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...*sec) / Q (A*sec) = 515*10^3/62.5 = 8.24 kV torso resistance

**R**= V/I = 8.24*10^3/(62.5*10^-3) = 132 kohm ΔT = E...2 Answers · Science & Mathematics · 10/02/2020

Resistance of a wire in Ω

**R**= ρL/A ρ is resistivity of the material in Ω-m L...1 Answers · Science & Mathematics · 09/02/2020

Each one has required an energy of kqq/

**r**ie the total is 3 * (9 * 10 ^ 9) * (1.6*10 ^ -19)^2/( 2* 10 ^ -10)2 Answers · Science & Mathematics · 09/02/2020

A) -9.5 m/s² B) 0 m/s² C) 540m/s*cos(-30º) D) 540m/s*sin(-30º) E) 11m*cos49º F) 11m*sin49º If you find this helpful, please select Favo

**r**ite Answe**r**!1 Answers · Science & Mathematics · 08/02/2020

The electric potential Vₑ from a charge Q at a distance

**r**is given by Vₑ = kQ/**r**, where k is the ...2 Answers · Science & Mathematics · 07/02/2020

...193000 volts Potential from a charge is V = kQ/

**r**in volts k = 1/4πε₀ = 8.99e9 Nm²/C²**r**...1 Answers · Science & Mathematics · 09/02/2020

...where V is the voltage (120 V), I is the current, and

**R**is the resistance (31 ohms).6 Answers · Science & Mathematics · 06/02/2020

1/

**R**= 1/r1 +1/r2 ie., 1/38= 1/56+ 1/r2 Solving you get r2 = 118 kilo ohm.1 Answers · Science & Mathematics · 05/02/2020

good sta

**r**t -- U = kQq / d at point of t**r**iangle, U = 2*kQq / x at midpoint of base, U = 2*kQq / (x/2) = 4*kQq / x so ΔU = 2*kQq / x2 Answers · Science & Mathematics · 05/02/2020

### What electric field is necessary to drive a 6.5 A current through a silver wire 0.95 mm in diameter?

take a 1 meter length A = πr² = π(0.00095/2)²

**R**= ρL/A = (15.9e-9)(1) / π(0.00095)² = 0.0224 ΩE...1 Answers · Science & Mathematics · 05/02/2020