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4 years ago

Name the main classes of polymer and define their characteristic properties

Engineering

1 answer:

Svetlanka [38]4 years ago

8 0

Answer:

Polymers are the naturally occurring or synthetic macromolecules that are composed of repeating subunits, called monomers.

The three main classes of polymers are: thermoplastic, thermosetting, and the elastomers.

Thermoplastic polymers have linear bonding. These polymers can be melted again and thus can recycled.

Thermosetting polymers have cross-linked bonding. These polymers decompose when heated and thus can not be remelted and recycled.

Elastomers have linear bonding with some cross-linking. These polymers extreme elastic extensibility and thus can revert back to its original shape after deformation, without causing any permanent damage.

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i know this isnt suppose to be here or any of that but this needs to be stop any where...ok now that i have your attention every

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I know this isnt suppose to be here or any of that but this needs to be stop any where...ok now that i have your attention everyone needs to know this people all over the world are burning bibles LETS STOP THIS because that is so disrepecful that people are doing that to the bible and thats hurting jesus and gods heart LETS STOP THIS BURNING BIBLE STUFF pls copy and paste this and share the word so we can stop this.

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3 years ago

Read 2 more answers

Air enters an adiabatic gas turbine at 1590 oF, 40 psia and leaves at 15 psia. The turbine efficiency is 80%, and the mass flow

melamori03 [73]

Answer:

a) 158.4 HP.

b) 1235.6 °F.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to set up an energy balance for the turbine's inlets and outlets:

$m_{in}h_{in}=W_{out}+m_{out}h_{out}$

Whereas the mass flow is just the same, which means we have:

$W_{out}=m_{out}(h_{out}-h_{in})$

And the enthalpy and entropy of the inlet stream is obtained from steam tables:

$h_{in}=1860.7BTU/lbm\\\\s_{in}= 2.2096BTU/lbm-R$

Now, since we assume the 80% accounts for the isentropic efficiency for this adiabatic gas turbine, we assume the entropy is constant so that:

$s_{out}= 2.2096BTU/lbm-R$

Which means we can find the temperature at which this entropy is exhibited at 15 psia, which gives values of temperature of 1200 °F (s=2.1986 BTU/lbm-K) and 1400 °F (s=2.2604 BTU/lbm-K), and thus, we interpolate for s=2.2096 to obtain a temperature of 1235.6 °F.

Moreover, the enthalpy at the turbine's outlet can be also interpolated by knowing that at 1200 °F h=1639.8 BTU/lbm and at 1400 °F h=174.5 BTU/lbm, to obtain:

$h_{out}=1659.15BUT/lbm$

Then, the isentropic work (negative due to convention) is:

$W_{out}=2500lbm/h(1659.15BUT/lbm-1860.7BUT/lbm)\\\\W_{out}=-503,875BTU$

And the real produced work is:

$W_{real}=0.8*-503875BTU\\\\W_{real}=-403100BTU$

Finally, in horsepower:

$W_{real}=-403100BTU/hr*\frac{1HP}{2544.4336BTU/hr} \\\\W_{real}=158.4HP$

Regards!

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4 years ago

A hollow steel tube with an inside diameter of 100 mm must carry a tensile load of 400 kN. Determine the outside diameter of the

Olegator [25]

Answer:

119.35 mm

Explanation:

Given:

Inside diameter, d = 100 mm

Tensile load, P = 400 kN

Stress = 120 MPa

let the outside diameter be 'D'

Now,

Stress is given as:

stress = Load × Area

also,

Area of hollow pipe = $\frac{\pi}{4}(D^2-d^2)$

Area of hollow pipe = $\frac{\pi}{4}(D^2-100^2)$

thus,

400 × 10³ N = 120 × $\frac{\pi}{4}(D^2-100^2)$

D² = tex]\frac{400\times10^3+30\pi\times10^4}{30\pi}[/tex]

D = 119.35 mm

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3 years ago

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dmitriy555 [2]

Answer:

age = int(input())

ticket = 100

total = 0

passengers = 1

while passengers <= 5:

if age < 3:

continue

else:

total += ticket

passengers += 1

print(total)

Explanation:

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2 years ago

Find the magnitude of the two pulling forces P and Q when their resultant is 50 N at 20° with Q. P 20°

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