Let L be a language defined over Σ = {a, b}. Let L˜ ⊆ {a, b} ∗ be the set of strings derived from strings of L by toggling the first letter. For example, if bbba ∈ L, then abba ∈ L˜. Λ ∈ L if and only if Λ ∈ L˜. For example, if L = aa∗ b ∗ , then L˜ = ba∗ b ∗ .
(a) Build a finite automaton for a ∗ b(aa ∪ b) ∗
(b) Show that regular languages are closed under the ~ operator. Do this by giving a general method that takes any finite automaton M that accepts a language L, and constructs a DFA or NFA that accepts the language L˜. Hint: create a new start state that has arrows with labels different from the original start state.
6. (20 pts) Let L be a language defined over Σ = {a,b}. Let L ≤ {a,b}*
C
be the set of strings derived from strings of L by t(c) Apply your construction on the automaton you built

Enterprise resource planning, or ERP, is the term used to describe the electronic flow of data between businesses across networks like the Internet. false

Electric(al) devices are tools that functionally rely on electricity to power their fundamental parts. They can be considered as being superior to traditional mechanical systems, which depend on diverse power sources like fuels or human physical strength. A certain type of electrical equipment called an electronic device is used more often to process data than to generate mechanical forces. To more clearly distinguish between the two types, electric gadgets that emphasize physical work are also known as electromechanical devices. Mechatronics demonstrates how the two sciences overlap.The study of electronic and electric devices, as well as their design, maintenance, and power supply, is known as electrical engineering.

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The entropy of 2 moles of and ideal gas expanding freely to 3 times it's initial volume is 18.3J/k

The following formula is required:

∆S = nx R x ℓn x (Vf/Vi)

Where

n = number of moles of gas (n = 2)

R = gas constant (R = 8.314 J/(mol * K))
Vf = final volume (Vf = 3.V1)

Vi = intial volume

Vi = 1L (Asumption )

∆S = 2 x 8.314 x 1.099

∆S =18.3 j/K

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Translation:

Calculate the entropy of 2 moles of an ideal gas expanding freely to three times its initial volume, use: ∆S =n・R・ℓn (Vf / Vi)

Answer:

10

Explanation:

Answer:

A uniform T-section beam is 100 mm wide and 150 mm deep with a flange thickness of 25 mm and a web thickness of 12 mm. If the limiting bending stresses for the material of the beam are 80 MN/m2 in compression and 160 MN/m2 in tension, find the maximum u.d.l. that the beam can carry over a simply supported span of 5 m.

Explanation:

hope this helps if not let me know

A 4-digit password may be set to prevent unauthorized operation of the transceiver without permission.

It is a system created to stop unapproved access to or from a private network. So, Option 4 is the right choice. A firewall can be put into place as either hardware, software, or a hybrid of the two.

programme firewall: You can help protect your computer from unauthorised data entry and exit by installing a software firewall on it. Securing only the machine on which it is installed is the main objective of a software firewall. Many antivirus scanners come with a software firewall as standard equipment.

taking or requesting to "borrow" another employee's badge. Following another person through access doors in an attempt to "piggyback." using a colleague's password or login information to access a computer.

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A NAAQS exceedance refers to temporary levels exceeding the established standard, while a NAAQS violation indicates a persistent or recurring non-compliance with the standard.

NAAQS (National Ambient Air Quality Standard), set by regulatory agencies to protect public health and the environment, establish maximum allowable levels for pollutants in the ambient air. The terms "exceedance" and "violation" are used to describe different scenarios of non-compliance:

1. NAAQS Exceedance: A NAAQS exceedance refers to a temporary event where pollutant concentrations surpass the standard. It may occur due to short-term spikes in pollution levels caused by localized sources, unusual weather conditions, or specific events. Exceedances are typically evaluated and addressed on a case-by-case basis and may not immediately trigger regulatory actions.

