Using conservation of energy, potential energy at the top = kinetic energy at the bottom. The total kinetic energy is the sum of translational and rotational kinetic energy. Thus, mgh = (1/2)mv^2 + (1/5)mv^2, leading to v = √(10gh/7).

Using conservation of energy, potential energy at the top = kinetic energy at the bottom. The total kinetic energy is the sum of translational and rotational kinetic energy. Thus, v = √(5gh/7).

Using the formula λ = v/f, where v is the speed and f is the frequency, we have λ = 340 m/s / 1700 Hz = 0.2 m.

Potential Energy (PE) = 1/2 * k * x^2 = 1/2 * 200 N/m * (0.1 m)^2 = 1 J

Potential Energy in Spring (PE) = 1/2 kx^2 = 1/2 * 500 N/m * (0.2 m)^2 = 10 J

Using the equation of motion: h = ut + 0.5gt². Solving for t gives t ≈ 4 seconds.

Using τ = Iα, we have α = τ/I = 15/3 = 5 rad/s².

According to Newton's second law for rotation, τ = Iα. If τ is doubled, α also doubles.

According to Newton's second law for rotation, τ = Iα, thus α = τ/I.

Using the formula: final velocity = initial velocity + acceleration * time. Final velocity = 0 + 2 * 10 = 20 m/s.

Time period (T) = 1 / frequency = 1 / 440 Hz ≈ 0.00227 s

Time period (T) = 1 / frequency = 1 / 440 Hz ≈ 0.00227 s.

The torque τ = Mg(L/2) and moment of inertia I = (1/3)ML². Using τ = Iα, we find α = 3g/2L.

Using conservation of energy, potential energy at the top is converted to rotational kinetic energy at the bottom. The angular velocity ω can be found using the relation ω = √(3g/L).

Using conservation of energy, potential energy at the top converts to rotational kinetic energy at the bottom. The angular speed ω = √(3g/L).

The speed of a wave is given by the formula v = fλ. Here, v = 50 Hz * 2 m = 100 m/s.

The speed of the wave can be calculated using the formula v = fλ. Here, v = 500 Hz * 0.5 m = 250 m/s.

The speed of the wave can be calculated using the formula v = fλ. Thus, v = 500 Hz * 0.68 m = 340 m/s.

Using the formula ω = ω₀ + αt, where ω₀ = 0, α = 2 rad/s², and t = 5 s, we get ω = 0 + 2*5 = 10 rad/s.

Using the formula ω = ω₀ + αt, we have ω = 10 rad/s + (2 rad/s²)(5 s) = 20 rad/s.

The total kinetic energy is the sum of translational and rotational kinetic energy. K.E. = (1/2)Mv² + (1/2)(Iω²) where I = (1/2)MR² for a solid cylinder.

The linear velocity v of the center of the wheel is given by v = Rω.

Stress = Force / Area. Halving the diameter increases the area by a factor of 1/4, thus stress quadruples.

Elongation is inversely proportional to the area. Doubling the diameter increases the area by a factor of 4, thus elongation becomes L/4.

Stress (σ) = Force (F) / Area (A). Halving the diameter increases the area by a factor of 4, thus stress quadruples.

Elongation is inversely proportional to the area. Halving the diameter reduces the area to a quarter, thus elongation quadruples.

Elongation (ΔL) = (F * L) / (A * Y), where A is the cross-sectional area and Y is Young's modulus.

Bernoulli's equation states that an increase in the speed of a fluid results in a decrease in pressure.

Bernoulli's principle states that an increase in the speed of a fluid occurs simultaneously with a decrease in pressure.

Bernoulli's principle states that an increase in the speed of a fluid occurs simultaneously with a decrease in pressure.