Question 1 of 3View PoliciesCurrent Attempt in ProgressCompound axial member ABC has a uniform diameter of d = 2.4 in. Segment (1) is an aluminum [E₁ = 10,000 ksi] alloy and segment (2)is a copper [E₂ = 17,000 ksi] alloy. The lengths of segments (1) and (2) are L₁ = 78 in. and L2 = 150 in., respectively. Determine the forceP required to stretch compound member ABC by a total of 0.35 in.APart 1292892374_46143....jpgAluminumL₁Answer:A = iBCalculate the cross-sectional area of the compound member.Save for LaterL2(2)CopperADin.2291131836_34905....png^Attempts: 0 of 1 used291761570 73033....png|^!!PsSubmit Answer292042643_58410....pngD4285767222_54900....pngENGUSSShow14/07/20 Part 1Calculate the cross-sectional area of the compound member.Answer:92374_46143....jpgA =iSave for Laterin.2Part 2The parts of this question must be completed in order. This part will be available when you complete the part above.Part 3The parts of this question must be completed in order. This part will be available when you complete the part above.^Attempts: 0 of 1 used291131836_34905....png291761570_73033....png^Submit Answer292042643_58410....pngA285767222 54900....png

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Question 1 of 3 View Policies Current Attempt in Progress Compound axial member ABC has a uniform diameter of d = 2.4 in. Segment (1) is an aluminum [E₁ = 10,000 ksi] alloy and segment (2) is a copper [E₂ = 17,000 ksi] alloy. The lengths of segments (1) and (2) are L₁= 78 in. and L₂ = 150 in., respectively. Determine the force - P required to stretch compound member ABC by a total of 0.35 in. L₁ Part 1 292892374_46143....jpg M Answer: A= Aluminum (1) ^ Calculate the cross-sectional area of the compound member. Save for Later O B L2 (2) Copper in.2 291131836_34905....png 13 291761570_73033....png Attempts: 0 of 1 used Submit Answer ^ D 292042643_58410....png 285767222_54900....png 0 ENG US ^ $4 Sho 14/07

Question 1 of 3 View Policies Current Attempt in Progress Compound axial member ABC has a uniform diameter of d = 2.4 in. Segment (1) is an aluminum [E₁ = 10,000 ksi] alloy and segment (2) is a copper [E₂ = 17,000 ksi] alloy. The lengths of segments (1) and (2) are L₁= 78 in. and L₂ = 150 in., respectively. Determine the force - P required to stretch compound member ABC by a total of 0.35 in. L₁ Part 1 292892374_46143....jpg M Answer: A= Aluminum (1) ^ Calculate the cross-sectional area of the compound member. Save for Later O B L2 (2) Copper in.2 291131836_34905....png 13 291761570_73033....png Attempts: 0 of 1 used Submit Answer ^ D 292042643_58410....png 285767222_54900....png 0 ENG US ^ $4 Sho 14/07

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Work and Heat Transfer

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The 200 x 200 x 1,250-mm oak [E-12 GPa] block (2) shown is reinforced by bolting two 5 x 200 x 1,250 mm steel [E - 200 GPa] plates (1) to opposite sides of the block. A concentrated load of 300 kN is applied to a rigid cap. Assume P-300 kN, L-1.25 m. Determine (a) the normal stresses in the steel plates (1) and the oak block (2). (b) the shortening of the block when the load is applied. B (1) Answers: A₁- A₂- i Part 2 Calculate the cross-sectional area of each steel plate, A₁, and the cross-sectional area of the oak block, Az. i Save for Later P CL (2) C Answer: F₂/F₁- CL Save for Later CL (1) mm² mm² Derive a compatibility equation using force-deformation relationships substituted into a geometry-of-deformation relationship. Express the compatibility equation here as the ratio of the force in the oak block, F₂, to the force in each steel plate, F₁. Attempts: 0 of 1 used Submit Answer Attempts: 0 of 1 used Submit Answer
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Part A Cables of negligible weight support the loading shown. (Figure 1) If W1 = 5.00 lb , W2 = 71.0 lb , yB = 4.50 ft , yc = 7.50 ft , YD = 2.50 ft , and rc = 2.00 ft , find æg- Express your answer numerically to three significant figures in feet. • View Available Hint(s) ΠV ΑΣφ It vec IB = ft Submit