2. NAAQS Violation: A NAAQS violation signifies a sustained or recurring non-compliance with the established standard. It occurs when pollutant levels consistently exceed the NAAQS over a specified timeframe, such as an averaging period (e.g., 24 hours or annual). Violations trigger regulatory consequences and the implementation of corrective measures, such as emission controls, enforcement actions, or mandated pollution reduction plans.

Differentiating between exceedances and violations is crucial in regulatory decision-making and prioritizing resources for air quality management. While exceedances may warrant investigation and localized actions, violations indicate the need for more significant and sustained efforts to achieve and maintain compliance with the NAAQS.

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The value of ΔS when 6.00 mol of Br2(l) is vaporized at 58.8 °C is approximately 559.2 J/K.

To calculate the change in entropy (ΔS) when 6.00 mol of Br2(l) is vaporized at 58.8 °C, we need to use the equation:

ΔS = n × ΔSvap

where:

n = number of moles

ΔSvap = molar entropy of vaporization

The molar entropy of vaporization (ΔSvap) can be found in reference sources and is typically given in units of J/(mol·K).

Let's assume that the value of ΔSvap for Br2 is 93.2 J/(mol·K). Using this value, we can calculate ΔS as follows:

ΔS = 6.00 mol × 93.2 J/(mol·K)

ΔS = 559.2 J/K

Therefore, the value of ΔS when 6.00 mol of Br2(l) is vaporized at 58.8 °C is approximately 559.2 J/K.

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Answer:

To divide the ultimate (or maximum) stress by the typical (or working) stress.

Explanation:

Hopes this helps :)

Answer:

10.203 Volts

Explanation:

For this problem, we need to understand that a series resistive circuit is simply a circuit with some type of voltage source and some resistors, in this case, R1 and R2.

First, we need to find the voltage in the circuit.  To do this, we need to find the total resistance of the circuit.  When two resistors are in series, you sum the resistance.  So we can say the following:

R_Total = R1 + R2

R_Total = 570 Ω + 560 Ω

R_Total = 1130 Ω

Now that we have R_Total for the circuit, we can find the voltage of the circuit by using Ohm's law, V = IR.

V_Total = I_Total * R_Total

V_Total = 17.9 mA * 1130 Ω

V_Total = 20.227 V

Now that we have V_Total, we can find the voltage drop across each resistor by using Ohm's law once more.  Note, that since our circuit is series, both resistors will have the same current (I.e., I_Total = I_1 = I_2).

V_Total = V_1 + V_2

V_Total = V_1 + I_2*R2

V_Total - I_2*R2 = V_1

20.227 V - (17.9 mA * 560 Ω) = V_1

20.227 V - (10.024 V) = V_1

10.203 V = V_1

Hence, the voltage drop across R1 is 10.203 Volts.

Cheers.

The common workplace for both science and math pathways, with engineering and technology in laboratories and ships. Thus, option A is correct.

A workplace is given as the location or the place at which the employee are performing their jobs and tasks.

The science or technology-mediated employees are related to research where they perform the tests in laboratories, as well as the ships are the location for the engineering and the math candidates for the analysis of the studies and to perform the job.

Hence, the location in the workplace common for both engineering and technology candidates has been the laboratories and the ships. Thus, option A is correct.

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Answer:

a. The time required for the tank to empty halfway is presented as follows;

\(t_1 = \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \left (\sqrt{2} -1 \right)\)

b. The time it takes for the tank to empty the remaining half is presented as follows;

\(t_2 = { \dfrac{ D_0^2 }{D} \cdot\sqrt{\dfrac{H}{g} }\)

The total time 't', is presented as follows;

\(t = \sqrt{2} \cdot \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} }\)

Explanation:

a. The diameter of the tank = D₀

The height of the tank = H

The diameter of the orifice at the bottom = D

The equation for the flow through an orifice is given as follows;

v = √(2·g·h)

Therefore, we have;

\(\dfrac{P_1}{\gamma} + z_1 + \dfrac{v_1}{2 \cdot g} = \dfrac{P_2}{\gamma} + z_2 + \dfrac{v_2}{2 \cdot g}\)

