Let ρAl represent the density of aluminum and ρFe that of iron. Find the radius of a solid aluminum sphere that balances a solid iron sphere of radius rFe on an equal-arm balance. (Use r_F for r Fe, rho_A for ρAl, and rho_F for ρ as necessary.)
Viron = 4/3 πr Fr^3
Valu = 4/3 πr Al^3
pFr 4/3πrfe^3 = pAl4/3πrAl^3
or
rAl = \(\sqrt[3]{pFe/pAl}\)
Hope this helps
Answer:
r = [3 (pAl/MAl)]/(4 pi)]^1/3
r = [3 (pFe / MFe)]/(4 pi)]^1/3
Explanation:
In the equilibrium state, aluminum and iron have the same mass. From the density equation and solving for the mass we have:
Mass = density/volume
MFe = pFe/V
MAl = pAl/V
In equilibrium, we have that MFe = MAl
Solving for the volume:
MFe = pFe/V
V = pFe/MFe
MAl = pAl/V
V = pAl/MAl
The equation for the volume of a sphere is equal to:
V = (4 pi r^3)/3
Replacing the volume of both iron and aluminum, we have:
V = (4 pi r^3)/3
r = [(3 V)/(4 pi)]^1/3
r = [3 (pAl/MAl)]/(4 pi)]^1/3
r = [3 (pFe/MFe)]/(4 pi)]^1/3