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anonymous
 2 years ago
For problem 1 of challenge problems for units and dimensional analysis, im confused on section 1.10 of the solution. How does Y = 1 and Z = 1?
anonymous
 2 years ago
For problem 1 of challenge problems for units and dimensional analysis, im confused on section 1.10 of the solution. How does Y = 1 and Z = 1?

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theEric
 2 years ago
Best ResponseYou've already chosen the best response.0Hi! I won't be on too much longer, but maybe I can help. I don't know what the question or solution are, though, since I'm not in the course. Feel free to provide that information, so that other people on OpenStudy can help!

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Problem 1: An ideal (nonviscous) liquid with a density of ! is poured into a cylindrical vessel with a crosssectional area of A1to a level at a height h from the bottom. The bottom has an opening with a crosssectional area A2. Find the time it takes the k=liquid to flow out. I'll post the link to the solution because copying it would take up too much space. http://ocw.mit.edu/courses/physics/801scphysicsiclassicalmechanicsfall2010/introductiontomechanics/unitsanddimensionalanalysis/MIT8_01SC_problems01_soln.pdf

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Hi, I just finished the set of questions. Well, to answer your question as to how is Y = 1 and Z = 1, Considering equation 1.10 and analyzing using dimensions, we obtain, [T] = [L]^y/[L]^(z) Which leads to y+z = 0 since there is no [L] dimension on the LHS Further implies y =z which implies that these quantities cancel out giving y=1 and z=1. This is how I have understood it, let me know what you think?

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Ah ha! That makes sense, I forgot to use dimensional analysis for that last equation. Thank you for clarifying :)

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Yes is it easy to miss :)
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