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 11 months 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?
 11 months 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
 11 months 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!

adev
 11 months 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

kimkelle
 11 months ago
Best ResponseYou've already chosen the best response.1Hi, 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?

adev
 11 months 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 :)

kimkelle
 11 months ago
Best ResponseYou've already chosen the best response.1Yes is it easy to miss :)
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