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

Using the lens maker's formula, the focal length is calculated to be 20 cm.

Using the lens maker's formula, f = R/(n-1) = 30/(1.5-1) = 30/0.5 = 60 cm.

Using the lens maker's formula, the focal length is calculated to be 15 cm.

Once the applied force exceeds the maximum static frictional force, the block will start moving.

The maximum angle for static friction is given by tan(θ) = μs. Here, θ = tan⁻¹(0.5) which is approximately 26.57 degrees, so the block can remain at rest at angles less than this.

The block will not slide if the component of gravitational force down the incline is less than or equal to the maximum static friction force, leading to θ = tan⁻¹(μs).

The presence of kinetic friction opposes the motion, thus decreasing the acceleration of the block compared to a frictionless incline.

Maximum static frictional force (Fs) = μs * N = μs * mg = 0.5 * 10 kg * 9.8 m/s² = 49 N, approximately 50 N.

Frictional force (f_k) = μ_k * N = μ_k * mg = 0.3 * 10 kg * 9.8 m/s² = 29.4 N, approximately 30 N.

Work done = Force × Distance = 10 N × 5 m = 50 J.

Work done = Force × Distance = 10 N × 5 m = 50 J.

Work done = Force × Distance = 10 N × 5 m = 50 J.

Work done = Force × Distance = 10 N × 5 m = 50 J.

Work done = Force × Distance = 10 N × 5 m = 50 J.

Work done = Force × Distance = 10 N × 5 m = 50 J.

Work done = Force × Distance = 10 N × 5 m = 50 J.

Using conservation of energy, potential energy at height = kinetic energy just before hitting the ground. mgh = 0.5mv². Solving gives v = sqrt(2gh) = sqrt(2 * 9.8 * 10) = 14 m/s.

Using conservation of energy, potential energy at height = kinetic energy just before hitting the ground. mgh = 0.5mv^2. Solving gives v = sqrt(2gh) = sqrt(2*9.8*10) = 14 m/s.

Using conservation of energy, potential energy at height = kinetic energy just before hitting the ground. mgh = 0.5mv². Solving gives v = √(2gh) = √(2 * 9.8 * 5) = 10 m/s.

The acceleration a = g * sin(θ) = 9.8 * sin(30°) = 9.8 * 0.5 = 4.9 m/s².

Using Newton's second law, F = ma, we have a = F/m = 10 N / 5 kg = 2 m/s².

Using Newton's second law, F = ma, we have a = F/m = 20 N / 5 kg = 4 m/s².

Using Newton's second law, F = ma, we have a = F/m = 10 N / 5 kg = 2 m/s².

Using Newton's second law, F = ma, we have a = F/m = 20 N / 5 kg = 4 m/s².

Frictional force (Ff) = μk * N = μk * mg = 0.3 * (5 kg * 10 m/s²) = 15 N.

Maximum static frictional force (Fs) = μs * N = μs * mg = 0.4 * (5 kg * 10 m/s²) = 20 N.

The period T is given by T = 2π√(m/k). Here, T = 2π√(2/200) = 2π√(0.01) = 2π(0.1) = 0.2π ≈ 0.63 s.

Angular frequency (ω) = 2πf = 2π(3) = 6 rad/s.

The period of a mass-spring system is T = 2π√(m/k). Halving the mass does not change the period since it is independent of mass in this case.