In nuclear reactions, both mass and charge are conserved, according to the law of conservation of mass-energy and charge.

The energy released when a nucleus is formed from its constituent nucleons is called binding energy.

The energy released in a nuclear reaction is referred to as nuclear energy, which is a result of changes in the binding energy of the nucleus.

The mass defect is the difference in mass between the reactants and products in a nuclear reaction, which is related to the binding energy.

'BAC' = 11*16^2 + 10*16^1 + 12*16^0 = 186 + 10 + 12 = 188.

When a p-n junction diode is forward biased, the depletion region narrows, allowing current to flow easily through the junction.

When a p-n junction diode is reverse-biased, the depletion region widens, preventing current flow.

When a p-n junction is forward biased, the depletion region narrows, allowing current to flow.

An electric field is formed at the p-n junction due to the diffusion of charge carriers.

The depletion region is the area around the p-n junction where charge carriers are depleted, creating an electric field.

The depletion region is the area around the p-n junction where charge carriers have recombined, leaving behind immobile ions.

In the p-type region of a p-n junction, holes are the majority charge carriers.

In a p-type region, holes are the predominant charge carriers.

Vertically opposite angles are equal. Therefore, if one angle measures 120 degrees, the other angle also measures 120 degrees.

The equivalent resistance for resistors in parallel is given by 1/R_eq = 1/R1 + 1/R2 + 1/R3. Thus, 1/R_eq = 1/2 + 1/3 + 1/6 = 1. Therefore, R_eq = 1Ω.

Using the formula for resistors in parallel, 1/R_eq = 1/R1 + 1/R2 + 1/R3, we find R_eq = 1 / (1/6 + 1/12 + 1/18) = 3 ohms.

1/R_eq = 1/R1 + 1/R2 + 1/R3 = 1/2 + 1/3 + 1/6 = 1. Therefore, R_eq = 3Ω.

The formula for equivalent resistance in parallel is 1/R_eq = 1/R1 + 1/R2 + 1/R3. Thus, 1/R_eq = 1/6 + 1/3 + 1/2 = 1/6 + 2/6 + 3/6 = 6/6 = 1, so R_eq = 1 ohm.

The equivalent resistance in parallel is given by 1/R_total = 1/R1 + 1/R2 + 1/R3. Thus, 1/R_total = 1/2 + 1/3 + 1/6 = 1Ω.

For parallel resistors, 1/R_eq = 1/R1 + 1/R2. Thus, 1/R_eq = 1/4 + 1/6 = 5/12, so R_eq = 12/5 = 2.4 ohms.

First, find the equivalent resistance: 1/R_eq = 1/4 + 1/6 => R_eq = 2.4 ohms. Then, I = V/R_eq = 12V / 2.4Ω = 5 A.

First, find the equivalent resistance: 1/R_eq = 1/4 + 1/6 => R_eq = 2.4 ohms. Then, I = V/R_eq = 12 V / 2.4 ohms = 5 A.

First, find the equivalent resistance: 1/R_eq = 1/4 + 1/12 = 3/12 + 1/12 = 4/12, R_eq = 3Ω. Then, I = V/R_eq = 24V / 3Ω = 8A.

Total current I = V/R_eq. First, find R_eq = 1/(1/4 + 1/12) = 3Ω. Then, I = 24V / 3Ω = 8A.

First, find the equivalent resistance: 1/R_eq = 1/4 + 1/6 => R_eq = 2.4Ω. Then, I = V/R_eq = 12V / 2.4Ω = 5A.

Using the formula for resistors in parallel, 1/R_eq = 1/R1 + 1/R2, we have 1/R_eq = 1/6 + 1/3 = 1/6 + 2/6 = 3/6. Therefore, R_eq = 6/3 = 2 ohms.

The total resistance in parallel is given by 1/R_total = 1/R1 + 1/R2. Thus, 1/R_total = 1/6 + 1/3 = 1/2, so R_total = 2 ohms.

First, find the equivalent resistance: 1/R_eq = 1/6 + 1/3 = 1/2, so R_eq = 2Ω. Then, I = V/R_eq = 12V / 2Ω = 6A.

Using the formula 1/R_eq = 1/R1 + 1/R2, we get 1/R_eq = 1/6 + 1/3 = 1/6 + 2/6 = 3/6, thus R_eq = 2Ω.

First, find the equivalent resistance: 1/R_eq = 1/8 + 1/4 = 1/8 + 2/8 = 3/8, so R_eq = 8/3Ω. Then, I = V/R_eq = 16V / (8/3)Ω = 16 * 3/8 = 6A.