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 problems: Group work: Write True or False if not 1. Deformation cannot be change size or shape but tensile deformation is changing of length. False. Deformation is* a change in size or shape. Tensile deformation is a specific type of deformation that involves a change in length. 2. Plastic and elastic deformations are the same. False. Elastic deformation is temporary and reversible, while plastic deformation* is permanent. 3. Stress and strain are force over area. False. Stress is force over area, but strain* is the ratio of change in length to original length, making it a dimensionless quantity. Work out 4. A force of 200 \, N acts over area, and stress formed by the force is 4000 \, Pa. Find Area? Step 1: Recall the formula for stress. = (F)/(A) Where is stress, F is force, and A is area. Step 2: Rearrange the formula to solve for area and substitute the given values. A = (F)/() = 200 \, N4000 \, Pa Step 3: Calculate the area. A = (200)/(4000) \, m^2 = 0.05 \, m^2 The area is 0.05 m^2. 5. A force of 100 \, N acts over 0.05 \, m^2 cross section area of a wire of length 5 \, m, if wire increase by 5 \, cm. Find (a) Stress (b) Strain a) Stress Step 1: Recall the formula for stress and substitute the given values. = (F)/(A) = 100 \, N0.05 \, m^2 Step 2: Calculate the stress. = 2000 \, N/m^2 = 2000 \, Pa The stress is 2000 Pa. b) Strain Step 1: Convert the increase in length from centimeters to meters. L = 5 \, cm = 0.05 \, m Step 2: Recall the formula for strain and substitute the values. = ( L)/(L_0) = 0.05 \, m5 \, m Step 3: Calculate the strain. = 0.01 The strain is 0.01. Send me the next one 📸