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An imperial fluid ounce is 1 ⁄ 20 of an imperial pint, 1 ⁄ 160 of an imperial gallon or exactly 28.4130625 mL. A US customary fluid ounce is 1 ⁄ 16 of a US liquid pint and 1 ⁄ 128 of a US liquid gallon or exactly 29.5735295625 mL, making it about 4.08% larger than the imperial fluid ounce. A US food labeling fluid ounce is exactly 30 mL.
The tun (Old English: tunne, Latin: tunellus, Medieval Latin: tunna) is an English unit of liquid volume (not weight), used for measuring wine, [1] oil or honey. Typically a large vat or vessel, most often holding 252 wine gallons, but occasionally other sizes (e.g. 256, 240 and 208 gallons) were also used. [2] The modern tun is about 954 litres.
Plastic tubing. A tube, or tubing, is a long hollow cylinder used for moving fluids (liquids or gases) or to protect electrical or optical cables and wires.. The terms "pipe" and "tube" are almost interchangeable, although minor distinctions exist — generally, a tube has tighter engineering requirements than a pipe.
Measurement of volume by displacement, (a) before and (b) after an object has been submerged. The amount by which the liquid rises in the cylinder (∆V) is equal to the volume of the object. In fluid mechanics, displacement occurs when an object is largely immersed in a fluid, pushing it out of the way and taking its place. The volume of the ...
A pipe is a tubular section or hollow cylinder, usually but not necessarily of circular cross-section, used mainly to convey substances which can flow — liquids and gases , slurries, powders and masses of small solids. It can also be used for structural applications; a hollow pipe is far stiffer per unit weight than the solid members.
[nb 2] Note that a 252-gallon tun of wine has a mass of approximately 2060 pounds, [4] between a short ton (2000 pounds) and a long ton (2240 pounds). The tun is approximately the volume of a cylinder with both diameter and height of 42 inches, as the gallon was originally a cylinder with diameter of 7 inches and height of 6.
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Much more work is needed to find the volume if we use disc integration.First, we would need to solve = () for x.Next, because the volume is hollow in the middle, we would need two functions: one that defined an outer solid and one that defined the inner hollow.