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Allushta [10]
3 years ago
9

Technician A says that you don’t need to use an exhaust extraction system when working on vehicles equipped with a catalytic con

verter. Technician B says that carbon monoxide effectively asphyxiates a person. Who is correct?
Engineering
1 answer:
pav-90 [236]3 years ago
8 0

Answer:

Technician B

Explanation:

When working on vehicles which are equipped with a catalytic coverter, the best practice of asphyxiating someone is to use carbon monoxide. This implies that you need an extraction sytem. Therefore, technician A is wrong by assuming no need of the system. In conclusion, technician B is correct.

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Proper handling of blueprints includes which of the following
marta [7]

Answer:

folding plans neatly after use

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A private plane pilot is what kind of individual transportation position? professional level mid-level entry-level EPA-certified
gayaneshka [121]

Answer:

A private plane pilot is a mid-level position.

Explanation:

The six pilot certifications in the US are as follows:

  1. Sport Pilot
  2. Recreational Pilot
  3. Private Plane Pilot
  4. Commercial Plane Pilot
  5. Flight Instructor
  6. Airline Transport Pilot

From this listing, it is evident that the Private Plane Pilot is in the mid of the line so it is a mid-level position.

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What are the three elementary parts of a vibrating system?
zhenek [66]

Answer:

the three part are mass, spring, damping

Explanation:

vibrating system consist of three elementary system namely

1) Mass - it is a rigid body due to which system experience vibration and kinetic energy due to vibration is directly proportional to velocity of the body.

2) Spring -  the part that has elasticity and help to hold mass

3) Damping - this part considered to have zero mass and  zero elasticity.

7 0
3 years ago
Considering only (110), (1 1 0), (101), and (10 1 ) as the possible slip planes, calculate the stress at which a BCC single crys
vredina [299]

Solution :

i. Slip plane (1 1 0)

Slip direction -- [1 1 1]

Applied stress direction = ( 1 0 0 ]

τ = 50 MPa    ( Here slip direction must be perpendicular to slip plane)

τ = σ cos Φ cos λ

$\cos \phi = \frac{(1,0,0) \cdot (1,1,0)}{1 \times \sqrt2}$

       $=\frac{1}{\sqrt2 }$

$\cos \lambda = \frac{(1,0,0) \cdot (1,-1,1)}{1 \times \sqrt3}$

       $=\frac{1}{\sqrt3 }$

  τ = σ cos Φ cos λ

∴ $50= \sigma \times \frac{1}{\sqrt2} \times \frac{1}{\sqrt3} $

  σ = 122.47 MPa

ii. Slip plane  --- (1 1 0)

   Slip direction -- [1 1 1]

  $\cos \phi = \frac{(1, 0, 0) \cdot (1, -1, 0)}{1 \times \sqrt2} =\frac{1}{\sqrt2}$

   $\cos \lambda = \frac{(1, 0, 0) \cdot (1, 1, -1)}{1 \times \sqrt3} =\frac{1}{\sqrt3}$

 τ = σ cos Φ cos λ

∴ $50= \sigma \times \frac{1}{\sqrt2} \times \frac{1}{\sqrt3} $

  σ = 122.47 MPa

iii. Slip plane  --- (1 0 1)

    Slip direction --- [1 1 1]

$\cos \phi = \frac{(1, 0, 0) \cdot (1, 0, 1)}{1 \times \sqrt2} =\frac{1}{\sqrt2}$

   $\cos \lambda = \frac{(1, 0, 0) \cdot (1, 1, -1)}{1 \times \sqrt3} =\frac{1}{\sqrt3}$

τ = σ cos Φ cos λ

∴ $50= \sigma \times \frac{1}{\sqrt2} \times \frac{1}{\sqrt3} $

  σ = 122.47 MPa

iv. Slip plane -- (1 0 1)

    Slip direction  ---- [1 1 1]

$\cos \phi = \frac{(1, 0, 0) \cdot (1, 0, -1)}{1 \times \sqrt2}=\frac{1}{\sqrt2}$

$\cos \lambda = \frac{(1, 0, 0) \cdot (1, -1, 1)}{1 \times \sqrt3} =\frac{1}{\sqrt3}$

τ = σ cos Φ cos λ

∴ $50= \sigma \times \frac{1}{\sqrt2} \times \frac{1}{\sqrt3} $

  σ = 122.47 MPa

∴ (1, 0, -1). (1, -1, 1) = 1 + 0 - 1 = 0

3 0
3 years ago
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