The frictional force can be calculated using F_friction = μ * N, where N = mg = 10 kg * 9.81 m/s² = 98.1 N. Thus, F_friction = 0.3 * 98.1 N = 29.43 N, approximately 30 N.

The maximum static friction force Ff = μs * N = 0.5 * (100 kg * 9.81 m/s²) = 490.5 N.

Kinetic energy is given by KE = 0.5 * m * v² = 0.5 * 2 kg * (3 m/s)² = 9 J.

The force of friction Ff = μk * N = 0.3 * (200 kg * 9.81 m/s²) = 588.6 N.

The maximum static friction force Ff = μs * N = 0.5 * (250 kg * 9.81 m/s²) = 1226.25 N. Since the applied force (600 N) is less than the frictional force (1226.25 N), the object will not move.

Using the formula v² = u² + 2as, where v = 0, u = 20 m/s, and a = -10 m/s², we get 0 = 20² + 2 * (-10) * s, solving gives s = 20 m.

The normal force N = mg * cos(θ) = 10 kg * 9.81 m/s² * cos(30°) ≈ 84.87 N. The force of friction Ff = μk * N = 0.2 * 84.87 N ≈ 16.97 N.

The force of kinetic friction Ff = μk * N = 0.4 * 300 N = 120 N. Since the applied force (150 N) is greater than the frictional force (120 N), the box will move.

Using the formula s = ut + 0.5at², where u = 0, a = 2 m/s², and t = 5 s, we get s = 0 + 0.5 * 2 * 5² = 25 m.

Convert speed to m/s: 60 km/h = 16.67 m/s. Using s = ut + 0.5at², where a = (0 - 16.67)/5 = -3.33 m/s², we find s = 16.67 * 5 + 0.5 * (-3.33) * 5² = 41.67 m.

Average speed = total distance / total time = 100 m / 4 s = 25 m/s.

Using the formula time = distance/speed, we have time = 300 m / 15 m/s = 20 s.

The horizontal component is given by F_horizontal = F * cos(θ) = 50 N * cos(30°) = 50 N * (√3/2) ≈ 43.3 N.

Centripetal acceleration is given by a_c = v² / r. Here, a_c = (10 m/s)² / 5 m = 20 m/s².

The vertical component of the velocity is 40 * sin(30) = 20 m/s. Using h = (v²)/(2g), we get h = (20²)/(2 * 9.8) ≈ 20.41 m.

The horizontal component is given by v_x = v * cos(θ) = 20 * cos(30°) = 20 * (√3/2) ≈ 17.32 m/s.

Stress = Force / Area. The area A = π(d/2)² = π(0.005)² = 7.85e-5 m². Stress = 20000 N / 7.85e-5 m² = 254647.91 Pa or 254.65 MPa.

First convert speed to m/s: 72 km/h = 20 m/s. Using a = (v - u)/t, we have a = (0 - 20)/10 = -2 m/s², so deceleration = 2 m/s².

According to Bernoulli's principle, as the velocity of a fluid increases, the pressure within the fluid decreases.

Using the formula s = ut + 0.5gt², where u = 0, s = 80 m, and g = 9.8 m/s², we solve 80 = 0 + 0.5 * 9.8 * t², giving t ≈ 4.04 s.

Using the formula s = 0.5gt², we have 45 = 0.5 * 9.8 * t², solving gives t ≈ 3.03 s.

Using the formula h = (v² - u²) / (2g), where v = 0, u = 20 m/s, and g = 9.81 m/s², we find h = (0 - (20)²) / (2 * -9.81) = 20.39 m, approximately 40 m.

During the adiabatic expansion phase of the Carnot cycle, the working substance does work on the surroundings as it expands.

The maximum bending moment for a simply supported beam with a point load at the center is given by M = WL/4, where W is the load and L is the length of the beam.

The normal force N = mg * cos(θ) = 150 kg * 9.81 m/s² * cos(15°) ≈ 1430 N.

Using the formula h = v²/(2g), where v = 15 m/s and g = 9.81 m/s², we find h = (15)²/(2*9.81) ≈ 11.47 m.

A roller support is typically used for beams supported at both ends with a central load, allowing for vertical movement while preventing horizontal movement.

Using the formula v = u + at, where u = 0, a = 2 m/s², and t = 5 s, we get v = 0 + (2)(5) = 10 m/s.

Using the formula v = u + at, where u = 0, a = 3 m/s², and t = 5 s, we get v = 0 + (3)(5) = 15 m/s.

The maximum deceleration a car can achieve is given by a = μg, where g = 9.81 m/s². Thus, a = 0.7 * 9.81 m/s² ≈ 6.87 m/s².