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Question 6: In the following figure, the magnification of this low carbon steel image is 500x....
25 Sub Question 5: In the following figure, the magnification of this image is 100x. The circle in the photograph has an inscribed circle with a diameter of 3.14 in with an enclosed area of 7.75 in? The grain size number obtained for this photomicrograph is 3.89. Can you confirm this result?
Problem 6: Link 2, shown in the figure, is 25 mm wide, has 12-mm diameter bearings at the ends, and is cut from low-carbon steel bar stock having a minimum yield strength of 165 MPa. The end- condition constants are C-1 and C-1.2 for buckling in and out of the plane of the drawing, respectively. a. Using a design factor na 4, find a suitable thickness for the link. b. Are the bearing stresses at O and B of any...
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HEAT TREATMENT OF STEELS EXPERIMENT 6 EXPERIMENT 6 HEAT TREATMENT OF STEELS THEORY The Effect of Cooling Rate One of the most convenient methods for controlling the properties of a given steel, i.e., a steel whose composition is already fixed, consists of austenizing the steel and ten cooling to room temperature at some predetermined rate. A variation of cooling rates...
Question I.5 Figure 1.5 shows a frame with loads at A and D. Select the closest value for the magnitude of the total reaction at B. Assume the weight of the frame is zero. 40 kN VE 96.2 kN (a) (Ь -40 kN 5 m (c) -87.5 kN 30 kN (d) 57 kN 4m ao 1 m (e) 50 kN Figure L.5 Low mass frame Question I.6 In the shear and bending moment equations for beams, which of the following...
summatize the following info and break them into differeng key points. write them in yojr own words
apartus
6.1 Introduction—The design of a successful hot box appa- ratus is influenced by many factors. Before beginning the design of an apparatus meeting this standard, the designer shall review the discussion on the limitations and accuracy, Section 13, discussions of the energy flows in a hot box, Annex A2, the metering box wall loss flow, Annex A3, and flanking loss, Annex...
Consider a cylindrical capacitor like that shown in Fig. 24.6. Let d = rb − ra be the spacing between the inner and outer conductors. (a) Let the radii of the two conductors be only slightly different, so that d << ra. Show that the result derived in Example 24.4 (Section 24.1) for the capacitance of a cylindrical capacitor then reduces to Eq. (24.2), the equation for the capacitance of a parallel-plate capacitor, with A being the surface area of...