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February 1, 2015

February 1, 2015

Posted by **Sara** on Saturday, April 10, 2010 at 6:30pm.

- Itroductory Physical Science -
**drwls**, Saturday, April 10, 2010 at 8:44pmLet D be the diameter in cm. The number of shot in the tray is approximately 100/D^2. The equation is not accurate unless D<<10 cm. With a hexagonal arrangement of spheres, you can get about 119/D^2 speheres in. It depends upon how you arrange the spheres and how much loose space remains at the edges.

The same number fills the cylinder up to 20 cm^3. Calculate the number of spheres of diameter D that can occupy that volume with hexagonal close packing. According to a theorem of Gauss, 74% of the volume can be occupied by solid spheres, whose volume is pi*D^3/6, so the number of spheres also equals

0.74*20/(pi*D^3/6) = 28.3/D^3

28.3/D^3 = 100/D^2 ; therefore

D = 0.28 cm

You can complete the other questions with that information.

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