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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

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

For z >>

**R**, E = qz / [4πε₀*(z²+R²)^(3/2)] ≈ qz / [4πε₀...2 Answers · Science & Mathematics · 04/09/2019

**R**= 37.2^2/9.80*sin 28.2 = 141.2*0.4726 = 66.73 m2 Answers · Science & Mathematics · 12/08/2019

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

... 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

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 nm2 Answers · Science & Mathematics · 14/06/2019

Power = V²/

**R**= (1.5+1.5)²/10Ω = 0.9W <----- (a) 2/3 * 0...1 Answers · Science & Mathematics · 09/06/2019

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

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 mA1 Answers · Science & Mathematics · 04/03/2019