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Ugo [173]
2 years ago
5

Determine the normal stress acting perpendicular to the seam, when the tube is subjected to an axial compressive force of 200

Engineering
1 answer:
astraxan [27]2 years ago
8 0

Normal Stress = 200 units / Area of the tube face

Normal Stress is equal to the force applied perpendicular to the surface area divided by the surface area. This force is also known as the axial load.

Therefore, we get the following formula:

Normal Stress = 200 units of Force / Area of the tube face

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For each topic, find the total number of blurts that were analyzed as being related to the topic. Order the result by topic id.
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Answer:

Explanation: see attachment below

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3 years ago
Carbon dioxide contained in a piston–cylinder device is compressed from 0.3 to 0.1 m3. During the process, the pressure and volu
Lunna [17]

Answer:

W = -6.5 Kpa m^6 (\frac{1}{0.1 m^3}- \frac{1}{0.3 m^3})= -6.5 Kpa m^6 (6.66 m^-3) =-43.33 Kpa m^3 = -43.33 KJ

Explanation:

For this case we know the following info :

V_i = 0.3 m^3 represent the initial volume

V_f = 0.1 m^3 represent the final volume

We know that the pressure and volume are related with the following expression:

P = aV^{-2}= \frac{a}{V^2}

Where a is a constant given a = 6.5 Kpa m^6

And we need to calculate the work associated to this process.

We have a compression, and by definition the work is defined with the following general expression:

W = \int_{V_i}^{V_f} P dV

If we replace the expression for P we got:

W = \int_{V_i}^{V_f} a V^{-2} dV

If we integrate we got:

W = -a (\frac{1}{V}) \Big|_{V_i}^{V_f}

Using the fundamental theorem of calculus we have:

W = -a (\frac{1}{V_f} -\frac{1}{V_i})

And replacing the values we got:

W = -6.5 Kpa m^6 (\frac{1}{0.1 m^3}- \frac{1}{0.3 m^3})= -6.5 Kpa m^6 (6.66 m^-3) =-43.33 Kpa m^3 = -43.33 KJ

5 0
4 years ago
5 kg of a wet steam has a volume of 2 m3
Verizon [17]
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3 years ago
A commercial facility has a three-phase, 277 volt, 1200 Amp service. What is the required nominal transformer size needed for th
alekssr [168]

Answer:

The answer is "Option e".

Explanation:

Given value:

\to v_{ph}=277  \ \ \ \ \ \ \  I_{ph} =1200\\\\

vA= v_{Ph} I_{ph}  \  \ \ \to  for \ single \ phase\\\\=3v_{ph} I_{ph}  \ \ \ \ \ \to  for \ 3 - \phi  \\\\s = 3\times 277 \times 1200 \\\\

  =997.200 \ K VA\\\\= 1000 \ k VA

7 0
3 years ago
I need answer to this question
blagie [28]

Answer:

1. Graph C

2. Friction

Explanation:

1. The line on all of the graphs shown represents velocity. The formula for velocity is v=\frac{d}{t} where d is distance and t is time. Focusing on the first lap, the starting point on the graph should be the origin and the "ending" point should be (20, 3). These requirements eliminate graph A as an answer because its "end" is not (20, 3). During the break, the student does not move, so the slope of the line should be completely horizontal. The break lasted for 5 minutes, so the correct graph should have a horizontal line between the points (20, 3) and (25, 3). This requirement eliminates graph B and D because their break is either not long enough (B) or too long (D).

2. Friction slows down the movement of objects. When an object is rough, it produces more friction which causes the object to be slowed more. When an object is smooth, friction slows it less than it would for a rough object.

6 0
3 years ago
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