PLEASE ANSWER QUICKLY Q8. The performance of an element can be written as p = fi(F).f:(G),f³(M).F, G and Min the performance equation representI. FunctionalRequirementII. GeometricШ. MaterialsIV. Materials IndexParametersPropertiesA. I, II and IIIВ. 1, and IVС. П, and IVD. I, II and IV

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The performance of an element can be written as p = fi(F).f:(G)fs(M). F, G and M in the performance equation represent I. Functional II. Geometric II. Materials IV. Materials Index Requirement Parameters Properties A. I, II and II B. I, III and IV С. П, I and IV D. I, II and IV

The performance of an element can be written as p = fi(F).f:(G)fs(M). F, G and M in the performance equation represent I. Functional II. Geometric II. Materials IV. Materials Index Requirement Parameters Properties A. I, II and II B. I, III and IV С. П, I and IV D. I, II and IV

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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Q8. The performance of an element can be written as p = fi(F).f2(G).f3(M). F, G and M in the performance equation represent I. Functional Requirement II. Geometric III. Materials IV. Materials Index Parameters Properties A. I, II and III В. I, I and IV С. П, Ш and IV D. I, II and IV
Temperature (°C) 1600 0 1500 1400 1300 1200 1100 1000 1085°C 0 (Cu) 1 20 20 Alloy Brass Steel Composition (at% Ni) 40 Liquid Liquidus line Aluminum Titanium B @x+L 60 CE 40 60 Composition (wt% Ni) 80 Yield Strength [MPa (ksi)] 345 (50) 690 (100) 275 (40) 480 (70) Solidus line 1453°C 80 100 2800 2600 Problem 5 A structural member 250 mm (10 in.) long must be able to support a load of 44,400 N (10,000 lb,) without experiencing any plas- tic deformation. Given the following data for brass, steel, aluminum, and titanium, rank them from least to greatest weight in accordance with these criteria. 2400 8.5 7.9 2.7 4.5 2200 2000 100 (Ni) Density (g/cm³) Temperature (°F)
CALCULATE THE FOLLOWING: We performed the experiment to measure MATERIAL 1 - BRASS the thermal conductivity of 2 materials (Brass & Steel) in the laboratory and measured the following tabulated values: Calculation for Brass Quantities Values Material 1- BRASS Calculated Power (Q') W (Diameter = 25mm) Area of cross section (A) m2 Power Difference in Temperature Temperature (°C) between two points (AT) °C Difference in distance between two points (Ax) m Q' 1 2 3 (W) Thermal conductivity of brass (kp) W/m°C 4 5 6 8 9. 14.65 79 77.4 76 50.1 46.5 42.4 36 34.5 33.6 MATERIAL 2 - STEEL Calculation for Steel Material 2 - STEEL %3D Quantities Values Calculated (Diameter = 25mm) %3D Power (Q') W Power Area of cross section (A) m2 Temperature (°C) Difference in Temperature between two points (AT) °C Q' 1 |(W) Difference in distance between two points (Ax) m 2 3 7 8 14.2 88.6 87.5 85 33.9 33.6 32.4 Thermal conductivity of steel (kg) W/m°C
A 30 wt% Ni-70 wt% Cu alloy (Animated Figure 9.3a) is slowly cooled from 1400°C (2550°F) to 1150°C (2100°F). (a) At what temperature does the first solid phase form? i ! °C (b) What is the composition of this solid phase? i %wt Ni (c) At what temperature does the last of the liquid solidify? i °℃ (d) What is the composition of this last remaining liquid phase? i %wt Ni
E and 5 v X O file:///C:/Users/Hp/Desktop/mm/OUTCOME%20NO.2%20and%205%20with%20Problems.pdf + O Fit to page D Page view A Read aloud 1 Add notes Problems 1. A brass rod of diameter 25 mm and length 250 mm is subjected to a tensile load of 50 kN and the extension of the rod is equal to 0.3 mm. Find the Young's Modulus or Modulus of Elasticity. 99+ to search hp delete prt sc 14 DI backs & 7 8 24 4 P EJR -0 J K H D. F
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