Answer:
(B) dimensions, tolerances, materials, and finishes of a component.
Explanation:
An engineering drawing :
An engineering drawing is a technical drawing which draws the actual component .
An engineering drawing shows
1. Materials
2.Dimensions
3.Tolerance
4.Finishes of a component
Engineering drawing does not shows any information about the cost of component.
So the option B is correct.
Answer:
Information such as tolerance and scale can be found in the <u>title block</u> of an engineering drawing
Explanation:
The title block of an engineering drawing can normally be found on the lower right and corner of an engineering drawing and it carries the information that are used to specify details that are specific the drawing including, the name of the project, the name of the designer, the name of the client, the sheet number, the drawing tolerance, the scale, the issue date, and other relevant information, required to link the drawing with the actual structure or item
Answer:
Answer for the question:
Use the Hurricanes data and via Multiple regression select the three input variables: Min-pressure, Gender, Category in order to predict the All-Deaths. Pick one of the three variables that has the best P-value and re-do the regression using ONLY this one variable to predict the All-deaths. Answer the questions. (Numerical answers are rounded so choose the answer that matches the best).
is explained in the attachment.
Explanation:
Answer:
the minimum expected elastic modulus is 372.27 Gpa
Explanation:
First we put down the data in the given question;
Volume fraction 
= 0.84
Volume fraction of matrix material 
= 1 - 0.84 = 0.16
Elastic module of particle 
= 682 GPa
Elastic module of matrix material 
= 110 GPa
Now, the minimum expected elastic modulus will be;
= ( × ) / ( + )
so we substitute in our values
= (682 × 110 ) / ( [ 682 × 0.16 ] + [ 110 × 0.84] )
= ( 75,020 ) / ( 109.12 + 92.4 )
= 75,020 / 201.52
= 372.27 Gpa
Therefore, the minimum expected elastic modulus is 372.27 Gpa
Answer:
a) 0.4393 mm
b) -0.141 mm
Explanation:
Cylindrical bar : Diameter = 20.3 mm , Length = 205 mm
Force that deforms bar of metal = 46400 N
elastic modulus = 66.9 GPa
Poisson's ratio ( u ) = 0.32
A) Determine the amount by which this specimen will elongate in the direction of applied stress
dl = 
where P = 46400 N
L = 205 mm
A = 
= 323.65
E = 66.9 GPa
dl = ( 46400 * 205 ) / ( 323.65 * 66.9 * 10^3 ) = 0.4393 mm
B) determine the change in diameter of the specimen
change in diameter( compressed due to elongation in length )
= - u * dl
= - 0.32 * 0.4393
= -0.141 mm
