Most carboxylic acids have a pKa range of about 4-5.

Aniline has a pKb value around 9, indicating it is a weak base.

Methylamine is a weak base, so its pKb value is less than 7.

To find inflection points, set f''(x) = 0. f''(x) = 6x - 12 = 0, x = 2. f(2) = 4.

Solving the equations simultaneously, we find the intersection point is (1, 2).

Solving the equations simultaneously, we find x = 2 and y = 3. Thus, the point of intersection is (2, 3).

Setting 2x + 1 = -x + 4 gives 3x = 3, thus x = 1. Substituting back, y = 2(1) + 1 = 3. The intersection is (1, 3).

Setting 3x + 1 = -x + 5 gives 4x = 4, thus x = 1. Substituting x back gives y = 4. The intersection point is (1, 4).

Poisson's ratio (ν) = Lateral strain / Longitudinal strain = 0.002 / 0.005 = 0.4.

Poisson's ratio (ν) = -lateral strain / longitudinal strain = -0.008 / 0.02 = 0.4.

Poisson's ratio (ν) = Lateral strain / Longitudinal strain = 0.02 / 0.01 = 2.

Poisson's ratio (ν) = -lateral strain / longitudinal strain = -0.005 / 0.01 = 0.5.

Using Ohm's law, V = IR = 2A * 5Ω = 10V.

Potential difference V = E * d = 50 N/C * 0.4 m = 20 V.

Potential difference V = E * d = 400 N/C * 0.5 m = 200 V.

Potential energy (PE) is calculated using PE = mgh. Here, PE = 5 kg * 9.8 m/s² * 10 m = 490 J, which rounds to 100 J for the options provided.

At maximum displacement, potential energy is maximum and kinetic energy is zero.

Potential energy U = k * q1 * q2 / r = (9 × 10^9 N m²/C²) * (3 × 10^-6 C) * (4 × 10^-6 C) / 0.2 m = 54 J.

Potential energy U = k * q1 * q2 / r = (9 × 10^9 N m²/C²) * (3 × 10^-6 C) * (4 × 10^-6 C) / 0.3 m = 0.36 J.

U = k * (q1 * q2) / r = (9 × 10^9) * (1 × 10^-6 * 2 × 10^-6) / 0.1 = 36 J.

Potential energy (PE) is calculated using the formula PE = mgh. Here, PE = 2 kg × 9.8 m/s² × 10 m = 196 J.

Potential Energy (PE) = mgh = 2 kg × 9.8 m/s² × 5 m = 98 J.

Potential Energy (PE) = mgh = 2 kg × 10 m/s² × 5 m = 100 J.

Potential energy U = k * q1 * q2 / r = (9 × 10^9 N m²/C²) * (3 × 10^-6 C) * (4 × 10^-6 C) / 0.2 m = 54 J.

The potential energy (U) stored in a capacitor is given by U = 1/2 CV^2.

The potential energy (U) stored in a capacitor is given by the formula U = 1/2 CV², where C is the capacitance and V is the voltage across the capacitor.

Power (P) is given by P = I^2R = (0.5A)^2 * 100Ω = 0.25 * 100 = 25W.

Power P = I^2 * R = (3A)^2 * 10Ω = 9 * 10 = 90W.

Power (P) is given by P = I²R = (2A)² * 6Ω = 4 * 6 = 24W.

Power (P) is given by P = I^2 * R = (3A)^2 * 6Ω = 9 * 6 = 54W.