Answer:
6032 g
Step-by-step explanation:
The battery can be modeled as a cylinder.
Volume of a cylinder
[tex]\sf V=\pi r^2 h[/tex]
where:
- r is the radius
- h is the height
Given:
Substitute the given values into the formula and solve for v:
[tex]\begin{aligned}\textsf{Volume of a cylinder} & =\sf \pi r^2 h \\ \implies \textsf{Volume of the battery} & = \sf \pi (4)^2(10)\\ v& = \sf 160 \pi\:\:cm^3\end{aligned}[/tex]
Density equation
[tex]\rho =\dfrac{m}{v}[/tex]
where:
- [tex]\rho[/tex] is density
- m is mass
- v is volume
Given:
[tex]\rho = 12 \dfrac{\text{g}}{\sf cm^3}[/tex]
[tex]v=160 \pi \sf \:\:cm^3[/tex]
Substitute the given values into the Density equation and solve for m:
[tex]\begin{aligned}\rho & =\dfrac{m}{v}\\\implies 12 \dfrac{\text{g}}{\sf cm^3} & = \dfrac{m}{160 \pi \sf \:\:cm^3}\\m & = 12 \dfrac{\text{g}}{\sf cm^3} \cdot 160 \pi \sf \:\:cm^3\\m & = 12 \cdot 160 \pi \:\text{g}\\ m & = 1920\pi\:\text{g}\\ m & = 6031.857895...\:\text{g}\end{aligned}[/tex]
Therefore, the total mass of the battery to the nearest gram is 6032 g.
