Question

A sample of unknown material weight 900N In air and and 400N when submerged in an alcohol solution with a density of 0.7 g/cm³.What is the density of the material ?

Answer 1

1.26 g/cm³

Step-by-step explanation

Step 1

given

`\begin{gathered} F_(g(air))=900\text{ N} \\ F_(g(alchodol))=400\text{ N} \\ \rho_(alcohol)=0.7\text{ }(g)/(cm^3) \end{gathered}`

unknown; the density of the material, so

`\begin{gathered} F_B=F_(g(air))-F_(g(alcohol)) \\ F_B=900\text{ N-400 N=500 N} \end{gathered}`

so, the proportion is

the ratio of the force equals the ratio of the density ,so

`\begin{gathered} (F_(g(air)))/(F_B)=(\rho_(material))/(\rho_(alcholol)) \\ replace \\ \frac{900\text{ N}}{500\text{ N}}=\frac{\rho_(material)}{0.7\text{ }(g)/(cm^3)} \\ mutliply\text{ both sides by 0.7}(g)/(cm^3) \\ \frac{900\text{N}}{500\text{N}}*0.7\text{ }(g)/(cm^3)=\frac{\rho_(mater\imaginaryI al)}{0.7\text{(g)/(cm^(3))}}*0.7(g)/(cm^3) \\ 1.26(g)/(cm^3)=\text{ density of the material } \\ \end{gathered}`

so, the density of the material is

1.26 g/cm³

I hope this helps you