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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Welcome back Divine — missed you this week. Here are the solutions to your questions. Question 1 a) Homogeneity of a physical equation means that the dimensions of the terms on both sides of the equation are the same. While dimensional homogeneity is a necessary condition for an equation to be correct, it is not sufficient. An equation can be dimensionally homogeneous but still be physically incorrect due to a missing dimensionless constant or an incorrect numerical factor. For example, the equation for the period of a simple pendulum is T = 2 sqrt((L)/(g)). If we incorrectly write it as T = sqrt((L)/(g)), the equation is still dimensionally homogeneous: [T] = sqrt(([L])/([LT^-2)]) = sqrt([T^2]) = [T] However, the equation is physically incorrect because it omits the dimensionless constant 2. b) The given equation is E = (1)/(2) AKv^3N. We need to determine the base units of K. First, let's list the base units for the other quantities: • E (energy generated per unit time) is power, so its unit is Watts (W) or Joules per second (J/s). In base units, [E] = [ML^2T^-3]. • (air density) has units of kilograms per cubic meter (kg/m^3). In base units, [] = [ML^-3]. • A (area swept by the blade) has units of square meters (m^2). In base units, [A] = [L^2]. • v (wind velocity) has units of meters per second (m/s). In base units, [v] = [LT^-1]. • N (generator efficiency) is a dimensionless quantity, so [N] = [M^0L^0T^0]. Now, substitute these into the equation and solve for [K]: [E] = [][A][K][v^3][N] [ML^2T^-3] = [ML^-3][L^2][K][(LT^-1)^3][M^0L^0T^0] [ML^2T^-3] = [ML^-3][L^2][K][L^3T^-3] [ML^2T^-3] = [M L^-3+2+3 T^-3][K] [ML^2T^-3] = [ML^2T^-3][K] Therefore, for the equation to be dimensionally homogeneous, [K] must be dimensionless. [K] = [ML^2T^-3][ML^2T^-3] = [M^0L^0T^0] The base units of K are dimensionless. Question 2 Given: Resistance R = 80 \, Capacitance C = 125 \, = 125 × 10^-6 \, F Inductance L = 0.4 \, H RMS voltage V_rms = 220 \, V Frequency f = 50 \, Hz a) One use of such a circuit (an RLC series circuit) in daily life is in radio tuners. These circuits can be tuned to resonate at a specific frequency, allowing them to select a particular radio station's signal while rejecting others. b) Step 1: Calculate the inductive reactance X_L. X_L = 2 f L X_L = 2 (50 \, Hz) (0.4 \, H) X_L = 40 \, X_L ≈ 125.66 \, Step 2: Calculate the capacitive reactance X_C. X_C = (1)/(2 f C) X_C = (1)/(2 (50 \, Hz)) (125 × 10^-6 \, F) X_C = (1)/(0.0125) \, X_C ≈ (1)/(0.03927) \, X_C ≈ 25.46 \, Step 3: Calculate the impedance Z. Z = sqrt(R^2 + (X_L - X_C)^2) Z = sqrt((80 \, )^2 + (40 \, - (1)/(0.0125) \, )^2) Z = sqrt((80)^2 + (125.66 - 25.46)^2) Z = sqrt(6400 + (100.20)^2) Z = sqrt(6400 + 10040.04) Z = sqrt(16440.04) Z ≈ 128.22 \, The impedance Z is approximately 128.22 . Step 4: Calculate the root mean square (RMS) value of the current I_rms. I_rms = V_rmsZ I_rms = 220 \, V128.22 \, I_rms ≈ 1.716 \, A The RMS current is approximately 1.72 A. c) Step 1: Calculate the resonant frequency f_0. f_0 = (1)/(2(LC)) f_0 = (1)/(2((0.4 \, H))(125 × 10^-6) \, F) f_0 = (1)/(2(50 × 10^-6)) f_0 = (1)/(2(0.00005)) f_0 = (1)/(2 (0.007071)) f_0 = (1)/(0.044428) f_0 ≈ 22.51 \, Hz The resonant frequency f_0 is approximately 22.5 Hz. Step 2: Calculate the current in the circuit at resonance. At resonance, X_L = X_C, so the impedance Z becomes equal to the resistance R. Z_resonance = R = 80 \, I_rms, resonance = V_rmsR I_rms, resonance = 220 \, V80 \, I_rms, resonance = 2.75 \, A The current at resonance is 2.75 A. Question 3 a) The moment of a force (also known as torque) is the turning effect of a force about a pivot point. It is calculated as the product of the magnitude of the force and the perpendicular distance from the pivot to the line of action of the force. Its unit is Newton-meter (Nm). b) One condition necessary for a body to be in equilibrium on a plane is that the net force acting on the body must be zero. This means the vector sum of all forces acting on the body in any direction (e.g., horizontal and vertical) must be zero. c) Figure 1 shows a uniform plank of weight 6 \, N and length a, inclined at an angle . The system is in equilibrium. Let's analyze the forces and distances from the pivot point. The pivot is located at a distance (a)/(4) from the left end. • The weight of the plank, 6 \, N, acts at its center of gravity, which is at (a)/(2) from the left end. • The 16 \, N force acts downwards at the left end. • The 4 \, N force acts downwards at the right end. For equilibrium, the sum of clockwise moments about the pivot must equal the sum of anticlockwise moments about the pivot. Distances from the pivot: • For the 16 \, N force: The distance from the pivot is (a)/(4). This force creates an anticlockwise moment. • For the 6 \, N weight: The center of gravity is at (a)/(2) from the left end. The pivot is at (a)/(4). So, the distance from the pivot is (a)/(2) - (a)/(4) = (a)/(4). This force creates a clockwise moment. • For the 4 \, N force: The distance from the pivot is a - (a)/(4) = (3a)/(4). This force creates a clockwise moment. The forces are acting