what are the turns of a univalve shell?
352430
The spiral shell allows the animal inside to grow.  As it grows, the sheel develops a wider turn at the open end and the animal grows into this space.
113217
There are two names for a seven sided shape, heptagon and septagon.  Both answers are equally correct
984859
internal rotation on shouldwer and elbow bent
359367
The parabolic structure is stronger than a circular strucure or dome.\n\n---Addition---\n\nLet's think in an egg. An egg has two different shapes in each end, one is more rounded like an hemisphere and the other is more like the shape of a paraboloid.\nIf you press the egg from both ends, the egg brakes always \nfrom the more rounded one, the hemisphere looking shape instead of the paraboloid looking end.
513263
Spiral, lenticular, elliptical and irregular\n\nSee the link
693221
It's a face that is not flat.  A cone has two faces: the base which is flat and the rest of the cone which is curved.
338257
a) vcenter = (m1*v1+m2+m2)/(m1+m2)\n\nb) v1 rel = v1 - v center\nv2 rel = v2 - v center\n\nremember velocity is a vector so you must keep track of magnitude and direction
672001
The moment of inertia for a spherical shell is 2/3*M*R2. You might imagine the spherical shell to be made up of a series of tiny mass elements the mass of each being its volume times its density r. The volume would be the thickness of the shell times the difference in latitude covered by the mass element times the difference in longitude covered by the mass element. Lets call the differential longitude dq and the differential latitude dl. Then the mass of each element is r*dq*dl. The moment of inertia of the whole shell is the sum of the moments of inertia of each of the little mass elements. The moment of inertia, i, for a mass particle is m*r2 where m is the particle mass and r is its distance from the axis or rotation. In our case i=r*dq*dl*R*cos(l) where R is the radius of the spherical shell and l is the latitude measured from the equator of the sphere. To sum up all the moments of inertia i we must integrate over all q from 0 to 2*p and all l from -p to p. Carrying out this integration gives us the result.
247069
A rolled plate is aprox a Archimedes spiral. For a spiral constant thickness or pitch you have that:\n\nLength = pi*(R(outer)^2-R(inner)^2)/ Pitch
740030
f = uN\n198 = u*(55*9.8)\nu = 0.367 for static\n\nf = uN\n175 = u*(55*9.8)\nu = 0.325 for kenetic
291300
Try this:\nhttp://flash.lakeheadu.ca/~yzhang/myText/dspGuide/CH32.PDF#search='laplace%20transform'
754180
Here's one simple animation: (I'll keep looking).\nhttp://janus.astro.umd.edu/javadir/orbits/ssv.html\n\nHere's a FANCY video:\nhttp://www.bbc.co.uk/science/space/solarsystem/3d_tour.shtml
829392
just a piont of view, but normally to divide something symmetrically you have to do it at the widest piont along the centre line. where as assymetricaly is off the centre line at a narrower piont and therefore not a symmetry,this only applies to poloygons and symmetric shapes.  hope this helps
1070589
Assuming you mean as 4 shapes equivalent to the original L shape, the answer is given in the attached image.
80065
Rotated counter-clockwise:\nD (-3,-2) and K (-2,1)\n\nRotated clockwise:\nD (3,2) and K (2,-1)
976121
Let us assume that we require a cone with d as bottom circle dia, h as vertical height which are the known factors. Let l be the slant height of the cone which is equal to Square root of (h^2+(d/2)^2). To arrive the cone we have to make a sector with l as radius and A as angle of the sector. \nl is already calculated. A is calculated as follows:\nA = pi()*d/l which is obtained in radians; \nTo convert it to degrees, multiply A with 180 and divide by pi();\nSo we get A in degrees and l, from which we can get a sector. By connecting the two slant heights together we can get a cone with dia d and height h.
757387
(a) Assuming the base is a half-circle and that there are no other forces acting on the mass (friction, drag, etc), then the kinetic energy at the base is equal to the potential energy at the top. Note that the change in potential energy is not affected by the semi-circular path the mass follows, but rather the net change in height. This net change is simply the radius of the half-circle container:\n\nKE = mgR = (.71 g * 1 kg/1000 g)(9.8 m/s²)(25 cm * 1 m/100 cm)\n\nKE = .00174 J\n\n(b) The kinetic energy associated with rotation is given as:\n\nKER = (1/2) I w², where w is the rotational velocity. Note that since the sphere rolls without slipping, the translational velocity of its cener of mass is simply:\n\nv = w * r\n\nSubstituting this into the expression for the rotational kinetic energy:\n\nKER = (1/2) (I / r²) v²\n\nNote that the expression for the translational kinetic energy is:\n\nKET = (1/2) m v²\n\nThus, the total kinetic energy of the ball at the base of the hemisphere is;\n\nKE = KER + KET = (v²/2)( I/r² + m)\n\nThe fraction of the rotational kinetic energy over the total is:\n\nf = KER / KE = (v²/2)( I/r²) / (v²/2)( I/r² + m)\n\nf = ( I/r²)/( I/r² + m)\n\nFinally, the moment of inertia, I, of a sphere is:\n\nI = (2/5)mr²\n\nSubstituting this into the expression for f:\n\nf = (2/5)m/( (2/5)m + m)\n\nf = (2/5)/( (2/5) + 1)\n\nf = 2/7\n\n(c) At the base of the hemisphere, there are two forces acting on the ball: the force of gravity and the centripetal force. At the base, these forces are coincident (act in the same direction). Thus the normal force, which opposes these two, is simply:\n\nN = Fgrav + Fcent\n\nN = mg + mv²/R\n\nTo find v, recall that KET / KE is 2/7. Using this, we can manipulate the terms to get v:\n\n(1/2) (I / r) v² = (2/7)(mgR)\n\n(1/2) ((2/5)m) v² = (2/7)(mgR)\n\n(1/5) v² = (2/7)(gR)\n\nv = sqrt((10/7)(gR))\n\nv = sqrt((10/7)(9.8 * 25 cm * 1m/100 cm))\n\nv = sqrt((10/7)(9.8 * 25 cm * 1m/100 cm))\n\nv = 1.871 m/s\n\nFinally, the normal force is computed as:\n\nN = mg + mv²/R\n\nN = (.71 g)(9.81 m/s²) + (.71 g)(1.871 m/s)²/(25 cm)\n\nN = (.71 g)(9.81 m/s²) + (.71 g)(1.871 m/s)²/(25 cm)\n\nN = .0169 Newtons\n\n\nGood Luck! :)