what volume of water will be displaced when a 10g of weight is dropped in a bucket of water?
470447
You need to know the specific gravity of the weight.  (Specific gravity is weight per unit of volume, typically expressed in grams per milliliter).  The specific gravity of water is generally accepted as 1 (meaning that it weighs 1 gram per milliliter).  The equation to determine volume of water displaced would be:\n\n[10g/(S.G. of the weight)] X 1 = Vol. of water displaced\n\nNotes:\nS.G. is short for Specific Gravity\n\nThe 1 in the equation is the S.G. of water.\n\nIn actuality, the S.G. of water varies with temperature, so you may need to adjust the S.G. of water to the correct value if the problem gives you that information.\n\nThe answer will be volume of water in grams.
1096402
If the bucket is swinging around you at a constant speed, the water will be flat against the bottom of the pail.  As the speed is changed, the water will lag behind or press ahead depending whether it is accelerating or decelerating.
1027854
In this following, I'll assume that the drop maintains a spherical shape.  Otherwise, the math will get horrendous, and one probably can't get an analytical solution.\n\nThe surface area of a sphere is given by:\n\nS = 4*pi*r^2\n\nThe mass of a sphere of density D is simply the volume times the density.  The volume of a sphere with radius r is given by:\n\nV = 4/3 * pi * r^3,\n\nso the mass is given by:\n\nM = 4/3 * pi * r^3 * D.\n\nWe know that the drop gains mass according to:\n\ndM/dt = k*S = k*4*pi*r^2\n\nSubstitute the expression for the mass as a function of radius into this differential equation:\n\n4/3*pi*D*d(r^3)/dt = k*4*pi*r^2\n\nSimplifying, we get:\n\nd(r^3)/dt = 3*k*r^2/D\n\nlet u = r^3\n\nthen\n\ndu/dt = 3*k*u^(2/3)/D\n\nu^(-2/3) du = 3*k/D dt\n\nIntegrate this to obtain\n\n3(u^1/3 -u0^1/3) = 3*k*(t-t0)/D\n\nwhere u0 and t0 are the constants of integration, and in this problem, we can let t0 = 0.\n\nSubstituting back for u and simplifying:\n\nr - r0 = k*t/D\n\nr(t) = k*t/D + r0\n\nThis gives the radius as a function of time, which is the desired result.\n\nIn the absence of friction and neglecting the change in gravitational acceleration as a function of height, the drop would have a constant acceleration of g (9.807 m/s^2).  If we include air friction (and continue to assume a spherical raindrop) then the net acceleration is:\n\na = g - 0.5 * C * D * A * v^2/M\n\nwhere C is the drag coefficient (= 1/2 for a sphere), \nA is the cross sectional area\nM is the mass\nv is the velocity\nD is the density\n\nSubstituting expressions for the mass and area in terms of the radius, we get:\n\na = g - (0.5 * C * D * pi * r^2 * v^2)/(4/3 * pi * r^3 * D)\na = g - 3/8 * C * v^2/r\n\nif C = 0.5 (for a sphere)\na(t) = dv(t)/dt = g - (3*v(t)^2)/(16*r(t))\n\nFrom here, one would need to substitute the expression derived previously for r as a function of t, solve the differential equation to find v(t), then take the derivative of v(t) with respect to t to get a(t) in terms of t.  This is straightforward, but involves some tedious algebra that I'm not inclined to do at the moment!
152879
No, it will not.  The other answers to this question are on the right track, in that they point out that an object's density is the key thing in most situations, since if it is less dense than the fluid it is in, it will displace fluid equal to its weight before it is completely submerged, and hence float.\n\nBut in *this* problem, we are told from the outset that there is not enough water to be displaced.  Therefore, the object will displace all the water, all 1 lb. of it, and still not have displaced enough water to equal its weight.  Therefore, it will rest on the bottom, i.e. not float.
