Answer: 3.21 N
[tex] Specific\hspace{1mm} gravity = \frac {Density\hspace{1mm}of\hspace{1mm}substance}{Density\hspace{1mm} of \hspace{1mm}water}[/tex]
[tex]\Rightarrow Density \hspace{1mm}of\hspace{1mm}substance= 2.32\times 1000\hspace{1mm} kg/m^3 = 2320\hspace{1mm}kg/m^3\\ Mass =Density\times volume\\ \Rightarrow 2320 \hspace{1mm} kg/m^3\times 8.64 \hspace{1mm}in^3\times \frac {1.64\times10^{-5} m^3}{1\hspace{1mm}in^3}=0.328 kg[/tex]
For weight, we will multiply by [tex] g=9.8 m/s^{-2}[/tex]
[tex]weight= 0.328\times9.8=3.21\hspace{1mm}N[/tex]
Hence, the rock would weigh 3.21 N.
