A standard gold bar weighs 12.5kg. The density of gold is 19.3g/cm^3. What is the diameter of a sphere made from the same gold bar? I have tried various versions of the question in WolframAlpha. WA will happily understand and return the diameter of a sphere of a certain volume. Is there a way to get it to return the answer in one question?
first calculate the volume from the mass and the density, then calculate the diameter of the sphere from the volume.
Yes that would work. I have no issue with doing this on paper even by hand without a calculator! If I type in Wolfram Alpha "diameter of sphere of volume 10" it will not only understand the question and interpret it but also return the answer.
If I type "Volume of a sphere of volume of a gold bar" it fails. As does
"Volume of a sphere of volume of a solid of 12.5 kg and a density of 19.3g/cm^3"
Can I get it to solve the problem in one pass?
solve 4/3 pi r^3=12.5 kg/19.3 g/cm^3 for r
I tried to get it to understand something like
weight=12.5 kg, density=19.3g/cm^3, volume=weight/density, solve 3/4 pi r^3=volume for r
but had no success
Solve[(4 Pi/3)* r^3 == m*d, r]