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

The radius of the solid aluminum sphere has to be
`r_(Al) =0.20m=20cm`

Step-by-step explanation:

Hi

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

the second step is to find Volume of the iron sphere (
`V_(sphere)=(4\pi r^(3)  )/(3)`) so,
`V_(Fe)=(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)=(m_(Fe))/(\rho_(Al) )` to find aluminum Volume hence its radius,
`(4\pi r^(3) _(Al))/(3) =(0.0904kg)/(2.7010kg/m^(3))`, as we clear
`r_(Al)` we find that,
`r^(3) _(Al)=(3)/(4\pi ) 0.0335m^(3) =0.008m^(3)`, then
`r _(Al)=\sqrt[3]{0.008m^(3)} =0.2m`.