Torque (τ) = F × r × sin(θ) = 50 N × 0.5 m × sin(60°) = 50 N × 0.5 m × (√3/2) = 43.3 Nm.

Torque (τ) = F × d × sin(θ) = 50 N × 1 m × sin(30°) = 50 N × 1 m × 0.5 = 25 Nm.

Torque (τ) = F × r × sin(θ) = 50 N × 2 m × sin(30°) = 50 N × 2 m × 0.5 = 50 N·m.

Horizontal component = Fcos(θ) = 50 N × cos(30°) = 50 N × (√3/2) ≈ 43.3 N.

Torque (τ) = F × r × sin(θ) = 50 N × 1 m × sin(60°) = 50 N × 1 m × (√3/2) ≈ 43.3 Nm.

Torque (τ) = F × r × sin(θ) = 50 N × 2 m × sin(60°) = 50 × 2 × (√3/2) = 43.3 Nm.

Horizontal component = F × cos(θ) = 50 N × cos(60°) = 50 N × 0.5 = 25 N.

Using F = ma, a = F/m = 50 N / 25 kg = 2 m/s².

Using F = ma, a = F/m = 50 N / 25 kg = 2 m/s².

Work = Force * Distance = 50 N * 4 m = 200 J.

Using F = mAω², we find A = F / (mω²) = 12 / (3*(2π*2)²) ≈ 0.2 m.

Total parts = 1 + 2 + 3 + 4 = 10. Third project's share = (3/10) * 5000 = $1500.

If the seller sells 25% of 120 apples, he sells 0.25 * 120 = 30 apples. Therefore, the number of apples left is 120 - 30 = 90.

Apples sold = 30% of 120 = 36. Apples left = 120 - 36 = 84.

The ratio of apples to oranges is 5:3. If there are 40 apples, we can set up the proportion: 5/3 = 40/x. Cross-multiplying gives us 5x = 120, so x = 120/5 = 24. Therefore, he has 24 oranges.

If the ratio of apples to oranges is 5:3, then for every 5 apples, there are 3 oranges. If there are 40 apples, we can set up the proportion: 5/3 = 40/x. Cross-multiplying gives us 5x = 120, so x = 24. Therefore, there are 24 oranges.

The HCF of 36 and 48 is 12, which is the maximum number of rows he can plant.

The largest number of bouquets is the HCF of 60 and 90, which is 30.

The minimum height is the LCM of 3 and 5, which is 15 feet.

The LCM of 4 and 6 is 12. Therefore, they will be watered together again in 12 days.

The RMS speed is proportional to the square root of the temperature. Therefore, at 600 K, the RMS speed will be 400 * sqrt(600/300) = 400 * sqrt(2) m/s.

The RMS speed is proportional to the square root of the temperature. Therefore, v_rms at 600 K = 500 * sqrt(600/300) = 500 * sqrt(2) ≈ 707 m/s.

Work done by the gas during expansion at constant pressure is W = PΔV. Here, ΔV = 5 L - 2 L = 3 L, so W = 1 atm * 3 L = 3 L·atm.

Work done (W) = P * ΔV = 100 kPa * (5 m³ - 2 m³) = 100 kPa * 3 m³ = 300 kJ.

Work done (W) = P * ΔV = P * (V2 - V1).

According to Boyle's Law, P1V1 = P2V2, thus P2 = P1(V1/V2).

In an isothermal process, the work done by the gas is equal to the heat absorbed. Therefore, work done = 600 J.

Work done (W) = P * ΔV = 100 kPa * (2 m³ - 1 m³) = 100 kPa * 1 m³ = 100 kJ = 10 kJ.

Work done (W) = nRT ln(Vf/Vi). Here, nRT = PVi = 100 kPa * 1 m³ = 100 kJ. W = 100 kJ * ln(2) ≈ 69.3 kJ.

Work done (W) = PΔV = 100 kPa * (2 m³ - 1 m³) = 100 kPa * 1 m³ = 100 kJ = 20 kJ.