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2 years ago

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At his review last year, Lucas was promised a 20 percent raise if he met his production goals. Raises were included in today’s p

aychecks, and although Lucas has met all of his goals, he received only a cost-of-living raise. In the future, Lucas’ ________ will probably be

1 answer:

Murrr4er [49]2 years ago

4 0

Answer: Instrumentality;low

Explanation: Instrumentality is the impact a person have or will be able to render to a given activity or his or her job. The Instrumentality of a person has been found to be proportional to the what outcome of the person's efforts. Especially if the person's explanations are meant.

When a person's expectations are not meant it will cause the person's Instrumentality to be low.

You might be interested in

Explain why failure of this garden hose occurred near its end and why the tear occurred along its length. Use numerical values t

alukav5142 [94]

Answer:

hoop stress
longitudinal stress
material used

all this could led to the failure of the garden hose and the tear along the length

Explanation:

For the flow of water to occur in any equipment, water has to flow from a high pressure to a low pressure. considering the pipe, water is flowing at a constant pressure of 30 psi inside the pipe which is assumed to be higher than the allowable operating pressure of the pipe. but the greatest change in pressure will occur at the end of the hose because at that point the water is trying to leave the hose into the atmosphere, therefore the great change in pressure along the length of the hose closest to the end of the hose will cause a tear there. also the other factors that might lead to the failure of the garden hose includes :

hoop stress ( which acts along the circumference of the pipe):

αh = $\frac{PD}{2T}$ EQUATION 1

and Longitudinal stress ( acting along the length of the pipe )

αl = $\frac{PD}{4T}$ EQUATION 2

where p = water pressure inside the hose

d = diameter of hose, T = thickness of hose

we can as well attribute the failure of the hose to the material used in making the hose .

assume for a thin cylindrical pipe material used to be

$\frac{D}{T}$ ≥ 20

insert this value into equation 1

αh = $\frac{20 *30}{2}$ = 60/2 = 30 psi

the allowable hoop stress was developed by the material which could have also led to the failure of the garden hose

8 0

2 years ago

Poles are values of Laplace transform variable, s, that make denominator of transfer function zero. Zeros are values of Laplace

Ostrovityanka [42]

Answer:

Zero 1 = -1

Zero 2 = -3

Pole 1 = 0

Pole 2 = -2

Pole 3 = -4

Pole 4 = -6

Gain = 4

Explanation:

For any given transfer function, the general form is given as

T.F = k [N(s)] ÷ [D(s)]

where k = gain of the transfer function

N(s) is the numerator polynomial of the transfer function whose roots are the zeros of the transfer function.

D(s) is the denominator polynomial of the transfer function whose roots are the poles of the transfer function.

k [N(s)] = 4s² + 16s + 12 = 4[s² + 4s + 3]

it is evident that

Gain = k = 4

N(s) = (s² + 4s + 3) = (s² + s + 3s + 3)

= s(s + 1) + 3 (s + 1) = (s + 1)(s + 3)

The zeros are -1 and -3

D(s) = s⁴ + 12s³ + 44s² + 48s

= s(s³ + 12s² + 44s + 48)

= s(s + 2)(s + 4)(s + 6)

The roots are then, 0, -2, -4 and -6.

Hope this Helps!!!

3 0

2 years ago

A three-point bending test was performed on an aluminum oxide specimen having a circular cross section of radius 3.5 mm (0.14 in

RideAnS [48]

To resolve this problem we have,

$R=3.5mm\\F_f1=950N\\L_1=50mm\\b=12mm\\L_2=40mm$

$F_{f2}$ is unknown.

With these dates we can calculate the Flexural strenght of the specimen,

$\sigma{fs}=\frac{F_{f1}L}{\pi R^3}\\\sigma{fs}=\frac{(950)(50*10^{-3})}{\pi 3.5*10^{-3}}\\\sigma{fs}=352.65Mpa$

After that, we can calculate the flexural strenght for the square cross section using the previously value.

