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Nina [5.8K]
3 years ago
8

Poly(cis-1,4-isoprene), or natural rubber (NR), has a tendency crystallize. The Tm of this polymer is slightly below room temper

ature, 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.
Engineering
1 answer:
Nonamiya [84]3 years ago
8 0

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.

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D. Leak-proof

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Using an "AND" and an "OR", list all information (Equipment Number, Equipment Type, Seat Capacity, Fuel Capacity, and Miles per
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Answer:

Explanation :

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Check the attached document for the solution.

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3 years ago
A shaft made of stainless steel has an outside diameter of 42 mm and a wall thickness of 4 mm. Determine the maximum torque T th
fiasKO [112]

Answer:

Explanation:

Using equation of pure torsion

\frac{T}{I_{polar} }=\frac{t}{r}

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T is the applied Torque

I_{polar} is polar moment of inertia of the shaft

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r is distance from center

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D_{0} = Outer Diameter

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I_{polar}=\frac{\pi (D_{o} ^{4}-D_{in} ^{4}) }{32}

Applying values in the above equation we get

I_{polar} =\frac{\pi(0.042^{4}-(0.042-.008)^{4})}{32}\\I_{polar}= 1.74 x 10^{-7} m^{4}

Thus from the equation of torsion we get

T=\frac{I_{polar} t}{r}

Applying values we get

T=\frac{1.74X10^{-7}X100X10^{6}  }{.021}

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3 years ago
(a) Consider a message signal containing frequency components at 100, 200, and 400 Hz. This signal is applied to a SSB modulator
MakcuM [25]

Answer:

Explanation:

The frequency components in the message signal are

f1 = 100Hz, f2 = 200Hz and f3 = 400Hz

When amplitude modulated with a carrier signal of frequency fc = 100kHz

Generates the following frequency components

Lower side band

100k - 100 = 99.9kHz\\\\100k - 200 = 99.8kHz\\\\100k - 400 = 99.6kHz\\\\

Carrier frequency 100kHz

Upper side band

100k + 100 = 100.1kHz\\\\100k + 200 = 100.2kHz\\\\100k + 400 = 100.4kHz

After passing through the SSB filter that filters the lower side band, the transmitted frequency component will be

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At the receive these are mixed (superheterodyned) with local ocillator frequency whichh is 100.02KHz, the output frequencies will be

100.02 - 100.1k = 0.08k = 80Hz\\\\100.02 - 100.2k = 0.18k = 180Hz\\\\100.02 - 100.4 = 0.38k = 380Hz

After passing through the SSB filter that filters the higher side band, the transmitted frequency component will be

100k, 99.9k, 99.8k\ \ and \ \99.6kHz

At the receive these are mixed (superheterodyned) with local oscillator frequency which is 100.02KHz, and then fed to the detector whose output frequencies will be

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3 0
3 years ago
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