U = 1794.005 × 10⁶ J. Explanation: Information provided indicates that the capacitance of the original capacitor is C = 1.27 F, and the potential difference applied to it is V = 59.9 kV, or 59.9 × 10³ V. The potential energy (U) for the capacitor is determined by the formula: U = (1/2) × C × V². Substituting the respective values, we find U = (1/2) × 1.27 × (59.9 × 10³)², resulting in U = 1794.005 × 10⁶ J.
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
This is somewhat misleading, and I encountered the same question in my homework. An electric field strength of 1*10^5 N/C is provided, along with a drag force of 7.25*10^-11 N, and the critical detail is that it maintains a constant velocity, indicating that the particle is in equilibrium and not accelerating.
<span>To solve, utilize F=(K*Q1*Q2)/r^2 </span>
<span>You'll want to equate F with the drag force, where the electric field strength translates to (K*Q2)/r^2; substituting the values results in </span>
<span>(7.25*10^-11 N) = (1*10^5 N/C)*Q1 ---> Q1 = 7.25*10^-16 C </span>
The frequency of a sound remains constant as it departs from the source. It does not alter.
The voices of swimmers do not modify in frequency when transitioning to or from the water. Only their speed and wavelength vary.
