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Solve for the length of a cylinderby lshine09
Tags: length of a cylinder 
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#1
Jan1310, 08:28 PM

P: 7

1. The problem statement, all variables and given/known data
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? 2. Relevant equations Volume of Cylinder= (PI)(R^2)(h) V=M/D 3. 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= 1.475155E7 m3 3. I converted the Radius from um into m 2.8 um * (1m/10E6 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)= 1.475155E7 m3 5. I then solved for h and got the answer of: 5.989E21 (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? 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution 


#2
Jan1310, 08:34 PM

HW Helper
P: 6,206

2.8μm=2.8x10^{6}m



#3
Jan1310, 08:38 PM

P: 7

ha THANKS! As always, a retarded mistake.



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