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  1. a) The force on q1 due to q3 (r=10cm = 0.1m) is: F13 = k.q1.q3/r² = 9*10^9 * (-3*10^-6) * (3... = 0.1² + 0.1² = 0.02 The force on q1 due to q2 (with r=d so r² = d² = 0.02) is: F12 = k.q1.q1/r²...

    1 Answers · Science & Mathematics · 22/06/2019

  2. Fg / r = GMm/r^3 Which is not particularly useful if you ask me, are you sure that's what you mean?

    3 Answers · Science & Mathematics · 11/07/2019

  3. For z >> R, E = qz / [4πε₀*(z²+R²)^(3/2)] ≈ qz / [4πε₀...

    2 Answers · Science & Mathematics · 04/09/2019

  4. R = 37.2^2/9.80*sin 28.2 = 141.2*0.4726 = 66.73 m

    2 Answers · Science & Mathematics · 12/08/2019

  5. P = V I = V^2 / R 3.4E3 = 240^2 / R R = 17 ohm to two sig figs When you get a good response, please consider giving a best answer. This is the only reward we get.

    2 Answers · Science & Mathematics · 24/04/2019

  6. ... what you are to solve for is V of the battery and R of the battery. When the resistors are in parallel the effective...

    6 Answers · Science & Mathematics · 27/02/2019

  7. change in energy = R(1/3² - 1/5²) = 1.55e-19 J and change in energy = hv / λ so λ = hv / change in energy = 6.626e−34J·s * 2.998e8m/s / 1.55e-19J λ = 1.282e-6 m = 1.282 µm = 1282 nm

    2 Answers · Science & Mathematics · 14/06/2019

  8. Power = V²/R = (1.5+1.5)²/10Ω = 0.9W <----- (a) 2/3 * 0...

    1 Answers · Science & Mathematics · 09/06/2019

  9. A/ torque τ = r x F = 0.10m * 10N = 1.0 N·m B/ For a solid cylinder...

    1 Answers · Science & Mathematics · 12/06/2019

  10. first get total R as seen by the battery 1 in parallel with 7.8 is 7.8/8.8 = 0.886... which produces current thru 7.8 Ω of I = E/R = 0.094/7.8 = 0.012 amp or 12 mA

    1 Answers · Science & Mathematics · 04/03/2019

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