When a block of volume 1.00 × 10–3 m3 is hung from a spring scale as shown in Figure A, the scale reads 10.0 N. When the same block is then placed in an unknown liquid, it floats with 2/3 of its volume submerged as suggested in Figure B. The density of water is 1.00 × 103 kg/m3.\nDetermine the mass of the block and the density of the unknown liquid?
133095
ok, i'm assuming that the block is being hung into water. otherwise the problem's too easy.\nso 1.00x10^-3 m3 of water will be 1 kg, or 9.8 N. thus block weighs 19.8 N or 2.02 kg.\nThen, 2/3*1.00x10^-3 = vol of unknown liquid displaced = 6.67x10^-4 m3. since its floating, we also know that amount of liquid weighs 19.8N, or 2.02 kg, so 2.02kg/6.67x10^-4 = 3.03x10^3 kg/m3\nI hope that's right.
200901
to calculate the density you don't need to worry about the temprature and pressure if you already know the volume, as you do.\n\ndensity = mass per unit volume   \n\nit's standard units are g/cm^3\nyou volume is in mL, but that is fine since 1mL=1cm\n\ndensity = 1.6/15000\n=1.07x10^-4 g/cm^3
548185
density=mass/volume\n\nd=125.6g/750cm^3=.1674g/cm^3\n\napproximately .167g/cm^3 (dont forget the cubed. if u forget it u'll get it wrong)
155450
look at column 7.  all of the densities appear there in grams/cm3.\n\nhttp://www.science.co.il/PTelements.asp
349307
Poiseuille's Law: dV/dt = (πR^4/8η)(∆P/L)\n\nWe're given\n\ndV = 490 cm^3 or 0.00049 m^3\ndt = 35 min or 2100 sec\nL = 2.5 cm or 0.025 m\n\nη - viscosity of water is 8.90 × 10-4 Pa•s or 8.90 × 10-3 dyne-sec/cm2at about 25°C.\n\nWe also have ∆p = ρg∆h\n\nρ is the density of pure water (1000 kg/m^3)\ng is the acceleration due to gravity (9.81 m/s^2)\n∆h is the height of the fluid column (we are given 0.8 m)\n\nHence, ∆p = (1000)(9.81)(0.8) = 7848 N/m^2\n\nNow we solve for D, the diameter of the needle, which is = 2R\n\nD = 2R = 2[(8ηLdV)/(π∆Pdt)]^-4\n\n= 2 [(8(8.90×10^-4)(0.025)(0.00049)) / ((3.147)(7848)(2100))]^(1/4)\n\n= 2(2.025 x 10^-4) = 4.05 x 10^-4 m or 0.405 mm
270055
density of pure water is 62.4 lbs/sq ft.\n\nSo, load = dAgh\n\nor A = 40000/(62.4*5*32) = 4.01 sq ft.
437535
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 :)
536076
Volume of water\n=volume of flask\n=mass/density\n=183.07/.99754\n=183.52g\n=.18352 kg\n1000gm\n=1kg\n=1lit.\n.18352kg\n=.18352lit
557939
mass =concentration\n-------\nvolume
543539
Looks like this is your homework. anyway, i'll only answer two of your questions.\n\n1. the density of water is almost 1 g/mL. and 1 liter = 1000 mL. hence the mass of 1 liter of water is 1000 g.\n\n2. since 1 liter of water is 1000 g or 1.00 kg, its weight is 1.00 * 9.8 = 9.80 N
67233
The water pressure t a given depth is give as the weight of the water above that depth divided by the area it is being applied.\n\nSo imagine a column of water extending from the top of the lake down the depth.  Calculate the volume of this column of water then using the density of water (1 grams / mL) find the mass.  The take the mass of the water and find its weight (W = mg).  The weight it is the force exerted by the water, but pressure is the force of the water divided by the area it is applied.  So divide the force by the area of the base.\n\nEDIT:\nI see you added the density of water (even though I think my density [1 g/cc] is better).  I will leave you to do your own homework.\n[I also refuse to work with such an ugly unit, I prefer metric]
160292
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.
361830
For water, it's easy: 1mL=1g. For everything else, you need to know the density. Density is g/mL. So, if you have 3g of something that has a density of 6g/mL, you have 0.5mL of stuff.
485592
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
148027
density is mass divided by volume - so if you know the volume and the density, then you multiply them to get the mass  (which is synonomous with weight)
62020
in reality, it is the surface tension (adherance of molecules to one another) and density of the fluid\nbecause the question does not state that the object is suspened in the fluid\n\notherwise the force acting to keep an object buoyant would need to equate and thus counteract the objects weight (which is the force of gravity pulling said object down)\n\nmass X volume = density which is a factor in determining a force acting upon the object; a force working to cause the object to disperse within the less dense fluid or conversely the fluid to compress and colapse the object\n\nSo the author of the question given is expecting the answer: weight
489698
The idea is that the increase in hydrostatic pressure of the water will cause a decrease in volume of the balloon. You are told that the temperature of the ballon is constant so that you can make the assumption PV = constant = Pa Va, where Pa Va are the pressure and volume at atmospheric conditions.\n\nThe bouyant force is simply the weight of water displaced, and therefore is it proportional to the volume of the balloon. You can assume that the water is incompressible and therefore constant density.\n\nSo, saying that the bouyant force is 1/2.4 of the surface value means that the volume has decreased to Va/2.4.\n\nPa Va = Pb Vb; Vb = Va/2.4 --> Pb = 2.4 Pa.\nRemember that this means absolute pressure.\nThe absolute hydrostatic pressure in the water is:\nPw = Pa + (rho) g h\nwhere h is the depth below the surface.\nSo: 2.4 Pa = Pa + (rho) g h\nh = (1.4 Pa)/((rho) g)\nPa = 101 000 N/m^2\nrho = 1000 kg/m^3\ng = 9.8 m/s^2\nh = 14.4 m