a glass prism must fit in a cylindrical tube that has an inside diameter of d inches.?
919521
Imagine the center of the triangle. Draw a line segment from it to a vertex, and then one from the center to a midpoint of a side. This makes a 30-60-90 triangle, so the line segment is (p/2) sqrt(3). This is a radius of the cylinder, so we multiply it by 2 to get the diameter.\n\nd = p sqrt(3)\n\nor, as the question wants it\np = d/sqrt(3) = d sqrt(3)/3
293478
both will have a miniscus and this is due to the adhesive property of water. the cohesive property keeps the water molecules togherter but the adhesive props allow for movement as in a cappilary tube. the gLass is clearer than the plastic so im thinking glass
237112
the cylinder shape and the contents.
796040
material needed=2pir(h+r)\nvolume=pir^2h\npir^2h=1000cc (given)\nh=1000/(pir^2)\nsubstituing h in the equation 2pir(h+r)\nmaterial=2pir(1000/pir^2+r)\n=2pir*1000/pir^2+2pir^2\ndA/dt=-2000/r^2+4pir\nsetting this to zero\n-2000+4pir^3=0\n4pir^3=2000\nr^3=2000/4pi\nr=cube root of 500/3.14\n=5.4 approx\nsubstituting\nh=1000/(3.14*5.4*5.4)\n=10.9 aapprox\nso the radius is 5.4 cm approx\nand the height 10.9 approx
805804
In the lab we used to enlarge the hole on micropipette tips when pipetting things like glycerol. Just cut off a little bit of the tip with a razor blade. With samples over 1 mL we were just patient because the plastic disposable pipettes tend to shatter if you try to trim them a little.
827251
V = πr²h\nA (top & bottom) = πr²\nEnd Area = 2πr²\nLateral Area = 2πrh\nCost of tube = 2πrhC, C being cost per unit area\nCost of ends = 2πr²C*1.2\nTotal cost, P = 2πC(1.2r² + rh)\nh = V/πr², so\nP = 2πC(1.2r² + rV/πr²)\nP = 2πC(1.2r² + V/πr)\ndP/dr = 2πC(2.4r - V/πr²) = 0 for max or min, so\n2.4r = V/πr²\nr³ = V/2.4π\nr³ = 5/2.4π\nr = 0.872"\nh = 5/π0.872²\nh = 2.093"\n\nCheck:\nC(2.4*π0.872² + 10/0.872) = 17.201050C\nC(2.4*π0.87² + 10/0.87) = 17.201444C (+0.000394)\nC(2.4*π0.874² + 10/0.874) = 17.201137C (+0.000087)
553954
Base times height gives the volume of the prism.\nVolume of prramid= 1/3 x base x height =1/3 x 24=8 cubic inches.
335828
The radius is half the diameter.\nThe diameter is twice the radius.\nd = 2r\n8 = 2r\nr = 8/2\nr = 4 inches
936045
Boy, you probably need to consider the liquid containing bottle's dimensions too. \nThe total average height of liquid will be 0.8 m + half the height in the bottle (say.\n\nAssuming the average height of liquid in the bottle = 100 mm = 0.1 m,\ntotal average height = 0.8 + 0.1 = 0.9 m\n\npressure exerted = potential energy= densityx acceleration due to gravityx height\n\nLength of needle can be neglected because they are usually installed horizontally and hardly add to the height. Also, it can be presumed that the flexible tubing and the needle were fiiled with the liquid at the comencement of the I.V drip. and the pressure difference if any due to this may be neglected.\n\nKinetic energy imparted = potential energy\n\n1/2 m v^2 = mgh\nv^2 = 2 gh\nv^2 = 2 x 9.8 x 0,9\nv= (2 x 9.8 x 0,9)^1/2\nv = 4.2 m/s= 420 cm/sec\n\nvolume to be injected/time in seconds\n= 490/ (35x60) = 1.4 cm3/ sec\n\nvolume flow (V) = speed (S)x cross section area\narea = V/S= 1.4/420 = 1/300 cm^2 \narea = 1/2 pi r^2\n1/300 = 1/2 x 22/7 x r^2\n\nr^2 = 1/300 x 2 x 7/22 cm2\n      = 7/ 3300 cm2\n\n  r    = .046cm = 0.46 mm\ndiameter (d) = 2r= 2 x 0.46= 0.92 mm................. Answer
100213
Diameter will be 1 units\n\nCircumference =piD\nif Circumference =pi\nthen D=1
760371
1. The ratio of h to r is not dependent on the can being made of aluminium. The surface area is just a concept - it cannot be aluminium.\n\n2. Is a cube (assuming that no aluminium is used in forming the edges) \n\n3. You need to do this bit.\n\nFurther Investigation  - you need to do this bit , it looks as though if you limit your investigation to regular prisms that will be O.K., investigating cones would be impressive. \n\nIn so far as the wastage is concerned. If you find out what the area+wastage is for the optimally sized circular cylinders and compare with optimally sized hexagonal prisms then this may be sufficient?
135114
you need to measure the outside diameter of your scope/dew shield, and then compare that to the listed sizes at this site:\nhttp://www.kendrickastro.com/astro/solarfilters.html
851866
Does it have to be a triangle??\n\nA cylinder is what comes to mind for me.  Like a soda can sliced in three equal parts.
419817
deviation= (refractive index -1)*angle of prism\n\nif prism angle is small...\nand\ndependence of rf with wavelength is given by cauchy's formula\n\nrf= A/wavelength....  A is constant\n\nso dependence (deviation) proportional to [(1/wavelength) -1)]
238652
Assuming you have a rectangular object, then L x H x W will work. For other shapes you need something else. See sources.
843492
depending on how big the glass is!!! do the math!!!
1087393
SPH means sphere, CYL means cylinder.  Just some instructions about how should the optician cut your glass.