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Sveta_85 [38]
2 years ago
10

The team needs to choose a primary view for the part drawing. Three team members make suggestions:

Engineering
1 answer:
dexar [7]2 years ago
6 0

Answer:

<u>Option 1</u>

Explanation:

As the team has already submitted the plans for the part drawing, the best way to proceed would be how it was given in the plans. Hence, the option to be selected :

  • <u>Team member 1 suggests an orthographic top view because that is how the plans for the part were submitted.</u>
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The provided length of the vertical curve is satisfactory for the reconstruction design speed of 60 mi/h

Explanation:

The explanation is shown on the first uploaded image

8 0
3 years ago
The link acts as part of the elevator control for a small airplane. If the attached aluminum tube has an inner diameter of 25 mm
aksik [14]

Answer:

Tmax=14.5MPa

Tmin=10.3MPa

Explanation:

T = 600 * 0.15 = 90N.m

T_max =\frac{T_c}{j}  = \frac{x}{y}  = \frac{90 \times 0.0175}{\frac{\pi}{2} \times (0.0175^4-0.0125^4)}

=14.5MPa

T_{min} =\frac{T_c}{j}  = \frac{x}{y}  = \frac{90 \times 0.0125}{\frac{\pi}{2} \times (0.0175^4-0.0125^4)}

=10.3MPa

7 0
3 years ago
What are the benefits of using the engineering design process? combining scientific knowledge and creativity in a rigid and stru
Trava [24]

Answer:

combining scientific knowledge, careful reasoning, and artistic invention in a flexible approach to problem-solving

Explanation:

7 0
3 years ago
Which kind of fracture (ductile or brittle) is associated with each of the two crack propagation mechanisms?
Nina [5.8K]

dutile is the correct answer

6 0
3 years ago
A circular specimen of MgO is loaded in three-point bending. Calculate the minimum possible radius of the specimen without fract
Hitman42 [59]

Answer:

radius = 9.1 × 10^{-3} m

Explanation:

given data

applied load = 5560 N

flexural strength = 105 MPa

separation between the support =  45 mm

solution

we apply here minimum radius formula that is

radius = \sqrt[3]{\frac{FL}{\sigma \pi}}      .................1

here F is applied load and  is length

put here value and we get

radius =  \sqrt[3]{\frac{5560\times 45\times 10^{-3}}{105 \times 10^6 \pi}}  

solve it we get

radius = 9.1 × 10^{-3} m

8 0
3 years ago
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