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3 years ago

Eutectic product in Fe-C system is called A. Pearlite B. Bainite C. Ledeburite D. Spheroidite

Engineering

1 answer:

zimovet [89]3 years ago

6 0

Answer:

Eutectic product in Fe-C system is called Ledeburite-C.

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Anne works with other engineers studying physical processes that involve the flow of particles. Which field of engineering would

slega [8]

Answer:

Option A

Chemical engineering

Explanation:

Chemical engineering mainly encompass the study of behavior of different particles such as petroleum, water, drugs and other products. When Anne is involved in a study with engineers who study flow of particles, the flow, viscosity and other properties are among the behavior that chemical engineers are involved in.

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3 years ago

The section of the area to be examined is shown circumscribed by broken lines with circles at

Aloiza [94]

This question is about Circle Geometry. it evaluates connected and broken lines with respect to circles.

<h3>What is Circle Geometry?</h3>

This refers to the body of knowledge in mathematics that has to do with the various problems associated with the Circle.

In real-world scenarios, circle geometry is used in technologies involving:

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Learn more about Circle Geometry at:
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2 years ago

Who play 1v1 lol unblocked games 76

nikklg [1K]

Answer:

I did not what is it about?

8 0

2 years ago

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Once you have chosen a topic, what should you do before beginning the research process? a. Find as many possible facts and detai

dlinn [17]

Answer:

The answer is C

Explanation:

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2 years ago

A cylindrical resistor element on a circuit board dissipates 0.6 W of power. The resistor is 1.5 cm long, and has a diameter of

Burka [1]

Answer:

a. 51.84Kj

b. 2808.99 W/m^2

c. 11.75%

Explanation:

Amount of heat this resistor dissipates during a 24-hour period

= amount of power dissipated * time

= 0.6 * 24 = 14.4 Watt hour

(Note 3.6Watt hour = 1Kj )

=14.4*3.6 = 51.84Kj

Heat flux = amount of power dissipated/ surface area

surface area = area of the two circular end + area of the curve surface

$=2*\frac{\pi D^{2} }{4} + \pi DL\\=2*\frac{\pi *(\frac{0.4}{100} )^{2} }{4} + \pi *\frac{0.4}{100} *\frac{1.5}{100}$

= 2.136 *10^-4 $m^{2}$

Heat flux = $\frac{0.6}{2.136 * 10^{-4} }$ = 2808.99 $W/m^{2}$

fraction of heat dissipated from the top and bottom surface

$=\frac{\frac{2*\pi D^{2} }{4} }{\frac{2*\pi D^{2}}{4} + \pi DL } \\\\=\\\frac{\frac{2*\pi *(\frac{0.4}{100} )^{2} }{4} }{\frac{2*\pi *(\frac{0.4}{100} )^{2} }{4} +\pi *\frac{0.4}{100} *\frac{1.5}{100} } \\\\=\frac{2.51*10^{-5} }{2.136*10^{-4} } \\\\\= 0.1175$

=11.75%

8 0

3 years ago

Read 2 more answers