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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Here are the solutions to the questions. 5.a.i) Voltage drop across the remainder of the circuit Step 1: Express the given voltages in phasor form. The supply voltage is v = 30(100 t), so its phasor is V = 30 0 rad. The voltage drop across one component is v_1 = 20(100 t - 0.59), so its phasor is V_1 = 20 -0.59 rad. Step 2: Convert V_1 to rectangular form. V_1 = 20((-0.59) + j(-0.59)) V_1 ≈ 20(0.8310 - j0.5564) V_1 ≈ 16.62 - j11.13 Step 3: Subtract V_1 from V to find the remainder voltage V_R. V_R = V - V_1 = (30 + j0) - (16.62 - j11.13) V_R = (30 - 16.62) + j(0 - (-11.13)) V_R = 13.38 + j11.13 Step 4: Convert V_R back to polar form A . Amplitude A = sqrt((13.38)^2 + (11.13)^2) = sqrt(179.0244 + 123.8769) = sqrt(302.9013) ≈ 17.404 V. Phase angle = ((11.13)/(13.38)) = (0.8318) ≈ 0.694 rad. The voltage drop across the remainder of the circuit is 17.40(100 t + 0.694) V. 5.a.ii) Supply frequency Step 1: Identify the angular frequency from the supply voltage equation v = 30(100 t). = 100 rad/s. Step 2: Use the relationship = 2 f to find the frequency f. 100 = 2 f f = (100)/(2) = 50 Hz. The supply frequency is 50 Hz. 5.a.iii) Periodic time of the supply Step 1: Use the relationship between periodic time T and frequency f. T = (1)/(f). Step 2: Substitute the frequency found in part (ii). T = (1)/(50) = 0.02 s. The periodic time of the supply is 0.02 s. 5.a.iv) r.m.s. value of the supply voltage Step 1: Identify the peak voltage V_m from the supply voltage equation v = 30(100 t). V_m = 30 V. Step 2: Use the formula for the RMS value of a sinusoidal voltage. V_rms = (V_m)/(sqrt(2)). V_rms = (30)/(sqrt(2)) = 30sqrt(2)2 = 15sqrt(2) V. V_rms ≈ 15 × 1.4142 = 21.213 V. The r.m.s. value of the supply voltage is 15sqrt(2) V (or approx. 21.21 V). 5.b) Solve for I_1, I_2, I_3 using determinants Step 1: Write the system of equations in matrix form AX = B. 2 & 3 & -4 \\ 1 & -5 & -3 \\ -7 & 2 & 6 I_1 \\ I_2 \\ I_3 = 26 \\ -87 \\ 12 Step 2: Calculate the determinant of the coefficient matrix D. D = 2 & 3 & -4 \\ 1 & -5 & -3 \\ -7 & 2 & 6 = 2((-5)(6) - (-3)(2)) - 3((1)(6) - (-3)(-7)) + (-4)((1)(2) - (-5)(-7)) D = 2(-30 + 6) - 3(6 - 21) - 4(2 - 35) D = 2(-24) - 3(-15) - 4(-33) = -48 + 45 + 132 = 129 Step 3: Calculate D_1 by replacing the first column with the constant terms. D_1 = 26 & 3 & -4 \\ -87 & -5 & -3 \\ 12 & 2 & 6 = 26((-5)(6) - (-3)(2)) - 3((-87)(6) - (-3)(12)) + (-4)((-87)(2) - (-5)(12)) D_1 = 26(-30 + 6) - 3(-522 + 36) - 4(-174 + 60) D_1 = 26(-24) - 3(-486) - 4(-114) = -624 + 1458 + 456 = 1290 Step 4: Calculate D_2 by replacing the second column with the constant terms. D_2 = 2 & 26 & -4 \\ 1 & -87 & -3 \\ -7 & 12 & 6 = 2((-87)(6) - (-3)(12)) - 26((1)(6) - (-3)(-7)) + (-4)((1)(12) - (-87)(-7)) D_2 = 2(-522 + 36) - 26(6 - 21) - 4(12 - 609) D_2 = 2(-486) - 26(-15) - 4(-597) = -972 + 390 + 2388 = 1806 Step 5: Calculate D_3 by replacing the third column with the constant terms. D_3 = 2 & 3 & 26 \\ 1 & -5 & -87 \\ -7 & 2 & 12 = 2((-5)(12) - (-87)(2)) - 3((1)(12) - (-87)(-7)) + 26((1)(2) - (-5)(-7)) D_3 = 2(-60 + 174) - 3(12 - 609) + 26(2 - 35) D_3 = 2(114) - 3(-597) + 26(-33) = 228 + 1791 - 858 = 1161 Step 6: Calculate I_1, I_2, I_3 using Cramer's Rule. I_1 = (D_1)/(D) = (1290)/(129) = 10 mA I_2 = (D_2)/(D) = (1806)/(129) = 14 mA I_3 = (D_3)/(D) = (1161)/(129) = 9 mA The currents are I_1 = 10 mA, I_2 = 14 mA, I_3 = 9 mA. 5.c) Acceleration and tensions for three masses Given: m_A = 4 kg, m_B = 3 kg, m_C = 1 kg, = 0.6, g = 10 m/s^2. Step 1: Draw free-body diagrams and set up equations of motion. Assume m_A moves down, m_B moves right, and m_C moves up. Let a be the acceleration. For