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 any variable or symbol stated above as necessary.)

Question

anonymous · Answer

in the form of  rAl= ( answer basically has to be a formula over a formula)

anonymous · Answer

$$\frac{4}{3} * \pi (rAL)^{3} *g= \frac{4}{3} * \pi (rFe)^{3} *g$$

anonymous · Answer

your balancing the m1*g=m2*g

anonymous · Answer

woops forgot the density, its volume * density *gravity .....on both sides

anonymous · Answer

$$ rAL=\sqrt[3]{rFE^{3} * \frac{\rho FE}{\rho Al}}$$

anonymous · Answer

Right on Point, Thank You

anonymous · Answer

you can simplify and write rFE on the outside of the cube root