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anyanavicka [17]
3 years ago
6

During a shrinkage limit test, a 19.3 cm3 saturated clay sample with a mass of 37 g was placed in a porcelain dish and dried in

the oven. The oven-dried sample had a mass of 28 g with a final volume of 16 cm3 . Determine the shrinkage limit and the shrinkage ratio.
Engineering
1 answer:
alexira [117]3 years ago
4 0

Answer:

shrinkae limit = 20.35%

shrinkage ratio = 1.45

Explanation:

1. to get the shrinkage limit we would first calculate the moisture content w.

w = (37-28)/28

= 9/28

= 0.3214

then the formula for shrinkage limit is

[w-\frac{(V-Vd}{wd} ]*100

w = 0.3214

V = 19.3

Vd = 16

Wd = 28

when we put these values into the formula:

[0.3214-\frac{(19.3-16)}{28} ]*100\\

= 20.35%

2. the shrinkage limit = Wd/V

= 28/19.3

= 1.45

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With reference to the NSPE Code of Ethics, which one of the following statements is true regarding the ethical obligations of th
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Answer: c. The VW engineers involved were ethically obligated to hold paramount the health, welfare and safety of the public even if their supervisors directed them to implement software and hardware that enabled cheating on the emissions testing software.

Explanation: The National Society of professional Engineers, NSPE define the code of ethics which must guide engineers in their duty. These codes act as principles of personal conduct, towards the public and their employers.  

One of the areas covered by these codes is overriding importance of the safety and health of the public to any other factor. In addition, engineers are to avoid deception and maintain the reputation of their profession. These cannot be sacrificed for the financial gain of their employers or explained away by saying they are following the direction of their employers. While they have certain responsibilities to their employers, the health welfare and safety of the public is more important.  

4 0
3 years ago
D
TEA [102]

Answer:

true

Explanation:

4 0
3 years ago
Two steel plates are to be held together by means of 16-mm-diameter high-strength steel bolts fitting snugly inside cylindrical
dusya [7]

Answer:

The outer diameter of the spacers that yields the most economical and safe design is 25.03 mm

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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

5 0
3 years ago
Problem 4.079 SI A rigid tank whose volume is 3 m3, initially containing air at 1 bar, 295 K, is connected by a valve to a large
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Answer:

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State 1

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u_1 = 210.49kJ/kg\\h_i = 295.17kJkg

State 2

P_2 =6bar\\T_2 = 296K\\T_f = 320K

For tables we know at T=320K

u_2 = 228.42kJ/kg

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m_1 = \frac{p_1V}{RT_1}

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Q_{cv} = -1007.86kJ

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Lorico [155]

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