subaruturbo_18

I have a question on one of my work sheets that im stuck with, i've tried looking on the internet but cannot find too much, i do not have the books i need yet as student loan hasn't come through

The question is ' a gas bubble from an explosion under water, oscillates with a period proportional to P (to the power of a) d (to the power of b) and E (to the power of c), where P is the static pressure, d is the density, and E is the total energy of the explosion. Find the values of a, b, and c.

So first i found the dimensions of each P, d, and E,

P = m . l-1 . t-2

d = m . l-3

E = m . T-2

which then gives me

[ma l-a t2a] [mb l-3b] [mc t-2c]

which when seperated gives me

m abc l-a+(-3b) t -2c+(-2c) = T

a + b + c = 0 (1)

-a + -3b = 0 (2)

-2a -2c = 1 (3)

so (im not 100% on this, correct me if im wrong) i need solve the above equations to find out the values of a b and c?

boxst

Sorry not very helpful, but I'm extremely glad I don't have to do that stuff anymore. I'm dreading when my daughters get to the age where I will have to dust off my brain to help them.

Steve

my06 ppp silver

iTrader: (2)

my head has just gone in to meltdown!!!

my06 ppp silver

iTrader: (2)

try a new post, DOES ANYONE KNOW CAROL VORDERMAN?

JPL

Thanks! You've just made my head explode!

subaruturbo_18

it's funny as i feel stupid in my class not knowing this, but yet here i feel superior for getting that far.

AndyC_772

iTrader: (2)

(1) + (2) => -2b + c = 0, c = 2b (4)

Sub into (3) => -2a - 4b = 1, -2a = 1 + 4b, a = -1/2 - 2b (5)

Sub (4) and (5) into (1): -1/2 - 2b + b + 2b = 0
-1/2 + b = 0, b = 1/2

Sub into (2): -a + (-3/2) = 0
a = -3/2

Sub into (1): -3/2 + 1/2 + c = 0
c = 1

Haven't checked that your dimensional analysis is correct, but if it is then there are your answers

RJMS

iTrader: (2)

From equation (3)
-2a-2c=1 so -2c=1+2a so c=-1/2-a

From equation (2)
-a-3b=0 so -3b=a so b=-a/3

Substitute into equation (1)
a + b + c = 0 so a-a/3-1/2-a = 0 so -a/3-1/2=0 -a/3=1/2 so a=-3/2

Substitute into equation (2)
3/2-3b=0 so 3/2=3b so b=1/2

Substitute into equation (1)

-3/2+1/2+c=0 so c=1

You'll notice the quote was from before you changed equation 3 - it was pretty much not solvable with the original version!

scud8

iTrader: (1)

I think your dimensions for energy are wrong - it should be m l2 t-2 (mass x velocity squared, or force x distance). See if that gives you a more plausible answer.

alcazar

iTrader: (2)

OK, from the second equation, -a-3b=0, so -a=3b, and b=-a/3, (call this (4))

If you now substitute this into equation 1, you get,

a-a/3+c=0, therefore 2a/3 +c =0, call this (5)


Now taking (5), and (3), we have:

2/3a + c = 0 (multiply this eq by 2, call the result (6))
-2a -2c = 1
4/3a +2c = 0 (6) (ADD these last two equations, (3) + (6))

-2/3a = 1, so 2/3a = -1
and, a = -1 (divided by) 2/3, = -3/2

You should now bwe able to substitute this value into the other ORIGINAL equations to get the other two values, starting with (2), then (1)


Does this help?

RJMS

iTrader: (2)

At least we're all coming up with the same answers

alcazar

iTrader: (2)

LOL, but are they correct?

It's actually NOT that advanced algebra. Are you reading this, Andy?

RJMS

iTrader: (2)

Thye are based onthe final three equations No idea before that - and no it's not really advanced at all, try working it with the original three equations though that's a real bugger (unless you want a,b & c to all be 0 which seemed unlikely!)

scud8

iTrader: (1)

The answer is a is -5/6, b is 1/2 and c is 1/3. See equation (3) of http://personalpages.manchester.ac.u...derwater-I.pdf for independent verification. You definitely need the right dimensionality for energy (m l2 t-2) to get the answer - see my post above - your original three equations are wrong.

(Carol Vorderman only got a third - I got a 2:1.)

chocolate_o_brian

iTrader: (22)

Just got in!

mr_D

iTrader: (12)

subaruturbo_18

Just out of curiosity, does anybody think this is a bit hard for a question on the first work sheet we have been given on the second week of a foundation year of engineering?

I am finding it very difficult. and don't really understand how anyone got to the answer they did.

scud8

iTrader: (1)

I don't think it was that hard - it was the kind of stuff I did in A level Applied Maths back in the early 80s. You seemed to have the right idea in your original post.

I was a bit thrown by the -5/6 when I worked it out, which was why I looked for an independent verification!

zeuss

iTrader: (2)

dreading my daughter asking me anything like this with homework I havent a clue

dunx

iTrader: (3)

Unless she goes for physics or engineering, don't worry !

LOL

dunx

P.S. I've just got in from work where ml^2t^-2 was a choc chip muffin and a coffee !

P.P.S. http://personalpages.manchester.ac.u...derwater-I.pdf that's way cooler than the "sinking ship" we did on my course

chocolate_o_brian

iTrader: (22)

Ahem, ask Jeff about that, as he's currently tutoring me in algebra etc, and I'm an engineer wannabe

wayne9t9

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