923764
a) F = ma\n    F = 0.0412 kg * 9.8m/s^2\n    F = 0.404 N\n\nb) F = 0.0370 kg * 9.8 m/s^2\n    F = 0.363 N\n\nc) The difference in "weight" is equivalent the weight of the displaced volume of water...  So...\n   0.404N - 0.363N = V * 1000 kg/m^3 * 9.8 m/s^2\n   0.041N = V * 9800N/m^3\n   V = 4.18 * 10^-6 m^3   or  4.18 cm^3\n\nd) Oh Juliet you are my density.... (bad joke!)\n     Density = m/V\n     D = 0.0412 kg / 4.18  * 10^-6 m^3\n     D = 9848 kg/m^3
257654
Say the anchor weighs 15 pounds in air and displaces 2 pounds of water.\nBefore the drop, the boat displaces its weight plus the 15 lb weight in air of the anchor. This is the maximum displacement situation, so the water is at maximum height.\nAfter the drop, if the anchor hits bottom, the boat displaces its weight only and the anchor displaces 2 lb of water, the minimum displacement situation, so the water is at minimum height.\nAfter the drop, if the anchor doesn't reach the bottom, the boat displaces its weight plus the 13 lb weight in water of the anchor. The anchor displaces 2 lb water. The total displacement is the same as the maximum displacement situation, so the water is at maximum height.
63147
Archimedes' principle states that the buoyant force experienced by a submerged object is equal to the weight of the liquid displaced by the object. Experimentally this appears in the fact that the submerged object apparently weighs less by an amount equal to the weight of the liquid displaced. The buoyant force can be expressed as:\n\nFb = W(air) - W(liquid) = dgv \n\nwhere d is the density of the liquid, g is the acceleration of gravity and v is the volume of the immersed object (or the immersed part of the body if it floats).
368357
The test liquid will sink to the bottom because it is more dense than water. Water has a density of 1.0 g/ml. There fore it can be simply thought of as "lighter."\n\nThere will be two visible separate layers.
835572
Gotta approach this using basic principles.\n\nEnergy is the integral of force applied over a distance.\nTheres two sets of forces working:\nfluid and gravitational.\n\nSo you make your control volume, and look at it.\nYou want to raise the water in it, thats mgh or rho*v*g*h\n\nYou also have to exert a pressure to push against the water ahead of it.  The downward pressure acting on your control volume is f/a so the force you must overcome is p*a and you are traversing distance h so the energy is p*a*h\n\nTherefore to raise the fluid vertically against a down pressure and against gravity the minimum energy (real world is lossy) is \nE = h*(mg+pa)
151746
Weigh it.  Each gallon weighs 10 pounds.
760373
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
823430
the buoyancy of the object and its volume.
613715
First find the volume of the object, i.e metal, by placing it in water and then u know the amount of water displaced.\nIf it displaces half the water than it's weight, then u know that it's specific gravity is twice that of water.
302446
I agree with the first answer:  Here is something that might help you with maths:  When you have a problem such as this make a 'rough estimate' of the answer.  In your problem almost all the tomatoes were water, nearly 100kg, the water was reduced by 1% - that's very little, it being 1/100 th. So you nearly have the same weight as you started with.  It cannot be 50kg as that would have meant that half the water 50% would have evaporated.  I hope you follow. Go back to my first point, the rough estimate should have been 'nearly 100kg but a little less'.  I hope I have made sense and that this gives you a good idea as to how you should proceed in the future.  Always work out a 'nearly right' answer first.....
781338
The density is 0.357 g/mL. This is because specific gravity is the ratio of the density of the unknown liquid to the density of water, 1.0 g/mL.\n\nIf, say, you wanted to know the density of the liquid in pounds per gallon, you would multiply 0.357 by the density of water, which is 8.320 lb/gal
951151
Weight the person has to lift = mass of the bucket * accelaration due to gravvity\nWeight the person has to lift = 24 * 9.8 J \nWork done = weight lifted * DIstace\n8*1000 J = 24 * 9.8 * Distance\nDistance = 24*9.8/8*1000 meters\nDistance = 3*9.8/1000 meters\nWell this is the answer. Right now i am feeling quite lazy to calculate all this. Hope you can manage with this answer. \nThanks
810641
Use Archimedes's buoyancy principal.  The weight outside represents the response of mass to the Earth's gravity.  The weight in water represents the displaced water's weight.  Use the ratio to get the answer.  HINT: if you convert to mass, the answer almost gives itself away.\nThe displaced water has a well known density.  That density will give the volume of the both the displaced water and the object displacing it.