Angular deceleration α = (final angular speed - initial angular speed) / time = (0 - 20 rad/s) / 5 s = -4 rad/s².

Using τ = Iα, we find α = τ/I. Assuming I = 1 kg·m², α = 5 rad/s². Time to stop = ω/α = 20 rad/s / 5 rad/s² = 4 s.

Using τ = Iα, where τ is torque, I is moment of inertia, and α is angular acceleration. Assuming I = 25 kg·m², α = τ/I = 5/25 = 0.2 rad/s².

Angular deceleration = (final angular velocity - initial angular velocity) / time = (0 - 15) / 3 = -5 rad/s², so the magnitude is 5 rad/s².

Angular acceleration α = Torque/I. Assuming I = 6 kg·m², α = 3 N·m / 6 kg·m² = 0.5 rad/s².

Angular deceleration α = (ω_f - ω_i) / t = (0 - 20 rad/s) / 5 s = -4 rad/s², so the magnitude is 4 rad/s².

The angular deceleration α = (ω_final - ω_initial) / time = (10 - 20) / 5 = -2 rad/s². Torque τ = Iα = 4 kg·m² * (-2 rad/s²) = -8 N·m, so the magnitude is 8 N·m.

Maximum possible value = measured value + uncertainty = 100 + 2 = 102 N.

Percentage uncertainty = (absolute uncertainty / measured value) * 100 = (1 / 50) * 100 = 2%.

Percentage uncertainty = (absolute uncertainty / measured value) * 100 = (2 / 50) * 100 = 4%.

Relative uncertainty = (absolute uncertainty / measured value) = 2 / 50 = 0.04 or 4%.

Torque (τ) = Force (F) × Distance (d) = 10 N × 0.5 m = 5 Nm.

Torque = Force × Distance = 10 N × 2 m = 20 Nm.

Torque = Force × Distance = 10 N × 2 m = 20 Nm.

Torque = Force × Distance × sin(60°) = 10 N × 2 m × (√3/2) = 17.32 Nm.

Using F = ma, a = F/m = 10 N / 2 kg = 5 m/s². Final velocity v = u + at = 0 + 5 * 5 = 25 m/s.

Net force = applied force - frictional force = 10 N - 4 N = 6 N.

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

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

Work done = F × d × cos(θ) = 15 N × 4 m × cos(30°) = 15 N × 4 m × (√3/2) = 60 J.

Torque = Force × Distance × sin(θ) = 15 N × 1 m × sin(60°) = 15 N × 1 m × (√3/2) = 12.5 Nm.

Work Done = F * d * cos(θ) = 15 N * 4 m * cos(60°) = 30 J

Net force = applied force - frictional force = 15 N - 5 N = 10 N.

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

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

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

Work done = F × d × cos(θ) = 15 N × 3 m × cos(60°) = 15 N × 3 m × 0.5 = 22.5 J.

Work done = Force × Distance = 15 N × 4 m = 60 J.

Torque = Force × Distance × sin(30°) = 20 N × 1 m × 0.5 = 10 Nm.

Torque = Force × Distance × sin(θ) = 20 N × 2 m × sin(60°) = 20 N × 2 m × (√3/2) = 20√3 Nm.