Solve for the length of a cylinder

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AI Thread Summary
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.
 
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