\(\left( \dfrac{P_1}{\gamma} -\dfrac{P_2}{\gamma} \right) + (z_1 - z_2) + \dfrac{v_1}{2 \cdot g} = \dfrac{v_2}{2 \cdot g}\)

Where;

P₁ = P₂ = The atmospheric pressure

z₁ - z₂ = dh (The height of eater in the tank)

A₁·v₁ = A₂·v₂

v₂ = (A₁/A₂)·v₁

A₁ = π·D₀²/4

A₂ = π·D²/4

A₁/A₂ = D₀²/(D²) = v₂/v₁

v₂ = (D₀²/(D²))·v₁ = √(2·g·h)

The time, 'dt', it takes for the water to drop by a level, dh, is given as follows;

dt = dh/v₁ = (v₂/v₁)/v₂·dh = (D₀²/(D²))/v₂·dh = (D₀²/(D²))/√(2·g·h)·dh

We have;

\(dt = \dfrac{D_0^2}{D} \cdot\dfrac{1}{\sqrt{2\cdot g \cdot h} } dh\)

The time for the tank to drop halfway is given as follows;

\(\int\limits^{t_1}_0 {} \, dt = \int\limits^h_{\frac{h}{2} } { \dfrac{D_0^2}{D} \cdot\dfrac{1}{\sqrt{2\cdot g \cdot h} } } \, dh\)

\(t_1 =\left[{ \dfrac{D_0^2}{D\cdot \sqrt{2\cdot g} } \cdot\dfrac{h^{-\frac{1}{2} +1}}{-\frac{1}{2} +1 } \right]_{\frac{H}{2} }^{H} =\left[ { \dfrac{D_0^2 \cdot 2\cdot \sqrt{h} }{D\cdot \sqrt{2\cdot g} } \right]_{\frac{H}{2} }^{H} = { \dfrac{2 \cdot D_0^2 }{D\cdot \sqrt{2\cdot g} } \cdot \left(\sqrt{H} - \sqrt{\dfrac{H}{2} } \right)\)

\(t_1 = { \dfrac{2 \cdot D_0^2 }{D^2\cdot \sqrt{2\cdot g} } \cdot \left(\sqrt{H} - \sqrt{\dfrac{H}{2} } \right) = { \dfrac{\sqrt{2} \cdot D_0^2 }{D^2\cdot \sqrt{ g} } \cdot \left(\sqrt{H} - \sqrt{\dfrac{H}{2} } \right)\)

\(t_1 = { \dfrac{\sqrt{2} \cdot D_0^2 }{D^2\cdot \sqrt{ g} } \cdot \left(\sqrt{H} - \sqrt{\dfrac{H}{2} } \right) = { \dfrac{D_0^2 }{D^2\cdot \sqrt{ g} } \cdot \left(\sqrt{2 \cdot H} - \sqrt{{H} } \right) =\dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \left (\sqrt{2} -1 \right)\)The time required for the tank to empty halfway, t₁, is given as follows;

\(t_1 = \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \left (\sqrt{2} -1 \right)\)

(b) The time it takes for the tank to empty completely, t₂, is given as follows;

\(\int\limits^{t_2}_0 {} \, dt = \int\limits^{\frac{h}{2} }_{0 } { \dfrac{D_0^2}{D} \cdot\dfrac{1}{\sqrt{2\cdot g \cdot h} } } \, dh\)

\(t_2 =\left[{ \dfrac{D_0^2}{D\cdot \sqrt{2\cdot g} } \cdot\dfrac{h^{-\frac{1}{2} +1}}{-\frac{1}{2} +1 } \right]_{0}^{\frac{H}{2} } =\left[ { \dfrac{D_0^2 \cdot 2\cdot \sqrt{h} }{D\cdot \sqrt{2\cdot g} } \right]_{0 }^{\frac{H}{2} } = { \dfrac{2 \cdot D_0^2 }{D\cdot \sqrt{2\cdot g} } \cdot \left( \sqrt{\dfrac{H}{2} } -0\right)\)

