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tensa zangetsu [6.8K]
3 years ago
5

In the engineering design process, testing is the least important step. True False

Engineering
2 answers:
Wittaler [7]3 years ago
7 0

Answer:

false

Explanation:

Romashka-Z-Leto [24]3 years ago
4 0

i think the answer is false

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Use the drop-down menus to complete the statements about using OneNote in Outlook meeting requests.
romanna [79]

Answer:

OneNote can be used to share  

✔ information

before the meeting.

OneNote integrates seamlessly with Outlook by placing a  

✔ link

in the meeting request.

OneNote can also be used to pass information along both  

✔ during

and after the meeting.

8 0
2 years ago
A cantilever beam AB of length L has fixed support at A and spring support at B.
atroni [7]

The force in the spring will be F =\dfrac{KPl^3}{3EI}.

The deflection of the beam will be \rho = 0.15(\dfrac{KPL^3}{3EI})

<h3>What is a cantilever beam?</h3>

A rigid, horizontally extending structural member known as a cantilever is supported at only one end. Typically, it extends from a solidly affixed flat vertical surface, such as a wall.

Given that:-

  • A cantilever beam AB of length L has fixed support at A and spring support at B.
  • The spring behaves in a linearly elastic manner with stiffness k. If a concentrated load P is applied at B.

The spring force will be calculated as:-

F = kx

Deflection will be given by:-

x = \dfrac{PL^3}{3EI}

The spring force will be calculated by:-

F = \dfrac{KPL^3}{3EI}

The deflection of the beam will be given as:-

\rho = \dfrac{0.15KPL^3}{3EI}

Therefore the force in the spring will be F =\dfrac{KPl^3}{3EI}..The deflection of the beam will be \rho = 0.15(\dfrac{KPL^3}{3EI})

To know more about Cantilever beam follow

brainly.com/question/16791806

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7 0
2 years ago
The A/C compressor will not engage when the A/C is turned on. The static refrigerant pressure is 75 psi and the outside temperat
VikaD [51]

In the case above,  poor connection at the pressure cycling switch  and also a faulty A/C clutch coil could be the cause.

<h3>What is likely the reason when an A/C compressor will not engage if A/C is turned on?</h3>

The cause that hinders the A/C Compressor from engaging are:

  • Due to low pressure lockout.
  • Due to a poor ground
  • Due to bad clutch coil.
  • Dur to an opening in the wire that links to the clutch coil.
  • Due to a blown fuse.

Note that the pressure switches is known to be one that control the on/off function of any kind of AC compressor and as such, if there is switch failure, it can hinder the AC compressor from functioning at all.

Therefore, technician A and B are correct.

Learn more about refrigerant pressure from

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3 0
2 years ago
A centrifugal pump is required to pump water to an open water link situated 4 km away from the location of the pump through a pi
11111nata11111 [884]

Answer:

P= 5.5 bar

Explanation:

Given that

L= 4000 m

d= 0.2 m

Friction factor(F) = 0.01

speed V= 2 m/s

Head = 5 m

Head loss due to friction

h_f=\dfrac{FLV^2}{2gd}

h_f=\dfrac{0.01\times 4000\times 2^2}{2\times 9.81\times 0.2}

h_f=40.77m

So the total head(H) = 5 + 40.77 + 10.3 =56.07

Where 10.3 m is the atmospheric head.

We know that

P=ρ g H

So total Pressure

P= 1000 x 9.81 x 56.07 Pa

P=5.5\times 10^5\ Pa

P= 5.5 bar

5 0
3 years ago
#5 Air undergoes an adiabatic compression in a piston-cylinder assembly from P1= 1 atm and Ti=70 oF to P2= 5 atm. Employing idea
otez555 [7]

Answer:

The work transfer per unit mass is approximately 149.89 kJ

The heat transfer for an adiabatic process = 0

Explanation:

The given information are;

P₁ = 1 atm

T₁ = 70°F = 294.2611 F

P₂ = 5 atm

γ = 1.5

Therefore, we have for adiabatic system under compression

T_{2} = T_{1}\cdot \left (\dfrac{P_{2}}{P_{1}}  \right )^{\dfrac{\gamma -1}{\gamma }}

Therefore, we have;

T_{2} = 294.2611 \times \left (\dfrac{5}{1}  \right )^{\dfrac{1.5 -1}{1.5 }} \approx 503.179 \ K

The p·dV work is given as follows;

p \cdot dV = m \cdot c_v \cdot (T_2 - T_1)

Therefore, we have;

Taking air as a diatomic gas, we have;

C_v = \dfrac{5\times R}{2} = \dfrac{5\times 8.314}{2} = 20.785 \ J/(mol \cdot K)

The molar mass of air = 28.97 g/mol

Therefore, we have

c_v = \dfrac{C_v}{Molar \ mass} = \dfrac{20.785}{28.97} \approx 0.7175 \ kJ/(kg \cdot K)

The work done per unit mass of gas is therefore;

p \cdot dV =W =   1 \times 0.7175 \times (503.179 - 294.2611) \approx 149.89 \ kJ

The work transfer per unit mass ≈ 149.89 kJ

The heat transfer for an adiabatic process = 0.

8 0
3 years ago
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