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.

In an inductor, the current lags the voltage by 90 degrees.

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 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 electrochemical cell, oxidation occurs at the anode.

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.

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.

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

Pioneer species are the first organisms to colonize previously uninhabited or disturbed environments.

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.

In electrolysis, oxidation occurs at the anode.

Chlorine gas is produced at the anode during the electrolysis of NaCl solution.

After fertilization, the ovary develops into a fruit, which contains the seeds.

The number of ways to award 3 prizes to 5 students is 5P3 = 60.

The number of arrangements of 4 distinct flags is 4! = 24.

The number of ways to choose 4 fruits from 10 is C(10, 4) = 210.

The number of ways to distribute 4 different prizes among 10 students is 10P4 = 5040.

The number of ways to distribute 4 prizes among 10 students is 10P4 = 5040.