Answer:
20 cm
Explanation:
The electric potential energy U is calculated with the formula U = kq₁q₂/r, where q₁ = 5 nC (5 × 10⁻⁹ C) and q₂ = -2 nC (-2 × 10⁻⁹ C) and r is determined as √(x - 2)² + (0 - 0)² + (0 - 0)² = x - 2. This leads to U = -0.5 µJ (-0.5 × 10⁻⁶ J), where k = 9 × 10⁹ Nm²/C².
Thus, solving for r gives us r = kq₁q₂/U
which leads to x - 2 = kq₁q₂/U
Then, rearranging gives x = 0.02 + kq₁q₂/U m
So, x = 0.02 + 9 × 10⁹ Nm²/C² × 5 × 10⁻⁹ C × -2 × 10⁻⁹ C/-0.5 × 10⁻⁶ J
Resulting in x = 0.02 - 90 × 10⁻⁹ Nm²/-0.5 × 10⁻⁶ J
This simplifies to x = 0.02 + 0.18 = 0.2 m, or 20 cm
Answer:
The energy delivered is E = 0.18 J
Explanation:
Given,
Battery voltage, V = 9 V
Charge in the circuit, Q = 20 mC
= 20 x 10³ C
Energy supplied in the circuit can be computed as
E = Q V
E = 20 x 10⁻³ x 9
E = 180 x 10⁻³
E equals 0.18 J.
The energy delivered in the circuit is therefore E = 0.18 J
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Answer:
A) The updated amplitude = 0.048 m
B) Period T = 0.6 seconds
Explanation: Please refer to the attached documents for the solution.
Answer:
Unfortunately, I do not possess the information
Explanation:
