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Answer:
La fuerza resultante sobre q₃ es -1.2245 × 10⁻¹⁵ i + -0.24 × 10⁻¹⁵ j
La magnitud de la fuerza resultante sobre q₃ es aproximadamente 1.25 × 10⁻¹⁵ N
Explanation:
q₁ = -45 μC = -45 × 10⁻⁶ C
r₁₂ = 30 mm = 30 × 10⁻³ m
q₂ = 25 μC = 25 × 10⁻⁶ C
r₂₃ = 50 mm = 50 × 10⁻³ m
q₃ = 20 μC = 20 × 10⁻⁶ C
k = 9×10⁻⁹ N·m²/C²
Por lo tanto;
r₁₃ = √(50² + 30²) = 10·√(34)
F₁₂ = 9×10⁻⁹ × (-45 × 10⁻⁶)×(25 × 10⁻⁶)/(30 × 10⁻³)² = -1.125 × 10⁻¹⁴
F₁₂ = -1.125 × 10⁻¹⁴ N
F₂₃ = 9×10⁻⁹ × (20 × 10⁻⁶)×(25 × 10⁻⁶)/(50 × 10⁻³)² = 1.8 × 10⁻¹⁵ j
F₁₃ = 9×10⁻⁹ × (-45 × 10⁻⁶)×(20 × 10⁻⁶)/(10·√34 × 10⁻³)² = -2.38× 10⁻¹⁵
Los componentes de F₁₃ son;
-2,38 × 10⁻¹⁵ × cos (arctan (30/50)) = -2,04 × 10⁻¹⁵ j
-2,38 × 10⁻¹⁵ × sin (arctan (30/50)) = -1,2245 × 10⁻¹⁵ i
La fuerza resultante sobre la carga q₃, =
+
∴ = 1.8 × 10⁻¹⁵ j + -1.2245 × 10⁻¹⁵ i + -2.04 × 10⁻¹⁵ j
La fuerza resultante sobre q₃ es = -1.2245 × 10⁻¹⁵ i + -0.24 × 10⁻¹⁵ j
La magnitud de la fuerza resultante sobre q₃,
= √((-1.2245 × 10⁻¹⁵)² + (-0.24 × 10⁻¹⁵)²) ≈ 1.25 × 10⁻¹⁵
La magnitud de la fuerza resultante sobre q₃, ≈ 1.25 × 10⁻¹⁵ N.
<h2>current flowing through the given circuit is 3 A</h2>
Explanation:
When the resistances are connected in series , the total resistance is the sum of individual resistances .
Thus R = R₁ + R₂
here R₁ is resistance of 10 ohms
and R₂ is resistance of 4 ohms .
Thus total resistance R = 10 + 4 = 14 ohm .
The potential difference applied = 42 volt
Thus current I = =
= 3 A
Thus current through the circuit is 3 Ampere