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3 years ago

The cost of hiring new employees outpaces the raises for established employees is

Engineering

1 answer:

GenaCL600 [577]3 years ago

4 0

Answer:

do you have to make the right decisions to be a person who has been a member who is not the first time a great person who has a job in a day of his life or

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Who can help me with electric systems for cars?

hoa [83]

Answer: i can see if i can what is the problem

Explanation:

7 0

3 years ago

A structural component in the form of a wide plate is to be fabricated from a steel alloy that has a plane-strain fracture tough

jeyben [28]

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 $= 98.9 MPa \sqrt{m}$

Yield strength Y = 860 MPa

Flaw detection apparatus = 3.0mm (12in)

y = 1.0

Let's use the expression:

$oc = \frac{K}{Y \sqrt{pi * a}}$

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

$a= \frac{1}{pi} * [\frac{k}{y*a}]^2$

Substituting figures in the expression above, we have:

$= \frac{1}{pi} * [\frac{98.9 MPa \sqrt{m}} {10 * \frac{860MPa}{2}}]^2$

= 0.0168 m

= 17mm

Therefore, since the flaw 17mm > 3 mm the critical flow is subject to detection

3 0

3 years ago

Read 2 more answers

Which of the following best distinguishes between superficial design improvements and functional

Contact [7]

Answer:

Superficial design improvements are typically only trivial changes to a design, while functional design improvements can change the way a product or process is used to significantly enhance performance.

Explanation:

As a PC board designer, I would sometimes spend a certain amount of time making traces have shorter routes, or fewer layer changes or bends. (I wanted to make the layout "pretty.") In some cases, these changes are superficial, affecting the appearance only. In some cases, they are functional, reducing crosstalk or emissions or susceptibility to interference.

I deal with a web site that seems to be changing all the time (Brainly). In many cases, the same information is rearranged on the page—a superficial change. In other cases, the information being displayed changes, or the way that certain information is accessed changes. These are functional changes. (Sometimes, they "enhance performance," and sometimes they don't, IMO.)

In short ...

<em>Superficial design improvements are typically only trivial changes to a design, while functional design improvements can change the way a product or process is used to significantly enhance performance.</em>

8 0

2 years ago

A water pump delivers 3 hp of shaft power when operating. If the pressure differential between the outlet and the inlet of the p

Natali [406]

Answer:

Mechanical Efficiency = 83.51%

Explanation:

Given Data:

Pressure difference = ΔP=1.2 Psi

Flow rate = $V=8ft^3/s\\$

Power of Pump = 3 hp

Required:

Mechanical Efficiency

Solution:

We will first bring the change the units of given data into SI units.

$P=1.2*6.895 = 8.274KPa\\V=8*0.00283=0.226 m^3/s\\P=3*0.746=2.238KW$

Now we will find the change in energy.

Since it is mentioned in the statement that change in elevation (potential energy) and change in velocity (Kinetic Energy) are negligible.

Thus change in energy is

$=(Mass * change in P)/density\\= \frac{M*P}{p}\\\\$

As we know that Mass = Volume x density

substituting the value

Energy = Volume * density x ΔP / density

Change in energy = Volumetric flow x ΔP

Change in energy = 0.226 x 8.274 = 1.869 KW

Now mechanical efficiency = change in energy / work done by shaft

Efficiency = 1.869 / 2.238

Efficiency = 0.8351 = 83.51%

5 0

3 years ago

Two containers initially contain dry air at 40 °C. Water is added to both containers so that the relative humidity is 30% in con

enot [183]

Answer:

A) Wet bulb temperature of #1 is less than that of #2

Explanation:

This can be gotten from pinpointing the states of the two containers on a psychometric chart.

8 0

3 years ago