#21
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Quote:
[6+(4 x 9)]/6 = y [6+ 36]/6 = y 42/6 = y 7 = y the instructions should say "simplify" not "solve" becuase saying y = [6+(4 x 3^2]/(7-1) is in fact already some said equation being solved in terms of y hope this helps Edit: Quote:
when they say changed by a factor of 4 they mean your mass or velocity (depending on what specifically they state in the problem) decreased (sometimes it could be increased that you have to read in the given problem) by a number divided by 4. For example, lets say you velocity initial was 20 m/s and they say velocity final decreased by a factor of 4 it would mean 20/4 which = 5m/s. It would be the same if they said a similar statement about your two masses. If this is still not clear give me the problem and ill help you solve it If you need help with your physics jalen feel free to pm me at anytime Mike~ Edit: Quote:
Ok so you have Vo = .505 and V = .301 so .505(x) = .301 x = .301/.505 x = .596 so you velocity changed by a factor of .596 ... check it .301 / (.596 your factor) is = to your Vo
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LETS GO YANKEES! CONGRATS ON #4 GIANTS! Last edited by Future Raptor Ace; 12-10-2010 at 11:44 PM.. Reason: Automerged Doubleposts |
#22
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I still didn't understand so I just put 4 down for everything,
I'll give the exact numbers later so you guys can try and explain it to me, but it was like freaking a factor of 1/3 and crap
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Be the leaf.
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#23
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I need help solving this problem, it is a fluid pressure problem and it is annoying the hell out of me.... if anyone here knows how to do this please help..... thanks in a million
The problem says a 3ft wide rectangular gate is pinned at is its center C. Determine the torque M that must be applied to its center to open the gate. The specific weight of water is 62.4 lb/ft^3 The dimensions are 12ft 3ft 3ft 2ft and 2ft if you cant read them..... thanks again
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LETS GO YANKEES! CONGRATS ON #4 GIANTS! Last edited by Future Raptor Ace; 12-17-2010 at 10:51 PM.. |
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Quote:
Don't know if you solved the problem or not. If not then PM me. Note: I can't read the dimensions because there is no drawing.
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--- pace e salute --- Last edited by Rev; 12-17-2010 at 11:47 PM.. |
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Rev or anyone please help me here im not understanding the derivation of a formula
Ok so Normal strain x = εx = σa/E - υσb/E where υ = Poisson's ratio (- εx/εy) (-lateral strain / axial strain) Normal strain y = εy = -υσa/E + σb/E where υ = Poisson's ratio (- εx/εy) (-lateral strain / axial strain) (a = x) and (b = y) E = the modulus of elasticity Now from the two equations above how does it come to these two equations below which are normal stress x and y ? σx = [E/(1-υ^2)](εx + υεy) (eq. 1) σy = [E/(1-υ^2)](εy + υεx) (eq. 2) I am solving for P ( a tensile force) so I know I need to solve for the two equations above and relate them to P/A. The only thing that is bothering me is I have no idea where (eq. 1) and (eq. 2) come from or how they are derived..... please help! Thanks! Is this a Morh's Circle im missing?
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LETS GO YANKEES! CONGRATS ON #4 GIANTS! Last edited by Future Raptor Ace; 04-10-2011 at 04:58 AM.. |
#26
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lolol seriusly bro that aint too hard. you shoulda just asked me instead of the peeps in here who dont study that stuff. its a shame tho that they arent as smart as us. i try to rub it in sometimes too hahah
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"i'm not conceited, i'm convinced. conceited people have faults, i have no faults." "life is a game, and I win" |
#27
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Your response did not help answer my question
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LETS GO YANKEES! CONGRATS ON #4 GIANTS! |
#28
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He's just jelly that he doesn't know how to type sigma
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#29
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The "I'm more awesome than..." thread is that-a-way.
My sig can be used as a visual. |
#30
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lol
If anyone still knows how to help please do, non of the classmates im friends with understand it either
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LETS GO YANKEES! CONGRATS ON #4 GIANTS! |
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