how do you find the mass and volume of a dog and a diamond ring?
963983
There are many ways to do this problem.  The best way to probably handle density, and perhaps mass even, is using Archimedes principle of buoyancy.  You can find a detailed discussion on www.wikipedia.com and search for Archimedes or buoyancy.  The main idea is that "the weight of the displaced fluid is directly proportional to the volume of the displaced fluid".  \n\ngood luck!
679198
Gold is measured by purity, not weight.  Pure gold is 24 carat but would be too soft for jewellery usage.  The most common used in US is 14 carat - meaning 14 parts pure gold and 10 parts of another alloy or metal.  In UK wedding rings are usually 18 carat whilst middle range jewellery is 9 carat - 9 parts pure gold and 15 parts another alloy or metal.\n\nThis 'carat' is not to be confused with the carat used in diamonds........that is a different proposition altogether and denotes weight..............
918429
use the formulas your given\nto find the volume of the cylinder\nthen pretend the semi hemi on  the top is a sphere find the volume and multiply by 1/2\n\nadd two totals to find volume
703001
To find the force P, its equal in size, but opposite in direction to the weight of the crate,resolved in the direction of the slope:\n\nP=Wcos60 or P=Wsin30\nP=(80*9.81)cos60=392.4N\n\nR is equal in size, but opposite in direction to the weight of the crate, resolved perpindicular to the slope\n\nR=Wcos30 or R=Wsin60\nR=(80*9.81)cos30=679.66N\n\nPlease note you need to use weight, not mass ( W=m*g)\n\nGo do the math!
1144840
mass divided by volume\nmeasured in kgm^-3
430786
The formula for the capacity or volume, V, of a cylinder is:\n\nV = Π(d^2)h/4\nwhere:\nΠ = 3.1416 (value up to 4 decimal places)\nd = diameter of the cylinder\nh = height of the cylinder\n\nSo, merely substitute the values that you have and you will get the capacity or volume of the cylinder. Just make sure to use similar unit of measures both for the diameter and the height.
735291
The best way to measure its volume is to get a measuring cylinder with water in, pop the ring in the water (making sure that is is completely covered) and then measuring the change in the water level.\n\nThat will give you the volume of the ring.
984991
Density = mass / volume\nTo find the volume of an irregular object submerge it in water the amount the water rises is the volume of the object.
74771
hecka!!  damn.  i miss california.\n\nwhat you have to do is realize that this question is asking about the volume of a rectangular shape that is 1 meter by 1 meter by X meters, where X is the height of the column.\n\nso, you know the density.. density is defined as mass divided by volume.  and you know the weight.. which is mass times acceleration.  here the acceleration is the acceleration of gravity "g" = 9.8 m/s^2.  \n\nso you can do a little algebra to get the equation volume = mass / density \nvolume = weight / (acceleration * density)\n\nyou know the weight (100,000 N), the acceleration (9.8 m/s^2) and the density (1.2 kg/m^3).  plug those numbers in and you get the VOLUME of the atmosphere, and since you know 2 of the dimensions of the box (it is 1 square meter across the bottom face, remember) you can figure out the 3rd pretty easily.\n\nsince this is your homework i wont just tell you the answer, but you should be pretty much able to crank it out from here.
359140
Use the equation:\n\n    v = u + at\n\nwhere:\n      v = 0 (because the dog comes to a stop)\n      u = 1.5 m/s\n      a = -.35 m/s^2\n\nSubstitute into the above and solve for t (answer is between 4.0 and 4.5 s)
16584
You can find the volume using a triple integral
1119363
Easy... fill the cone with some liquid.... then pour this liquid into a measuring cylinder and measure the amount of liquid... you'll get the volume of the cone.\n\nMathematically it is 1/3 the volume of Cylinder\nthat is 1/3(pie*r*r*h)
1087435
well, first start with the molecular weight of gold (197 by definition), then divide it by the number of atoms in a mole (6.2*10^23), which gives you a mass of 3.18*10^-22 g.
961619
PART 1:\nIf the ratio of the diameters is 1 : 2 : 3, then that means the diameters are 18 : 36 : 54.  You can solve this with similar ratios:\n\n1/18 = 2/x\nx = 36\n\n1/18 = 3/y\ny = 54\n\nSo the respective diameters are 18", 36" and 54".  (The radii would be half this or 9", 18", 27").\n\nHowever, the answer is to be expressed in feet so divide by 12:\n54 inches / 12 inches/ft = 4.5 feet.\n\nPART 2:\nVolume of the *middle* sphere.  The radius is 18" (one half of 36"), so just plug this into the formula:\nV = (4/3)(pi)(18)^3\nV = 7,776 pi (cubic inches)\n\nPART 3:\nThe volume of the *small* sphere = (4/3)(pi)(9)^3\nThe volume of the *middle* sphere = (4/3)(pi)(18)^3\n\nTo get the ratio between these two, just divide smallest / biggest\n= (4/3)(pi)(9^3) / (4/3)(pi)(18^3)\nThe 4/3 and pi cancel out\n= 9^3 / 18^3\n= 729 / 5,832\n= 1 : 8\n\nThis makes sense because the ratio of lengths is linear (1 : 2 : 3), the ratio of areas is squared (1 : 4 : 9) and the ratio of volumes is cubic (1 : 8 : 27).\n\nPART 4:\nWe have the 3 diameters (18", 36", 54").  the total height is the sum:\n18 + 36 + 54 = 108"\n\nConvert this to feet (divide by 12):\n108 inches / 12 inches/ft = 9 feet.
658673
As someone said, this is a mass not a volume but that doesn't mean you can't do it : this is all about conversions.\n\n1) Gas : The only thing you need to know is that a gram mole of any gas at standard conditions occupies 22.41 liters. You can work out the equivalent volumen for a kilogram mole for yourself (hint, 1000 time as much) and don't forget to change the molar volume (use the ideal gas law) for the conditions if they are not standard. This will be to a first approximation the same volume for a kmol of any gas (providing its not at not too extreme conditions)\n\n2) liquid : you need to work out the molar volume from the density and the molecular weigh. If you don't have that data, it can't be done. Unlike the gas, this cam be very differnent for different liquids, and also chages with conditions but in a manner that cannot be calculated with a simple formula \n\nWhe you come to do real work with any of these things, you usually find that volumes are given in standard and actual conditions. Be careful to use the conversion factors at the appropraite conditions or you will get it wrong\n\nEnjoy.
651714
Homework?\nOK, only a hint. Consider a cylinder made of the circle and with a height dh. What is the volume of this cylinder? \n\nNow displace the cylinder along the path over dh, etc.
1003850
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.