What can you tell me about a Dyson Sphere?
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A Dyson Sphere is basically a globe that you would build around a sun.  All of the energy put out by the sun would fall on the inside of the Sphere.  Thus (I love saying that word) you would control all the energy of the sun.\nThe problems would be having enough materials to make it.  Using all of the metals in our solar system, a Dyson Sphere placed near the orbit of the earth would only be a few molecules thick... thus (he he he) very fragile.\nAlso, you wouldn't be able to approach the sphere safely from the outside because there would be very little light on the outside.\nAlso, where would you stand as you dismantled our solar system... a bit of a problem.\nIn the Larry Niven book, "Ring World" some of these problems are solved, because the created a 'dyson ring' instead of a sphere.  Check it out!
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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.
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http://scienceworld.wolfram.com/physics/YoungsModulus.html\n\nhttp://physics.uwstout.edu/Univphys1/labs/ymodul.htm\n\nTry them..
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Don't know what grade you're in, but looks like this is gonna involve some integration.\nLet the intersection point of the plane be z=h(=0.5 in this case).\n\nConsider a thin disc of thickness dz in the volume to be evaluated, at a distance z from the origin. Obviously, z>h. It would have a radius of \nr = sqrt(R^2 - z^2). R is the radius of the sphere, =1 in this case.\n\nHence, the volume of this disc would be\ndV = 2*pi*r^2*dz\n   = 2*pi*(R^2-z^2)*dz\n\nHence, for total volume, this is going to have to be integrated between z=h to z=R.\n\nI believe this works out to \nV = 2*pi*((2R^3/3)-R^2h+h^3/3) here. Put in the values mentioned above.\n\nWhew!
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A sphere is the shape that forms where all points are as close to the source of gravity as possible. If it was a cube, the corners would "fall" toward the center, forming a sphere.
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Let's look at the general case. It turns out that for even dimensional spheres, the ratio of the 'surface area' to the 'volume' of the next lower dimensional sphere is irrational in even dimensions and rational in odd dimensions. Your cases are just special cases of this general result.
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Can you design an ultra-light, ultra strong chamber (sphere?) from carbon nanotubes (and resin) large enough yet strong enough that when evacuated to near perfect vacuum it would float?  Can you collect such chambers (honeycomb?) into a dirigible that would raise weights and fly?  Good luck.
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Gravity shapes objects.  If an objects mass is over a certain threshold then gravity will be strong enough to overcome molecular forces to "mold" an object.  Most asteriods have masses that are too low for this however planets and most moons are massive enough.  \n\nWhy a sphere?  This is due to the inverse radial dependance of the gravitational force.  In order words at a given distance from the center of mass distribution the force of gravity is the same.  This type of shape is a sphere.
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The first is a sphere of radius a centered at (0,0,a) and the second is a sphere of radius a/2 centered at (0,0,a/2). Both spheres are above the xy-plane and are tangent to each other at the origin. The smaller sphere is contained inside the larger.\n\nIn spherical coordinates, the equations reduce to\nrho=2a*cos(phi) and\nrho=a*cos(phi).\n\nHence the limits are\nrho: from a*cos(phi) to 2a*cos(phi)\nphi: from 0 to pi/2\ntheta: from 0 to 2pi.\n\nI am using the mathematician's convention where phi is the angle down from the z-axis and theta is the angle from the xz-plane. Physicists use the reverse convention, so be careful!\n\nDon't forget the rho^2 sin(phi) !
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The "foamy" spacetime would look like a complex turbulent storm-tossed sea. Some physicists theorize the formation of wormholes therein; speculation arising from this includes the possibility of hyperspatial links to other universes.\n\nQuantum foam is theorized to create masses of virtual particles. These virtual particles make their existence known by the Casimir effect. The Casimir effect is a physical force exerted between separate objects.\n\nDutch physicist Hendrik B. G. Casimir first proposed the existence of the force, and he formulated an experiment to detect it in 1948 while participating in research at Philips Research Labs. The classic form of his experiment used a pair of uncharged parallel metal plates in a vacuum, and successfully demonstrated the force to within 15% of the value he had predicted according to his theory.\n\nSome physicists theorize that having two sets of these charged metal plates will create/connect a worm  hole between them.\n\nIf you want to take it a step further. Have a pair of plates on earth and another on a space craft moving close to the speed of light. Jumping in between the plates entering/the worm hole will allow you to travel backwards in time!
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well...lets call your known volume V.  so...what kind of shape are we dealing with...eh, lets claim we are dealing with a sphere...we know the volume of the sphere is determined by v=(4/3)*pi*r^3 so we could solve for r (the radius).\n\nso....\n\nV=(4/3)*pi*r^3\nr=((3V)/(4pi))^(1/3)\n\nwe can now substitute that r into our surface area of a sphere equation....which is SA=4*pi*r^2, so we would have\n\nSA=4*pi*((3V)/(4pi))^(2/3)\n\na similar series of actions would occur for any other shape\n\nhope this helps,\n\nMatttlocke
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http://hyperphysics.phy-astr.gsu.edu/hbase/class/phscilab/dens.html\n\ntry this site
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Besides a tasty treat for the physicists...\n\nThe effervescence of the bubbles has been pondered by many a physicists.\n\nhttp://www.stanford.edu/dept/news/relaged/911111Arc1079.html
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I like the joke!  To play along, I define:\nr = metric r  and \nd = metric d\n\nV = (pi/3)*(d^2)*(3r - d)\n\nThis will work for the tank being any fraction of a sphere, not just 1/2
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u can get a clear picture of it in howstuffworks.com or else u can find similar info on webopedia.com
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from calculus\ndivide the sphere up into circular slices of width dx.\nthe disk at a distance of x from the origin has a radius of sqrt(r^2-x^2) and so an area of pi(r^2-x^2) and a volume of pi(r^2-x^2)dx\nsumming up the volumes of all the disks gives the volume of the whole sphere = int pi(r^2-x^2)dx from x=-r to r\n=pi.r^2.x-pi/3.x^3 from x=-r to r\n=(pi.r^3-pi/3.r^3)-(-pi.r^3--pi/3.r^3)\n=4/3.pi.r^3
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nothing. even the vacuum of space has energy flowing through it. So, in theory, a perfect vacuum is nothing.