The moment of inertia of a disk is I = (1/2)MR^2. If the radius is doubled, the new moment of inertia becomes I' = (1/2)M(2R)^2 = 4I.

The absolute error is given as 2 m.

Using the lens formula 1/f = 1/v - 1/u, we have 1/(-10) = 1/v - 1/5. Solving gives v = -3.33 cm.

Torque (τ) = Force (F) × Distance (r) = 20 N × 0.8 m = 16 Nm.

Torque (τ) = F × d = 20 N × 1 m = 20 Nm.

Torque = Force × Distance = 50 N × 1.2 m = 60 Nm.

Torque (τ) = Force (F) × Distance (r) = 50 N × 1 m = 50 Nm.

Using the lens maker's formula, R = 2f(n-1) = 2*10*(1.5-1) = 20 cm.

Using the lens maker's formula, f = R/(n-1), we find f = 20/(1.5-1) = 40 cm.

For a drop to remain in equilibrium, the upward force due to surface tension must balance the downward gravitational force.

The spherical shape of a liquid drop is due to surface tension, which minimizes the surface area for a given volume.

The drop of oil spreads on water because the surface tension of water is higher than that of oil, causing the oil to spread out.

A drop of water is spherical due to surface tension, which minimizes the surface area.

For total internal reflection to occur in a fiber optic cable, the core must have a higher refractive index than the cladding.

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

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

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.

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

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

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

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.

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