Circumference = 2πr = 2π(2) = 4π m. Speed = Distance/Time = 4π/4 = π m/s ≈ 3.14 m/s.

Centripetal acceleration (a_c) = v²/r. First find v = 2πr/T = 2π(4)/8 = π m/s. Then a_c = (π)²/4 = 2.5 m/s².

Angular momentum L = mvr; if v is doubled, L also doubles.

Work done = Force × Distance = 12 N × 3 m = 36 J.

Work done = Force × Distance = 3 N × 6 m = 18 J.

Final kinetic energy = Initial kinetic energy + Work done = 100 J + 50 J = 150 J.

Angular momentum is constant as the particle moves with constant velocity.

The angular momentum is non-zero and depends on the distance from the point to the line of motion.

Angular momentum L = mvr, where r is the perpendicular distance from the line of motion to point P.

Final velocity (v) = u + at = 3 + (2 * 5) = 13 m/s.

Using the formula: final velocity = initial velocity + acceleration * time. Final velocity = 2 + 4 * 5 = 22 m/s.

Final velocity = initial velocity + (acceleration * time) = 10 m/s + (2 m/s² * 5 s) = 20 m/s.

The moment of inertia of a point mass about an axis is given by I = mr^2.

The moment of inertia of a point mass is given by I = mr^2.

Angular momentum L = mvr, where v is the linear speed and r is the radius.

The magnetic force acting on a charged particle moving in a magnetic field is given by F = qvB sin(θ), where θ is the angle between the velocity vector and the magnetic field vector.

There are 26 choices for each letter (3 letters) and 10 choices for the first digit and 9 for the second. Total = 26^3 * 10 * 9 = 17576.

Frequency f = number of oscillations / time = 20 / 40 s = 0.5 Hz.

Time period (T) = 2π√(L/g) = 2π√(1/9.8) ≈ 2 s

Relative uncertainty = (Uncertainty / Measured value) * 100 = (0.1 / 2.0) * 100 = 5%.

Time period (T) = 2π√(L/g) ≈ 2π√(1/9.8) ≈ 2 s.

Using conservation of energy, potential energy at the top = kinetic energy at the bottom. mgh = 0.5mv². Solving gives v = sqrt(2gh) = sqrt(2 * 9.8 * 1) = 4.4 m/s.

The period of a simple pendulum is given by T = 2π√(L/g). If L is doubled, T becomes T' = 2π√(2L/g) = √2 * T ≈ 1.41 s.

The potential energy of a pendulum is maximum at the highest point of its swing, where its kinetic energy is minimum.

Using conservation of energy, potential energy at the top = kinetic energy at the bottom. mgh = 0.5mv². Solving gives v = √(2gh) = √(2 * 9.8 * 2) = 4 m/s.

Using conservation of energy, potential energy at the top = kinetic energy at the bottom. mgh = 0.5mv^2. Solving gives v = sqrt(2gh) = sqrt(2*9.8*5) = 10 m/s.

Time period T = 2π√(l/g) = 2π√(1/10) = 2π/√10 ≈ 2 s.

Time period T = 2π√(l/g) = 2π√(2/10) = 2π√(0.2) ≈ 2.0 s.

Frequency (f) = 1 / (2π√(L/g)) = 1 / (2π√(1/9.8)) ≈ 1 Hz.

The period T of a simple pendulum is given by T = 2π√(L/g). Here, L = 1 m and g = 9.8 m/s². Thus, T = 2π√(1/9.8) ≈ 2.0 s.