the linear density of a 24 m long rod is 36/ (x+1)^2 kg /m, where x is measured in meters from one end of the?
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use average value of integral 1/(b-a)*integral f(x) dx where f(x) is the given density function with limits 0 to 24\n \n1. factor out the 36 and use a u substitution w/u=x+1 dont forget to change limits of integration to u values\n\n2. 1.44 kg/m^3 is approx. answer
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T=2pi*sqrt(L/g) for pedulum with center of mass at the end, so rod length is twice that, so T=2pi*sqrt(L_rod/2/g), the answer is L=g*T^2/2/pi^2=0.243m.\n\nThe length is not 12.2cm because the center of mass of the rod is half distance of the length of the rod
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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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If the density is 'd' kg/m^3 then the weight is '9.81d' Newton. [remember weight is force]
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If as described the entire meter stick is in balance, then the center of its mass/gravity would be where the support is, at the 22.4cm mark\n\naha ... here we go then,\n\n142.3 * (22.4-9)  = 134 * (x - 22.4)\n\nsolve this for x to get the answer\n( think of the entire stick as a ball placed at position x)
101463
that depends on the material's specific gravity\nweight in kg = specific gravity * volume in ltrs
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Density is mass over volume.\n\np = m/v\n1.74 = m/43.5\n1.74 * 43.5 = 75.69 (mass)\n\n75.69 g
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The same way that Archimedes did -- determine the density.\n\nMeasure the mass of the coin, then determine its volume by water displacement.\n\nDensity = mass/volume\n\nThe density of pure (24 karat) gold is 19.3 grams/cm3\n\nFor other alloys and karat ratings of gold, check this table:\nhttp://www.24carat.co.uk/densityofgoldandothermetalsframe.html\n\nMost U.S. gold coins in current production are 22 karat gold.\n\nYou may need to do some more research to determine the expected gold content of your coin, and then you can compare it to the chart for a density value.
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the weight will be according 2 density of the material \n\nfor example for water 1  liter wighs 1 kg
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2.36 meters  = 236 cm \n2.54 cm = 1 inch\n\nso 236 /  2.54 = inches\n\n92.91 inches
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D = M/V\n\nDensity= Mass/Volume
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Density is mass/volume.  Find its mass by weighing it. Find its volume by immersing it in water and measuring how much water it displaces.  That is its volume.\nDivide mass by volume and you've got density.\nP.
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look at column 7.  all of the densities appear there in grams/cm3.\n\nhttp://www.science.co.il/PTelements.asp
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Density = mass / Volume\n\nYou are given the mass of the cyllinder (m = 4 kg).\nYou are given the dimentions of of the cyllinder which you can then use the find the volume.\n\nThe volume of a cyllinder = (pi*r^2)*h\nwhere pi is the constant pi (the ratio of the circumfrance of a circle to its diameter, ~3.1415), r is the radius, and h is the height.\nUsing this formula, I calculate the volume of the cyllinder to be about 9425 cm^3.\n\nD = m / v\nD = 4000 g / 9425 cm^3\nD = 4.24 E-1 g/cm^3 or 424 kg/m^3\nThis is a low density by comparison to many other things, for example, water has a density of about 1 g/cm^3.\n\nNext Time, do your own home work.
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I answered this question for jc yesterday!  Search for questions to find the answer.\n\n\nJust integrate p(x) along the rod.
280201
Density = mass/volume
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PV = nRT\nsince n = g/MM\nMM = gRT/PV\n\nso...\n\nMM = 1.097g * (62.3637 L·Torr/Kmol) * 372.15K / 743 torr * 0.256L\nMM = 133.85 g/mol