Average induced emf = -ΔΦ/Δt = -(0.2 - 0.5)/2 = 0.15 V = 150 V.

An induced EMF is generated when there is a change in magnetic flux through a coil.

If the mass of the Earth doubles, the acceleration due to gravity also doubles, as it is directly proportional to mass.

Angular momentum L = Iω, so if I is doubled and ω remains constant, L also doubles.

Distance (d) = Potential (V) / Electric field (E) = 5 V / 2 V/m = 2.5 m.

The energy stored in a capacitor is given by U = 1/2 CV^2. If V is halved, the energy becomes U' = 1/2 C(V/2)^2 = 1/8 CV^2, which is halved.

The charge (Q) stored in a capacitor is directly proportional to the potential difference (V) across it, given by Q = CV.

Work done (W) = Charge (Q) × Potential difference (V) = 2 C × 10 V = 20 J.

Linear velocity v = rω, so if r is halved and ω remains constant, v also halves.

The focal length (f) of a concave mirror is given by f = R/2, where R is the radius of curvature. Thus, f = 20 cm / 2 = 10 cm.

The focal length f of a lens is given by f = R/2. Therefore, f = 20 cm / 2 = 10 cm.

The focal length (f) of a mirror is given by f = R/2. For a convex mirror, R is positive, so f = 20 cm / 2 = 10 cm.

According to Ohm's Law, if the resistance (R) is doubled and the voltage (V) remains constant, the current (I) will be halved (I = V/R).

The peak voltage (V_peak) is calculated as V_peak = V_rms × √2 = 100 V × √2 ≈ 141.42 V.

According to Ohm's law, if voltage increases and resistance remains constant, current will also increase proportionally.

Inductive reactance (X_L) is given by X_L = V/I. Here, X_L = 100 V / 2 A = 50 Ω.

The electric fields due to both charges at the midpoint cancel each other out, resulting in a net electric field of 0 N/C.

Total resistance R = 4Ω + 12Ω = 16Ω. Voltage V = I * R = 2A * 16Ω = 32V.

Using the formula for parallel resistors, 1/R_eq = 1/R1 + 1/R2 = 1/4 + 1/6 = 5/12, R_eq = 12/5 = 2.4Ω.

In series, R_total = R1 + R2 = 5Ω + 10Ω = 15Ω.

Increasing the plate area while keeping the distance constant increases the capacitance, as capacitance is directly proportional to the area of the plates.

Power P = V^2/R = 5^2/10 = 25/10 = 2.5W.

Total resistance R = 10Ω + 5Ω = 15Ω. Voltage V = I * R = 2A * 15Ω = 30V.

Total resistance R = 4Ω + 4Ω = 8Ω. Current I = V/R = 12V/8Ω = 1.5A.

The voltage drop across the resistor is equal to the battery voltage, which is 15V.

The total resistance is 15Ω. The current I = V/R = 15V/15Ω = 1A. Voltage drop across 10Ω = I * R = 1A * 10Ω = 10V.

Power (P) can be calculated using P = V^2 / R. Here, P = 24V^2 / 6Ω = 96W.

Power P = V^2/R = 24^2/8 = 576/8 = 72 W.

Total resistance R = 3Ω + 6Ω = 9Ω. Current I = V/R = 9V/9Ω = 1A.

Total resistance R_total = 3Ω + 6Ω = 9Ω. Current I = V/R_total = 9V/9Ω = 1A. Voltage across 6Ω = I * R = 1A * 6Ω = 6V.