What is the area of the gold leaf and the length of the gold fiber?

[digink]

I just don't get how to solve this problem:

3. Gold, which has a mass of 19.32 g for each cubic centimeter of volume, is the most ductile metal and can be pressed into a thin leaf or drawn out into a long fiber.

(a) If a sample of gold, with a mass of 33.16 g, is pressed into a leaf of 2.400 µm thickness, what is the area of the leaf?
___m2
(b) If, instead, the gold is drawn out into a cylindrical fiber of radius 1.000 µm, what is the length of the fiber?
___m

I don't understand what to use please help.

 

[digink]

does anyone know?

 

[wire2]

Your sample is 33.16 ÷ 19.32 = 1.716356108 cc's
1 cc = 1 000 000 sq cm at 1 µm thick
so at 2.4 µm = 416666.6667 sq cm

volume of cylinder is pi r² h
so 3.1416 x .5 x .5 x h = 1,716,356.108
so h = 2185327.359 cm

I think...

 

[digink]

neither of those are right, also could you please explain a little. This is in the beginning chapter of my physics book and I completely forgot/dont know what the hell they are talking about.

any help would be great.

 

[MathStudent]

Hint:
They gave you the density which = mass/volume.
You know the mass of the sample of gold... Thus you can figure out what volume it should have.

 

[digink]

they didnt give me density.. am I missing something?

 

[futb0l]

a mass of 19.32 g for each cubic centimeter of volume

That's density. Mass/Volume.

 

[dextercioby]

So u've got the leaf's mass,the leaf's density and now use the definition for density to find the volume.
For point (a),u can assume the leaf to have a parallelipipedic shape.U're asked for the area of one side,being given the thickness...

For point (b),u can assume the fiber to be a rectangular circular cylinder and,this time,you're asked for the height/length,knowing the base(circular) surface and its volume...

Daniel.

 

[wire2]

>neither of those are right, also could you please explain a little.

ok, I'm curious why but anyway,
the sample is 33.16g, 1 cc = 19.32g so to divide will give the # of cc of the sample;
>>Your sample is 33.16 ÷ 19.32 = 1.716356108 cc's

>>1 cc = 1 000 000 sq cm at 1 µm thick
yes, no, anyone?
if so, we have 1,716,356.108 sq cm at 1 µm thick

then I divided by the nominal thickness of 2.4 µm = 416666.6667 sq cm

D'oh! that should have been 715148.3783 sq cm

>>volume of cylinder is pi r² h
h being the height (length) of the rod
>>so 3.1416 x .5 x .5 x h = 1,716,356.108 (volume of sample in µ cc)
then I transposed h to solve
>>so h = 2185327.359 cm
It's been 40 years since my physics classes so I *may* have forgotten something.