Question

One cubic meter (1.00 m) of aluminum has a mass of 2.70 10 kg, and the same volume of iron has a mass of 7.8610'kg. Find the radus of a solid aluminum sphere that will balance a sod irn sphere o radu 14 squal-a balano rm

Answer 1

Answer:

The radius of the solid aluminum sphere has to be $r_{Al} =0.20m=20cm$

Explanation:

Hi

First of all, we're going to find the density of both materials, to do that we use ρ$=\frac{mass}{Vol}=\frac{m}{V}$. Then $\rho_{Al} =2.7010\frac{kg}{m^{3} }$ and $\rho_{Fe} =7.8610\frac{kg}{m^{3} }$.

the second step is to find Volume of the iron sphere ($V_{sphere}=\frac{4\pi r^{3}  }{3}$) so, $V_{Fe}=\frac{4\pi (0.14)^{3}  }{3}=0.0115m^{3}$.

The third step is to find the mass ($m=V\rho$) on this iron sphere so, $m_{Fe}=0.0115kg^{3}*7.861kg/m^{3} =0.0904kg$

The fourth step, we are going to use $V_{Al}=\frac{m_{Fe}}{\rho_{Al} }$ to find aluminum Volume hence its radius, $\frac{4\pi r^{3} _{Al}}{3} =\frac{0.0904kg}{2.7010kg/m^{3}}$, as we clear $r_{Al}$ we find that, $r^{3} _{Al}=\frac{3}{4\pi } 0.0335m^{3} =0.008m^{3}$, then$r _{Al}=\sqrt[3]{0.008m^{3}} =0.2m$.