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...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

...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

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

...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

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

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

... (270°) Thus q must be negative.

**k**= 8.988e9 q = -3²*2824.5/**k**= -2.83µ**C**or about -2.8µ**C**with 2 s.f. 2. ½mv² = 3.2e-16...1 Answers · Science & Mathematics · 01/09/2019

... 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

... 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/C²1 Answers · Science & Mathematics · 31/08/2019

...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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