mass A: m_A g - T_1 = m_A a 4(10) - T_1 = 4a 40 - T_1 = 4a (Eq. 1) For mass C: T_2 - m_C g = m_C a T_2 - 1(10) = 1a T_2 - 10 = a (Eq. 2) For mass B: Normal force N = m_B g = 3(10) = 30 N. Friction force f_k = N = 0.6 × 30 = 18 N. The equation of motion for B (moving right): T_1 - T_2 - f_k = m_B a T_1 - T_2 - 18 = 3a (Eq. 3) Step 2: Solve the system of equations. From (Eq. 1): T_1 = 40 - 4a From (Eq. 2): T_2 = 10 + a Substitute T_1 and T_2 into (Eq. 3): (40 - 4a) - (10 + a) - 18 = 3a 40 - 4a - 10 - a - 18 = 3a 12 - 5a = 3a 12 = 8a a = (12)/(8) = 1.5 m/s^2 Step 3: Calculate the tensions T_1 and T_2. T_1 = 40 - 4a = 40 - 4(1.5) = 40 - 6 = 34 N T_2 = 10 + a = 10 + 1.5 = 11.5 N The acceleration of the masses is 1.5 m/s^2. The tension in the string between A and B is 34 N. The tension in the string between B and C is 11.5 N. 5.d) Atwood machine Given: m_1 = 4 kg, m_2 = 3 kg, g = 9.81 m/s^2. Released from rest. Step 1: Set up equations of motion for each mass. Let T be the tension in the string and a be the acceleration. For m_1 (moving downwards): m_1 g - T = m_1 a 4(9.81) - T = 4a 39.24 - T = 4a (Eq. 1) For m_2 (moving upwards): T - m_2 g = m_2 a T - 3(9.81) = 3a T - 29.43 = 3a (Eq. 2) Step 2: Solve for acceleration a. Add (Eq. 1) and (Eq. 2): (39.24 - T) + (T - 29.43) = 4a + 3a 9.81 = 7a a = (9.81)/(7) ≈ 1.4014 m/s^2 i) The acceleration of each mass is 1.4014 m/s^2. (4 kg mass downwards, 3 kg mass upwards). Step 3: Calculate the speed after 2 seconds. Using the kinematic equation v = u + at, with initial velocity u=0. v = 0 + (1.4014)(2) v = 2.8028 m/s ii) Their speed after 2 seconds is 2.8028 m/s. 5.e) Regression analysis Given data: Electrical (x): 3, 6, 4, 5, 7, 8, 9, 7, 10, 9 Electronics (y): 5, 6, 5, 6, 7, 9, 7, 8, 9, 6 Step 1: Calculate the necessary sums. n = 10 x = 68 y = 68 x^2 = 510 y^2 = 482 xy = 485 x = (68)/(10) = 6.8 y = (68)/(10) = 6.8 i) Scatter diagram for the data: A scatter diagram would plot the given points (x, y) on a Cartesian plane. The x-axis would represent Electrical marks and the y-axis would represent Electronics marks. The points would be: (3,5), (6,6), (4,5), (5,6), (7,7), (8,9), (9,7), (7,8), (10,9), (9,6). Visually, these points would show a general upward trend, indicating a positive correlation between Electrical and Electronics marks. ii) Equation of best fit and prediction: To predict the Electrical Installation mark (x) given the Electronics mark (y), we need the regression line of x on y, which has the form x = a + by. Calculate the slope b: b = (n xy - ( x)( y))/(n y^2 - ( y)^2) b = (10(485) - (68)(68))/(10(482) - (68)^2) = (4850 - 4624)/(4820 - 4624) = (226)/(196) ≈ 1.1531 Calculate the y-intercept a: a = x - by a = 6.8 - (1.1531)(6.8) = 6.8 - 7.8411 = -1.0411 The equation of best fit (x on y) is x = -1.0411 + 1.1531y. Predict the Electrical Installation mark (x) for a student who scored 7 in Electronics (y=7): x = -1.0411 + 1.1531(7) x = -1.0411 + 8.0717 = 7.0306 The predicted Electrical Installation mark is 7.03. iii) Regression line of y on x: The regression line of y on x has the form y = a' + b'x. Calculate the slope b': b' = (n xy - ( x)( y))/(n x^2 - ( x)^2) b' = (10(485) - (68)(68))/(10(510) - (68)^2) = (4850 - 4624)/(5100 - 4624) = (226)/(476) ≈ 0.4748 Calculate the y-intercept a': a' = y - b'x a' = 6.8 - (0.4748)(6.8) = 6.8 - 3.2286 = 3.5714 The equation of the regression line of y on x is y = 3.5714 + 0.4748x. To draw this line on the scatter diagram, you would plot two points from this equation (e.g., for x=3, y ≈ 5.0; for x=10, y ≈ 8.3) and draw a straight line connecting them. This line represents the best linear fit for predicting Electronics marks based on Electrical marks. 3 done, 2 left today. You're making progress.