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Montano1993 [528]
3 years ago
10

Since the passing of the Utah GDL laws in 1999:

Engineering
1 answer:
wlad13 [49]3 years ago
7 0
The answer is b, I hope this helps you
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spetember 7th

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Flank wear data were collected in a series of turning tests using a coated carbide tool on hardened alloy steel at a feed of 0.3
Paladinen [302]

Answer:

A) n =  0.6143, c ≈ 640m/min

B) n = 0.6143 , c = 637.53m/min

Explanation:

using the given data

A) A plot of flank wear as a function of time and also A plot for tool when

Flank wear is 0.75 and cutting edge speed is 100m/min, Time of cutting edge is said to be 20.4 min  also for cutting edge speed of 155m/min , time for cutting edge is 10 min

is attached below

calculate for the constant N from the second plot

note : the slope will be negative because cutting speed decreases as time of cutting increase

V1 = 100m/min , V2 = 155m/min,  T1 = 20.4 min, T2 = 10 min

= - N = \frac{In(V2) - In(V1)}{In(T2)-ln(T1)}

therefore  - N = \frac{5.043 - 4.605}{2.302 -3.015}

                       = - 0.6143

THEREFORE  ( N ) = 0.6143

Determine for the constant C from the second plot as well

note : C is the intercept on the cutting speed axis in 1 min tool life

connecting the two points with a line and extend it to touch the cutting speed axis and measure the value at that point

hence   C ≈ 640m/min

B) Calculate the values of  N and C in the Taylor equation solving simultaneous equations

using the above cutting speed and time of cutting values we can find the constant N via Taylor tool life equation

Taylor tool life equation = vT = C ------------- equation 1

cutting speed = v = 100m/min and 155m/min

tool life = T = 20.4 min and 10 min

also constant  n and c are obtained from the previous plot

back to taylor tool life equation = 100 * 20.4 = C

therefore C = (100)(20.4)^n  ---------------- equation 2

also using the second values of  v and T

taylor tool life equation = 155 * 10 = C

therefore C = ( 155 )(10)^n ----------------- equation 3

Equate equation 2 and equation 3 and solve simultaneously

(100)(20.4)^n = (155)(10)^n

To find N

take natural log of both sides of the equation

= In ((100)(20.4)^n) = In((155)(10)^n)

= In (100) + nIn(20.4) = In(155) + nIn(10)^n

= n(3.0155) - n (2.3026) = 5.043 - 4.605

= 0.7129 n = 0.438

therefore n = 0.6143

To find C

substitute 0.6143 for n in equation 2

C = (100)(20.4) ^ 0.6143

C = 637.53 m/min

Attached are the two plots for solution A

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How does the start function make your programs easier to understand?
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A network address of 172.16.0.0 /12 has been given. Which of the following accurately describes this network? (select one or mor
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Answer:

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In IP version 4 there are four following type of classes

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4)Class D(224-239)

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Generally class A,B,C and D are used.

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In a parallel circuit, as more resistances are added, what happens to the total circuit current?
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Answer:

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Explanation:

hope this helps

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