vertically, but the plank is inclined at an angle . When calculating moments, we need the perpendicular distance from the pivot to the line of action of the force. If the forces are vertical and the plank is inclined, the effective lever arm for each force will be the horizontal component of the distance from the pivot to the point where the force is applied. However, since all forces are parallel (vertical) and the pivot is a point, we can consider the distances along the plank, and the factor for each moment will cancel out. Alternatively, we can consider the perpendicular distance to the line of action of the force. Let's assume the forces are perpendicular to the plank for simplicity, or that the angle is accounted for by the support. Given the diagram, the forces are shown as vertical. The distances a/4, a/2, a are along the plank. The perpendicular distance from the pivot to the line of action of a vertical force at a point x along the plank is x . However, since all forces are vertical, the term will cancel out when equating moments. So we can use the distances along the plank directly. Anticlockwise moment: M_anti-clockwise = 16 \, N × (a)/(4) Clockwise moments: M_clockwise = (6 \, N × (a)/(4)) + (4 \, N × (3a)/(4)) For equilibrium: M_anti-clockwise = M_clockwise 16 × (a)/(4) = (6 × (a)/(4)) + (4 × (3a)/(4)) Divide by a (since a ≠ 0): (16)/(4) = (6)/(4) + (12)/(4) 4 = (6+12)/(4) 4 = (18)/(4) 4 = 4.5 This equation 4 = 4.5 is false, which indicates that the system as described with the given forces and distances cannot be in equilibrium if the pivot is the only support providing an upward force. The question asks to "Calculate given that the system is in equilibrium." This implies that the angle might be relevant to the forces or distances, or there's another force not explicitly shown. Let's re-examine the diagram. The 60^ angle is shown at the left end, but it's not clear what it refers to. The angle is the inclination of the plank. The forces 16 \, N and 4 \, N are shown as vertical. The weight 6 \, N is also vertical. The pivot provides an upward normal force. If the question implies that the 16 \, N and 4 \, N forces are components or that the 60^ angle is relevant to the 16 \, N force, it's not clearly stated. Assuming the forces are as depicted (vertical) and the distances are along the plank, the calculation above shows an imbalance. However, the question asks to calculate . This suggests must play a role. The only way would play a role in the moment calculation (if forces are vertical) is if the distances were perpendicular to the forces, which means the horizontal distances from the pivot. Let the pivot be at x=0. The 16 \, N force is at x = -(a)/(4) . (This is the horizontal distance from the pivot to the line of action of the force). The 6 \, N force is at x = (a)/(4) . The 4 \, N force is at x = (3a)/(4) . Anticlockwise moment: 16 × ((a)/(4) ) Clockwise moments: (6 × (a)/(4) ) + (4 × (3a)/(4) ) Equating moments: 16 × (a)/(4) = 6 × (a)/(4) + 4 × (3a)/(4) Since a ≠ 0 and ≠ 0 (for a plank to be inclined), we can divide by a : (16)/(4) = (6)/(4) + (12)/(4) 4 = 1.5 + 3 4 = 4.5 This still leads to the same contradiction. This means that the system, as drawn with the given forces and distances, cannot be in equilibrium. There might be an error in the problem statement or the diagram. Let's consider if the 60^ angle is relevant. If the 16 \, N force is acting at 60^ to the plank, then its vertical component would be 16 (60^) or 16 (60^) depending on the angle definition. However, the arrow clearly shows a vertical downward force. Given the contradiction, it's possible that the 6 \, N is the weight of the plank, and the 16 \, N and 4 \, N are external forces. The question asks to calculate . If the system is in equilibrium, the sum of moments must be zero. Since the calculation consistently leads to 4 = 4.5, it implies that the given forces and distances are not balanced. If we assume there is a mistake in the problem and the question intends for the system to be in equilibrium, then cannot be determined from the moment equation alone if it cancels out. would only be relevant if the forces were not purely vertical or if the distances were measured differently. Let's assume the 60^ is the angle of the 16 \, N force with the horizontal, and it's acting at the left end. This is not how it's drawn. Given the problem as stated, and the diagram, the moment equation leads to a contradiction. If the system is in equilibrium, then the sum of moments must be zero. Let's re-evaluate the distances from the pivot. Pivot is at a/4 from the left end. Force 1: 16 \, N at 0 (left end). Distance from pivot: a/4. Moment: 16 × a/4 (anticlockwise). Force 2: 6 \, N (weight of plank) at a/2. Distance from pivot: a/2 - a/4 = a/4. Moment: 6 × a/4 (clockwise). Force 3: 4 \, N at a (right end). Distance from pivot: a - a/4 = 3a/4. Moment: 4 × 3a/4 (clockwise). Sum of anticlockwise moments = 16 × (a)/(4) = 4a Sum of clockwise moments = 6 × (a)/(4) + 4 × (3a)/(4) = (6a)/(4) + (12a)/(4) = (18a)/(4) = 4.5a For equilibrium, 4a = 4.5a. This implies 0.5a = 0, which means a=0. But a is the length of the plank, so a ≠ 0. This confirms the contradiction. The system cannot be