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  1. ... to get a Kelvin value of 0 corresponding to absolute zero. K = C+273. 4). Rearrange part 3 algebraically. K = C+273 Subtract...

    1 Answers · Science & Mathematics · 11/07/2011

  2. Celsius [°C] = [K] − 273.15 [K] = [°C] + 273.15 Fahrenheit [°F] = [K] × 9⁄5 − 459.67 [K...

    1 Answers · Science & Mathematics · 10/02/2014

  3. ... radioactive decay, you have : -------------------------------------------- dC/dt = ( - k ) ( C ) dC/C = ( - k ) dt ln ( C / C0 ) = ( - k ) t C...

    1 Answers · Science & Mathematics · 20/03/2015

  4. –32°F = –36ºC = 238 KC] = ([°F] − 32) × 5⁄9 [°F] = [°C] × 9⁄5 + 32 [K...

    1 Answers · Science & Mathematics · 02/03/2011

  5. For a monatomic ideal gas : ----------------------------------------------------- k = C sub P / C sub V = 1.667 Q sub const V = ( m...

    2 Answers · Science & Mathematics · 07/01/2011

  6. k(-3q)/r + k(C)/(r/sqrt(2)) + k(4q)/r = 0 -3q + sqrt(2)C + 4q = 0 C = -(sqrt(2)/2)q

    1 Answers · Science & Mathematics · 26/01/2009

  7. ...have no more kinetic energy left to transfer is___B______ b: 0 k a: -273 k c: 0 C b: 0 k D: 273 K At absolute zero -273.15C...

    1 Answers · Science & Mathematics · 21/10/2008

  8. ...this Equation 1 where T(t) = temperature at any time "t" k & c are constants Ta = ambient temperature <&...

    3 Answers · Science & Mathematics · 06/02/2009

  9. T = deltaE / deltaS delta E = 3e-21 delta S = k ln Q a) 159.45 K b) 201.99 K c) C = deltaE / deltaT = (3e-21) / (201.99-159.45) = 7.05219e-23 / 2 = 3.52609e-23

    2 Answers · Science & Mathematics · 24/04/2012

  10. F = (9/5)C + 32 K = C + 273.15 C = K – 273.15 substituting: F = (9/5)(K...

    2 Answers · Science & Mathematics · 30/09/2012

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