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3 years ago

Identify the types of bonding that occur within a polymer system and how this affects its properties

Engineering

1 answer:

Vinil7 [7]3 years ago

7 0

Answer:

Covalent bond.

Explanation:

As we know that the main component pf polymer is carbon.The carbon element present in the left side of the periodic table and carbon makes covalent bond.That is why in the polymer covalent bonding present.Polymer is the long chain of covalent bond.Covalent bond is also known as polar bond.We know that in the covalent bond balance electron are shared by both elements and to get stability.

The properties of covalent bond given as

1. Poor conductor of electricity and heat

2. Low melting point

3. Low boiling point

4.Generally brittle

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What is the magnitude of the maximum stress that exists at the tip of an internal crack having a radius of curvature of 3.5 × 10

babunello [35]

Answer:

The magnitude of the maximum stress that exists at the tip of the internal crack is 3900.183MPa (565703.38 psi).

Explanation:

The magnitude of the maximum stress that exists at the tip of an internal crack can be estimated using fracture mechanics. If a material that have internal crack is loaded with tensile stress, the crack tends to widen due to the tensile force created by the stress, and this leads to a concentration of stress near the tip of the crack, thereby expanding the crack length and finally leading to failure of the material.

Fracture mechanics involves the study of propagation of cracks in a material. It is used to predict the conditions at which failure of a material will likely occur.

A crack may appear in the surface of the material (surface crack) or at the interior of the material (internal crack), but the same formula can be used to estimate the magnitude of the maximum stress at the tip of the crack.

The formula is given as Πm=2Πo√(a/r)

Where,

Πm is the maximum stress

Πo is the tensile stress

r is the radius of curvature

a is the crack length.

The major difference between the surface crack and the internal crack is the value for their crack length.

Crack length can be understood as the length of a crack at which the crack becomes unstable under certain applied stress.

For surface crack, crack length=a

For internal crack, crack length=2a because internal cracks forms two surfaces of a complete sphere, while the surface crack forms one surface of a sphere.

Given in the question,

Radius of curvature, r=3.5×10^(-4) mm=0.00035mm

Crack length, a=5.5×10^(-2) mm

For internal crack, a=[(crack length)/2]={[5.5×10^(-2)]/2}=0.0275mm

Tensile stress, Πo=220Mpa

Maximum stress=Πm

Using,

Πm=2Πo√(a/r)

=(2×220)√(0.0275/0.00035)

=440√78.571

=3900.183MPa (565703.38 psi)

4 0

3 years ago

8. Two 40 ft long wires made of differing materials are supported from the ceiling of a testing laboratory. Wire (1) is made of

san4es73 [151]

Answer:

Material K has a modulus of elasticity E=3.389× 10¹¹ Pa

Material H has a modulus of elasticity E=1.009 × 10⁹ Pa

Material K has higher value of modulus of elasticity than material H

Material K is stiffer.

Explanation:

Wire 1 material H

Length=L = 40 ft =12.192 m

Diameter= 3/8 in = 0.009525 m

Area= A= πr²,where r=0.009525/2 =0.004763

A=3.142*0.004763² =0.00007126 m²

Force, F= 225 lb= 225*4.45 =1001.25 N

Change in length =Δ L= 0.10 in = 0.00254

To find modulus of elasticity apply'

E=F*L/A*ΔL

E=1001.25*12.192/(0.004763*0.00254)

E= 1009027923.58 Pa

E=1.009 × 10⁹ Pa

For Wire 2 material K

Length=L= 40 ft =12.192 m

Diameter = 3/16 in = 0.1875 in = 0.004763 m

Area= πr² = 3.142 * (0.004763/2)² = 0.00000567154 m²

Force, F= 225 lb= 225*4.45 =1001.25 N

Change in length =Δ L= 0.25 in =0.00635 m

To find modulus of elasticity apply'

E=F*L/A*ΔL

E= (1001.25*12.192)/(0.00000567154 * 0.00635 )

E=338955422575 Pa

E=3.389× 10¹¹ Pa

Material K has a greater modulus of elasticity

The material with higher value of E is stiffer than that with low value of E.The stiffer material is K.

8 0

3 years ago

Question 1 An ice making machine operates on the ideal vapour-compression cycle using refrigerant R134a. The refrigerant enters

-BARSIC- [3]

Answer:

check the attached image for the correct solution

Explanation:

In an ideal vapor-compression refrigeration cycle, the refrigerant enters the compressor as a saturated vapor and is cooled to the saturated liquid state in the condenser. It is then throttled to the evaporator pressure and vaporizes as it absorbs heat from the refrigerated space.

the correct workings/calculation is in the attached image

4 0

4 years ago

The process of hazard recognition, evaluation, and control is the foundation of an effective___program.

Alona [7]

Answer: A. Safety program

7 0

3 years ago

In a river reach, the rate of inflow at any time is 350 cfs and the rate of outflow is 285 cfs. After 90 min, the inflow and out

Semenov [28]

Answer:

change in storage = -310,500 ft^3

intital storage= 3.67 acre ft

Explanation:

Given data:

Rate of inflow = 350 cfs

Rate of outflow = 285 cfs

After 90 min, rate of inflow = 250 cfs

Rate of outflow = 200 cfs

final storage = 10.8 acre-ft

calculating the average inflow and outflow

average inflow $= \frac{(350+250)}{2} = 300 cfs$

average outlow $= \frac{(285+200)}{2} = 242.5 cfs$

total amount of water drain during the period of one hour

= (average outflow - average inflow) *60*90

= (242.5 - 300)*60*90 = -310,500

change in storage is calculate as

= -310,500 ft^3

in cubic meter

= -310500/35.315 = 8792.30 cm^3

in acre-ft

= -310,500/43560 = 7.13 acre ft

initial storage = 10.8 - 7.13 = 3.67 acre ft

3 0

3 years ago