Using the formula v² = u² + 2as, where v = 0, u = 20 m/s, and a = -g = -10 m/s², we find 0 = (20)² + 2(-10)s, giving s = 20 m.

Using v = u - gt, where v = 0, u = 20 m/s, and g = 9.8 m/s², we find t = u/g = 20/9.8 ≈ 2.04 s.

Using the formula h = v²/(2g) = (20 m/s)² / (2 * 10 m/s²) = 20 m.

Using the formula v² = u² + 2as, where v = 0, u = 20 m/s, and a = -g = -10 m/s², we get 0 = (20)² + 2(-10)s, solving gives s = 20 m.

Using the formula h = v² / (2g), h = (15 m/s)² / (2 × 9.8 m/s²) = 11.5 m.

Using the formula v² = u² + 2as, where v = 0, u = 20 m/s, and a = -g = -10 m/s². 0 = (20)² + 2(-10)h, solving gives h = 20 m.

Using the formula h = v²/(2g), we have h = (20 m/s)² / (2 * 10 m/s²) = 20 m.

Using the equation v² = u² + 2as, where v = 0, u = 30 m/s, and a = -9.8 m/s² (acceleration due to gravity), we have 0 = (30)² + 2*(-9.8)*s. Solving gives s = 45.92 m, approximately 45 m.

Using conservation of energy, initial kinetic energy = potential energy at maximum height. 0.5mv² = mgh. Solving gives h = v²/(2g) = (30)²/(2 * 9.8) = 45.9 m.

At the highest point, the tension can be zero if the centripetal force is provided entirely by the weight.

At the highest point, the centripetal force is provided by the weight, so T + mg = mv²/r, T = 0.

At the highest point, T + mg = mv²/r. 2 N + 3 N = mv²/1. v² = 5, v = √5 ≈ 2.24 m/s.

Using conservation of energy, mgh = (1/2)mv^2 + (1/2)(2/5)mv^2. Solving gives the moment of inertia I = (2/5)m(10^2).

The potential energy at the top of the ramp is given by PE = mgh.

Using conservation of energy, the final velocity v at the bottom is given by v = √(2gh).

Using conservation of energy, potential energy mgh converts to kinetic energy (1/2 mv^2). Thus, v = √(2gh).

The relationship is given by v = Rω, where v is the linear speed, R is the radius, and ω is the angular speed.

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

The moment of inertia of a solid sphere is (2/5)MR^2. If the radius is doubled, the moment of inertia increases by a factor of 4.

The moment of inertia for a solid sphere is (2/5)MR^2. If the radius is doubled, the moment of inertia increases by a factor of 4.

For rolling without slipping, the relationship is v = ωR, where v is linear velocity, ω is angular velocity, and R is the radius.

Using Snell's law, sin(θ2) = sin(60)/1.5, we find θ2 = 30 degrees.

Using Snell's law, n1 * sin(θ1) = n2 * sin(θ2). Here, n1 = 1 (air), θ1 = 45 degrees, n2 = 1.5 (prism). Solving gives θ2 ≈ 30 degrees.

Critical angle θc = sin⁻¹(n2/n1) = sin⁻¹(1.33/2.42) ≈ 36.9°.

To determine if total internal reflection occurs, we first find the critical angle using sin(θc) = 1/n = 1/1.5, which gives θc ≈ 41.8°. Since 60° > 41.8°, total internal reflection will not occur.

Since the angle of incidence (60°) is greater than the critical angle (approximately 41.8° for glass to air), total internal reflection occurs.

Critical angle θc = sin⁻¹(1/2.42) ≈ 24.4°.

For a prism, the angle of minimum deviation D is given by D = A(n - 1), where A is the angle of the prism. Here, D = 60(1.5 - 1) = 30 degrees.

For a prism, the angle of minimum deviation (D) can be approximated as D = A for small angles, thus D = 60 degrees.

Using the lens formula, 1/f = 1/v - 1/u; here, f = 15 cm and u = -30 cm. Thus, 1/v = 1/15 + 1/30 = 1/10, giving v = 10 cm.