Unit and Dimensional Analysis: Challenge Problem 1
Is it supposed that the information provided in this module is sufficient on its own to be able to do the accompanying challenge questions? I have done first year university physics and algebra but no physics for a long time and felt that these challenge questions require more knowledge then is obtained in the module (for example in Problem 1 what the hell is 'b'?). It's a bit disappointing after getting all excited about studying physics.
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You’re not alone in asking this… The variable b is a numerical constant and g is the speed of the liquid. It looks like they are beginning their analysis by writing down all the things they can think of which might be involved... could depend on the density, could depend on the area of hole 1, could depend on the area of hole 2, could depend on the speed of the liquid, could depend on the height of the standing liquid, and of course could have a numerical constant involved. Don't know that it depends on the first power of each of those, so throw in powers. I think the point is if you guess wrong, the power will end up being 0, or you will not be able to get a solution
Thanks for the answer. I understand that we are meant to have a go but it did seem a bit of a sudden leap. I'll keep going with it and see if I can't get into the swing of things.
p.s. I meant to say that I've done first year uni calculus and linear algebra. I definitely haven't done physics
I'm a bit stuck on this one too. I feel like I now have to go away and learn some mechanics on my own before I can move on.
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Gremlin - I think this is a relatively minor part pf physics - I wouldn't give up now! If you go a little further I think things will make much more sense.
I still don't get it. Why is "0 = -3V +W + X + 2(Y + Z)" equal to zero? and why did they put "X" instead of "-2X"? Please, can somebody help me on this? Thanks.
In that equation he is looking only at the powers of L
On the left side, if L is there its power must be 0 since anything to the power 0 is 1
On the right side, the first term has L to the power -3V, the second term has L to the power W, the third term has L to the power X, the fourth term has L to the power 2Y and the last has L to the power 2Z... sine these are powers, the "combined" power of L on the right side is -3V + W + X + 2(Y + Z)... Equating the power of L on the left to the power of L on the right gives their equation