Poly(cis-1,4-isoprene), or natural rubber (NR), has a tendency crystallize. The Tm of this polymer is slightly below room temperature, although lightly-crosslinked NR can partially crystallize at room temperature when stretched. Apparently, Tm is elevated upon stretching which allows for crystallization above the Tm of the unstretched polymer. Explain.

Answers

Answer 1

Answer:

Explanation:

Crystalline melting temperature (Tm) is termed as the temperature required for a  crystalline polymer to change to a fluid or glasslike crystalline spaces of a semi-crystalline polymer liquefy (expanded sub-atomic movement).  

Crystallization of polymers is an interaction related with incomplete arrangement of their atomic and molecular chains. These chains crease together and structure requested districts called lamellae, which form bigger spheroidal designs named spherulites. Polymers can solidify after cooling from melting, mechanical extending, or dissolvable dissipation. Crystallization influences the optical, mechanical, and synthetic chemical properties of the polymer.  

For a crystalline polymer, a required polymer chain is present in or goes along a few crystalline and amorphous zones. The crystalline zones are comprised of intermolecular & intramolecular arrangements or deliberate and thus firmly stuffed plan of atoms or chain fragments, and an absence of it brings about the development of amorphous zones.  

The mechanical property boundary, for example, shear modulus expansions in the temperature of perception for polymer material framework.  

The temperature reaction of direct linear polymers might be seen as partitioned into three particularly separate fragments:  

1. Above Tm: The polymer stays as fluid whose consistency & viscosity would rely upon atomic molecular weight and temperature.  

2. Between Tm and Tg: This area may go between close to 100% crystalline & 100% amorphous chain atomic bunches relying upon the polymer underlying consistency. The amorphous part carries on similar to supercooled fluid in this section. The generally actual conduct of the polymer in this moderate portion is similar to an elastic rubber.  

3.Below Tg: The polymer material saw as glass is hard and inflexible, showing and emanating a predetermined coefficient of thermal extension. The glass is more like a crystalline strong than the fluid in personal conduct standard regarding mechanical property boundaries. In regard to the molecular atomic request, in any case, the glass all the more intently takes after the fluid. There is little contrast between the direct linear and cross-connected polymer beneath Tg.


Related Questions

Leland wants to work in a Production career operating heavy machinery. Which type of education or training should Leland seek?

a bachelor’s degree then a master’s degree
vocational school certificate or master’s degree
on-the-job training or vocational school certificate
associate’s degree then a bachelor’s degree

Answers

Answer:

it is indeed C

Explanation:

Answer:

c

Explanation:

How does distribution add value to goods and services being sold,
including intellectual property?

Answers

Answer:

Distribution (or its more sophisticated counterpart, supply chain management) can add value to goods and services by making them more easily and conveniently available to consumers. ... This means that you need good wholesalers and good transportation systems to get your products to the retailers.

Explanation:

Distribution (or its more sophisticated counterpart, supply chain management) can add value to goods and services by making them more easily and conveniently available to consumers. ... This means that you need good wholesalers and good transportation systems to get your products to the retailers.

Answer:

Distribution (or its more sophisticated counterpart, supply chain management) can add value to goods and services by making them more easily and conveniently available to consumers.

Explanation:

hope it helps <33

A 4 stroke over-square single cylinder engine with an over square ratio of 1.1,the displacement volume of the engine is 245cc .The clearance volume is 27.2cc the bore of this engine is ?

Answers

Answer:

10.007

Explanation:

Assuming we have to find out the compression ratio of the engine

Given information

Cubic capacity of the engine, V = 245 cc

Clearance volume, V_c = 27.2 cc

over square-ratio = 1.1

thus,

D/L = 1.1

where,

D is the bore

L is the stroke

Now,

Volume of the engine V =[tex]\frac{\pi}{4} D^2L[/tex]

plugging values we get

245 = [tex]\frac{\pi}{4} D^3/1.1[/tex]

Solving we get D =7 cm

therefore,  L= 7/1.1 =6.36 cm

Now,

the compression ratio is given as:

r =(V+V_c)/V_c

on substituting the values, we get

r = (245+27.2)/27.2 =10.007

Hence, Compression ratio = 10.007

4. What element is missing from construction drawings?
A. Physical arrangement of specific electrical equipment
B. Electrical layout
C. Electrical connections
D. Side elevation views

Answers

Answer:

C Electrical Connections

Explanation:

In reading says . However, electrical

connections aren’t shown in construction drawings.

QUESTION 6
Which of the following is NOT a resume format?
01. Chronological
O2. Portfolio
3. Functional
04. Combination

Answers

It would be 2 Portfolio

A shaft is made from a tube, the ratio of the inside diameter to the outside diameter is 0.6. The material must not experience a shear stress greater than 500KPa. The shaft must transmit 1.5MW of mechanical power at 1500 revolution per minute. Calculate the shaft diameter

Answers

Answer:

shaft diameter = [tex]\sqrt[3]{0.3512}[/tex] mm = 0.7055 mm

Explanation:

Ratio of inside diameter to outside diameter ( i.e. d/D )= 0.6

Shear stress of material ( Z ) ≤ 500 KPa

power transmitted by shaft ( P ) = 1.5MW of mechanical power

Revolution ( N ) = 1500 rev/min

Calculate shaft Diameter

Given that: P = [tex]\frac{2\pi NT}{60}[/tex]  ---- 1

therefore; T = ( 1.5 *10^3 * 60 ) / ( 2[tex]\pi[/tex] * 1500 )  = 9.554 KN-M

next

[tex]\frac{T}{I_{p} } = \frac{Z}{R}[/tex]

hence ; T = Z[tex]_{p} *Z[/tex]

attached below is the remaining part of the solution

A 5-m-long, 4-m-high tank contains 2.5-m-deep water when not in motion and is open to the atmosphere through a vent in the middle. The tank is now accelerated to the right on a level surface at 2 m/s2. Determine the maximum gage pressure in the tank. Mark that point at the interior bottom of the tank. Draw the free surface at this acceleration.

Answers

Answer: hello your question lacks the required diagram attached below is the diagram

answer :  29528.1  N/m^2

Explanation:

Given data :

dimensions of tank :

Length = 5-m

Width = 4-m

Depth = 2.5-m

acceleration of tank = 2m/s^2

Determine the maximum gage pressure in the tank

Pa ( pressure at point A )  = s*g*h1

    = 10^3 * 9.81 * 3.01

    = 29528.1  N/m^2

attached below is the remaining part of the solution

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