Question-208999

Question Number 208999 by Spillover last updated on 30/Jun/24

Answered by mr W last updated on 30/Jun/24

Commented by mr W last updated on 30/Jun/24

F=mg−B m=Vρ_(Iron) ⇒V=(m/ρ_(Iron) ) B=((Vρ_(Oil) g)/2)=((mρ_(Oil) g)/(2ρ_(Iron) )) ⇒F=(1−(ρ_(Oil) /(2ρ_(Iron) )))mg =(1−((0.9)/(2×7.8)))×0.36×10=3.39N

$${F}={mg}−{B} \\ $$$${m}={V}\rho_{{Iron}} \:\Rightarrow{V}=\frac{{m}}{\rho_{{Iron}} } \\ $$$${B}=\frac{{V}\rho_{{Oil}} {g}}{\mathrm{2}}=\frac{{m}\rho_{{Oil}} {g}}{\mathrm{2}\rho_{{Iron}} } \\ $$$$\Rightarrow{F}=\left(\mathrm{1}−\frac{\rho_{{Oil}} }{\mathrm{2}\rho_{{Iron}} }\right){mg} \\ $$$$\:\:\:\:\:\:\:\:=\left(\mathrm{1}−\frac{\mathrm{0}.\mathrm{9}}{\mathrm{2}×\mathrm{7}.\mathrm{8}}\right)×\mathrm{0}.\mathrm{36}×\mathrm{10}=\mathrm{3}.\mathrm{39}{N} \\ $$

Commented by Spillover last updated on 30/Jun/24