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Lana71 [14]
3 years ago
5

Complete the following sentence.

Engineering
1 answer:
Lyrx [107]3 years ago
6 0

Answer:

bottom

Explanation:

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The following median grain size data were obtained during isothermal liquid phase sintering of an 82W-8Mo-8Ni-2Fe alloy. What is
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Answer:

The probable grain-coarsening mechanism is : Ideal grain growth mechanism

( d^{2}- d_{0} ^{2} = kt )

Explanation:

The plot attached below shows the time dependence of the growth of grain.

The probable grain-coarsening mechanism is : Ideal grain growth mechanism

the ideal growth follows this principle = d^{2} - d^{2} _{0}  = kt

d = final grain size

d_{0} = initial grain size

k = constant ( temperature dependent )

t = 0

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4 years ago
A mason requires 12 mason-hours per 100 ft2 for the bottom 4 ft of wall, and a hod carrier requires 14.5 helper hours to mix the
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3 years ago
A roadway with a rough-asphalt pavement has a cross slope of 2%, a longitudinal slope of 2.5%, a curb height of 8 cm, and a 90-c
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b

Explanation:

5 0
3 years ago
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°

6 0
3 years ago
In a TDM communication example, 15 voice signals are badlimited to 5kHz and transmitted simultaneously using PAM. What is a prel
MA_775_DIABLO [31]

Answer:

Option D

160 kHz

Explanation:

Since we must use at least one synchronization bit, total message signal is 15+1=16

The minimum sampling frequency, fs=2fm=2(5)=10 kHz

Bandwith, BW required is given by

BW=Nfs=16(10)=160 kHz

5 0
4 years ago
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