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Svet_ta [14]
3 years ago
9

How do we infer that there is

Engineering
1 answer:
Katena32 [7]3 years ago
6 0

Answer:

Not too far removed from Collingwood’s concern with the elimination of physical and moral force via social civilization are accounts of civilized society concerned with the management of violence, if only by removing it from the public sphere. Such a concern is extended in Zygmunt Bauman’s account of civilization to the more general issue of producing readily governable subjects. The “concept of civilization,” he argues, “entered learned discourse in the West as the name of a conscious proselytising crusade waged by men of knowledge and aimed at extirpating the vestiges of wild cultures” (1987, 93).

This proselytizing crusade in the name of civilization is worth considering further. Its rationale is not too difficult to determine when one considers Starobinski’s (1993, 31) assertion: “Taken as a value, civilization constitutes a political and moral norm. It is the criterion against which barbarity, or non-civilization, is judged and condemned.” A similar sort of argument is made by Pagden (1988, 33), who states that civilization “describes a state, social, political, cultural, aesthetic—even moral and physical—which is held to be the optimum condition for all mankind, and this involves the implicit claim that only the civilized can know what it is to be civilized.” It is out of this implicit claim and the judgments passed in its name that the notion of the “burden of civilization” was born. And this, many have argued, is one of the less desirable aspects and outcomes of the idea of civilization

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An energy system can be approximated to simply show the interactions with its environment including cold air in and warm air out
Elenna [48]

Answer: The energy system related to your question is missing attached below is the energy system.

answer:

a) Work done = Net heat transfer

  Q1 - Q2 + Q + W = 0

b)  rate of work input ( W ) = 6.88 kW

Explanation:

Assuming CPair = 1.005 KJ/Kg/K

<u>Write the First law balance around the system and rate of work input to the system</u>

First law balance ( thermodynamics ) :

Work done = Net heat transfer

Q1 - Q2 + Q + W = 0 ---- ( 1 )

rate of work input into the system

W = Q2 - Q1 - Q -------- ( 2 )

where : Q2 = mCp T  = 1.65 * 1.005 * 293 = 485.86 Kw

             Q2 = mCp T = 1.65 * 1.005 * 308 = 510.74 Kw

              Q = 18 Kw

Insert values into equation 2 above

W = 6.88 Kw

5 0
3 years ago
A ________ can be installed in a cast-iron block to repair a worn or cracked cylinder. Question 24 options:
VLD [36.1K]

Answer:

Cylinder sleeve

Explanation:

7 0
2 years ago
You are working as an electrical technician. One day, out in the field, you need an inductor but cannot find one. Looking in you
telo118 [61]

Answer:

a) the inductance of the coil is 6 mH

b) the emf generated in the coil is 18 mV  

Explanation:

Given the data in the question;

N = 570 turns

diameter of tube d = 8.10 cm = 0.081 m

length of the wire-wrapped portion l =  35.0 cm = 0.35 m

a) the inductance of the coil (in mH)

inductance of solenoid

L = N²μA / l

A = πd²/4  

so

L = N²μ(πd²/4) / l

L = N²μ(πd²) / 4l

we know that μ = 4π × 10⁻⁷ TmA⁻¹

we substitute

L = [(570)² × 4π × 10⁻⁷× ( π × (0.081)² )] / 4(0.35)

L =  0.00841549 / 1.4

L = 6 × 10⁻³ H    

L = 6 × 10⁻³ × 1000 mH

L = 6 mH

Therefore, the inductance of the coil is 6 mH

b)

Emf ( ∈ ) = L di/dt

given that; di/dt = 3.00 A/sec

{∴ di = 3 - 0 = 3 and dt = 1 sec}

Emf ( ∈ ) = L di/dt

we substitute

⇒ 6 × 10⁻³ ( 3/1 )

= 18 × 10⁻³ V

= 18 × 10⁻³ × 1000

= 18 mV  

Therefore, the emf generated in the coil is 18 mV  

7 0
2 years ago
Turn on your____
storchak [24]

Answer:

b

Explanation:

5 0
3 years ago
Read 2 more answers
A three-point bending test was performed on an aluminum oxide specimen having a circular cross section of radius 5.6 mm; the spe
ankoles [38]

Answer:

F =  8849 N

Explanation:

Given:

Load at a given point = F =  4250 N

Support span = L = 44 mm

Radius = R = 5.6 mm

length thickness of tested material = 12 mm

First compute the flexural strength for circular cross section using the formula below:

σ_{fs} = F_{f} L / \pi  R^{3}

σ = FL / π R³

Putting the given values in the above formula:

σ = 4250 ( 44 x 10⁻³ ) / π  ( 5.6 x 10⁻³ ) ³

  = 4250 ( 44 x 10⁻³ )  / 3.141593 ( 5.6 x 10⁻³ ) ³

  = 4250 (44 x 1 /1000 )) / 3.141593 ( 5.6 x 10⁻³ ) ³

  = 4250 ( 11 / 250  ) / 3.141593 ( 5.6 x 10⁻³ ) ³

  = 187 / 3.141593 ( 5.6 x 1 / 1000 ) ³

  = 187 / 3.141593 (0.0056)³

  = 338943767.745358

  = 338.943768 x 10⁶

σ = 338 x 10⁶ N/m²

Now we compute the load i.e. F from the following formula:

F_{f} = 2 σ_{fs} d³/3 L

F = 2σd³/3L

  = 2(338 x 10⁶)(12 x 10⁻³)³ / 3(44 x 10⁻³)

  = 2 ( 338 x 1000000 ) ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)

  = 2 ( 338000000 ) ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)

  = 676000000 ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)

  = 676000000 ( 12  x  1/1000  )³ / 3 ( 44 x 10⁻³)

  = 676000000 (  3  / 250  )³ / 3 ( 44 x 10⁻³)

  = 676000000 (  27  / 15625000 )  / 3 ( 44 x 10⁻³)

  = 146016  / 125 / 3 ( 44 x 1 / 1000  )

  = ( 146016  / 125 ) /  (3 ( 11 /  250 ))

  =  97344  / 11

F =  8849 N

4 0
3 years ago
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