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Sagot :
1.
First, let's calculate the volume of the cube:
[tex]V_1=6^3=216\text{ cm^^b3}[/tex]Now, let's calculate the volume of the cylinder drilled:
[tex]\begin{gathered} V_2=\frac{\pi d^2h}{4}\\ \\ V_2=\frac{3.14159\cdot2.5^2\cdot6}{4}\\ \\ V_2=29.45\text{ cm^^b3} \end{gathered}[/tex]So the volume of the cube after being drilled is:
[tex]\begin{gathered} V=V_1-V_2\\ \\ V=216-29.45\\ \\ V=186.55\text{ cm^^b3} \end{gathered}[/tex]If the weight is 6.3 N, let's find the mass:
[tex]\begin{gathered} W=m\cdot g\\ \\ 6.3=m\cdot9.8\\ \\ m=\frac{6.3}{9.8}\\ \\ m=0.643\text{ kg} \end{gathered}[/tex]And the density of the metal is:
[tex]\begin{gathered} d=\frac{m}{V}\\ \\ d=\frac{0.643}{186.55}\\ \\ d=0.0034468\text{ kg/cm^^b3}\\ \\ d=3.4468\text{ g/cm^^b3} \end{gathered}[/tex]2.
To find the weight of the cube before being drilled, let's use the following rule of three:
[tex]\begin{gathered} volume\rightarrow weight\\ \\ 216\text{ cm^^b3}\rightarrow x\text{ N}\\ \\ 186.55\text{ cm^^b3}\rightarrow6.3\text{ N}\\ \\ \\ \\ \frac{216}{186.55}=\frac{x}{6.3}\\ \\ x=\frac{216\cdot6.3}{186.55}\\ \\ x=7.295\text{ N} \end{gathered}[/tex]
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