Using the formula for surface area and volume, the optimal dimensions are found to be 5 cm height and approximately 15.87 cm radius.

Using the formula for surface area and volume, the optimal dimensions are found to be radius = 5 and height = 20.

Using the formula for volume and surface area, the optimal dimensions are found to be radius = 5 cm and height = 20 cm.

The dipole moment (p) is defined as p = qd, where q is the charge and d is the separation distance.

The torque (τ) experienced by a dipole in a uniform electric field is given by τ = pE sin(θ), where p is the dipole moment and θ is the angle between p and E.

Rotational kinetic energy K.E. = (1/2)Iω². For a disc, I = (1/2)MR², thus K.E. = (1/4)MR²ω².

The moment of inertia of a solid disc about its central axis is given by I = (1/2)MR².

The moment of inertia I of a disc about its axis is (1/2)MR². Therefore, angular momentum L = Iω = (1/2)MR²ω.

The disk has a smaller moment of inertia than the ring, resulting in greater acceleration down the incline.

A diverging lens always forms a virtual and upright image when the object is placed in front of it.

For a fixed area, the minimum perimeter occurs when the rectangle is a square. Thus, dimensions are approximately 14 m by 14.28 m.

For maximum area, the rectangle should be a square. Each side = 100/4 = 25 m.

From Newton's second law for rotation, τ = Iα, thus α = τ/I.

The rotational kinetic energy is given by K.E. = (1/2)Iω^2.

Using the formula α = (ω - ω₀)/t, we have α = (0 - 10)/5 = -2 rad/s², so the deceleration is 2 rad/s².

Using the formula α = (ω - ω₀)/t, we have α = (0 - 30 rad/s) / 10 s = -3 rad/s².

Angular deceleration α = (final angular velocity - initial angular velocity) / time = (0 - 10) / 5 = -2 rad/s².

Rotational kinetic energy K.E. = (1/2)Iω² = (1/2)(2)(10)² = 100 J.

Using F = ma, mass m = F/a = 10 N / 2 m/s² = 5 kg.

Work done = Force × Distance = 10 N × 5 m = 50 J

Work done = Force × Distance = 10 N × 5 m = 50 J

Work done = Force × Distance = 10 N × 5 m = 50 J

Work done (W) = Force (F) × Distance (d) × cos(θ). Here, θ = 0° (force and displacement are in the same direction). W = 10 N × 5 m × cos(0°) = 10 N × 5 m = 50 J.

Work done = Force × Distance = 10 N × 5 m = 50 J.

Acceleration = Force / Mass = 100 N / 50 kg = 2 m/s².

Using F = ma, a = F/m = 100 N / 50 kg = 2 m/s².

Using F = ma, acceleration a = F/m = 15 N / 3 kg = 5 m/s².

Work done (W) = F × d × cos(θ) = 20 N × 4 m × cos(60°) = 20 × 4 × 0.5 = 40 J.

Using F = ma, a = F/m = 30 N / 3 kg = 10 m/s².

Using F = ma, we find a = F/m = 30 N / 6 kg = 5 m/s².