ENGINEERING COLLEGE

Although the viscoelastic response of a polymer can be very complex (time-dependent stress cycling for instance), two special loading scenarios are fairly simple to describe mathematically. _refers to scenarios for which the stress applied to a polymer must decay over time in order to maintain a constant strain. Otherwise, over time, the polymer chains will slip and slide past one another in response to a constant applied load and the strain will increase (in magnitude).
_refers to scenarios for which a polymer will permanently flow over time in response a constant applied stress.

Answers

Answer 1

Answer:

Answer:

Viscoelastic stress relaxation

Viscoelastic creep

Explanation:

Viscoelastic stress refers to the reduction of tensile stress with the relaxation time that occurs when the body is kept at a certain length under tensile stress. The purpose of this study was to demonstrate viscoelastic stress relaxation
Therefore, when the viscous tension stress occurs, the pressure decreases due to the steady-state stress of the phases.
When under constant pressure, the viscoelastic material experiences a time-dependent increase in pressure. This phenomenon is called viscoelastic creep.
Therefore the phase constant pressure decreases when there is relaxation in the viscoelastic stress

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In an adiabatic process, the temperature of the system remains constant. a)- True b)- False

Answers

Answer:

A. True

Explanation:

For some transformation having kinetics that obey the Avrami equation (Equation 10.17), the parameter n is known to have a value of 2.1. If, after 146 s, the reaction is 50% complete, how long (total time) will it take the transformation to go to 86% completion?

Answers

Answer:

t = 25.10 sec

Explanation:

we know that Avrami equation

$Y = 1 - e^(-kt^n)$

here Y is percentage of completion of reaction = 50%

t is duration of reaction = 146 sec

so,

$0.50 = 1 - e^(-k^146^2.1)$

$0.50 = e^(-k306.6)$

taking natural log on both side

ln(0.5) = -k(306.6)

$k = 2.26* 10^(-3)$

for 86 % completion

$0.86 = 1 - e^{-2.26* 10^(-3) * t^(2.1)}$

$e^{-2.26* 10^(-3) * t^(2.1)} = 0.14$

$-2.26* 10^(-3) * t^(2.1) = ln(0.14)$

$t^(2.1) = 869.96$

t = 25.10 sec

Consider the products you use and the activities you perform on a daily basis. Describe three examples that use both SI units and customary units for measurement.

Answers

Three activities that I can do on a daily basis that involve both metric units (SI units) and customary units are: measuring the length of a door with a tape measure, which includes both SI units and customary units (like feet, inches, and centimeters); baking a cake that calls for one teaspoon (customary unit) of baking soda, which can also be converted to four grams (SI unit); and weighing myself on a weighing scale, which can be measured in pound and kilogram (metric unit).

Answer: Three examples of activities that I can perform on a daily basis that involves both metric units (SI units) and customary units include: measuring the length of a door using a tape measure, which includes both SI units and customary units (like feet, inches, and centimeters); baking a cake that requires one teaspoon (customary unit) of baking soda, which could also be converted into four grams (SI unit); weighing myself on a weighing scale, which can be measured by pounds (customary unit) or kilograms (metric unit).

Explanation:I big brain :) (Not Really I Just Wanted To Help) I hope this helped! ;)

Shear modulus is analogous to what material property that is determined in tensile testing? (a)- Percent reduction of area (b) Yield strength (c)- Elastic modulus (d)- Poisson's ratio

Answers

Answer:

(c)- Elastic modulus

Explanation:

We know that in tensile test we measure the properties of the material like yield strength,ultimate tensile strength ,Poisson ratio.

In tensile test

σ = ε E

Where σ is the stress

ε is the strain.

E is the elastic modulus.

Now for shear tress

τ = Φ G

Where τ the shear stress

Φ is the shear strain.

G is the shear modulus.

So we can say that Shear modulus is analogous to Elastic modulus.

A 1-m3 tank containing air at 10°C and 350 kPa is connected through a valve to another tank containing 3 kg of air at 35°C and 150 kPa. Now the valve is opened, and the entire system is allowed to reach thermal equilibrium with the surroundings, which are at 19.5°C. Determine the volume of the second tank and the final equilibrium pressure of air. The gas constant of air is R = 0.287 kPa·m3/kg·K.

Answers

Answer:

V₂=1.76 m³

P=222.03 KPa

Explanation:

Given that

For tank 1

V₁=1 m³

T₁= 10°C = 283 K

P₁=350 KPa

For tank 2

m₂=3 kg

T₂=35°C = 308 K

P₂=150 KPa

We know that for air

P V = m R T

P=pressure ,V= Volume,R= gas constant ,T= temperature ,m =mass

for tank 2

P₂ V₂ = m₂ R T₂

By putting the values

150 x V₂ = 3 x 0.287 x 308

V₂=1.76 m³

Final mass = m₁+m₂

m =m₁+m₂

The final volume V= V₂+V₁

V= 1.76 + 1 m³

V= 2.76 m³

The final temperature T= 19.5°C

T= 292.5 K

$m=(PV)/(RT)$

$m_1=(P_1V_1)/(RT_1)$

$m_1=(350* 1)/(0.287* 283)$

$m_1=4.3\ kg$

m =m₁+m₂

m =4.3 + 3 = 7.3 kg

Now at final state

P V = m R T

P x 2.76 = 7.3 x 0.287 x 292.5

P=222.03 KPa

For steels, we can assume the endurance limit measured by rotating beam tests is the 50% of the ultimate tensile strength (UTS) as long as the UTS is equal to or less than 200 ksi (1400 MPa).True/False

Answers

Answer:

True

Explanation:

For steels, we can assume the endurance limit measured by rotating beam tests is the 50% of the ultimate tensile strength (UTS) as long as the UTS is equal to or less than 200 kpsi (1400 MPa).

Also, It is a simplistic rule of thumb that, for steels having a UTS less than 160 kpsi, the endurance limit for the material will be approximately 45 to 50% of the UTS.

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