\(t_2 = { \dfrac{ D_0^2 }{D} \cdot\sqrt{\dfrac{H}{g} }\)

The time it takes for the tank to empty the remaining half, t₂, is presented as follows;

\(t_2 = { \dfrac{ D_0^2 }{D} \cdot\sqrt{\dfrac{H}{g} }\)

The total time, t, to empty the tank is given as follows;

\(t = t_1 + t_2 = \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \left (\sqrt{2} -1 \right) + t_2 = { \dfrac{ D_0^2 }{D} \cdot\sqrt{\dfrac{H}{g} } = \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \sqrt{2}\)

\(t = \sqrt{2} \cdot \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} }\)

Answer:

as all the people should go near stratosphere

Based on the information given, we can use the principle of work and energy to solve for the velocity of the collar when it reaches point B. Since the force acting on the cord is constant, the work done on the collar can be expressed as:

W = Fd

where W is the work done, F is the force acting on the cord (300N), and d is the distance travelled by the collar. We can express d in terms of the angle θ through which the pulley rotates:

d = 2πrθ

where r is the radius of the pulley. Neglecting the size of the pulley means that we can assume r is constant. We can then express θ in terms of the distance travelled by the collar:

θ = d/r

Substituting this expression for θ into the expression for d, we get:

d = 2πr(d/r) = 2πd

which gives us:

d = 2πr/(1-2π)

Next, we can use the work-energy principle to relate the work done on the collar to its change in kinetic energy:

W = ΔK

where ΔK is the change in kinetic energy of the collar. Initially, the collar is at rest, so its initial kinetic energy is zero. When it reaches point B, its final kinetic energy is:

Kf = (1/2)mv^2

where m is the mass of the collar (15kg) and v is its velocity. Therefore, the work done on the collar can be expressed as:

W = (1/2)mv^2 - 0

Substituting the expression for W, we get:

300d = (1/2)mv^2

Substituting the expression for d, we get:

300(2πr/(1-2π)) = (1/2)(15)v^2

Solving for v, we get:

v = √(600πr/(1-2π*15))

Since the radius of the pulley was not given, we cannot provide a numerical answer. However, this expression gives us the velocity of the collar in terms of the radius of the pulley. We can see that neglecting the size of the pulley does not affect the expression for v, since the radius cancels out.

In conclusion, the velocity of the collar when it reaches point B is given by the expression:

v = √(600πr/(1-2π*15))

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The straightforward response is that PJM is not profitable. Federal law mandates that PJM operate in a profit-neutral manner, which requires that overall income and expenses balance out over the long run.

The movement of wholesale electricity in all or parts of Delaware, Illinois, Indiana, Kentucky, Maryland, Michigan, New Jersey, North Carolina, Ohio, Pennsylvania, Tennessee, Virginia, West Virginia, and the District of Columbia is coordinated by the regional transmission organization (RTO) known as PJM Interconnection.

Electricity is purchased by the PJM Energy Market to satisfy both immediate and future consumer demand. It incorporates the sale or purchase of energy in the five-minute Real-Time Energy Market and the day-ahead Market of PJM (one day forward). The Learning Center has further information on the energy industry.

More than 1,379 electric generators and more than 85,103 miles of high-voltage transmission lines are monitored and coordinated by PJM. We are not equipment owners, just like air traffic controllers. Electrical plants and power lines belong to third parties.

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The complete question is:

Why is PJM, a regional network operator that includes PECO, asking consumers to find ways to safely save electricity until 10:00 AM tomorrow?

False. Fire extinguishing equipment cannot be frozen according to Ref. 213/91.

According to Ref. 213/91, fire extinguishing equipment cannot be frozen. Fire extinguishers are essential safety devices designed to combat fires effectively. They contain pressurized agents that are specifically formulated to extinguish different types of fires. Freezing temperatures can significantly impair the functionality of fire extinguishers and render them ineffective in emergency situations.

