Question

1. challenge: find the density of a soda can with a radius of 3.25 cm, a height of 12.2 cm, and a mass of 40 g.

Explanation:

Step1: Recall the formula for the volume of a cylinder

The volume \( V \) of a cylinder is given by \( V=\pi r^{2}h \), where \( r \) is the radius and \( h \) is the height.
Given \( r = 3.25\space cm\) and \( h=12.2\space cm\), we substitute these values into the formula:
\( V=\pi\times(3.25)^{2}\times12.2 \)
First, calculate \( (3.25)^{2}= 3.25\times3.25 = 10.5625\)
Then, \( V=\pi\times10.5625\times12.2\approx3.1416\times10.5625\times12.2\)
\( 3.1416\times10.5625 = 33.1831\)
\( 33.1831\times12.2\approx404.834\space cm^{3}\)

Step2: Recall the formula for density

Density \(
ho \) is given by the formula \(
ho=\frac{m}{V} \), where \( m \) is the mass and \( V \) is the volume.
Given \( m = 40\space g\) and \( V\approx404.834\space cm^{3}\)
Substitute these values into the formula:
\(
ho=\frac{40}{404.834}\approx0.0988\space g/cm^{3}\) (rounded to a reasonable number of decimal places)

Answer:

The density of the soda can is approximately \( 0.10\space g/cm^{3}\) (or more precisely \( \approx0.0988\space g/cm^{3}\))