In summary, Mike has gathered data that suggests a perfect inverse relationship between time and water level in a 2 liter bottle, with a constant of 0.0169. He is trying to determine what the 0.0017 is so he can derive a formula for the time it takes to empty a bottle depending on initial volume, bottle radius, and slot area.
#36
Cyrus
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16
Looking at your data, your equation matches very well to what vanesche gave.
Wait, which equation, the one provided on the site or the one i have just posted?
#38
Cyrus
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The one on the site. I don't know what that mess you posted is, sorry.
#39
gtabmx
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So you manipulated the equation ont he site into something useful where velocity is not included or did you use the equation as is?
#40
gtabmx
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How can that be, I plug 25seconds into that equationa nd get -2.3meters?
#41
Cyrus
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what do you mean velocity, where do you see velocity anywhere?
#42
gtabmx
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"The symbol u stands for the fluid velocity"
on site
#43
Cyrus
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Yes, and where do you see it in the final equation?
Think about conservation of energy.
#44
gtabmx
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Nowhere, ok so your are using the last equation, I thought you rederived another equation from Bernouilli's equation. Yes, I am using that last equation and it does not fit, perhaps my values for lambda are worng. I solved for them and got around 4.0E-3. Correct?
#45
Cyrus
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I got your graph to match, you better redo those calculations boy, or I am going to smack you!
#46
gtabmx
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OKOKOK i will gimme a sec
#47
gtabmx
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OK, i get lambda is either equal to 9.251E-3 or 3.526E-3 with inconsistent units, correct?
#48
Cyrus
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Nope. Check your work some more. You are close though.
#49
gtabmx
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Oh god, look I know you are trying to do this for my own good but I looked over it and found nothing wrong and I have fired up maple to triple check and here.
{x = 0.003525914211}, {x = 0.009251617087}
#50
Cyrus
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Can you write me your equation your using to solve for lambda?
It might be that its rounding off, but I want to be sure its not what I think it might be.
#51
gtabmx
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0
0.0251 = 3062.500000 x - 39.13118960 x + 0.125
0.0251 = 3062.500000 x - 39.13118960 x + 0.125
"->"
{x = 0.003525914211}, {x = 0.009251617087}
#52
Cyrus
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16
No, this is not correct.
Where did you get these numbers?
Intercept is ok, but the rest are wrong.
Where did this come from?
0.0251?
3062.500000 ??
39.13118960 ??
#53
gtabmx
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I used the equation and i plugged 25s into time and the corresponding height 0.0251m from my results and i simplified.
#54
Cyrus
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16
Ah, this is fine.
BUTTT, you should use data points as far as possible, i.e. use 35.
You will find that you get:
0.0035822
0.00551068
#55
gtabmx
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ok, so wat a couple of decmals :tongue2: so, wat are the units of lumbda?
and now wat? there must be a way to find the values before even doing the experiment?
#56
Cyrus
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You tell me.
Not a couple of decimals, a factor of 2.
#57
gtabmx
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It was a joke, i understand.
#58
gtabmx
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No units, makes me feel better i think
#59
gtabmx
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So now i know lambda has no units and u must do the experient to find it, that's not taht good since its not a general formula yet
#60
gtabmx
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Wait, lambda is not consistent, it has not units is in first term and units in second
#61
Cyrus
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Are you sure?
#62
gtabmx
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m = m/s^2 * ?^2 * s^2 is first term so m = m * ? no units
m = m^(1/2)/s * ? * m^(1/2) * s is second term so m = ? units
#63
Cyrus
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What is this?
Do a proper dimensional analysis, and take your time.
#64
gtabmx
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I DID IT 5 TIMES adn I was about to take a picture of it with my phone until i found out the qualiy sucks
#65
Cyrus
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Then do it a 6th time, but TAKE YOUR TIME!
Show me some work, try to make it cleaner.
#66
gtabmx
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same thing again, look, why can't u understand that i get the same thing?
#67
gtabmx
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WAIT 1/m 1/m no, oh now I'm even more confused
#68
Cyrus
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Here:
[tex] \frac{m}{s^2} s^2 * \lambda = m[/tex]
Lambda is units of meter
[tex] \sqrt{ \frac{m}{s^2} {m}} * s *\lambda = m[/tex]
lambda is in units of meter, and it better be.
come on, this is algebra 1. I warned you to TAKE YOUR TIME.
#69
gtabmx
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You know I have a feeling that's not right look at the final equations, each term has to be in meters after all the steps are carried out now the first temr is Y^2 times gravity times time squared which is m = Y^2 * m*s(-2) * s^2 so then s^(-2) outs s^2 and you have m = Y^2 * m no?
#70
gtabmx
56
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ohh
wait
wait wait
heres my deal, i made a stupid mistake probably due to the fact that I've been up till 2AM adn woke up at 6 for the past 3 days so I'm sorry
also, i get consistent, but NO UNITSyup sextuple checked no units as simplified its m = m * y^2 - m * y + m