A hollow plastic ball floats in a tub of jello (the flavor is irrelevant, but I’ll tell you: it’s key lime). The jello sets, and when the ball is removed, the saucer-shaped depression in the jello measures 6cm across and 1cm deep. If the density of the plastic is twice that of jello, what is the average thickness of the plastic shell?
(In reply to
solution by Charlie)
... continued
(btw, in the previous post, make ((√10/2)(√10) = 5 read ((√10)/2)(√10) = 5)
The volume of the plastic in the ball is (4/3)π(R³-r³) where r is the inner radius. We set this equal to half the volume of the depression (spherical cap) as the density of the plastic is twice that of the jello and we need to make the volumes match. So
(4/3)π(125-r³)=(14π/3)/2). Solving this for r, gives ³√(125-7/4) or 4.97656. Subtracting this from the outer radius, 5, gives .02344 cm as the thickness of the plastic shell, or 234.4 micrometers or microns.
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Posted by Charlie
on 2003-04-04 11:00:19 |