Critical angle θc = sin⁻¹(n2/n1) = sin⁻¹(1.4/1.5) ≈ 42.0°.

Critical angle θc = sin⁻¹(n2/n1) = sin⁻¹(1.5/1.6) ≈ 38.7°.

For total internal reflection, the core must have a higher refractive index than the cladding, so it must be greater than 1.45.

The cladding has a lower refractive index than the core, ensuring that light is kept within the core through total internal reflection.

The cladding has a lower refractive index than the core, ensuring that light remains within the core by total internal reflection.

Angular momentum is conserved; it remains constant, but her angular velocity increases.

By conservation of angular momentum, if the moment of inertia decreases, the angular velocity must increase.

By conservation of angular momentum, pulling arms in decreases moment of inertia, thus increasing angular velocity.

Pulling her arms in decreases her moment of inertia, causing her rotational speed to increase to conserve angular momentum.

Angular momentum remains the same; however, her angular velocity increases due to a decrease in moment of inertia.

The correct form is 'A flock of birds was flying over the lake.' The subject 'flock' is singular, so the verb should be 'was'.

Using Poiseuille's law, the flow rate Q = (π * r^4 * ΔP) / (8 * η * L). Assuming L = 1 m, Q = (π * (0.05)^4 * 1000) / (8 * 0.1 * 1) = 0.01 m³/s.

Shear stress = viscosity × (velocity/radius) = 0.1 × (1/0.05) = 2 Pa.

Using the formula for shear stress, τ = (4 * η * Q) / (π * r^3), we find τ = 0.4 Pa.

Using the equation ω_f = ω + αt, where α = τ/I, we get ω_f = ω + (τ/I)t.

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

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.

First convert rpm to rad/s: 1000 rpm = (1000 * 2π)/60 rad/s. Then use α = (ωf - ωi)/t = (0 - (1000 * 2π)/60) / 10.

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

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 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.

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.

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

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.