# Question a6330

Sep 1, 2015

The sheet of lead has a thickness of $1.55 \cdot {10}^{- 4} \text{cm}$

#### Explanation:

You can think of your sheet of lead as being a rectangular prism with a very, very small thichness, $h$. The volume of a rectangular prism of width $w$, length $l$, and thickness $h$ is given by

$\textcolor{b l u e}{V = l \cdot w \cdot h}$

The idea here is that you can determine the volume of the sheet of lead by using its density and its mass. Since density is defined as mass per unit of volume, you can use the known mass to figure out the volume that much lead would occupy.

color(blue)(rho = m/V implies V = m/(rho)

V = (0.627color(red)(cancel(color(black)("g"))))/(11.34color(red)(cancel(color(black)("g")))/"cm"^3) = "0.05529 cm"""^3#
$V = l \cdot w \cdot h \implies h = \frac{V}{l \cdot w}$
$h = \left(\text{0.05529 cm"^color(red)(cancel(color(black)(3))))/(15.5color(red)(cancel(color(black)("cm"))) * 23.0color(red)(cancel(color(black)("cm")))) = color(green)(1.55 * 10^(-4)"cm}\right)$