The RMS voltage is V_rms = V0/√2 = 100/√2 = 70.7 V.

The peak voltage is given by the coefficient of sin, which is 100√2.

RMS voltage V_rms = V_peak / √2 = 120 / √2 = 84.85 V.

The peak voltage in the equation V(t) = V0 sin(ωt) is V0.

The peak voltage V0 is given directly in the equation as 10V.

In an AC generator, the induced EMF is maximum when the coil is positioned at 90 degrees to the magnetic field, as this is when the rate of change of magnetic flux is greatest.

In an adiabatic process, the relationship is given by PV^n = constant, where n is the heat capacity ratio.

In an adiabatic process, as the temperature decreases, the pressure also decreases due to the ideal gas law.

In an adiabatic process, there is no heat exchange with the surroundings, meaning no heat is absorbed or released.

In an adiabatic process, there is no heat exchange, so the internal energy remains constant for an ideal gas.

In an isochoric process, the volume of the system remains constant.

In an isochoric process, the volume remains constant.

According to Boyle's Law, P1V1 = P2V2. If V2 = 2V1, then P2 = P1/2.

For an ideal gas, the internal energy depends only on temperature. Since the temperature remains constant in an isothermal process, the internal energy also remains constant.

In an isothermal process, the temperature of the system remains constant.

Net reactance = XL - XC = 10Ω - 5Ω = 5Ω.

Net reactance X = XL - XC = 30Ω - 20Ω = 10Ω.

Calculate XL = 2πfL = 31.42Ω and XC = 1/(2πfC) = 159.15Ω. Z = √(R² + (XL - XC)²) = √(10² + (31.42 - 159.15)²) = 15.81Ω.

Total impedance Z = √(R² + (XL - XC)²) = √(6² + (8 - 4)²) = √(36 + 16) = √52 ≈ 10Ω.

The quality factor Q is inversely proportional to resistance. Increasing resistance decreases Q.

If the inductive reactance (XL) is greater than the capacitive reactance (XC), the circuit is inductive.

The resonant frequency f0 = 1/(2π√(LC)) = 1/(2π√(0.1 * 100e-6)) = 159 Hz.

The resonant frequency f0 = 1/(2π√(LC)) = 1/(2π√(0.1 * 100e-6)) = 159 Hz.

Using f = 1/(2π√(LC)), we find f = 1/(2π√(0.1 * 100e-6)) ≈ 159 Hz.

Using Z = √(R² + (XL)²), solve for XL.

Using Z = √(R² + X²), we find X = √(Z² - R²). Here, Z = 20Ω, R = 15Ω.

Using Z = √(R² + X²), we find X = √(Z² - R²) = √(50² - 30²) = √(2500 - 900) = √1600 = 40 ohms.

1 mole of Cu²⁺ deposits 63.5 g of copper.

Using the formula: moles = (current × time) / (n × F), where n=1 for electrons and F=96500 C/mol.

In electrolysis, oxidation occurs at the anode.