Find Area of flattened gold with limited info.

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To find the area of a flattened gold leaf with a mass of 1.708 g and a thickness of 6.811 μm, the volume is calculated using the formula V = M/D, yielding a volume of 0.088 cm³. The area is then determined by rearranging the volume formula to A = V/H, where H is the thickness converted to centimeters. An initial error in converting micrometers to centimeters was identified, which led to an incorrect area calculation. After correcting the conversion, the area was recalculated to be 1.29e-3 m². The discussion highlights the importance of accurate unit conversions in solving volume and area problems.
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Homework Statement


Gold, which has a density of 19.32 g/cm3, 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 1.708 g, is pressed into a leaf of 6.811 μm thickness, what is the area (in m2) of the leaf? (b) If, instead, the gold is drawn out into a cylindrical fiber of radius 2.100 μm, what is the length (in m) of the fiber?


Homework Equations


D = M/V
V = LxWxH(thickness)

The Attempt at a Solution


So I have been working on this for quite a while now. Here is my attempt:

I first solved D=M/V for V: V=M/D

Then I plugged 1.708g/19.32g/cm^3 to find my volume of .088cm^3

Then using Volume = L x W x H(thickness) I plugged in for volume and height and solved the equation.

When I converted micrometers to centimeters my result was 6.811e-3cm.

So (.088cm^3)/.006811cm = L x W (Area).

After I found the area in cm^2 which the above result yields, I converted to meters^2 and got an answer of 1.29e-3m^2.

It says it is wrong, where did I go wrong?
 
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Your conversion for the thickness is wrong.
1 um is 10^-3mm so 10^-4cm
 
Alright thanks. That made all the difference, problem solved.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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