Solve for the length of a cylinder

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To find the length of a cylindrical fiber made from a 2.850 g sample of gold with a density of 19.32 g/cm3, the volume is calculated using the formula V = M/D, yielding approximately 1.475155E-7 m3. The radius is converted from micrometers to meters, resulting in 2.8 x 10^-6 m. The volume of the cylinder is then expressed as V = (PI)(R^2)(h), and upon solving for height (h), the result is approximately 5.989E-21 m. Despite repeated calculations yielding this answer, the online homework system WileyPlus indicates it is incorrect, suggesting a possible error in the setup or calculations. The discussion highlights the importance of unit conversions and careful application of formulas in physics 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.If a sample of gold with a mass of 2.850 g is drawn out into a cylindrical fiber of radius 2.800 μm, what is the length (in m) of the fiber?

Homework Equations



Volume of Cylinder= (PI)(R^2)(h)
V=M/D

The Attempt at a Solution



1. I converted the density from g/cm3 in g/m3.
D= 19320000 g/m3

2. I use the equation V=M/D to solve for the Volume of the cylinder.
V=(2.850g)/(19320000 g/m3)
V= 1.475155E-7 m3

3. I converted the Radius from um into m
2.8 um * (1m/10E-6 um)= 2800000

4. I set the Volume of the cylinder equal to (PI)(R^2)(h) and plugged in the radius in m
(PI)(2800000^2)(h)= 1.475155E-7 m3

5. I then solved for h and got the answer of: 5.989E-21 (rounded to 4 SF)





I have gotten that answer over and over but our online physics homework system WileyPlus says it is incorrect. Any ideas where I went wrong?
 
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2.8μm=2.8x10-6m
 
ha THANKS! As always, a retarded mistake.
 
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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