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Density of a platinum-iridium cylinder |
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| Sep2-08, 07:49 PM | #1 |
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Density of a platinum-iridium cylinder
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 |
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| Sep2-08, 08:07 PM | #2 |
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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 |
| Aug28-09, 02:15 PM | #3 |
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That answer is correct but be careful with units. 21464 is in millimeters. so it would be 21.464 kg/m^3
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| Aug28-09, 08:19 PM | #4 |
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Recognitions:
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Density of a platinum-iridium cylinder
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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