Help solving this problem.

I just dont 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

does anyone know?

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...

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.

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.

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
Homework Helper
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.

>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.

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