SOLUTION: Hi there! I am not very good at math and I have been stuck on this question for 20 minutes now. The question is: calculate the volume of a box to hold a sphere-shaped beaker with a

Algebra ->  Bodies-in-space -> SOLUTION: Hi there! I am not very good at math and I have been stuck on this question for 20 minutes now. The question is: calculate the volume of a box to hold a sphere-shaped beaker with a      Log On


   

Question 953226: Hi there! I am not very good at math and I have been stuck on this question for 20 minutes now. The question is: calculate the volume of a box to hold a sphere-shaped beaker with a volume of 413cm^3. The second part is : how much space is being occupied by the beaker and how much is being wasted, in percents.
Thank you so much, I honestly have no idea what to do.
Found 2 solutions by macston, josmiceli:
Answer by macston(5194) About Me  (Show Source):
You can put this solution on YOUR website!
Volume of sphere=%284%2F3%29%28pi%29%28r%5E3%29
413cm%5E3=%284%2F3%29%28pi%29%28r%5E3%29 Multiply each side by 3%2F%28%284%29%28pi%29%29
%28%28413cm%5E3%29%283%29%29%2F%284%28pi%29%29=r%5E3
%281239cm%5E3%29%2F12.57=r%5E3
98.6=r%5E3 Find the cube root of each side.
4.62cm=r
The cubic box to hold the sphere must have edges=2r (diameter of the sphere)
The volume of the cube =%28edge%29%5E3=%282r%29%5E3=9.24%5E3cm%5E3
Volume of cube=789 cm^3
Volume wasted=vol(cube)-vol(sphere)=789cm^3-413cm^3=376cm^3
Percent of cube volume wasted=vol(wasted)/vol(cube)=(376cm^3/789cm^3)(100%)=47.7%

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
The volume of a sphere is:
+V%5Bs%5D+=+%284%2F3%29%2Api%2Ar%5E3+
where +r+ is the radius
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You are given that:
+V%5Bs%5D+=+413+ cm3
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+413+=+%284%2F3%29%2Ar%5E3+
Multiply both sides by +3+
+1239+=+4%2Api%2Ar%5E3+
Divide both sides by +4%2Api+
+309.75+%2F+pi+=+r%5E3+
+r%5E3+=+98.5966+
+r+=+4.6198+
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check:
+V%5Bs%5D+=+%284%2F3%29%2Api%2Ar%5E3+
+V%5Bs%5D+=+%284%2F3%29%2Api%2A4.6198%5E3+
+V%28s%29+=+1.333%2A3.14159%2A98.5983+
+V%28s%29+=+412.997+
This looks close enough
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If this sphere is tightly enclosed in a cube,
the volume of the box is +V%5Bc%5D+=++%282r%29%5E3+
This is because +2r+ is equal to a side
of the box
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+V%5Bc%5D+=+%28+2%2A4.6198+%29%5E3+
+V%5Bc%5D+=+9.2396%5E3+
+V%5Bc%5D+=+788.7866+
----------------------
+V%5Bc%5D+-+V%5Bs%5D+=+788.7866+-+413+
+V%5Bc%5D+-+V%5Bs%5D+=+375.7866+ ( wasted volume in cm3 )
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You are asked for:
+%28+V%5Bc%5D+-+V%5Bs%5D+%29+%2F+V%5Bc%5D+%29%2A100+ which is a %
+%28+375.7866+%2F+788.7866+%29%2A100+
+.4764%2A100+=+47.64+
47.64% is wasted space inside the cube
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Hope I got it -tough calculations