Recent content by x.users

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    Achieving High Value LC Circuits: What's Possible?

    Re look you know frecuancy = 1/( 2 byi (LC)^1/2) in resonance circute . A its any number and its unit Henri . Farad
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    Achieving High Value LC Circuits: What's Possible?

    Hello mm so sory bro about wrong i mean L C ( Inductor and capasitor ) = A x 10^(-78) . Cuz i need it to make balance to produce H2 from H2O by waves . thank you so much . can you add me Zkharove.laser@msn.com
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    Achieving High Value LC Circuits: What's Possible?

    Hello bro I'm so sorry to some wrong but i was meaning this facts : i try to obtaind a LC = A x 10^(-78) cuz i need it to make balance to produce H2 from H2O by Waves thank you so much .
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    Achieving High Value LC Circuits: What's Possible?

    Can we ...! Hello all I want to ask you immportant question ... Can we get High value of LC . i mean A x 10^(-45) and if we get it where we can use it ? i mean type of circute . thank you so much
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    What Determines the Thermal Conductivity of Gases?

    I need Info...! Homework Statement I need information on the thermal conductivity of gas . thank you so much . Homework Equations The Attempt at a Solution
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    Solving Integrals by Substitution: Tips and Tricks

    and and respect to second probelm int[1/(e^x - 1)] put 1 = e^x/e^x its become int[ e^x/(e^2x - e^x)] put u = e^x subtitution du = e^x dx du = u dx so dx = du/u int [u du/(u^2 - u ) u] and u =! 0 because u = e^x and exponantioal never will be zero...
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    Solving Integrals by Substitution: Tips and Tricks

    mm i think you must use fraction partial ...and in denominator adding +1 and - 1 . to use fraction partial . 1/(x^4 +1 +1 -1 ) = 1/(x^4 - 1 )+2 1/[(x^2-1)(x^2+1)+2] 1/[(x-1)(x+1)((x^2 -1)+2) +2] 1/[(x-1)(x+1)((x-1)(x+1)+2)+2] and you can use this web...
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    How to Solve a Second Order Differential Equation with a Trigonometric Function?

    x" = cosx its x-missing v = x' v' = x" = v dv/dx v dv/dx = cosx vdv = cosx dx v^2 = 2 sinx (x')^2 = 2 sinx x' = ( 2 sinx )^1/2 dx / (sinx )^1/2 = 2^1/2 dy Q = 2^1/2 y + C such that Q = http://integrals.wolfram.com/index.jsp
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    Antiderivative of Cotangent and Arcsine - Explanation Welcome

    mcmw mm you can visit this web to know http://en.wikipedia.org/wiki/Antiderivative you are welcome .
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