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  1. ...point at a distance (r) is given by: E = kQ / r² k is the coulomb constant = 8.99×10⁹ N·m²·C⁻² The field is away from a positive point charge and towards a...

    1 Answers · Science & Mathematics · 05/09/2019

  2. ...m2 /r²) {1/r² cancels out} (k.q1.q2 ) / (G.m1.m2 ) k = 8.99×10⁹ N·m²·C⁻² q1 = q2 = -1.27 x 10^6 C G = 6.67 × 10^-11 N·m²...

    2 Answers · Science & Mathematics · 05/09/2019

  3. electrostatic force F = kQq / d² where k = 8.99e9 N·m²/C² first condition: -0... = 1.87e-12 C² Q' = 1.367e-6 C and so the total charge present is Q_tot = 2...

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

  4. ...and so the magnitude of the field at the midpoint is E = k*(q1 + q2) / (d/2)² E = 8.99e9N·m²/...½*0.151m)² E = 9.90e5 N/C Hope this helps!

    1 Answers · Science & Mathematics · 05/09/2019

  5. energy = k*Q*q / d where k is some constant so adcb then dacb If you find this helpful, please award Best Answer!

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

  6. The unrounded answer from part a) is 3.2344x10^17 N/C. It's best to use this unrounded value in part b). The force...

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

  7. ... (270°) Thus q must be negative. k = 8.988e9 q = -3²*2824.5/k = -2.83µC or about -2.8µC with 2 s.f. 2. ½mv² = 3.2e-16...

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

  8. ... of all four charges. So to PE' we need to add k*q4*[q1/0.200m + q2/0.200m + q3/0.200m] = -4.05e-5 for q4 = -3.0e-9 C nd so PE" = 8.20e-4 J = 0.820 mJ = 820 µJ...

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

  9. ... Q = 5•0.36² / (9e9)(3.4) = 2.12e-11 C or 21.2 pC since the force is attractive, Q...force in newtons r is separation in meters k = 8.99e9 Nm²/C²

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

  10. ...286 m²/s²/K for air T = absolute temperature (273.15 K + °C) = 298.15 K v(s) = speed of sound = to be determined v...

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

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