The solid sphere will have a greater translational speed because it has a smaller moment of inertia.

The solid sphere will hit the ground first because it has a smaller moment of inertia, allowing it to roll down faster.

The solid sphere will have a greater linear speed because it has a smaller moment of inertia, allowing it to convert more potential energy into translational kinetic energy.

The solid sphere has a smaller moment of inertia compared to the hollow sphere, allowing it to accelerate faster down the incline.

Total angular momentum L = Iω, where I = (2/5)MR^2 for a solid sphere.

The moment of inertia of a solid sphere about its center is I = 2/5 MR^2.

The moment of inertia of a solid sphere about an axis through its center is I = 2/5 MR^2.

The acceleration of the center of mass of a rolling object is given by a = g sin(θ) / (1 + k^2/r^2). For a solid sphere, k^2/r^2 = 2/5, thus a = g sin(θ) / (1 + 2/5) = g sin(θ) / (7/5) = (5/7)g sin(θ).

The acceleration of the center of mass of a solid sphere rolling down an incline is given by a = g sin(θ) / (1 + (2/5)) = g sin(θ) / (7/5) = (5/7) g sin(θ).

Using conservation of energy, the final velocity of the solid sphere at the bottom is v = √(5gh/7).

Using conservation of energy, potential energy at the top (mgh) converts to kinetic energy (1/2 mv^2 + 1/2 Iω^2). For a solid sphere, I = (2/5)mr^2 and ω = v/r. Solving gives v = √(2gh).

The total kinetic energy is the sum of translational and rotational kinetic energy. For a solid sphere, the ratio of translational to total kinetic energy is 2:5, which simplifies to 2:3.

The total kinetic energy is the sum of translational and rotational kinetic energy. For a solid sphere, the ratio of translational to total kinetic energy is 2:3.

For a solid sphere, the ratio of translational kinetic energy to total kinetic energy is 2:3.

The total mechanical energy at the top is purely potential energy, which is mgh.

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

The total kinetic energy of a rolling sphere is the sum of translational and rotational kinetic energy, which is (1/2)mv^2 + (2/5)mv^2.

The angular speed ω of a rolling sphere is given by ω = v/R.

The moment of inertia of a thin rod about an end is I = 1/3 ML^2.

The moment of inertia of a thin rod about an end is I = 1/3 ML^2.

Using τ = Iα, we have α = τ/I = 10 Nm / 2 kg·m² = 5 rad/s².

Using τ = Iα, we have α = τ/I = 10 Nm / 2 kg·m² = 5 rad/s².

Using τ = Iα, we have α = τ/I = 10 Nm / 5 kg·m² = 2 rad/s².

Torque (τ) = F × r; therefore, F = τ / r = 10 Nm / 0.2 m = 50 N.

Using τ = Iα, we have α = τ/I = 10 N·m / 2 kg·m² = 5 rad/s².

Force = Torque / Distance = 12 Nm / 0.4 m = 30 N.

Torque (τ) = r × F, thus F = τ / r = 12 Nm / 0.4 m = 30 N.

Using τ = F × r, we have F = τ / r = 12 Nm / 0.4 m = 30 N.

Force = Torque / Distance = 12 Nm / 0.4 m = 30 N.

Force = Torque / Distance = 12 Nm / 4 m = 3 N.