# Thread: 2nd Conversion Problem / Question

1. ## 2nd Conversion Problem / Question

Thank You for the Help!

-qbkr21

2. Originally Posted by qbkr21
Thank You for the Help!

-qbkr21

Doesn't it seem odd that the flat sheet of metal measures to have a thickness of almost one meter?

1.You first need to find the volume of the sheet using the weight and density.
2.Then you need to use the given dimensions to find the thickness.
Conversions must be used throughout.

I can try to help if you have problems with one of those steps, but I'm no expert. I got 0.0014151310796295390612405333391297 mm for the answer.

3. area of the sheet ...
A = (2.4 ft*12 in/ft*2.54 cm/in)*(1*12*2.54) = 2229.67296 cm^2

volume of gold ...
V = density/mass = (19.32 g/cm^3)/(.200 g) = 0.01035... cm^3 {edit: please excuse my "dyslexic" fractions ... should be mass/density. the result is correct}

volume = area of sheet * thickness

thickness = V/A = 4.64 x 10^(-6) cm = 4.64 x 10^(-8) m = 46.4 nm

4. Originally Posted by skeeter
area of the sheet ...
A = (2.4 ft*12 in/ft*2.54 cm/in)*(1*12*2.54) = 2229.67296 cm^2

volume of gold ...
V = density/mass = (19.32 g/cm^3)/(.200 g) = 0.01035... cm^3

volume = area of sheet * thickness

thickness = V/A = 4.64 x 10^(-6) cm = 4.64 x 10^(-8) m = 46.4 nm
Isn't volume = mass/density ?

5. Originally Posted by qbkr21
Isn't volume = mass/density ?
Yes, $\displaystyle V=\frac{m}{\delta}$

So you should get $\displaystyle V=\frac{.2~g}{19.32~\tfrac{g}{cm^3}}\approx\color{ red}\boxed{.01035~cm^3}$

--Chris

6. ## RE:

RE:

7. Originally Posted by skeeter
area of the sheet ...
A = (2.4 ft*12 in/ft*2.54 cm/in)*(1*12*2.54) = 2229.67296 cm^2

volume of gold ...
V = density/mass = (19.32 g/cm^3)/(.200 g) = 0.01035... cm^3

volume = area of sheet * thickness

thickness = V/A = 4.64 x 10^(-6) cm = 4.64 x 10^(-8) m = 46.4 nm
Why doesn't the area have to be in cm^3 too?

8. Originally Posted by qbkr21
Why doesn't the area have to be in cm^3 too?
Area has square units [since you multiply two terms that contain the same units], but volume has cubed units.

So area won't have the same units as volume...area, in this case, will have the units $\displaystyle cm^2$, whereas volume will have $\displaystyle cm^3$

--Chris