What is the volume of the bubble?
108410
You can do this one yourself,\n\nPV = nRT
730124
1. volume = h*2*pi*r^2 =1000 so r^2= 1000/2/pi/h\n\nsurface area = 2*pi*(h+2*r^2)\nsurface area / volume = (h+2*r^2)/h/r^2 \n=1/r^2+2/h  = 2*pi*h/1000+2/h  which is a minimum when h is about 17.84  [(1000/pi)^0.5]
102106
Volume is equal to legnth times width times height. The dimensions you gave are the length, width, and height. So all you do is multiply the dimensions together.\n\nVolume=6*10*4=60*4=240
427529
The air in the bubble is enclosed by the smallest possible surface area -- a sphere.
499283
Hmm...\nThe formula for a cylinder volume is:\nArea * height\nThe area is a circle so it's pi * r * r\n\nVol = pi * r * r * h\n\nTherefore\n\nh = Vol / (pi * r * r)\nh = 121 / pi * r * r\nh = 38.515496228238671256069870736148 * r * r\nor h = 38.52 * (r square)
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.
966711
The volume of the water in the cup at any moment is:\n\nV = (1/3) pi [ (Rw)^2 + (1)Rw + (1)^2 ] \n   = (1/3) pi [ (Rw)^2 + Rw + 1 ] h\n\nRw = Radius of upper part of water\nh = height of water\n\nRw / h = 2 / 6\n\n6 Rw = 2 h\n\nRw = 1/3 h\n\nSubstitute in to the volume equation\n\nV = (1/3) pi [ (1/9) h^2 + (1/3) h + 1] h\n = (1/3) pi [ (1/9) h^3 + (1/3) h^2 + h ]\n\ndV / dt  = (1/3) pi [ (1/3) h^2 + (2/3) h + 1] (dh / dt)\n\nSolve for dh /dt  and plug in the values h=4 and dV/dt = 3
665350
As the volume is constant,\n(P/T)initial = (P/T) final\n1/(25+273.15) = P/(50+273.15)\nP = 1.084 atm\n\n(Taking the pressure of the initial gas was 1 atm.
526091
i'm forevever blowing bubbles
466950
Volume = Area x height\nArea = r x r x pi\npi is approximated by the fraction 22/7\nTherefore Volume of cylinder = 4 x4 x 22/7 x7 = 352 units^3
934849
291.13 L is the correct volume.\n\nCheck your algebra when you are solving for V1.
625137
Volume is L * W * H ... if the Volumen is 1000^3 and it's a cube, then L, W and H all = 1000cm\n\nDuh.
125839
I'm not sure. But the volume of a sphere is (4/3)(pi)(r)^(3). \n2) You know that the diameter is double the radius. So your radius is 1.\n3) Then, your volume is (4/3)(pi). \n4)Density=(mass/Volume)\n5)Density of water is 1.00 mL=1.00g/m^3\n6)MASS=(Density)(Volume)=((1.0g/m^3)(4/3)pi). Note that cubic meters cancel. The mass is (4/3)pi grams.\n\nHope I helped.
556558
Not the surface area of the water in the bowl, the vacuum in the bottle, or partial vacuum and the height of water in the bowl. The bottle will let air from the bowl in as soon as the water in the bowl drops...see the bubbles?
150532
The bubbles are steam, or water vapor.  They are NOT from hydrogen and oxygen.
760290
Popcorn ball. . . . \n\nGoogle search:  volume sphere (should have used "volume sphere," but it turned out not to matter)\n\nhttp://www.aaamath.com/geo79-volume-sphere.html\n\n4/3 (pi) r^3\n\n4/3 times pi times (6 to the 3rd power) \n\nequals 4/3 times pi times 216. . . and work from there.\n\nGoogle search:  volume cylinder\n\nhttp://www.321know.com/exp79_x5.htm\n\n(area of the base) * height =(pi r squared times height)\n\nPi times (16 squared) times 275. . . work from there.
144053
Pay more attention in class! And assuming it is an ideal gas, use PV=nRT. Since there is not change in volume you can basically drop it from the equation and make P=nRT. If you like algebra you can leave it in but it will cancel out eventually. You could solve for 'n' first using initial P and T, and then plug and chug.