Using F = ma, acceleration a = F/m = 200 N / 50 kg = 4 m/s².

Using v² = u² + 2gh, v² = 0 + 2 * 10 m/s² * 20 m = 400, v = √400 = 20 m/s.

Kinetic Energy (KE) = 0.5 * m * v² = 0.5 * 50 kg * (4 m/s)² = 400 J.

Momentum p = mv = 50 kg * 5 m/s = 250 kg·m/s.

Using F = ma, we find a = F/m = 200 N / 50 kg = 4 m/s².

If the object moves with constant velocity, the frictional force equals the applied force, which is 200 N.

Using F = ma, a = F/m = 18 N / 6 kg = 3 m/s².

Using Ohm's law, I = V/R = 9V / 3Ω = 3A.

Using the equation of motion for free fall: h = (1/2)gt². 20 m = (1/2)(9.8)t². Solving gives t ≈ 2 s.

Time to reach maximum height = initial velocity / g = 20 / 10 = 2 s.

Time to reach maximum height = initial velocity / g = 25 / 10 = 2.5 s.

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

Using the formula: t = v/g = 20/10 = 2 s.

Using the formula: height = (initial velocity^2) / (2 * g), where g = 9.8 m/s². Height = (15^2) / (2 * 9.8) = 11.25 m.

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

The angle for the first minimum in single-slit diffraction is given by a sin θ = λ, hence θ = λ/a.

According to the diffraction principle, as the slit width decreases, the width of the central maximum increases.

The angle of the first minimum in single-slit diffraction is given by sin(θ) = λ/a, where a is the width of the slit.

Work done = mgh = 2 kg * 9.8 m/s² * 5 m = 98 J.

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

Using conservation of energy, potential energy at the top = kinetic energy at the bottom. mgh = 0.5mv². Thus, v = sqrt(2gh) = sqrt(2 * 9.8 * 10) = 14 m/s.

Using conservation of energy, potential energy at the top = kinetic energy at the bottom. mgh = 0.5mv². Thus, v = sqrt(2gh) = sqrt(2 * 9.81 * 5) = 10 m/s.

Using conservation of energy, potential energy at the top (mgh) converts to kinetic energy at the bottom (1/2 mv²). Thus, v = sqrt(2gh) = sqrt(2 * 9.8 * 5) ≈ 10 m/s.

Net force = Applied force - Frictional force = 20 N - 5 N = 15 N. Using F = ma, a = F/m = 15 N / 5 kg = 3 m/s².

Net force = Applied force - Frictional force = 20 N - 5 N = 15 N. Using F = ma, a = F/m = 15 N / 5 kg = 3 m/s².

Net force = Applied force - Frictional force = 20 N - 5 N = 15 N. Acceleration = Net force / Mass = 15 N / 5 kg = 3 m/s².

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

Acceleration a = F/m = 15 N / 5 kg = 3 m/s². Velocity after 3 seconds = a × t = 3 m/s² × 3 s = 9 m/s.

Using F = ma, acceleration a = F/m = 15 N / 5 kg = 3 m/s².

Acceleration a = g sin(θ) = 9.8 m/s² × sin(30°) = 4.9 m/s².