Answer:
w = √ 1 / CL
This does not violate energy conservation because the voltage of the power source is equal to the voltage drop in the resistence
Explanation:
This problem refers to electrical circuits, the circuits where this phenomenon occurs are series RLC circuits, where the resistor, the capacitor and the inductance are placed in series.
In these circuits the impedance is
X = √ (R² + (
-
)² )
where Xc and XL is the capacitive and inductive impedance, respectively
X_{C} = 1 / wC
X_{L} = wL
From this expression we can see that for the resonance frequency
X_{C} = X_{L}
the impedance of the circuit is minimal, therefore the current and voltage are maximum and an increase in signal intensity is observed.
This does not violate energy conservation because the voltage of the power source is equal to the voltage drop in the resistence
V = IR
Since the contribution of the two other components is canceled, this occurs for
X_{C} = X_{L}
1 / wC = w L
w = √ 1 / CL
Answer:

Explanation:
You can consider that the force that acts over the proton is the same to the force over the electron. This is because the electric force is given by:


where E is the constant electric field between the parallel plates, and is the same for both electron and proton. Also, the charge is the same.
by using the Newton second law for the proton, and by using kinematic equation for the calculation of the acceleration you can obtain:

(it has been used that vp^2 = v_o^2+2ad) where d is the separation of the plates, ap the acceleration of the proton, vp its velocity and mp its mass.
By doing the same for the electron you obtain:

we can equals these expressions for both proton and electron, because the forces qE are the same:

Answer:
The magnitude of the electric field at a point equidistant from the lines is 
Explanation:
Given that,
Positive charge = 24.00 μC/m
Distance = 4.10 m
We need to calculate the angle
Using formula of angle



We need to calculate the magnitude of the electric field at a point equidistant from the lines
Using formula of electric field

Put the value into the formula



Hence, The magnitude of the electric field at a point equidistant from the lines is 
The solution to your problem is as follows:
2.2Kg*9.8m/s = 21.56N
<span>
21.56N*1.25m = 26.95J </span>
<span>We're only concerned with the work done against gravity, lifting the books to 1.25 meters. the distance walked has no effect on the problem, unless you take into account the wind resistance and the force needed to overcome it. Also, lowering the books onto the shelf doesnt count, because gravity does the work on the books.</span>