The characteristic equation has a repeated root r = 2. The general solution is y = (C1 + C2x)e^(2x).

The geometry is tetrahedral when there are four bonding pairs and no lone pairs around the central atom.

A molecule with four bonding pairs and no lone pairs has a tetrahedral geometry.

The geometry is linear when there are two bonding pairs and no lone pairs on the central atom.

The complex [Ag(NH3)2]+ has a linear geometry due to the presence of two ligands around the silver ion.

g = G * M / r² = (6.67 × 10^-11) * (100) / (10²) = 6.67 × 10^-10 N/kg

Weight = mass * g = 50 kg * 9.8 m/s² = 490 N

Weight W = m * g = 50 kg * 9.8 m/s² = 490 N.

F = G * (m1 * m2) / r² = (6.67 × 10^-11) * (1 * 1) / (1²) = 6.67 × 10^-11 N

Using the formula F = G * (m1 * m2) / r², we have F = (6.67 × 10^-11) * (5 * 10) / (2²) = 1.67 × 10^-10 N.

The gravitational force is given by Newton's law of gravitation: F = G * (m1 * m2) / r^2.

Gravitational potential energy (PE) is given by PE = mgh. Here, PE = 10 kg * 9.81 m/s² * 10 m = 981 J.

Gravitational potential energy (PE) = m * g * h = 10 kg * 9.81 m/s² * 5 m = 490.5 J.

Gravitational potential energy (PE) is given by the formula PE = mgh. Here, PE = 10 kg * 9.8 m/s² * 10 m = 980 J, which rounds to approximately 1000 J.

Gravitational potential energy (PE) is given by PE = mgh. Here, m = 10 kg, g = 9.8 m/s², and h = 5 m. PE = 10 * 9.8 * 5 = 490 J.

Gravitational potential energy (PE) = mgh = 2 kg × 9.8 m/s² × 10 m = 196 J.

Gravitational potential energy (PE) = m * g * h = 5 kg * 9.8 m/s² * 10 m = 490 J.

Alkali metals are located in group 1 of the periodic table.

Halogens are located in group 17 of the periodic table.

Heat required Q = mcΔT = (2000 g)(0.9 J/g°C)(50°C) = 90000 J.

In [Ni(CO)4], the nickel atom undergoes sp3 hybridization to accommodate four CO ligands.

In [Cr(NH3)6]3+, the hybridization is d2sp3 due to the octahedral arrangement of six ligands.

Impedance Z = √(R² + X²) = √(4² + 3²) = √(16 + 9) = √25 = 5 ohms.

Impedance Z = √(R² + X²) = √(4² + 3²) = √(16 + 9) = √25 = 5Ω.

Impedance Z = √(R² + (XL)²), where XL = 2πfL = 31.42Ω. Thus, Z = √(10² + 31.42²) = 15.71Ω.

Impedance Z = √(R² + (XL)²), where XL = 2πfL = 31.42Ω. Thus, Z = √(10² + 31.42²) = 15.71Ω.

Impedance Z = √(R² + (1/(ωC))²) where ω = 2πf. Here, R = 4Ω, C = 2μF, f = 50Hz. Calculate Z.

At resonance, the impedance is equal to the resistance, which is 4Ω.

The impedance Z in a series RL circuit is given by Z = √(R² + (2πfL)²).

The impedance Z in a series R-L circuit is given by Z = √(R² + (ωL)²).