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  1. ...for both charges. Sum the horizontal components: Ex = (k*4e-6/18) * cos45 + k*4e-6/18 *cos135 = k*4e-6/18 *[0.7071-0.7071] = 0 Sum...

    1 Answers · Science & Mathematics · 01/09/2019

  2. ...the field due to the other two corner charges: kQ / (2L²) = 2*k*10nC*cos45º / L² which leads directly to Q...

    1 Answers · Science & Mathematics · 01/09/2019

  3. ...with q1 in place and then you bring in q2, the new potential energy is PE = k*q1*q2 / (0.200m*√2) and if you then bring in q3, you have PE' = PE...

    1 Answers · Science & Mathematics · 01/09/2019

  4. ... is force in newtons r is separation in meters k = 8.99e9 Nm²/C²

    1 Answers · Science & Mathematics · 31/08/2019

  5. ... is force in newtons r is separation in meters k = 8.99e9 Nm²/C²

    1 Answers · Science & Mathematics · 29/08/2019

  6. ... vector arithmetic. for the left charge Ex = 2µk/r² pointed right Ey = 0 for the right charge... in Newtons/coulomb OR volts/meter k = 1/4πε₀ = 8.99e9 Nm²/C²

    1 Answers · Science & Mathematics · 27/08/2019

  7. ...standard temperature and pressure R = gas constant = 286 m²/s²/K for air T = absolute temperature (273.15 K + °C) = 298.15 K...

    3 Answers · Science & Mathematics · 24/08/2019

  8. ... = QE in Newtons/coulomb OR volts/meter k = 1/4πε₀ = 8.99e9 Nm²/C² re the other problem, just use...

    2 Answers · Science & Mathematics · 23/08/2019

  9. Potential energy of the spring at max compression is kx^2 / 2. This gets turned to gravitational potential energy at the top of the incline with no friction h = kx^2 / (2mg)

    1 Answers · Science & Mathematics · 23/08/2019

  10. ... following equation is used to solve this problem. F = k * Q1 * Q2 ÷ d^2 I use 9 * 10^9 for k. The unit of the...

    2 Answers · Science & Mathematics · 16/08/2019

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