a solid brass statue weighs 190 N in air. what is its volume?
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Use Archimedes Principal to answer this question:\n\n"The buoyant force is equal to the weight of the displaced fluid."\n\nThus, in air, the net force is equal to the statue's true weight minus the buoyant force.\n\n190 N = W - Fb\n\nThe statue's weight, W, is simply the density of the brass times its volume times gravity, and the buoyant force is equal to the density of the air times the statue's volume times gravity:\n\n190 N = d-brass * g * V - d-air * g * V\n\nRe-writing this expression provides the volume:\n\nV = 190 N / (d-brass * g - d-air * g)\n\nFrom Google, I found that the following typical densities (air is assumed to be taken at sea-level):\n\nd-brass = 8400 kg/m^3\nd-air = 1.2255 kg/m^3\nd-water = 1000 kg/m^3\n\nSubstituting these values provides:\n\nV = 190 N / (8400 * 9.8 - 1.2255 * 9.8)\n\nV = .00231 m^3\n\nWhen the statue is submerged in water, we again use Archimedes Principal to find its apparent weight:\n\nW = d-brass * g * V - d-water * g * V\n\nW = (.0231)*(9.8)(8400 - 1000)\n\nW = 167.4 N\n\nGood Luck :)
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2000N is the weight of the object...\nOr is that too obvious?
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Volume = 1/48 * 1*1 = 1/48 cu ft\ndensity = 480 pounds per cu ft\n\nmass = density*volume = 480* 1/48 lb = 10 lb
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309grams.\nThe dry mass is 58grams.
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1. The bouyant force of the water on the object is the difference in the reading, i.e. BF=(.0554-.0370)g\n                                   =0.0184gN\n\nBased on Archimedes' Principle, BF =Volume*Density of water displaced.\n \n0.018g=Volume*1000g\nVolume=.018/1000\n            =1.8*10^-5m3\n\n2. Density of Object=Mass/Volume\n                                =.0554/1.8*10^-5\n                                =3078kg/m3
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Mass is a constant; weight varies by gravitational pull.
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an ore sample weighs 15N (newton) in air. when the sample is suspended by a light cord and totally immersed in water, the tension in the cord is 10.80 N. find the volume and the total density of the sample.. \n\n\nApparent loss in weight=(15 -10.8 )N\nMass of water displaced=(1.50-1.08)kg\n                                        =0.42 kg\nVolume=0.42 l(since cm3= l)\n1 kg= 1 l (litre)\n\nTherefore,volume =0.42 litre\n                             =21/50000 m3\n\n\nsince density equals to mass over volume;\ndensity=0.42 (divide by) 21/50000\n           =100000 kg m-3\n\n\n\n\n\nThis is called Archimedes principle\n\nBelow is a table that will help you to change units from Wikipedia. \n\n\n\nName of unitSymbolDefinitionRelation to SI units\n\ncubic metre (SI unit)\nm³≡ 1 m × 1 m × 1 m\nlitre\nL≡ 1 dm³\nlambda\nλ≡ 1 mm³= 1 μL\ndrop (metric)\n ≡ 1/20 mL= 0.05 mL\nminim (Imperial)\nmin≡ 1/480 fl oz (Imp) = 1/60 fl dr (Imp)≈ 0.059 193 880 208 333 mL\nminim (U.S.)min≡ 1/480 US fl oz = 1/60 US fl dr= 0.061 611 519 921 875 mL\ndrop (U.S.) (alt)gtt≡ 1/456 US fl oz≈ 0.064 854 231 mL\ndrop (Imperial) (alt)gtt≡ 1/1824 gi (Imp)≈ 0.077 886 684 mL\ndrop (U.S.)gtt≡ 1/360 US fl oz= 0.082 148 693 229 1(6) mL\ndrop (medical) ≡ 1/12 mL= 0.08(3) mL\ndrop (Imperial)gtt≡ 1/288 fl oz (Imp)= 0.098 656 467 013 (8) mL\ndash (U.S.) ≡ 1/96 US fl oz = ½ US pinch= 0.308 057 599 609 375 mL\ndash (Imperial) ≡ 1/384 gi (Imp) = ½ pinch (Imp)= 0.369 961 751 302 08(3) mL\npinch (U.S.) ≡ 1/48 US fl oz = 1/8 US tsp= 0.616 115 199 218 75 mL\npinch (Imperial) ≡ 1/192 gi (Imp) = 1/8 tsp (Imp)= 0.739 923 502 604 1(6) mL\nfluid scruple (Imperial)\nfl s≡ 1/24 fl oz (Imp)= 1.183 877 604 1(6) mL\nfluid drachm (Imperial)\nfl dr≡ 1/8 fl oz (Imp)= 3.551 632 812 5 mL\nfluid dram (U.S.); U.S. fluidramfl dr≡ 1/8 US fl oz= 3.696 691 195 312 5 mL\nteaspoon (Canadian)\ntsp≡ 1/6 fl oz (Imp)≈ 4.735 510 416 667 mL\nteaspoon (U.S.)tsp≡ 1/6 US fl oz= 4.928 921 595 mL\nteaspoon (metric) ≡ 5 mL\nteaspoon (Imperial)tsp≡ 1/24 gi (Imp)= 5.919 388 020 8(3) mL\ndessertspoon (Imperial)\n ≡ 1/12 gi (Imp)= 11.838 776 041 (6) mL\ntablespoon (Canadian)\ntbsp≡ ½ fl oz (Imp)= 14.206 531 25 mL\ntablespoon (U.S.)tbsp≡ ½ US fl oz= 14.786 764 782 5 mL\ntablespoon (metric) ≡ 15 mL\ncubic inch\ncu in≡ 1 in³= 16.387 064 mL\ntablespoon (Imperial)tbsp≡ 5/8 fl oz (Imp)= 17.758 164 062 5 mL\npony ≡ 3/4 US fl oz= 22.180 147 171 875 mL\nfluid ounce (Imperial)\nfl oz (Imp)≡ 1/160 gal (Imp)= 28.413 062 5 mL\nfluid ounce (U.S.)fl oz (US)≡ 1/128 gal (US)= 29.573 529 562 5 mL\nshot ≡ 1 US fl oz= 29.573 529 562 5 mL\njigger ≡ 1 ½ US fl oz= 44.360 294 343 75 mL\ngill (U.S.)\ngi (US)≡ 4 US fl oz= 118.294 118 25 mL\ngill (Imperial); Noggingi (Imp); nog≡ 5 fl oz (Imp)= 142.065 312 5 mL\ncup (Canadian)\nc (CA)≡ 8 fl oz (Imp)= 227.3045 mL\ncup (U.S.)c (US)≡8 US fl oz ≡ 1/16 gal (US)= 236.588 236 5 mL\ncup (metric)c≡ 250 mL\nbreakfast cup ≡ 10 fl oz (Imp)= 284.130 625 mL\npint (U.S. fluid)\npt (US fl)≡ 1/8 gal (US)= 473.176 473 mL\npint (U.S. dry)pt (US dry)≡ 1/4 bu (US lvl)= 0.550 610 471 357 5 L\npint (Imperial)pt (Imp)≡ 1/8 gal (Imp)= 0.568 261 25 L\nfifth ≡ 1/5 US gal= 0.757 082 356 8 L\nquart (U.S. fluid)\nqt (US)≡ 1/4 gal (US fl)= 0.946 352 946 L\nquart (U.S. dry)qt (US)≡ 1/32 bu (US lvl) = ¼ gal (US dry)= 1.101 220 942 715 L\nquart (Imperial)qt (Imp)≡ ¼ gal (Imp)= 1.136 522 5 L\npottle; quartern ≡ ½ gal (Imp) = 80 fl oz (Imp)= 2.273 045 L\nboard-foot\nfbm≡ 144 cu in= 2.359 737 216 L\ngallon (U.S. fluid; Wine)\ngal (US)≡ 231 cu in= 3.785 411 784 L\ngallon (U.S. dry)gal (US)≡ 1/8 bu (US lvl)= 4.404 883 770 86 L\ngallon (Imperial)gal (Imp)≡= 4.546 09 L\nbeer gallonbeer gal≡ 282 cu in= 4.621 152 048 L\npeck (U.S. dry)\npk≡ ¼ US lvl bu= 8.809 767 541 72 L\npeck (Imperial)\npk≡ 2 gal (Imp)= 9.092 18 L\nbucket (Imperial)\nbkt≡ 4 gal (Imp)= 18.184 36 L\ntimber foot\n ≡ 1 cu ft= 28.316 846 592
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1 mass\n2 magnitude of separation
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the buoyancy of the object and its volume.
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The density of gold is 19.3*10^3 kg/m^3. So to calculate the statue's volume\n115 kg * (1 m^3/19.3*10^3 kg) = .00596 m^3. That mass of that volume of water \n.00596 m^3 * (1 kg/1*10^3 m^3) = 5.96 kg.\n\nThe weight of the statue: 115 kg*9.8 m/s^2 = 1127 Newtons\nThe weight of the 5.96 kg water: 5.96 kg*9.8 m/s^2 = 58.4 Newtons. This is the amount of bouyancy provided by the water.\n\nSo the tension in the cable while the statue is still under water is 1127 - 58.4 Newtons =1069 Newtons.\nAfter the statue is out of the water, tension is 1127 Newtons.
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It has no mass. But it has relative mass of ≤ 1. It is a matter.
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Conversion Calculator here\nhttp://education.yahoo.com/reference/weights_and_measures/index?p=volume
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Find the weight the object (with a precision scales) and its volume (put it in water and look at the amount displaced).\n\nCalculate the density = weight (in gram) / volume (in milliliter)\n\nYou will find a number lying between 19.3 (density of pure gold) and 8.92 (density of pure copper). Say you found 18 gram/ml\n\nThen the amount of gold is   (18-8.92)/(19.3-8,92) = 87.5% gold
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Designed by Silvio Gazzaniga and produced by Bertoni, Milano, it stands 36.5 cm (14.4 in) tall and is made of 5 kg (11 lbs.) of 18 carat (75%) solid gold with a base (13 cm [5.1 in] in diameter) containing two layers of malachite. The trophy, which weighs 6.175 kg total, depicts two human figures holding up the Earth.
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the amount of pressure exerted on an object
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Lift of WHAT?   Crane, Ballon, Brassier, Airplane wing....???\nGive us a CLUE
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this is in the guinness world records\nit is 1399.92 pounds