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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a) i) Tensile stress Step 1: Identify the given values. Force F = 200 \, N Cross-sectional area A = 2 × 10^-6 \, m^2 Step 2: Use the formula for tensile stress. = (F)/(A) Step 3: Substitute the values and calculate. = 200 \, N2 × 10^-6 \, m^2 = 100 × 10^6 \, N/m^2 = 1.0 × 10^8 \, Pa The tensile stress is 1.0 × 10^8 \, Pa. ii) Tensile strain Step 1: Identify the given values and convert units. Original length L_0 = 2.0 \, m Extension L = 4 \, mm = 4 × 10^-3 \, m Step 2: Use the formula for tensile strain. = ( L)/(L_0) Step 3: Substitute the values and calculate. = 4 × 10^-3 \, m2.0 \, m = 2 × 10^-3 The tensile strain is 2 × 10^-3. iii) Young's modulus Step 1: Use the calculated values for tensile stress and tensile strain. Tensile stress = 1.0 × 10^8 \, Pa Tensile strain = 2 × 10^-3 Step 2: Use the formula for Young's modulus. Y = ()/() Step 3: Substitute the values and calculate. Y = 1.0 × 10^8 \, Pa2 × 10^-3 Y = 0.5 × 10^11 \, Pa Y = 5.0 × 10^10 \, Pa Young's modulus is 5.0 × 10^10 \, Pa. iv) State two applications of elasticity of materials in industry and daily life. • Vehicle Tires • Suspension systems v) Using a well labelled diagram, describe a simple experiment to demonstrate Hooke's Law using springs and weights. To demonstrate Hooke's Law, set up a spring vertically from a rigid support. Attach a ruler alongside the spring to measure its length. Measure the initial length of the spring (L_0) when no weight is attached. Then, attach known weights (e.g., 100g, 200g, 300g) one by one to the free end of the spring. For each weight, measure the new length of the spring (L). Calculate the extension ( L = L - L_0) for each weight. Plot a graph of force (weight) against extension. Hooke's Law states that the extension is directly proportional to the applied force, so the graph should be a straight line passing through the origin, within the elastic limit of the spring. 5) Two springs are connected in series with spring constant of 150 \, N/m and 200 \, N/m respectively. A weight of 50 \, N is attached at the end of the spring, calculate total extension in cm. Step 1: Calculate the equivalent spring constant for springs connected in series. Given spring constants: k_1 = 150 \, N/m, k_2 = 200 \, N/m. For springs in series, the reciprocal of the equivalent spring constant is the sum of the reciprocals of individual spring constants: (1)/(k_eq) = (1)/(k_1) + (1)/(k_2) (1)/(k_eq) = (1)/(150 \, N/m) + (1)/(200 \, N/m) (1)/(k_eq) = (4)/(600 \, N/m) + (3)/(600 \, N/m) (1)/(k_eq) = (7)/(600 \, N/m) k_eq = (600)/(7) \, N/m k_eq ≈ 85.714 \, N/m Step 2: Calculate the total extension using Hooke's Law. Force F = 50 \, N F = k_eq × x x = (F)/(k_eq) x = 50 \, N(600)/(7) \, N/m x = (50 × 7)/(600) \, m x = (350)/(600) \, m x = (7)/(12) \, m x ≈ 0.5833 \, m Step 3: Convert the total extension to centimeters. x = 0.5833 \, m × 100 \, cm/m x = 58.33 \, cm The total extension is 58.33 \, cm. Send me the next one 📸