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  1. I'll express the force as F = kQq / d² where k = 1/4πε₀ = 8.99e9 N·m²/C² Both q1 and q2 will...

    1 Answers · Science & Mathematics · 27/09/2020

  2. ... m2/r^2) b is the correct answer. 2. You can equate h nu= k _b T. k _b is the Boltzmann constant, T is the...

    1 Answers · Science & Mathematics · 28/09/2020

  3. ... on the threshold of slipping, then Ff = 254 N = µ k *m*g*cosΘ = µs * 43kg * 9.8m/s² * sin37º means...

    1 Answers · Science & Mathematics · 24/09/2020

  4. ...net force = applied force horizontal - friction force m*a = Fx - µ k *(Fy + m*g) 8.67kg * -0.90m/s² = 20N - µ k *(25N...

    1 Answers · Science & Mathematics · 23/09/2020

  5. using kinematics, a = g*(sinΘ + µ k *cosΘ) a = 9.81m/s²*(sin15.0º + 0.18*cos15.0º) = 4.24...

    2 Answers · Science & Mathematics · 23/09/2020

  6. 1.381e-23J/ K * T = 1.6e-19 J T = 11,586 K So one might conclude that 1 eV of energy is quite a bit for a single particle. If you find this helpful, please select Favorite Answer!

    1 Answers · Science & Mathematics · 22/09/2020

  7. When periodic, assume a sin wave type solution. s= a sin kt. k is the time constant. s is the strain. s1=0.0032 = a sin kt1, s2 = a sin...

    3 Answers · Science & Mathematics · 22/09/2020

  8. a. intensity I = p*v where pressure p = B* k *A*sin(kx - ωt) and velocity v = A*ω*sin(kx - ωt) (see citation) so...

    1 Answers · Science & Mathematics · 20/09/2020

  9. ... * 20K = 896 J since for CHANGES in temperature, 1 K = 1 ºC 8) ΔQ = 3kg * 4186J/kg·ºC * ...

    1 Answers · Science & Mathematics · 20/09/2020

  10. type of steel is needed. I'll guess at 12e-6 / K ∆L/L = α∆T 0.19/L = 12e-6(42– –9) L = 0...

    3 Answers · Science & Mathematics · 18/09/2020

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