meters, what is the volume of solid A? Solids A and B are similiar.
133383
height is one dimensional, volume is 3 dimensional.\n\nLet ratio of height of A and B be r:R\n\n2.5 r = R\n\n250 / R^3 = volume of A / r^3\n\n= 250 * r^3/R^3\n\n= 250 * 2.5^3\n\n= 16
14470
Let No of square is N\nPerimeter = 2xN + 2 \n\nAssumptions:\nThey are arranged in a row.\nThey are of same size.\nThey are of UNIT size.
231037
piece a and piece b\n\na = 3b\n\na + b = 56\n\n4b = 56\n\nb = 14\n\na = 42\n\nANSWER : One piece was 14 cm and the other was 42 cm.
673698
The volume of the sphere is 4/3*pi*r^3\n\nThe volume of the cylinder is pi*r^2*h\n\nSince the cylinder's h is the same as 2r, then the volume becomes pi*r^2*2r which is the same as 2*pi*r^3.\n\nThe fraction of the space taken up by the sphere is the same as:\n\n4/3*pi*r^3\n-------------\n2*pi*r^3\n\nwhich is the same as:\n\n4/3\n----\n2\n\nwhich reduces down to 2/3
273441
Both are Same Weight , 1 Kg\nbut Volume is different
196305
Since they r similar,\n\nDivide the Area of the larger cylinder by the smaller one,\n49(3.142)/25(3.142) to receive proportionality\n\n= 1.96\n\nTherefore 1.96 * volume smaller cylinder,\n\n1.96*50(3.142) = 307.8 = 98(3.142) [The Volume of the larger cylinder]
163368
100\n\n(10mm=1cm; 10x10=100)
342354
density is mass per unit volume, so part (a) is correct.\n\nthe coefficient of volume expansion (I'll represent with the letter B) is:\n\nB = (1/V)*dV/dT\nwhere dV is the change in volume and dT is the change in temp.\n\nIn your problem we need to find the volume at the new temp.\nso rearranging this equation:\ndV = B*V*dT\nit might be easier if we write dV as the difference between final volume and initial volume, likewise for dT:\n\nV2 - V1 = B*V1*(T2-T1)\nso we need to find V2:\nV2 = V1*[1+B*(T2-T1)]\nV2 = 1.17 * [1+ [1.26*10^-3]*(20-0)] = 1.2 (rounded)\n\nd = m/v = 825 / V2 = 688 kg/m^3\n\nthis is a more reasonable answer than what you have.
497175
GB >= 0.6*(CS + GB)\n\n"cushion" variable U introduced : \n\nU >= 0 \nGB + U = 0.6 * (CS + GB)
179495
If the spheres are of same volume, their apparent loss of mass will be the same.\n\nIf the spheres are of same mass, Aluminium sphere's apparent loss of mass will be more.
448430
first blank=gas\nsecond blank=liquid\nthird blank=solids\nfourth blank=congruent diameter
431386
T1=m1*a, no matter are there m2 & m3 or not.\nT2=(m1+m2)*a, no matter whether m1 & m2 are connected with a string or are in one lump, the presence of m3 being unimportant too.
347616
What would what be in centimeters and meters?
228401
Unfortunately, this is wrong.  (And I do believe that I am the first one here with the correct answer...hint, hint)\n\nHere is how it should be solved algebraically:\n\nThe areas of the sides are:\nL*W = 80 \nL*H = 20 \nW*H = 16 \n\nTo find the lengths of the sides use the equations for the areas:\n\nH = 16/W\n\nsubstitute for H\nL*16/W = 20\n\nW = 80/L\nsubstitute for W\n\nL*16/(80/L) = 20\nL^2 = 1600/16 \nL = sqrt(100) = 10\n\nW= 80/10 = 8 \n\nH = 160/ 80 = 2\n\nV = L*W*H = 160 [in^3...don't forget the units!]\n\nTo check:\n\nMultiply:\n\nL*H = 10 * 2 = 20\n\nL*W = 10 * 8 = 80\n\nW*H = 8 * 2 = 16\n\n\nYou could have also solved it geometrically be realizing that:\n\nV = L*W*H \n\nV^2 = L*L*W*W*H*H = L*H*L*W*W*H = 20*80*16 = 25600\n\nV = sqrt(25600) = 160 in^3\n\nEither way you get the same answer for volume.  The algebraic way, however, gives you more information about the solid (ie, the lengths of the sides) and could be useful for any follow-up questions one might expect.
541498
Density is mass/volume\n\nYou've given 3 dimensions of an object, does that mean that it's a rectangular solid?  If so, then vol = L*H*D.
494784
if u are referring to expansion due to heating than the order of expansion of the substances from lowest to largest is solid, liquids, than gases. But a solid's expansion is extremely low, almost impossible to measure unless its a metal.
266438
Well, first of all, this isn't geometry.\n\nThe first brick has a volume of 0.2m * 0.1m * 0.05m = 0.001 cubic metres.  Its density is mass divided by volume, so 1.2kg/0.001 = 1200 kg/cubic metre.\n\nThe second brick's volume is 0.25m * 0.15m * 0.04m = 0.0015 cubic metres.  \n\n1200kg/cubic metre * 0.0015 cubic metres = 1.8kg.