Business Driven Information Systems
6th Edition
ISBN: 9781260004717
Author: Paige Baltzan
Publisher: MCGRAW-HILL HIGHER EDUCATION
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Chapter 2, Problem 6CBT
To determine
The point of view on Watson’s powerful services by IBM, whether we will have any access to this service in favourite service engine and its use in college career.
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Correct answer and complete fbd only. I will upvote.
A flanged bolt coupling consists of two concentric rows of bolts. The inner row has 6 nos. of 16mm diameterbolts spaced evenly in a circle of 250mm in diameter. The outer row of has 10 nos. of 25 mm diameter bolts spaced evenly in a circle of 500mm in diameter. If the allowable shear stress on one bolt is 60 MPa, determine the torque capacity of the coupling. The Poisson’s ratio of the inner row of bolts is 0.2 while that of the outer row is 0.25 and the bolts are steel, E =200 GPa.
Correct answer and complete fbd only. I will upvote.
10: The constant wall thickness of a steel tube with the cross sectionshown is 2 mm. If a 600-N-m torque is applied to the tube. Use G = 80 GPa forsteel.1. Find the shear stress (MPa) in the wall of the tube.2. Find the angle of twist, in degrees per meter of length.
Execute the code in a C++ program and send a
picture of the implementation from within the
program
> weight
;" :cout > volume
Calculate density //
; density = weight volume
Classify materials based on density //
} if (density 7) {
cout << "The material is Metal" <<
cout << "The material is Wood" <<
; endl
} else {
; endl
{
; return 0
{
Chapter 2 Solutions
Business Driven Information Systems
Ch. 2 - Prob. 1OCQCh. 2 - Prob. 2OCQCh. 2 - Prob. 3OCQCh. 2 - Prob. 4OCQCh. 2 - Prob. 5OCQCh. 2 - Prob. 6OCQCh. 2 - Prob. 1RQCh. 2 - Prob. 2RQCh. 2 - Prob. 3RQCh. 2 - Prob. 4RQ
Ch. 2 - Prob. 5RQCh. 2 - Prob. 6RQCh. 2 - Prob. 7RQCh. 2 - Prob. 8RQCh. 2 - Prob. 9RQCh. 2 - Prob. 10RQCh. 2 - Prob. 11RQCh. 2 - Prob. 12RQCh. 2 - Prob. 13RQCh. 2 - Prob. 14RQCh. 2 - Prob. 15RQCh. 2 - Prob. 16RQCh. 2 - Prob. 17RQCh. 2 - Prob. 1CCOCh. 2 - Prob. 2CCOCh. 2 - Prob. 3CCOCh. 2 - Prob. 4CCOCh. 2 - Prob. 5CCOCh. 2 - Prob. 6CCOCh. 2 - Prob. 1CCTCh. 2 - Prob. 2CCTCh. 2 - Prob. 3CCTCh. 2 - Prob. 1CBTCh. 2 - Prob. 2CBTCh. 2 - Prob. 3CBTCh. 2 - Prob. 4CBTCh. 2 - Prob. 5CBTCh. 2 - Prob. 6CBTCh. 2 - Prob. 7CBTCh. 2 - Prob. 8CBTCh. 2 - Prob. PIAYKBPCh. 2 - Prob. PIIAYKBPCh. 2 - Prob. PIIIAYKBPCh. 2 - Prob. PIVAYKBPCh. 2 - Prob. PVAYKBPCh. 2 - Prob. PVIAYKBPCh. 2 - Prob. PVIIAYKBPCh. 2 - Prob. PVIIIAYKBPCh. 2 - Prob. PIXAYKBPCh. 2 - Prob. PXAYKBPCh. 2 - Prob. PXIAYKBP
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- Execute the code in MATLAB and send me a picture of the implementation from within the program MATLAB code to analyze material shape % using computer vision image = imread('material_image.jpg'); % Read the image gray Image = rgb2gray(image); % Convert the image to grayscale BW = imbinarize (gray Image); % Convert the image to binary (black and white) Extract geometric properties (e.g., area % and bounding box) stats = regionprops (BW, 'Area', 'BoundingBox') Classify material based on shape % if stats. Area > 500 ; 'material = 'Plastic else ; 'material = 'Wood end ; disp(['The material is: ', material])arrow_forwardCORRECT ANSWER WITH COMPLETE FBD ONLY. I WILL UPVOTE. A torque wrench is used to tighten the pipe shown.Dimensions: S1 = 400 mm; S2 = 250 mm; S3 = 100 mmModulus of Rigidity G = 78 GPa1. The diameter of the solid pipe is 20 mm. How much is themaximum force P (N) that can be applied based on theallowable shear stress of 60 MPa?2. For a hollow pipe with 50 mm outside diameter and is 6 mmthick, compute for the maximum force P (kN) that can beapplied such that the angle of twist at A does not exceed 5degrees.3. The torque applied to tighten the hollow pipe is 200 N-m.Given: Pipe outside diameter = 50 mm Pipe thickness = 6 mmSolve for the resulting maximum shear stress (MPa) in the pipe.arrow_forwardUsing D flip-flops, design a synchronous counter. The counter counts in the sequence 1,3,5,7, 1,7,5,3,1,3,5,7,.... when its enable input x is equal to 1; otherwise, the counter count 0. Present state Next state x=0 Next state x=1 Output SO 52 S1 1 S1 54 53 3 52 53 S2 56 51 0 $5 5 54 S4 53 0 55 58 57 7 56 56 55 0 57 S10 59 1 58 58 S7 0 59 S12 S11 7 $10 $10 59 0 $11 $14 $13 5 $12 S12 $11 0 513 $15 SO 3 S14 $14 S13 0 $15 515 SO 0 Explain how to get the table step by step with drawing the state diagram and finding the Karnaugh map.arrow_forward
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