# Question a5b2a

Dec 25, 2017

The foil is only $0.0145 \text{mm}$ thick! This is much less than $\frac{1}{1000}$ of an inch!

#### Explanation:

You have to convert to a consistent set of units. We shall use grams for the mass unit and centimeters for the length unit.

The area of the foil is then

(75"ft"^2)×(30.48"cm"/{1"ft"})^2=69680"cm"^2

The mass isj

(9.6"oz")×(28.35"g/oz")=272.2"g"

Now the volume of the foil is the mass divided by the density, and the density is given to only three significant figures. So we render the volume accordingly:

(272.2"g")/(2.70"g"/{"cm"^3})=101"cm"^3

And then the thickness is the volume divided by the area:

(101"cm"^3)/(69680"cm"^2)=0.00145"cm"=0.0145"mm"

Dec 25, 2017

$1.45 \times {10}^{- 2}$mm

#### Explanation:

The critical calculation here is the volume, after all of the conversions. The conversions are more tedious and a potential source of errors than anything relevant to the real question. We want a final answer in metric units (mm), so it may be best to first put all of the data into metric units at the millimeter level.

$75 f {t}^{2} \times 92903.04 = 6967728 m {m}^{2}$
http://www.endmemo.com/convert/area.php

$1 c {m}^{3} = 1000 m {m}^{3}$
$2.70 \frac{g}{c {m}^{3}} \times \frac{0.001 c {m}^{3}}{m {m}^{3}} = 0.0027 \frac{g}{m {m}^{3}}$

$9.6 o z \times 28.35 = 272.16 g$
http://www.endmemo.com/convert/weight%20and%20mass.php
NOW we have:
Density: $0.0027 \frac{g}{m m} ^ 3$
Mass: $272.16 g$
Area: $6967728 m {m}^{2}$

The desired thickness calculation is:
T = "Mass"/("Density" xx "Area")# = $\frac{272.16}{0.0027 \times 6967728}$ = $1.45 \times {10}^{- 2}$