Answer:
7.9 g/cm³
Step-by-step explanation:
To find the density of the iron, we first need to find the volume of the solid cylinder.
[tex]\boxed{\begin{array}{l}\underline{\textsf{Volume of a Cylinder}}\\\\V=\pi r^2 h\\\\\textsf{where:}\\\phantom{ww}\bullet\;\textsf{$V$ is the volume.}\\\phantom{ww}\bullet\;\textsf{$r$ is the radius of the circular base.}\\\phantom{ww}\bullet\;\textsf{$h$ is the height.}\end{array}}[/tex]
The radius of a circle is half its diameter. Since the diameter of the circular base of the given cylinder is 18 cm, its radius is 9 cm. Therefore, the values to substitute into the volume formula are:
Substituting these values into the formula gives:
[tex]V=\pi \cdot 9^2 \cdot 3.5\\\\V=\pi \cdot 81 \cdot 3.5\\\\V=283.5\pi \rm \; cm^3[/tex]
To find the density of the iron given that the mass of the cylinder is 7.04 kg, we can use the formula for density:
[tex]\boxed{\begin{array}{l}\underline{\textsf{Density}}\\\\\rho=\dfrac{m}{V}\\\\\textsf{where:}\\\phantom{ww}\bullet\;\textsf{$\pho$ is the density.}\\\phantom{ww}\bullet\;\textsf{$m$ is the mass.}\\\phantom{ww}\bullet\;\textsf{$V$ is the volume.}\\\end{array}}[/tex]
In this case, the density should be in grams per cubic centimeter. Therefore, we need to convert the mass from kilograms to grams by multiplying it by 1000. Therefore:
- m = 7.04 kg = 7040 g
- V = 283.5π cm³
Substituting these values into the formula gives:
[tex]\rho=\dfrac{7040}{283.5\pi}\\\\\\\rho=7.9044148103...\\\\\\\rho=7.9\; \rm g/cm^3\;(2\;s.f.)[/tex]
Therefore, the density of the iron correct to 2 significant figures is:
[tex]\LARGE\boxed{\boxed{7.9\; \rm g/cm^3}}[/tex]
