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F = (k)(q_{1})(q_{2})/r² = [(9 x 10^{9} N-m²/Coul²)(2 Coul)(3 Coul)]/(0.3 m)²
F = 6 x 10^{11} N
q_{2} = (F)(r²)/(k)(q_{1}) = (3.5 N)[(1.2 m)²]/[(9 x 10^{9} N-m²)(3.4 Coul)]
q_{2} = 1.65 x 10^{-10} Coul
q² = (F)(r²)/(k) = (1.2 N)[(0.4 m)²]/(9 x 10^{9} N-m²)
q = 4.62 x 10^{-6} Coul
E = (k)(q)/r² = (9 x 10^{9} N-m²/Coul²)(1.8 Coul)/(1.5 m)² = 7.2 x 10^{8} N/Coul
E = (k)(q)/r² = (9 x 10^{9} N-m²/Coul²)(-1.8 Coul)/(2.5 m)²
E = -2.59 x 10^{9} N/Coul
V = (k)(q)/r = (9 x 10^{9} N-m²/Coul²)(1.8 Coul)/(1.5 m)
E = 1.08 x 10^{10} V
V = (k)(q)/r = (9 x 10^{9} N-m²/Coul²)(-1.8 Coul)/(2.5 m)
E = -6.48 x 10^{9} V
V = (k)(q)/r; V = V_{1} + V_{2}
V = (9 x 10^{9} N-m²/Coul²)(3 Coul)/(4.5 m) + (9 x 10^{9} N-m²/Coul²)(4.5 Coul)/(2.5 m)
E = 6 x 10^{9} + 1.6 x 10^{10} = 2.22 x 10^{10} V
q = (C)(V) = (15 x 10^{-6} F)(1.2 V) = 1.8 x 10^{-5} Coul
ρ = R(A/L) = (1.5 Ω)(0.002 m²)/(0.005 m) = 0.6 Ω-m
A = (ρ)(L)/R = (4 x 10^{-5} Ω-m)(0.001 m)/(300 Ω) = 1.3 x 10^{-10} m²
I = V/R = (6 V)/(2 Ω) = 3 A
V = IR = (2 A)(32 Ω) = 64 V
P = I²R = (2 A)²(32 Ω) = 128 W
I = V/R = (5 V)/(45 Ω) = 0.11 A
P = V²/R = (5 V)²/(45 Ω) = 0.56 W
R = V/I = (3 V)/(9 A) = 0.33 Ω
P = VI = (3 V)(9 A) = 27 W
I = P/V = (60 W)/(120 V) = 0.5 A
E = Pt = (60 W)(3600 sec) = 216 kJ
E = Pt = (100 W)(3600 sec) = 360 kJ
E = Pt = VIt = (120 V)(7.5 A)(120 sec) = 108 kJ
R = V/I = (120 V)/(7.5 A) = 16 Ω
R = V/I = (120 V)/(0.33 A) = 364 Ω
P = VI = (120 V)(0.33 A) = 39.6 W
E = Pt = VIt = (120 V)(0.33 A)(180 sec) = 7128 J
V = [(P)(R)]^{½} = [(60 W)(240 Ω)]^{½} = 120 V
I = V/R = (120 V)/(240 Ω) = 0.5 A
E = Pt = (60 W)(180 sec) = 10.8 kJ
R_{s} = R_{1} + R_{2} + R_{3} = 3R ⇒ R = (12 Ω)/3 = 4 Ω
1/R_{p} = 1/R_{1} + 1/R_{2} + 1/R_{3} = 3/R ⇒ R = (3)(0.667 Ω) = 2 Ω
F_{mag} = (I)(ΔL)(B_{⊥}) = (2 A)(40 m)(0.1 T) = 8 N, toward the West
F_{mag} = (I)(ΔL)(B_{⊥}) = (2 A)(40 m)(0.4 T)(cos 30°) = 27.7 N, Downward
B_{⊥} = F_{mag}/[(I)(ΔL)] = (5 N)/[(2 A)(40 m)] = 0.0625 T, Downward
I = F_{mag}/[(ΔL)(B_{⊥})] = (3 N)/[(100 m)(2 T)] = 0.015 T
ΔL = F_{mag}/[(I)(B_{⊥})] = (4 N)/[(1.5 A)(6 T)] = 0.444 m
F_{mag} = qVB_{⊥} = (1.6 x 10^{-19} Coul)(2 x 10^{8} m/sec)(0.0002 T)] = 6.4 x 10^{-15} N
B_{loop} = 2πIk/r = (2)(3.142)(2.4 A)(10^{-7} N/A²)/(0.5 m) = 3.02 x 10^{-6}
Φ = B_{⊥}A = (0.07 Wb/m²)(cos 25)(0.04 m)² = 1.02 x 10^{-4} Wb
E = ΔΦ/Δt = (1.02 x 10^{-4} Wb)/(0.15 sec) = 6.77 x 10^{-4} V
Mass Difference = (7)(1.0072 u) + (6)(1.0087 u) - 13.005739 u = 0.096861 u
Binding Energy = (931 MeV/u)(0.096861 u) = 90.2 MeV