Postby Guest » Thu Jan 09, 2014 8:45 pm
find the height of a cone with a half sphere on top (3-D object) given the volume is 0.65m cubed and a base of both cone and sphere with a diameter of 1 m
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To solve this will require use of two basic equations.
The volume of a standard cone
1/3Pi*(r 2)*h or ((Pi*d 2)*h)/12 = volume of cone
and the volume of a sphere
4/3Pi*(r 3) or (Pi*d 3)/6 = volume of sphere
Note: The above two equations for each shape are mathematically the same: one uses radius the other diameter and an integer divisor is used instead of a fractional multiplier. This is common in some engineering applications
Note also that a cone has two (2) parameters: diameter and height; and a sphere has only one (1) which is diameter.
The diameter is given in the question and is fixed at 1 meter, so the sphere is a fixed volume. The only variable we can control is the height of the cone.
The total volume (cone volume plus sphere volume) is 0.65 meters cubed (0.65 3).
A one (1) meter diameter sphere has a volume of (Pi*1 3)/6= 0.52 cubic meters.
Because it is a half-sphere take half the volume: 0.26. Subtract this value from the total volume: 0.65-0.26 = 0.39 cubic meters. A cone with a diameter 1 meter and volume of 0.39 Cubic Meters
Rearrange the volume of a cone equation to solve for its height.
((Pi*d 2)*h)/12 = volume of cone
((Pi*d 2)/v)/12 = height = 12v/(Pi(d 2))
h=(12*(0.39m 2))/(Pi(1.00 2)) = ??? meters high.
Don't forget to use units and significant digits
~~D~~
+330 
