Question 36675

Sep 16, 2016

The thickness of the foil is 0.014 mm.

Explanation:

Step 1. Convert the area of the foil to square centimetres.

${\text{A" = 75 color(red)(cancel(color(black)("ft")))^2 × ((12 color(red)(cancel(color(black)("in"))))/(1 color(red)(cancel(color(black)("ft")))))^2 × (("2.54 cm")/(1 color(red)(cancel(color(black)("in")))))^2 = 6.97 × 10^4color(white)(l) "cm}}^{2}$

Step 2. Convert the mass of the foil to grams.

$\text{Mass" = 9.6 color(red)(cancel(color(black)("oz"))) × (1 color(red)(cancel(color(black)("lb"))))/(16 color(red)(cancel(color(black)( "oz")))) × "454 g"/(1 color(red)(cancel(color(black)("lb")))) = "272 g}$

Step 3. Calculate the volume of the foil.

The formula for the density of an object is

color(blue)(bar(ul(|color(white)(a/a) ρ = m/V color(white)(a/a)|)))" "

where

ρ = density of the object
$m$ = mass of the object
$V$ = volume of the object

We can rearrange the formula to get

V = m/ρ

V = 272 color(red)(cancel(color(black)("g"))) × ("1 cm"^3)/(2.70 color(red)(cancel(color(black)("g")))) = "101 cm"^3

Step 4. Calculate the thickness of the foil.

The formula for the volume of a rectangular solid is

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} V = A h \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

where

$V$ = volume of the solid
$A$ = area of the solid
$h$ = height (or thickness) of the solid.

We can rearrange this formula to

$h = \frac{V}{A}$

h = (101 stackrelcolor(blue)("cm")(color(red)(cancel(color(black)("cm"^3)))))/(6.97 × 10^4 color(red)(cancel(color(black)("cm"^2)))) = 1.45 × 10^"-3"color(white)(l) "cm"

Step 5. Convert the thickness to millimetres.

1.45 × 10^"-3" color(red)(cancel(color(black)("cm"))) × "10 mm"/(1 color(red)(cancel(color(black)("cm")))) = "0.014 mm"#