Kinetic Energy (KE) = 0.5 × m × v² = 0.5 × 75 kg × (15 m/s)² = 0.5 × 75 × 225 = 8437.5 J.

Height (h) = PE / (mg) = 3000 J / (75 kg * 9.8 m/s²) = 4.08 m.

Using F = ma, F = 70 kg * 2 m/s² = 140 N.

Distance = speed * time = 10 m/s * 5 s = 50 m.

Distance = speed * time. Time in seconds = 2 * 60 = 120 s. Distance = 15 m/s * 120 s = 1800 m.

Distance = speed * time. Time in seconds = 2 minutes = 120 seconds. Distance = 15 m/s * 120 s = 1800 m.

Distance = speed * time = 15 m/s * 20 s = 300 m.

Distance = speed * time = 15 m/s * 20 s = 300 m.

For rolling without slipping, a_cm = (2/3)g sin(θ).

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.

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.

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.

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

Work = Force * Distance = 50 N * 4 m = 200 J.

Work done by the gas during expansion at constant pressure is W = PΔV. Here, ΔV = 5 L - 2 L = 3 L, so W = 1 atm * 3 L = 3 L·atm.

Work done (W) = P * ΔV = 100 kPa * (5 m³ - 2 m³) = 100 kPa * 3 m³ = 300 kJ.