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Dmitriy789 [7]
3 years ago
5

Oleg is using a multimeter to test the circuit branch you just installed. After turning off the current to the circuit at the se

rvice panel , hell first attach one pole to the ______________ at the end of the run? A) WHITE WIRE B) BOX C) BLACK WIRES) GROUND WIRE
Engineering
1 answer:
irakobra [83]3 years ago
7 0

Answer:

Option D, Ground Wire

Explanation:

Oleg is a mustimeter or multi tester or a basic voltage tester which is used to test the ground. The main purpose of the tester is to ensure that the outlet is connected to the ground of the circuit, thereby ensuring its proper functioning.

Hence, option D is correct

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c. Both Technicians A and B

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The hydrofoil boat has an A-36 steel propeller shaft that is 100 ft long. It is connected to an in-line diesel engine that deliv
Nataliya [291]

The question is incomplete. The complete question is :

The hydrofoil boat has an A-36 steel propeller shaft that is 100 ft long. It is connected to an in-line diesel engine that delivers a maximum power of 2590 hp and causes the shaft to rotate at 1700 rpm . If the outer diameter of the shaft is 8 in. and the wall thickness is $\frac{3}{8}$  in.

A) Determine the maximum shear stress developed in the shaft.

$\tau_{max}$ = ?

B) Also, what is the "wind up," or angle of twist in the shaft at full power?

$ \phi $ = ?

Solution :

Given :

Angular speed, ω = 1700 rpm

                              $ = 1700 \frac{\text{rev}}{\text{min}}\left(\frac{2 \pi \text{ rad}}{\text{rev}}\right) \frac{1 \text{ min}}{60 \ \text{s}}$

                              $= 56.67 \pi \text{ rad/s}$

Power $= 2590 \text{ hp} \left( \frac{550 \text{ ft. lb/s}}{1 \text{ hp}}\right)$

          = 1424500 ft. lb/s

Torque, $T = \frac{P}{\omega}$

                 $=\frac{1424500}{56.67 \pi}$

                 = 8001.27 lb.ft

A). Therefore, maximum shear stress is given by :

Applying the torsion formula

$\tau_{max} = \frac{T_c}{J}$

        $=\frac{8001.27 \times 12 \times 4}{\frac{\pi}{2}\left(4^2 - 3.625^4 \right)}$

      = 2.93 ksi

B). Angle of twist :

     $\phi = \frac{TL}{JG}$

         $=\frac{8001.27 \times 12 \times 100 \times 12}{\frac{\pi}{2}\left(4^4 - 3.625^4\right) \times 11 \times 10^3}$

         = 0.08002 rad

         = 4.58°

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High strength steels are being used to reduce weight on cars. Explain why using a high strength steel would allow you to reduce
gulaghasi [49]

Engaging the frequently tough requirements of vehicle safety, weight reduction for combustible economy, and manufacturability has influenced the steel industry to create a unique variety of 'super steels' for the automobiles of the future.

<h3><u>Explanation</u>:</h3>

• That steel, though, is far more advanced than the materials of just a few years ago.

• At the forefront of these is Advanced High Strength Steel, AHSS, developed by World Auto Steel’s member companies, which is demonstrating to be something of a vision in automobile production.  

• The standard engineering trade-off in steel preference involves considering the need for ultimate strength against flexibility and work-ability – stronger steels tend to be stiffer and less ductile, making them more difficult to develop into cars and more laborious to weld.

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• Where a typical mild steel might have a tensile strength of 300MPa, AHSS can exceed 1500MPa while retaining a highest elongation of 25%, compared to about 40% for mild steel. The intrigue is in the micro-structure, containing a martensite, bainite, austenite phase rather than ferrite, pearlite, or cementite.

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It will limit the materials for your bridge which could cause damage to your bridge

Explanation:

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