This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
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QUESTION 1: GENERAL 1.1 Resultant force The resultant force is the single force that represents the combined effect of all individual forces acting on an object. It is the vector sum of all forces. 1.2 Elastic limit The elastic limit is the maximum stress a material can withstand without undergoing permanent deformation. If the stress exceeds this limit, the material will not return to its original shape after the load is removed. 1.3 Strain Strain is a measure of the deformation of a material, defined as the ratio of the change in dimension to the original dimension. It is a dimensionless quantity. 1.4 Hydrostatic pressure Hydrostatic pressure is the pressure exerted by a fluid at rest due to gravity. It increases with depth and is calculated as P = gh, where is the fluid density, g is the acceleration due to gravity, and h is the depth. 1.5 Second moment of area The second moment of area (or area moment of inertia) is a geometric property of a cross-section that quantifies its resistance to bending. It is calculated by integrating the square of the distance from an axis to each infinitesimal area element over the entire cross-section. QUESTION 2: KINEMATICS 2.1 Given: Initial position P_0 = 60 km North East from OR Tambo. Velocity v = 96 km/h North for t = 6 hours. Step 1: Determine the initial displacement vector from OR Tambo. North East implies an angle of 45^ from the East (or North). d_0 = (60 45^, 60 45^) km = (60 × 0.7071, 60 × 0.7071) km ≈ (42.426, 42.426) km Step 2: Determine the displacement vector during the flight. The aircraft flies North, so the displacement is purely in the North direction. d_f = (0, 96 km/h × 6 h) = (0, 576) km Step 3: Calculate the total displacement vector from OR Tambo. D = d_0 + d_f = (42.426 + 0, 42.426 + 576) km = (42.426, 618.426) km Step 4: Calculate the magnitude of the total displacement. |D| = sqrt((42.426)^2 + (618.426)^2) = sqrt(1799.96 + 382450.5) = sqrt(384250.46) ≈ 619.88 km Step 5: Calculate the direction of the total displacement. The angle from the positive x-axis (East) is: = ((618.426)/(42.426)) ≈ (14.576) ≈ 86.06^ The direction can be expressed as 86.06^ North of East, or as a bearing from North: 90^ - 86.06^ = 3.94^ East of North. The position (displacement) is 619.88 km at 86.06^ North of East. 2.2 Given: Hoist K descends at v_K = 6.48 km/h. Hoist L ascends at v_L = 5.04 km/h. Assume upward direction is positive. v_K = -6.48 km/h v_L = +5.04 km/h 2.2.1 The velocity of hoist K relative to the velocity of hoist L in magnitude and direction. Step 1: Calculate the relative velocity v_K/L = v_K - v_L. v_K/L = (-6.48 km/h) - (5.04 km/h) = -11.52 km/h The velocity of hoist K relative to hoist L is 11.52 km/h downwards. 2.2.2 The velocity of hoist L relative to the velocity of hoist K in magnitude and direction. Step 1: Calculate the relative velocity v_L/K = v_L - v_K. v_L/K = (5.04 km/h) - (-6.48 km/h) = 5.04 km/h + 6.48 km/h = 11.52 km/h The velocity of hoist L relative to hoist K is 11.52 km/h upwards. 2.3 Given: Initial velocity v_0 = 42 m/s. Angle = 28^ to the horizontal. Assume g = 9.81 m/s^2. Step 1: Resolve the initial velocity into horizontal and vertical components. v_0x = v_0 = 42 m/s × (28^) = 42 × 0.8829 ≈ 37.082 m/s v_0y = v_0 = 42 m/s × (28^) = 42 × 0.4695 ≈ 19.719 m/s 2.3.1 The maximum height that the stone reaches. Step 2: Use the kinematic equation v_y^2 = v_0y^2 + 2a_y y. At maximum height, v_y = 0. 0^2 = (19.719 m/s)^2 + 2(-9.81 m/s^2) H_max 0 = 388.839 - 19.62 H_max H_max = (388.839)/(19.62) ≈ 19.818 m The maximum height reached by the stone is 19.82 m. 2.3.2 The horizontal displacement of the stone. Step 1: Calculate the time to reach maximum height (t_up). Using v_y = v_0y + a_y t_up: 0 = 19.719 m/s + (-9.81 m/s^2) t_up t_up = (19.719)/(9.81) ≈ 2.010 s Step 2: Calculate the total time of flight (T). For projectile motion returning to the same height, T = 2 t_up. T = 2 × 2.010 s = 4.020 s Step 3: Calculate the horizontal displacement (range). R = v_0x T = 37.082 m/s × 4.020 s ≈ 149.17 m The horizontal displacement of the stone is 149.17 m. QUESTION 3: ROTATIONAL MOTION 3.1 Define angular velocity. Angular velocity is the rate at which an object rotates or revolves relative to another point, i.e., the rate of change of angular displacement. It is a vector quantity, typically measured in radians per second (rad/s). 3.2 Given: Diameter D = 1.76 m, so radius r = D/2 = 0.88 m. Linear velocity v = 396 km/h. Step 1: Convert linear velocity from km/h to m/s. v = 396 kmh × 1000 m1 km × 1 h3600 s = 110 m/s 3.2.1 The rotational frequency of the wheel in revolutions per minute. Step 2: Calculate the angular velocity in rad/s using v = r. = (v)/(r) = 110 m/s0.88 m = 125 rad/s Step 3: Convert angular velocity from rad/s to r/min (revolutions per minute). Frequency (r/min) = 125 rads × 1 rev2 rad × 60 s1 min Frequency (r/min) = (125 × 60)/(2) = (7500)/(2) ≈ 1193.66 r/min The rotational frequency of the wheel is 1193.66 r/min. 3.2.2 The angular velocity of the wheel in rad/s. The angular velocity was calculated in Step 2 of 3.2.1. = 125 rad/s The angular velocity of the wheel is 125 rad/s. 3.2.3 The number of revolutions made by the wheel during 36 minutes. Step 1: Convert time from minutes to seconds. t = 36 min × 60 s/min = 2160 s Step 2: Calculate the total angular displacement in radians. = t = 125 rad/s × 2160 s = 270000 rad Step 3: Convert angular displacement from radians to revolutions. Revolutions = ()/(2) = 270000 rad2 rad/rev ≈ 42971.83 revolutions The number of revolutions made by the wheel is 42971.83 revolutions. What's next?