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Ede4ka [16]
2 years ago
13

Draw the six principal views of

Engineering
1 answer:
stepan [7]2 years ago
5 0

Explanation:

See what Challenge I completed on Ablo https://ablo.live/s/1/yIwE1xnX88zg0liSeC7njA

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Which of the following refers to a full-scale version of a product used to validate performance?
kompoz [17]
I’m thinking it would be c sorry if it’s wrong .
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2 years ago
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Another focus of effective communication, according to Stephen Covey, is ensuring that:
svetoff [14.1K]

Answer:

Stephen Covey believes this principle is the key to effective interpersonal communication. Seek first to understand, then to be understood. This habit is about communicating with others. It's about developing the habit of listening carefully and really understanding the other person BEFORE giving your thoughts.

Explanation:

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3 years ago
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The autorotation spin characteristics of a straight-wing aircraft are induced by Group of answer choices
NemiM [27]

Answer:

More Drag on the down going wing and More Lift on the up going wing

Explanation:

The autorotation spins of blades used in airborne wind energy technology sectors help drive and move the winds and water propeller-type turbines or shafts of generators to produce electricity at altitude and transmit the electricity to earth through conductive tethers.

Sometimes autorotation takes place in rotating parachutes, kite tails. Etc.

As a result, more Drag usually induces the autorotation spin characteristics of a straight-wing aircraft on the downgoing wing and More Lift on the up-going wing.

7 0
3 years ago
At a point on the free surface of a stressed body, the normal stresses are 20 ksi (T) on a vertical plane and 30 ksi (C) on a ho
victus00 [196]

Answer:

The principal stresses are σp1 = 27 ksi, σp2 = -37 ksi and the shear stress is zero

Explanation:

The expression for the maximum shear stress is given:

\tau _{M} =\sqrt{(\frac{\sigma _{x}^{2}-\sigma _{y}^{2}  }{2})^{2}+\tau _{xy}^{2}    }

Where

σx = stress in vertical plane = 20 ksi

σy = stress in horizontal plane = -30 ksi

τM = 32 ksi

Replacing:

32=\sqrt{(\frac{20-(-30)}{2} )^{2} +\tau _{xy}^{2}  }

Solving for τxy:

τxy = ±19.98 ksi

The principal stress is:

\sigma _{x}+\sigma _{y} =\sigma _{p1}+\sigma _{p2}

Where

σp1 = 20 ksi

σp2 = -30 ksi

\sigma _{p1}  +\sigma _{p2}=-10 ksi (equation 1)

\tau _{M} =\frac{\sigma _{p1}-\sigma _{p2}}{2} \\\sigma _{p1}-\sigma _{p2}=2\tau _{M}\\\sigma _{p1}-\sigma _{p2}=32*2=64ksi equation 2

Solving both equations:

σp1 = 27 ksi

σp2 = -37 ksi

The shear stress on the vertical plane is zero

4 0
3 years ago
4. The instant the ignition switch is turned to the start position,
geniusboy [140]

Answer:

D. Both pull-in and hold-in windings are energized.

Explanation:

The instant the ignition switch is turned to the start position, "Both pull-in and hold-in windings are energized." This is because the moment the ignition switch is turned to the start position, voltage passes through to the S terminal of the solenoid.

The hold-in winding is attached to the case of the solenoid. Similarly, the pull-in winding is also attached to the starter motor. Thereby, the current will move across both windings by getting energized to generate a strong magnetic field.

4 0
2 years ago
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