface area of a cylinder that is 35diameter and 40 ft tall. or \n((3.14)17.5^2)40)
542238
a very slender graduated cylinder with markings for individual mL.
693858
Difficult without paper, but the principle is to work out a formula for the total area of aluminium sheet which still maintaining 1000cm3 volume.\n\nIn the following, r = radius of cylinder and h = height of can.\n\nArea of aluminium = 2*pi*r^2 [the top and bottom] + 2*pi*r*h [because 2*pi*r is the circumference of the can]\n\nBecause pi*r^2*h = 1000, you can substitute h = 1000/pi*r^2 into the Area equation.\n\nThen differentiate with respect to r, and when dA/dr = 0, that's the most economical way to make the can.\n\nIt should turn out at close to r = 5.4; h = 10.9
501658
a British imperial capacity measure (liquid or dry) equal to 2 gallons
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
752805
The first thing you need to do is figure out which of the three liquids is going to be the one that limits the maximum quantity of the mixture that can be made.  Since the ratios are 5:2:1 you need 5 parts x, 2 parts y, and 1 part z.  You have 25 gallons of x and, if that's 5 'parts' then 1 'part' must be 5 gallons.  Now you need 2 'parts' (or 10 gallons) of y, and you do have that much in stock.  Lastly 1 'part' (5 gallons) of z is required and you have that much in stock as well.  So you'll mix 25 gallons (5 'parts') of x, 10 gallons (2 'parts') of y, and 5 gallons (1 'part') of z for a total of 25+10+5 = 40 gallons of the mixture.\n\nHope that helps.\n\n\nDoug