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Slav-nsk [51]
1 year ago
10

Summarize key

Engineering
1 answer:
BlackZzzverrR [31]1 year ago
7 0

Answer:

what are is ethiopia cultural ?

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Where’s the silicone rubber are made ? In a machine, vulcanization… explain please !!!!! I need help
Vaselesa [24]

Answer: This is done by heating a large volume of quartz sand to temperatures as high as 1800˚C. The result is pure, isolated silicon, which is allowed to cool and then ground into a fine powder. To make silicone, this fine silicon powder is combined with methyl chloride and heated once again.

Explanation:

4 0
3 years ago
Which of the following are not related to a materials structure? (Mark all that apply) a)- Atomic bonding b)- Crystal structure
belka [17]

The answer is c) atomic number

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3 years ago
1. For ball bearings, determine: (a) The factor by which the catalog rating (C10) must be increased, if the life of a bearing un
ElenaW [278]

Answer:

(b) Given the Weibull parameters of example 11-3, the factor by which the catalog rating must be increased if the reliability is to be increased from 0.9 to 0.99.

Equation 11-1: F*L^(1/3) = Constant

Weibull parameters of example 11-3: xo = 0.02 (theta-xo) = 4.439 b = 1.483

Explanation:

(a)The Catalog rating(C)

   Bearing life:L_1 = L , L_2 = 2L

   Catalog rating: C_1 = C , C_2 = ? ,

From given equation bearing life equation,

F\times\frac{1}{3} (L_1)  = C_1   ...(1)   \\\\    F\times\frac{1}{3} (L_2)  =C_2...(2)

we Dividing eqn (2) with (1)

\frac{C_2}{C_1} =\frac{1}{3}  (\frac{L_2}{L_1})\\\\ C_2 = C*(\frac{2L}{L})\frac{1}{3} \\\\   C_2 = 1.26 C

The Catalog rating increased by factor of 1.26

(b) Reliability Increase from 0.9 to 0.99

R_1 = 0.9 , R_2 = 0.99

Now calculating life adjustment factor for both value of reliability from Weibull parametres

a_1 = x_o + (\theta - x_o){ ln(\frac{1}{R_1} ) }^{\frac{1}{b}}

= 0.02 + 4.439{ ln(\frac{1}{0.9} ) }^{\frac{1}{1.483}} \\\\     = 0.02 + 4.439( 0.1044 )^{0.67}\\\\a_1 = 0.9968

Similarly

a_2 = x_o + (\theta - x_o){ ln(\frac{1}{R_2} ) }^{\frac{1}{b} }\\\\   = 0.02 + 4.439{ ln(1/0.99) }^{\frac{1}{1.483} }\\\\     = 0.02 + 4.439( 0.0099 )^{0.67}\\\\a_2 = 0.2215

Now calculating bearing life for each value

L_1 = a_1 * LL_1 = 0.9968LL_2 = a_2 * LL_2 = 0.2215L

Now using given ball bearing life equation and dividing each other similar to previous problem

\frac{C_2}{C_1}  = (\frac{L_2}{L_1} )^{\frac{1}{3} }\\\\   C_2 = C* (\frac{0.2215L }{0.9968L}  )^{1/3}\\\\   C_2 = 0.61 C

Catalog rating increased by factor of 0.61

6 0
3 years ago
In order to give a gradual increase in section modulus, frame reinforcement plates must be:Group of answer choicesrectangular.cu
Nataliya [291]

Answer:

Technician A says that when fifth wheel brackets are bolted to frame rails, the section modulus over that section of the frame is increased.

Explanation:

hope i helped

6 0
3 years ago
Trevor typically works on the highways doing maintenance. Who is most likely his employer? a large private company himself the g
Naddika [18.5K]

Answer: The Government.

Explanation: Highways are usually a roads where people travel on. They are usually owned mainly by the government some are owned by private entities. Of the above options the most possible employer of Trevor is the government as majority of the highways are usually owned by the government.

The government provide several services on the highways such as roads maintenance, security, accidents and emergency cares etc

3 0
3 years ago
Read 2 more answers
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