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Eva8 [605]
3 years ago
10

Will Give Brainliest

Engineering
1 answer:
Tom [10]3 years ago
8 0

Answer:

A hole within a hole

Explanation:

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When the grounded conductor is to be spliced in a 12-inch by 12-inch junction box, there shall be at least ? of free conductor l
slamgirl [31]

There should be a  6" free conductor left for splicing

Why there should be 6" splicing?

You must leave the junction box with at least six inches of free conductor wiring when running electrical cables from the box to the box for connecting needs.

Actually, 6-8 inches can be left when running the electric cables, additionally even if the grounded conductor is to be spliced in a 12-inch by 12-inch.

Conductor splicing: A splice, which can be finished using either the crimping or soldering method, is the joining of two or more conductors in a way that produces a permanent electrical termination and mechanical link.

Hence we can conclude that 6 inches is fine for the conductor splicing

To learn more about splicing
brainly.com/question/7668842
#SPJ4

7 0
1 year ago
What is the critical path and time duration?
kvasek [131]

The critical path is A-B-C, with a duration of 15 minutes.

<u>Explanation</u>:

  • The critical path is A-B-C, with a term of 15 minutes. You don't need to be knowledgeable in computer lingo to make sense of this one (as I made sense of it absent a lot of data on that half of the condition). Here's the manner by which I made sense of this.  
  • The main thing I saw was that "predecessor" was utilized, which implies something precedes the other thing. That provided me the primary insight that A precedes B and C, and that B must precede C. Then, I just included the term times for the three ways, which allowed me 15 minutes.
3 0
3 years ago
In a shear box test on sand a shearing force of 800 psf was applied with normal stress of 1750 psf. Find the major and minor pri
ryzh [129]

Answer:

The major and minor stresses are as 2060.59 psf, -310.59 psf and 1185.59 psf.

Explanation:

The major and minor principal stresses are given as follows:

\sigma_{max}=\dfrac{\sigma_x+\sigma_y}{2}+\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}

\sigma_{min}=\dfrac{\sigma_x+\sigma_y}{2}-\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}

Here

  • \sigma_x is the normal stress which is 1750 psf
  • \sigma_y is 0
  • \tau_{xy} is the shear stress which is 800 psf

So the formula becomes

\sigma_{max}=\dfrac{\sigma_x+\sigma_y}{2}+\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}\\\sigma_{max}=\dfrac{1750+0}{2}+\sqrt{\left(\dfrac{1750-0}{2}\right)^2+(800)^2}\\\sigma_{max}=875+\sqrt{\left(875)^2+(800)^2} \\\sigma_{max}=875+\sqrt{765625+640000}\\\sigma_{max}=875+1185.59\\\sigma_{max}=2060.59 \text{psf}

Similarly, the minimum normal stress is given as

\sigma_{min}=\dfrac{\sigma_x+\sigma_y}{2}-\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}\\\sigma_{min}=\dfrac{1700+0}{2}-\sqrt{\left(\dfrac{1700-0}{2}\right)^2+(800)^2}\\\sigma_{min}=875-\sqrt{(875)^2+(800)^2}\\\sigma_{min}=875-\sqrt{765625+640000}\\\sigma_{min}=875-1185.59\\\sigma_{min}=-310.59 \text{ psf}

The maximum shear stress is given as

\tau_{max}=\dfrac{\sigma_{max}-\sigma_{min}}{2}\\\tau_{max}=\dfrac{2060.59-(-310.59)}{2}\\\tau_{max}=\dfrac{2371.18}{2}\\\tau_{max}=1185.59 \text{psf}

5 0
3 years ago
Scientists learn more about the natural world through investigation. This involves identifying a problem, researching related in
mars1129 [50]

Answer:

This shows that the technological designs are similar to the scientific investigation processes.

Explanation:

Scientists carry out scientific investigations and discoveries. They learn from their environment by investigating the natural world. The scientist investigates by problem identification, research the related information, designing and conceptualizing, investigating and analyzing the results and deriving out conclusions.

Similarly the engineers follow the same step and provide solutions to the problems of the society through their products and technological designs. The engineers first identify the need of the product, design and implement the design to provide solution and then evaluate the solution. These are the similar steps that an engineer follow in a technological design of a product or a solution.

3 0
3 years ago
A normal shock wave takes place during the flow of air at a Mach number of 1.8. The static pressure and temperature of the air u
Darina [25.2K]

Answer:

The pressure upstream and downstream of a shock wave are related as

\frac{P_{1}}{P_{o}}=\frac{2\gamma M^{2}-(\gamma -1)}{\gamma +1}

where,

\gamma= Specific Heat ratio of air

M = Mach number upstream

We know that \gamma _{air}=1.4

Applying values we get

\frac{P_{1}}{100kPa}=\frac{2\times 1.4\times 1.8^{2}-(1.4 -1)}{1.4 +1}\\\\\frac{P_{1}}{100kPa}=3.61\\\\\therefore P_{1}=361.33kPa(Absloute)

Similarly the temperature downstream is obtained by the relation

\frac{T_{1}}{T_{o}}=\frac{[2\gamma M^{2}-(\gamma -1)][(\gamma -1)M^{2}+2]}{(\gamma +1)^{2}M^{2}}

Applying values we get

\frac{T_{1}}{423}=\frac{[2\times 1.4\times 1.8^{2}-(1.4-1)][(1.4-1)1.8^{2}+2]}{(1.4+1)^{2}\times 1.8^{2}}\\\\\therefore \frac{T_{1}}{423}=1.53\\\\\therefore T_{1}=647.85K=374.85^{o}C

The Mach number downstream is obtained by the relation

M_{d}^{2}=\frac{(\gamma -1)M^{2}+2}{2\gamma M^{2}-(\gamma -1)}\\\\\therefore M_{d}^{2}=\frac{(1.40-1)\times 1.8^{2}+2}{2\times1.4\times 1.8^{2}-(1.4-1)}\\\\\therefore M_{d}^{2}=0.38\\\\M_{d}=0.616

3 0
4 years ago
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