Complete question:
A structural component in the form of a wide plate is to be fabricated from a steel alloy that has a plane strain fracture toughness of 98.9 MPa root m (90 ksi root in.) and a yield strength of 860 MPa (125,000 psi). The flaw size resolution limit of the flaw detection apparatus is 3.0 mm (0.12 in.). If the design stress is one-half of the yield strength and the value of Y is 1.0, determine whether or not a critical flaw for this plate is subject to detection.
Answer:
Since the flaw 17mm is greater than 3 mm the critical flaw for this plate is subject to detection
so that critical flow is subject to detection
Explanation:
We are given:
Plane strain fracture toughness K 
Yield strength Y = 860 MPa
Flaw detection apparatus = 3.0mm (12in)
y = 1.0
Let's use the expression:
We already know
K= design
a = length of surface creak
Since we are to find the length of surface creak, we will make "a" subject of the formula in the expression above.
Therefore
![a= \frac{1}{pi} * [\frac{k}{y*a}]^2](https://tex.z-dn.net/?f=%20a%3D%20%5Cfrac%7B1%7D%7Bpi%7D%20%2A%20%5B%5Cfrac%7Bk%7D%7By%2Aa%7D%5D%5E2%20)
Substituting figures in the expression above, we have:
![= \frac{1}{pi} * [\frac{98.9 MPa \sqrt{m}} {10 * \frac{860MPa}{2}}]^2](https://tex.z-dn.net/?f=%20%3D%20%5Cfrac%7B1%7D%7Bpi%7D%20%2A%20%5B%5Cfrac%7B98.9%20MPa%20%5Csqrt%7Bm%7D%7D%20%7B10%20%2A%20%5Cfrac%7B860MPa%7D%7B2%7D%7D%5D%5E2)
= 0.0168 m
= 17mm
Therefore, since the flaw 17mm > 3 mm the critical flow is subject to detection
Answer:
Both of them are wrong
Explanation:
The two technicians have given the wrong information about the wires.
This is because firstly, a higher rating of AWG means it is smaller in diameter. Thus, the diameter of a 18 AWG wire is smaller than that of a 12 AWG wire and that makes the assertion of the technician wrong.
Also, the higher the resistance, the smaller the cross sectional area meaning the smaller the diameter. A wire with bigger cross sectional area will have a smaller resistance
So this practically makes the second technician wrong too
Answer:
The outer diameter of the spacers that yields the most economical and safe design is 25.03 mm
Explanation:
For steel bolt
Stress = 210 MPa or 210 N/mm2
Pressure = Stress* Area
Pbolt = 210 N/mm2 * 16^2 *(pi)/4
Pbolt = 210 N/mm2 * 200.96 mm^2 = 42201.6 N
For Brass spacer
Pressure = 42201.6 N
Area of Brass spacer = Pressure/Stress
Area of Brass spacer = 42201.6 N/145 N/mm^2 = 291.044 mm^2
Area of Brass spacer = (pi) (d^2 - 16^2)/4 = 291.044 mm^2
d^2 - 16^2 = 291.044 mm^2* 4/(pi) = 370.758
d^2 = 370.758 + 16^2
d^2 = 626.758
d = 25.03 mm
The outer diameter of the spacers that yields the most economical and safe design is 25.03 mm
Answer:
Farm equipment
Explanation:
Most people have heard claims that the US government pays farmers not to grow crops. The Agricultural Adjustment Act is the legislation that started this program. It was the first “Farm Bill.” The current farm bill provides for the following:
Subsidies for farmers
Insurance for farmers
Price supports
Food assistance for economically challenged Americans (the largest portion of the Farm Bill)
Forestry conservation programs
Alternative energy programs.
Answer:
The metal Harvey should use for reflective coatings in telescope mirror is;
C. Aluminum
Explanation:
In a telescope, the mirror of the telescope is usually made of a temperature-resistant and strong glass, while the reflective coating of the mirrors is made usually from aluminum. A protective coating often composed of silicon dioxide is placed n the top of the reflective coating
Aluminum is used in a telescope, rather than the silver, which is the most reflective material, because silver requires special conditions to be able to work, and silver tarnishes when exposed to air containing sulfur, and silver is susceptible to corrode.
