Using h = 0.5 * g * t^2, we have 45 = 0.5 * 10 * t^2. Solving gives t^2 = 9, so t = 3 s.

Using h = 0.5 * g * t^2, rearranging gives t = sqrt(2h/g) = sqrt(2 * 45 / 10) = sqrt(9) = 3 s.

Using h = 0.5 * g * t^2, rearranging gives t = sqrt(2h/g) = sqrt(2 * 45 / 10) = sqrt(9) = 3 s.

Using the formula: h = 0.5 * g * t^2. Rearranging gives t = sqrt(2h/g) = sqrt(2 * 45 / 10) = sqrt(9) = 3 s.

Using the formula: h = 0.5 * g * t^2. Rearranging gives t = sqrt(2h/g) = sqrt(2 * 80 / 10) = sqrt(16) = 4 s.

Probability of failing = 1 - Probability of passing = 1 - 0.7 = 0.3.

For destructive interference in a thin film with a higher refractive index below, the condition is 2t = (m + 1/2)λ/n, where n is the refractive index of the film.

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.

First convert speed to m/s: 72 km/h = 20 m/s. Using the formula: a = (v - u) / t = (0 - 20) / 10 = -2 m/s².

First convert speed to m/s: 72 km/h = 20 m/s. Using the formula: acceleration = (final velocity - initial velocity) / time = (0 - 20) / 5 = -4 m/s².

First convert speed to m/s: 72 km/h = 20 m/s. Deceleration = (final velocity - initial velocity) / time = (0 - 20) / 5 = -4 m/s².

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

Using the transformer equation Vp/Vs = Np/Ns, we find Vs = Vp × (Ns/Np) = 240 V × (50/200) = 60 V.

Using the transformer equation Vp/Vs = Np/Ns, we have 240/120 = 100/Ns, giving Ns = 50 turns.

Period T = 1/f = 1/440 Hz ≈ 0.00227 s.

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

The time period T is the reciprocal of frequency f. Thus, T = 1/f = 1/440 ≈ 0.00227 s.

Wavelength λ is given by λ = v/f. Thus, λ = 340 m/s / 440 Hz ≈ 0.77 m.

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

Time period T = 1/f = 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).

Using conservation of energy, potential energy converts to rotational kinetic energy. Angular speed ω = √(3g/L).

Using conservation of energy, potential energy converts to rotational kinetic energy. Angular speed ω = √(3g/L).

Average speed = total distance / total time = 100 m / 5 s = 20 m/s.

Distance = speed * time = 60 km/h * 2 h = 120 km.