$\sigma{fs}=\frac{F_{f2}L}{\pi R^3}\\(352.65*10^6)=\frac{3Ff(40*10^{-3})}{2(12*10^{-10})}\\F_{f2}=\frac{352.65*10^6}{34722.22}\\F_{f2}=10156.32N\\F_{f2}=10.2kN$

6 0

2 years ago

The 8-mm-thick bottom of a 220-mm-diameter pan may be made from aluminum (k = 240 W/m ⋅ K) or copper (k = 390 W/m ⋅ K). When use

Artemon [7]

Answer:

For aluminum 110.53 C

For copper 110.32 C

Explanation:

Heat transmission through a plate (considering it as an infinite plate, as in omitting the effects at the borders) follows this equation:

$q = \frac{k * A * (th - tc)}{d}$

Where

q: heat transferred

k: conduction coeficient

A: surface area

th: hot temperature

tc: cold temperature

d: thickness of the plate

Rearranging the terms:

d * q = k * A * (th - tc)

$\frac{d * q}{k * A} = th - tc$

$th = \frac{d * q}{k * A} + tc$

The surface area is:

$A = \frac{\pi * d^2}{4}$

$A = \frac{\pi * 0.22^2}{4} = 0.038 m^2$

If the pan is aluminum:

$th = \frac{0.008 * 600}{240 * 0.038} + 110 = 110.53 C$

If the pan is copper:

$th = \frac{0.008 * 600}{390 * 0.038} + 110 = 110.32 C$

7 0

2 years ago

Methane and oxygen react in the presence of a catalyst to form formaldehyde. In a parallel reaction, methane is oxidized to carb

Nezavi [6.7K]

Answer:

$y_{CH_4}^2=\frac{5mol/s}{100mol/s}=0.05\\y_{O_2}^2=\frac{3mol/s}{100mol/s}=0.03\\y_{H_2O}^2=\frac{47mol/s}{100mol/s}=0.47\\y_{HCHO}^2=\frac{43mol/s}{100mol/s}=0.43\\y_{CO_2}^2=\frac{2mol/s}{100mol/s}=0.02$

Explanation:

Hello,

a. On the attached document, you can see a brief scheme of the process. Thus, to know the degrees of freedom, we state the following unknowns:

- $\xi_1$ and $\xi_2$ : extent of the reactions (2).

- $F_{O_2}^2$ , $F_{CH_4}^2$ , $F_{H_2O}^2$ , $F_{HCHO}^2$ and $F_{CO_2}^2$ : Molar flows at the second stream (5).

On the other hand, we've got the following equations:

- $F_{O_2}^2=50mol/s-\xi_1-2\xi_2$ : oxygen mole balance.

- $F_{CH_4}^2=50mol/s-\xi_1-\xi_2$ : methane mole balance.

- $F_{H_2O}^2=\xi_1+2\xi_2$ : water mole balance.

- $F_{HCHO}^2=\xi_1$ : formaldehyde mole balance.

- $F_{CO_2}^2=\xi_2$ : carbon dioxide mole balance.

Thus, the degrees of freedom are:

$DF=7unknowns-5equations=2$

It means that we need two additional equations or data to solve the problem.

b. Here, the two missing data are given. For the fractional conversion of methane, we define:

$0.900=\frac{\xi_1+\xi_2}{50mol/s}$

And for the fractional yield of formaldehyde we can set it in terms of methane as the reagents are equimolar:

$0.860=\frac{F_{HCHO}^2}{50mol/s}$

In such a way, one realizes that the output formaldehyde's molar flow is:

$F_{HCHO}^2=0.860*50mol/s=43mol/s$

Which is equal to the first reaction extent $\xi_1$ , therefore, one computes the second one from the fractional conversion of methane as:

$\xi_2=0.900*50mol/s-\xi_1\\\xi_2=0.900*50mol/s-43mol/s\\\xi_2=2mol/s$

Now, one computes the rest of the output flows via:

- $F_{O_2}^2=50mol/s-43mol/s-2*2mol/s=3mol/s$

- $F_{CH_4}^2=50mol/s-43mol/s-2mol/s=5mol/s$

- $F_{H_2O}^2=43mol/s+2*2mol/s=47mol/s$

- $F_{HCHO}^2=43mol/s$

- $F_{CO_2}^2=2mol/s$

The total output molar flow is:

$F_{O_2}+F_{CH_4}+F_{H_2O}+F_{HCHO}+F_{CO_2}=100mol/s$

Therefore the output stream composition turns out into:

$y_{CH_4}^2=\frac{5mol/s}{100mol/s}=0.05\\y_{O_2}^2=\frac{3mol/s}{100mol/s}=0.03\\y_{H_2O}^2=\frac{47mol/s}{100mol/s}=0.47\\y_{HCHO}^2=\frac{43mol/s}{100mol/s}=0.43\\y_{CO_2}^2=\frac{2mol/s}{100mol/s}=0.02$

Best regards.

7 0

2 years ago