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Density of a platinum-iridium cylinder

  1. Sep 2, 2008 #1
    1. The problem statement, all variables and given/known data

    The standard kilogram is a platinum-iridium cylinder 39.0 mm in height and 39.0 mm in diameter. What is the density of the material?

    2. Relevant equations

    now I know that to find volume you have to use v=h(pi)r^2
    and I also know you have to convert it to meters. When I do I get this
    (0.039)(pi)(0.0195)^2
    which gives me 4.6589033654573236278091385713189e-5
    so this is what I get over all
    1 / (39.0 * pi * (19.5^2)) = 2.14642786 × 10-5
    yet the answer is 2.15*10^4k/m^3
    can someone tell me what im doing wrong, and if im suppose to convert the 2.14

    3. The attempt at a solution
     
  2. jcsd
  3. Sep 2, 2008 #2

    Astronuc

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    Staff: Mentor

    Well, 2.1464 is rounded up to 2.15 if one is using three significant digits in the final answer.

    One's solution is correct, the density is about 21464 or approximately 21500 kg/m3
     
  4. Aug 28, 2009 #3
    That answer is correct but be careful with units. 21464 is in millimeters. so it would be 21.464 kg/m^3
     
  5. Aug 28, 2009 #4

    rl.bhat

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    Homework Helper

    1 / (39.0 * pi * (19.5^2)) = 2.14642786 × 10-5

    This step is wrong. It should be
    1 / (0.0390 * pi * (0.195^2)) = 2.14642786 × 10^4
     
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