Ask question

Ask question

All categories

3 years ago

6

Briefly explain how each of the following influences the tensile modulus of a semicrystalline polymer and why:(a) molecular weig

ht(b) degree of crystallinity(:1 deformation by drawing(d) annealing ofan undefonned material(e) annealing of a drawn material

1 answer:

marin [14]3 years ago

5 0

Answer:

(a) Increases

(b) Increases

(c) Increases

(d) Increases

(e) Decreases

Explanation:

The tensile modulus of a semi-crystalline polymer depends on the given factors as:

(a) Molecular Weight:

It increases with the increase in the molecular weight of the polymer.

(b) Degree of crystallinity:

Tensile strength of the semi-crystalline polymer increases with the increase in the degree of crystallinity of the polymer.

(c) Deformation by drawing:

The deformation by drawing in the polymer results in the finely oriented chain structure of the polymer with the greater inter chain secondary bonding structure resulting in the increase in the tensile strength of the polymer.

(d) Annealing of an undeformed material:

This also results in an increase in the tensile strength of the material.

(e) Annealing of a drawn material:

A semi crystalline material which is drawn when annealed results in the decreased tensile strength of the material.

You might be interested in

Steam enters a turbine at 8000 kPa, 440oC. At the exit, the pressure and quality are 150 kPa and 0.19, respectively.

levacccp [35]

Answer:

$\dot W_{out} = 3863.98\,kW$

Explanation:

The turbine at steady-state is modelled after the First Law of Thermodynamics:

$-\dot Q_{out} -\dot W_{out} + \dot m \cdot (h_{in}-h_{out}) = 0$

The specific enthalpies at inlet and outlet are, respectively:

Inlet (Superheated Steam)

$h_{in} = 3353.1\,\frac{kJ}{kg}$

Outlet (Liquid-Vapor Mixture)

$h_{out} = 890.1\,\frac{kJ}{kg}$

The power produced by the turbine is:

$\dot W_{out}=-\dot Q_{out} + \dot m \cdot (h_{in}-h_{out})$

$\dot W_{out} = -2.93\,kW + (1.57\,\frac{kg}{s} )\cdot (3353.1\,\frac{kJ}{kg} - 890.1\,\frac{kJ}{kg} )$

$\dot W_{out} = 3863.98\,kW$

8 0

3 years ago

A search will start from a visual lead<br> true<br> false

BlackZzzverrR [31]

True

An organized searching process will need to start from the visual lead area. Eye focus and eye movements from the path of travel in an organized pattern describes a visual search process.

4 0

2 years ago

Read 2 more answers

A 0.25in diameter copper rod BC is securely attached between two identical 1in diameter steel rods (AB and CD).

sashaice [31]

Explanation:

Below is an attachment containing the complete question and solution

5 0

3 years ago

Lumber jacks use cranes and giant tongs to hoist their goods into trucks for transport. Fortunately, smaller versions of these d

Tems11 [23]

Complete Question

Lumber jacks use cranes and giant tongs to hoist their goods into trucks for transport. Fortunately, smaller versions of these devises are available for weekend warriors who want to play with their chain saws. Let us model the illustrated tongs as a planar mechanism that carries a log of weight 210 N. Given the following dimensions: 35 mm 10 mm 40 mm 230 mm 85 mm 45 mm 10 mm 35 mm 345 mm determine the force in N and moment in Nm that our worker exerts on the tongs. Also determine the pinching force magnitude in N that the tongs exert on the log; i.e. determine the horizontal force that the tong's teeth exert on the log. Assume the point E is centered between the tong's teeth.

The diagram for this question is shown on the first uploaded image

Answer:

The force P is $P= 210 N$

and the moment M is $M = -48.3N \cdot m$

The horizontal force that the tong teeth exerts is $F_T =89.67N$

Explanation:

First let denote the dimension to corresponding to the diagram

$a=35mm , b= 10mm, c= 40mm, d= 230mm, e= 85mm,f= 45mm,\\g= 10mm,h=35mm,i=345mm.$

Next looking at the diagram let us consider the vertical direction

At equilibrium

$\sum F_{vertical} =0$

This mean that

$P+ W = 0$

Since they are acting in opposite direction the equation becomes

$P - W = 0$

=> $P= W$

=> $P= 210 N$

At Equilibrium Moment about F gives

$\sum M_f = 0$

=> $F_T * (e +f + g+ h+ i) - F_T * (e+ f+g+ h+i) - W *d -M =0$

=> $M = -W *d$

=> $M = -210 * 0.230$

=> $M = -48.3N \cdot m$

Here $F_T$ is the horizontal force that the tong teeth exerts

Now let consider the part BAF of the system as shown on the second uploaded image

Now the angle $\theta$ is mathematically given as

$tan \theta = \frac{g+h}{a}$

=> $\theta = arctan \frac{g+h}{a}$

$= arctan (\frac{10+35}{35} )$

$=52.125^o$

Now at equilibrium the moment about A is

$\sum M_A = 0$

=> $P * (c+d) +M + F_{BC} cos \theta *f+F_{BC}sin\theta *(a+b) =0$

$210 * (0.040 + 0.230)-48.3+F_{BC} cos (52.125^o)*0.045+ F_{BC}sin(52.125^o)$ $* (0.035 +0.010) =0$

=> $10.29 +F_{BC} (0.02763+0.03552) =0$

=> $F_{BC} =\frac{10.29}{0.06315}$

=> $F_{BC} = - 162.925 N$

Looking at the forces acting on the teeth as shown on the third uploaded image

At Equilibrium the moment about D is

$\sum M_D = 0$

=> $\frac{W}{2} *d - F_T *(i+h) -F_{BC} cos \theta *h -F_{BC} sin \theta * (b+c) =0$

=> $\frac{210}{2} * 0.230 -F_T (0.345 +0.035) - (-162.925)cos(52.125^o) *0.035\\-(-162.925)sin(52.125^o)(0.010 +0.040) =0$

=> $34.081 = F_T(0.345 +0.035)$

=> $F_T =89.67N$

7 0

4 years ago

1. In a deck of cards, Kings, Queens and Jacks are called the face cards. If 4 cards are

VLD [36.1K]

The answer would be a standard of 26 to 2 because I did the same quiz

4 0

3 years ago