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mr Goodwill [35]
4 years ago
8

Matt just finished his first conversation with an international student from Rwanda. He found that things went smoothly without

his having to think about what he was saying even though he was a bit nervous about the interaction before it happened. Matt's experience illustrates _____. Group of answer choices
Engineering
1 answer:
ira [324]4 years ago
7 0

Answer:

Matt's experience illustrates

Unconscious competence

Explanation:

  • Competence is the ability of an individual or group of individuals to perform a task efficiently.
  • There are four stages of competence which are: unconscious incompetence, unconscious incompetence, conscious competence and unconscious competence.
  • Unconscious Competence is the last stage of competence in which a person has attain such competence in something that it has become his or her habit or characteristic. For example, if a person has achieved unconscious competence in a particular work then this work will become his or her second nature.
  • Similarly, Matt has experience illustrates us the unconscious competence as his interaction went smooth unintentionally.  

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What is the product of current and Resistance A power B voltage inductance D conductance
Sveta_85 [38]

Answer:

B. Voltage

Explanation:

Ohm's law: current times resistance equals voltage.

V = IR

5 0
3 years ago
A rectangular beam having b=300 mm and d=575 mm, spans 5.5 m face to face of simple supports. It is reinforced for flexure with
katovenus [111]

Answer:

provide 180 mm spacing

Explanation:

GIVEN DATA

rectangular beam: (b) = 300 mm, (d) = 575 mm

reinforced for flexure = 4Ф32 bars

WD = 30 kN /m,  WL = 45 kN/m

Wu = 1.4 * 30 + 1.6 * 45 = 114 kN/m

i) concrete shear stress ( vc)

100 Ac / bd = (100 * u * \frac{\pi }{4} *  32^2) / 300 * 575 = 1.865

from table 3.8

when:  100 Ac / bd = 1.865  then Vc = 0.778 N/mm^2

Ultimate shear force = (114 *5.5) / 2 = 313.5 kN

design shear stress =  V / bd = (313.5 * 10^3) / (300 * 575) = 1.82 N/mm^2

v < 0.8\sqrt{22}   =      1.82 < 3.75

design link provided  according to

Asv / sv =  b(v-vc) / 0.87 fy  = 300(1.82 - 0.778) / 0.87 (420)

ASv / Sv = 0.855

From table 3.13 :the value of Asv / sv can be calculated as

\frac{0.855 - 0.785}{0.897 - 0.785}  = \frac{x - 200}{175 - 200}

x = (-25) [ 0.625] + 200 = 184.375 mm

provide 180 mm spacing

8 0
4 years ago
Suppose we are managing a consulting team of expert computer hackers, and each week we have to choose a job for them to undertak
lina2011 [118]

Answer:

if number == 1

  then

  tempSolution= max(l[number],h[number])

else if number == 2 then

  tempSolution= max(optimalPlan(1, l, h)+ l[2], h[2])

else

  tempSolution= max(optimalPlan(number − 1, l, h) + l[number], optimalPlan(number − 2, l, h) + h[number])

end if

return Value

FindOptimalValue(number, l, h)

for itterator = 1 ! number do

  tempSolution[itterator] = 0

end for

for itterator = 1 ! number do

  if itterator == 1 then

      tempSolution[itterator] max(l[itterator], h[itterator])

  else if itterator == 2 then

      tempSolution[itterator] max(tempSolution[1] + l[2], h[2])

  else

      tempSolution[itterator] max(tempSolution[itterator − 1] + l[itterator], tempSolution[itterator − 2] + h[itterator])

  end if

end for

return Value[number]

OPtimalPlan(number, l, h, Value)

for itterator = 1 ! number do

  WeekVal[itterator]

end for

if tempSolution[number] − l[number] = tempSolution[number − 1] then

  WeekVal[number] ”Low stress”

  OPtimalPlan(number-1, l, h, Value)

else

  WeekVal[number] ”High stress”

  OPtimalPlan(number-2, l, h, Value)

end if

return WeekVal

7 0
3 years ago
Routers cannot be used to cut through material.<br><br> True<br> False
mario62 [17]

Answer:

Yes a router can be used to cut right through wood and sometimes it makes sense to do so. It leaves nice clean edges, can cut sharp curves and can follow a template

Explanation:

hope thats right

7 0
3 years ago
Read 2 more answers
Air flows at 45m/s through a right angle pipe bend with a constant diameter of 2cm. What is the overall force required to keep t
HACTEHA [7]

Answer:

b)1.08 N

Explanation:

Given that

velocity of air V= 45 m/s

Diameter of pipe = 2 cm

Force exerted by fluid  F

F=\rho AV^2

So force exerted in x-direction

F_x=\rho AV^2

F_x=1.2\times \dfrac{\pi}{4}\times 0.02^2\times 45^2

F=0.763 N

So force exerted in y-direction

F_y=\rho AV^2

F_y=1.2\times \dfrac{\pi}{4}\times 0.02^2\times 45^2

F=0.763 N

So the resultant force R

R=\sqrt{F_x^2+F_y^2}

R=\sqrt{0.763^2+0.763^2}

R=1.079

So the force required to hold the pipe is 1.08 N.

3 0
4 years ago
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