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[H] impossible conversion problems of death

Blogs > SaveYourSavior
Post a Reply
SaveYourSavior
Profile Blog Joined July 2008
United States1071 Posts
October 14 2008 05:29 GMT
#1
Hello there I am taking an earthquakes class expecting an easy peasy college class. Not only is it boring as hell unless you have a hardon for geology, it has some difficult labs. I have to do some conversion problems but i have no idea on how to convert some of these figures. My professor hasn't really taught me on how to convert these things and my book has nothing on this besides the following formulas for P waves and S waves:

The formulas for P waves (alpha) and S waves ( beta)

a = square root of ( ( k + 4/3(u) ) / p )

B= square root of (u/p)



I have to find p which has to be in units of grams per cubic centimeters ( g/cm^3) and i can use either of these 2 formulas:

a = square root of ( ( k + 4/3(u) ) / p )

B= square root of (u/p)

given values for #1

k= (27 x 10^10 ) dynes / cm^2
a = 4.8 km/s
u = (2.6 x 10^11) dynes/ cm^2
b= 3.0 km/s

i use b = square root of (u/p) because it is easier:

(3.0) km/s = square root of ( (2.6 x 10 ^11 dynes/cm^2)/ p )

here i get stuck, I have to square both sides but I am unsure of what happens when you square 3.0 km/s. Does it become 9 km^2/s^2? or 9 km/s^2?

I assume its 9 km/s^2 and the equation becomes:

p(9.0 km/s^2) = 2.6 x 10^11 dynes/ cm^2

p = (2.6 x 10^11 dynes/cm^2)/ (9.0 km/s^2)

i dont know how to convert dynes/cm^2 into g/ cm^3 although g/cm^2 exists, also i don't how the hell to convert km/seconds^2 into grams/ cm^3

i am completely lost and have 17 hours to do 6 of these problems

if you know how to do these things, can you please explain to me on how to do it? Thanks a lot. I assure you there will be something in for you besides lending me your knowledge and time. Thanks again!

a
HeavOnEarth
Profile Blog Joined March 2008
United States7087 Posts
Last Edited: 2008-10-14 06:02:43
October 14 2008 05:51 GMT
#2

i don't how the hell to convert km/seconds^2 into grams/ cm^3


Whoa. how'd you go from Dx/Dt to Dx/Dx O_O .... erm maybe use ... N (newton) = kg·m/s2 ... then ... gah what the fuck how does seconds go into grams or cm^3
thats ridiculous D:

1 (km / (seconds^2)) (grams / (cm^3)) = 1 000 000 m-2 kg s-2 is all i can do
that's an evil class
"come korea next time... FXO house... 10 korean, 10 korean"
LiAlH4
Profile Joined October 2007
New Zealand111 Posts
Last Edited: 2008-10-14 06:05:04
October 14 2008 06:03 GMT
#3
You've asked things in a big messed up mess, but I'll try to help you...

When you square 3 km/s it becomes 9 km^2/s^2 not 9km/s^2
... the way to multiply units is just to consider them as algebraic terms, so e.g. (x/y)*(x/y) = x^2/y^2

I didn't know this, but according to the great god of all knowledge (wikipedia), 1 dyne = 1 g.cm/s^2

So you have an equation that looks like this...

b km/s = (( u dynes/cm^2) / (p g/cm^3))^.5

now, by substituting the units of a dyne in, it becomes

b km/s = ((u g/(cm.s^2))/(p g/cm^3))^.5

now, by dividing the units of u by p...

b km/s = ((u/p)cm^2/s^2)^.5

b km/s = (u/p)^.5 cm/s

So now you just need to convert from cm to km

b km/s = (u/p)^.5 cm/s / (10^5 cm/km)

So that's how you work out what to do (hopefully I didn't stuff up my brackets or algebra or anything, and it's clear to follow). So if I did everything right, all you need to do in terms of calculations is just to take the squareroot of u/p and divide by 100000
SaveYourSavior
Profile Blog Joined July 2008
United States1071 Posts
Last Edited: 2008-10-14 08:24:37
October 14 2008 07:41 GMT
#4
ohhhh I see now


sorry about the confusion I gave upon you


thanks a lot for the help really appreciate it


edit: actually i still have trouble finding p dang
a
LiAlH4
Profile Joined October 2007
New Zealand111 Posts
Last Edited: 2008-10-14 08:50:41
October 14 2008 08:50 GMT
#5
oh, in my above post when i said what you need to do, that was for finding b not p. (ooops)

So to find p you'd do....

p = u/((b/100000)^2)

Does that help?
If not, I might have rearranged things wrong... I'm not sure.
Please log in or register to reply.
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