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Alik [6]
2 years ago
5

Airtight plugs are installed in the ends of large sizes of PVC conduit during bending to_____________.

Engineering
1 answer:
Lana71 [14]2 years ago
7 0

Airtight plugs are installed in the ends of large sizes of PVC conduit before bending to  A. prevent the conduit from collapsing when heated.

<h3>What is Airtight plugs?</h3>

Airtight plugs is necessary in the  PVC conduit so as to be able to avoid the collapse of conduit.

In this case, Airtight plugs are installed in the ends of large sizes of PVC conduit before bending to  A. prevent the conduit from collapsing when heated.

Learn more about Airtight plugs on:

brainly.com/question/20704252

#SPJ1

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Since the application can be the prospective employers first impression, it is important to a. always answer questions truthfull
loris [4]

Answer:

Definitely D

Explanation:

You should do all of those things in an application.

7 0
4 years ago
Read 2 more answers
A good visual lead is .... seconds from the front of the vehicle, focusing in the center of the path of travel. Searching 20 to
Tomtit [17]

A good visual lead is 20-30 seconds from the front of the vehicle, focusing in the center of the path of travel. Searching 20 to 30 seconds ahead, gives you time to assess within the next ..12-15 seconds, actions you may need to take to control an approaching risk.

Discussion:

Keeping eye focus centered in a path of travel at an interval of 20 to 30 seconds away from the vehicle is critical to gaining enough info. as possible in the driving scene. Good targeting sets up good sight lines for referencing and good peripheral fields for observing changes.

  • It is important to look ahead 12-15 seconds into your target area as one drives. Compromise. space by giving as much space to the greater of two hazards.

Read more on driving visual leads:

brainly.com/question/7067386

3 0
3 years ago
I am trying to create a line of code to calculate distance between two points. (distance=[tex]\sqrt{ (x2-x1)^2+(y2-y1)^2}) My li
k0ka [10]

Answer:

point_dist = math.sqrt((math.pow(x2 - x1, 2) + math.pow(y2 - y1, 2))

Explanation:

The distance formula is the difference of the x coordinates squared, plus the difference of the y coordinates squared, all square rooted.  For the general case, it appears you simply need to change how you have written the code.

point_dist = math.sqrt((math.pow(x2 - x1, 2) + math.pow(y2 - y1, 2))

Note, by moving the 2 inside of the pow function, you have provided the second argument that it is requesting.

You were close with your initial attempt, you just had a parenthesis after x1 and y1 when you should not have.

Cheers.

6 0
3 years ago
State and explain briefly four (4) functions that a machine foundation should fulfill.
MrMuchimi

Answer:

1 The weight of the foundation block should be enough to withstand vibrations

2 The foundation must be so dimensional that the resulting pressure due to weight of the device as well as the weight of foundation moves through the center of gravity of a base impact area in order to avoid the risk of specific settlement.

3.The base should be sufficiently rigid to have the requisite rigidity

4 a distance should be created all across  machine foundation to separate it from the adjacent parts of the building

Explanation:

 1. The weight of the foundation block should be enough to withstand vibrations and to avoid friction between device and the surrounding soil as well. This can be done by increasing the base block weight in supporting with engine power

2.The foundation must be so dimensional that the resulting pressure due to weight of the device as well as the weight of foundation moves through the center of gravity of a base impact area in order to avoid the risk of specific settlement.

3.The base should be sufficiently rigid to have the requisite rigidity as the slightest misdirection of foundation may cause significant bearing disorders.

4.To avoid propagation of vibration from a device to the adjacent parts of the building, a distance should be created all across  machine foundation to separate it from the adjacent parts of the building. 

8 0
4 years ago
Consider a steady isentropic airflow through a convergent channel with an inletto-exit cross section area ratio of 2. If the inl
Nina [5.8K]

Answer:

for m1=0.2

M2=0.4

for m1=2.5

M2=5

1=  channel input

2.channel output

Explanation:

Hello!

To solve this problem follow the steps below!

1.As it is an isentropic system, it means that the temperature of the fluid remains constant, so the volumetric flow is the same, remember that this is calculated by calculating the product between the velocity and the area.

V1A1=V2A2

1=  channel input

2.channel output

We can divide both sides of the equation by the speed of sound(V)

\frac{V1}{V}A1= \frac{V2}{V}A2

2. The mach number is defined as the ratio between the speed of a fluid and the speed of sound,

we rearrange the equation

M1A1=M2A2

\frac{M1A1}{A2} =M2

3.Now considering that the channel is convergent and that the ratio between the areas is 2, we have the following equation (A1 / A2 = 2)

M1(2)=M2

4.Finally we solve for the number of mach indicated (0.2 and 2.5)

for m1=0.2

0.2(2)=M2=0.4

for m1=2.5

2.5(2)=M2=5

8 0
3 years ago
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