Time constant (τ) = m/c, thus c = m/τ = 1/3 = 0.333 kg/s. Using c = 2ζ√(mk), we find ζ = 0.5.

Critical angle θc = sin^(-1)(1/n) = sin^(-1)(1/2.42) ≈ 24.4°.

The dipole moment p is defined as p = q * d.

The dipole moment p = q * d.

Dipole moment p = q * d = (1 × 10^-6 C) * (0.1 m) = 1 × 10^-7 C m.

Dipole moment p = q * d, if d is doubled, p also doubles.

Dipole moment p = q * d = (2 × 10^-6 C) * (0.1 m) = 2 × 10^-7 C m.

Torque τ = p × E = pE sin θ, where θ is the angle between p and E.

Angular momentum L = Iω, where I = (1/2)MR^2 for a disc.

Kinetic energy K = (1/2)Iω^2, where I = (1/2)MR^2 for a disc.

The moment of inertia I of a disc about its axis is (1/2)MR^2. Therefore, the kinetic energy K.E. = (1/2)Iω^2 = (1/2)(1/2)MR^2ω^2 = (1/4)MR^2ω^2.

The relationship between linear speed and angular speed for rolling without slipping is ω = v/R.

The disk has a lower moment of inertia compared to the ring, thus it reaches the ground first.

The disk has a lower moment of inertia than the ring, allowing it to convert more potential energy into translational kinetic energy.

The disk has a smaller moment of inertia compared to the ring, hence it will reach the bottom first.

The disk has a lower moment of inertia than the ring, allowing it to convert more potential energy into translational kinetic energy.

The disk will reach the bottom first because it has a smaller moment of inertia compared to the ring.

Using the formula ω = ω₀ + αt, we have ω = 10 + 2*5 = 20 rad/s.

Angular momentum L = Iω = (1/2)MR^2ω, where I = (1/2)MR^2 for a disk.

The moment of inertia I of a disk about its axis is (1/2)MR^2. Therefore, the kinetic energy K.E. = (1/2)Iω^2 = (1/2)(1/2)MR^2ω^2 = (1/4)MR^2ω^2.

Angular momentum L = Iω, where I = (1/2)MR^2 for a disk.

At the bottom, the total energy is converted into translational and rotational energy. The translational energy is 2/3 of the total energy.

At the bottom, the total kinetic energy is split equally between translational and rotational for a disk, hence the ratio is 1:1.

Using conservation of energy, potential energy at height h converts to kinetic energy at the bottom. The speed is √(2gh).

The relationship between linear speed and angular speed for rolling without slipping is given by ω = v/R.

Angular momentum L = Iω. If the radius is doubled, the moment of inertia increases by a factor of 4, thus L = 4Iω.

Angular momentum L = Iω, and if the radius is doubled, the moment of inertia I becomes 4I, thus L = 4Iω.

To conserve angular momentum, if the radius is doubled, the angular velocity must be halved.

The linear velocity v = rω. If the radius is doubled, to maintain the same v, the angular velocity must remain ω.

The moment of inertia I of a disk is proportional to r^2, so if the radius is doubled, I becomes 4I. Thus, angular momentum L = Iω becomes 4Iω.