Which of the following is the unit for electrical conductivity?Select one:O (Q-m) -1O (Q-mm)O m/SO Q-mIn a stress strain curve of mild steel, the elastic limit will be before the limit of proporype here to search99+hpfs 米f5f6f71-f8fg&7V36.8.9/64+%24%23

Answered: Which of the following is the unit for…

Engineering

Mechanical Engineering

Which of the following is the unit for electrical conductivity? Select one: O (2-m)-1 O (2-mm) O m/S O Q-m

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

See similar textbooks

Similar questions

. Data was collected from the tensile test demonstration for Aluminum 6061-T6. Below are the resulting stress-strain diagrams based on the tensile test we performed. o (MPa) Al 6061-T6 Stress-Strain Curve o (MPa) 350 320 300 250 200 150 100 50 0 Estimate 0 0.05 0.1 0.15 ε (mm/mm) 300 270 250 220 200 150 140 100 a) The modulus of Elasticity, E b) The yield stress, d, using the 0.2% offset method c) The strain at failure, & failure d) The modulus of resilience (approximately!) 50 0 0 Al6061-T6 Stress-Strain Curve, Magnified N 0.0032 0.002 0.004 0.006 0.008 € (mm/mm) The right diagram shows the magnified elastic region of stress-strain diagram to enhance the details.
4. We want to draw a 7.5 mm diameter copper wire having a yield strength of 140 MN.m²² into a 6 mm diameter wire. Will the drawn wire break during the drawing process? Strength (MN.m-2) 600 500 400 300 200 100 0 0 Tensile strength 20 Yield strength % Elongation 60 40 Percent cold work 80 100 90 80 70 60 50 40 30 20 10 100 % Elongation
Determine the percentage of ductility of a metal alloy having the following tensile stress-strain diagram. 600 500E 400 500 300 400 300 200 100 100 0.000 0.002 0.004 0.006 Strain 0.00 0.04 0.08 0.12 0.16 0.20 Strain Select one: O 0.6% O 19% O 4% 300% O 10% Stress (MPa) O O O O Stress (MPa)
Review Data taken from a stress-strain test for a ceramic are No elements selected o(ksi) given in the table. The curve is linear between the origin and the first point. 50 40 30 Figure 1 of 1 20 10 σ (ksi) e (in./in.) ex 10-*(in./in.) 0.5 1.0 1.5 2.0 25 33.2 0.0006 45.5 0.0010 Press (ENTER to select this element. Press (ESc to return to the main menu. Press (CTRL+Q) to quit the application. 49.4 0.0014 51.5 0.0018 53.4 0.0022
When stress is applied to a liquid over an application time tp the liquid needs a relaxation time tc to accordingly deform. Relaxation time tc is a physical property of liquids. The Deborah number is De= tdt, . The lower the Deborah number of a stress applied to a material the more.. : Select one: O The material behaves like a solid O The material behaves like visco-elastic O The material behaves like a liquid
Q1- The engineering tensile stress-strain data were obtained for AA6061-T6 aluminum alloy shown: note stress unit is (MPa). 314 294 266 226 287 325 329 337 334 a.(stress) 0 52 159 E.(strain) 0 1.07 1.70 2.23 3.07 4.80 7.13 9.13 11.33 12.97 14.40 16.10 Plot the stress-strain curves then determine the mechanical properties as followings, ay, ou, elastic zone, plastic zone, Young modulus, fracture stress and ductility if you know L - 30 mm, and Lf - 33.7 mm.
Ql- mm ,length 40 mm. The final diameter 18 mm and length 45 mm. Draw the force-extension diagram and determine the following : a) Ultimate tensile strength, c) Modulus of elasticity, d) Percent elongation and e) Percent reduction in area. The following data collected from tensile test of a metal bar diameter 20 b) The fracture strength, Force - kN 75 90 | 105 120 131 125 110 25 50 4.7 Extension-mm 0.02 0.04 0.06 0.20 - 0.60 1.60 3.00 4.0 Fracture
Determine the percentage of ductility of a metal alloy having the following tensile stress-strain diagram. Stress-Strain Diagram 1200 1000 800 1000 600 800 600 400 400 200 200 0.005 0.01 0015 0.02 0.025 003 0.1 0.0s 0.15 0.2 025 Strain (mm/mm) Select one: O 0.2% O 19% O 3% 25% O 0.5% Stress (Mpa)
The following data were obtained during a tension test of an aluminum alloy. The initial diameter of the test specimen was 0.505 in. and the gage length was 2.0 in. Load (lb) -O 2 310 4 640 6 950 9 290 11 600 12 600 Elongation (in.) O 0.00220 0.00440 0.00660 0.00880 0.0110 0.0150 Load (lb) 14 000 14 400 14 500 14 600 14 800 14 600 13 600 Elongation (in.) 0.020 0.025 0.060 0.080 0.100 0.120 Fracture determine the Plot the stress-strain diagram and (a) proportional following mechanical properties: limit; (b) modulus of elasticity; (c) yield point; yield strength at 0.2% offset; and (f) rupture strength. (d) (e) ultimate strength;
For the stress-strain curves below, which statements are correct? Stress (MPa) 60 50 40 30 20 10 0 a. 0 b. O c. e. f. A 1 2 The mechanical behavior shown in C is that of an elastomer with low elastic modulus Polymer B and C have an elastic behavior whereas A has a plastic behavior While the deformation in polymer A is fully reversible until breakage, polymer B and C only display reversible deformation until 0.1X and 5x elongations respectively Od. The polymer with brittle nature is the one that possesses the highest elastic modulus The mechanical behavior shown in B is that of an elastomer with low elasticity The polymer with the necking behavior has the highest strength B с 3 4 5 6 7 8 Strain
Determine the tensile yield strength (0.2% offset) and the maximum strength of a metal alloy having the following tensile stress-strain diagram. 600 500 400 500 300 400 300E 200 3200 100 100E 0.000 0.002 0.004 0.006 Strain ol 0.00 0.16 0.20 0.08 Strain 0.04 0.12 Select one: O The tensile yield strength Sy = 290 Mpa and the maximum strength Smax = 500 Mpa. O The tensile yield strength Sy = 170 Mpa and the maximum strength Smax = 400 Mca. G The tensile yield strength Sy = 200 Mpa and the maximum strength Smax = 500 Moa. O Tne tensile yield strength Sy = 390 Mpa and the maximum strength Smax= 500 Mpa. The tensile yed strength Sy = 150 Mpa and the maximum strength Smax = 250 Mpa. Stress (MPa) Stress (MPa)
A polypropylene has the following tensile creep compliance measured at 40 °C: D(t) = 1.2 x t0.1 GPa-1 where t is expressed in seconds. It is subjected to the following tensile stress sequence at 40 °C. Find the tensile strain at 3200 sec. Stress (MPa) 1.5 1.0 0.5 1000 2000 3000 Time (s)