When fire extinguishing equipment freezes, several issues can arise. First, the contents of the extinguisher may expand as they freeze, potentially leading to ruptures or leaks in the container. This can cause the extinguisher to malfunction or become hazardous when used. Second, freezing temperatures can affect the performance of the extinguishing agent itself. Certain agents, such as water-based solutions, can solidify or lose their effectiveness when exposed to extreme cold.

It is crucial to store fire extinguishers in suitable environments that are above freezing temperatures. This ensures that the equipment remains in optimal condition and is ready for immediate use during emergencies. Regular inspections and maintenance are also essential to identify any signs of damage or deterioration that may compromise the functionality of fire extinguishers.

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Answer: 3/5

Explanation:

The probability of an event is

(number of ways an event can happen)/(number of ways anything can happen)

There are 6 ways to choose a card from U to Z. You can draw U,V,W,X,Y, or Z.

There are 10 total cards. So, the probability is 6/10=3/5.

OSHA regulations outline the procedures that employers must follow to safeguard their workers against dangers. OSHA rules fall under the following four categories: general industry, construction, maritime.

With the Occupational Safety and Health Act of 1970, Congress founded the Occupational Safety and Health Administration (OSHA) to promote safe and healthful working conditions for workers by creating and enforcing standards and by providing training, outreach, information and assistance.

You are entitled to a secure workplace. To stop workers from being murdered or suffering other types of harm at work, the Occupational Safety and Health Act of 1970 (OSH Act) was passed. The law mandates that employers give their staff unrestricted working conditions.

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Answer:

Foot

Explanation:

It is. trust me

Answer:

A: TURBINE

B: THERMODYNAMICS

C: THERMODYNAMICS

D: DIFFUSER

E: DIFFUSER

F: HEAT EXCHANGER

G: THROTTLE THROTTLE

Explanation:

hope it helps

Espionage is  one of the following is an example of a disclosure threat.

What is Espionage ?

Espionage is the act of obtaining confidential or sensitive information by unauthorized means, typically through spying or covert means. This information may be related to national security, business secrets, scientific research, or other sensitive information. Espionage is often carried out by individuals working for governments, intelligence agencies, or criminal organizations for the purpose of gaining strategic advantage or economic benefit.

Espionage can take many forms, including hacking into computer systems, intercepting communications, recruiting insiders, or using physical means to access information.

The correct answer is c. Espionage, as it is an example of a disclosure threat.

A disclosure threat is a security risk that results in the unauthorized release of confidential or sensitive information. Espionage refers to the act of obtaining confidential or sensitive information by unauthorized means, typically through spying or covert means. This constitutes a threat to the confidentiality of the information, as it may be disclosed to unauthorized parties who can use it for malicious purposes.

The other options (alteration, denial, and destruction) are not examples of disclosure threats. Alteration refers to unauthorized changes to information, denial refers to the prevention of access to information, and destruction refers to the complete deletion or destruction of information. These are different types of security risks that have different impacts on the confidentiality, availability, and integrity of information.

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The question asks for the lowest possible elevation of the bottom of the overpass to ensure sufficient stopping sight distance at 60 mi/hr. Stopping sight distance is the distance required for a driver to bring their vehicle to a complete stop in an emergency.
To determine the stopping sight distance, we can use the following formula:

Stopping Sight Distance = Perception Distance + Reaction Distance + Braking Distance

Perception distance can be calculated as follows:

Perception Distance = 1.47 (t + 1.5V)

where t is the perception-reaction time (1.5 seconds) and V is the design speed (60 mi/hr).

Perception Distance = 1.47 (1.5 + 1.5 x 60) = 220.5 ft

Reaction distance can be calculated as follows:

Reaction Distance = V x t

Reaction Distance = 60 x 1.5 = 90 ft

Braking distance can be calculated as follows:

Braking Distance = (V² / 30f) + (V/2g)(a + f)

where f is the coefficient of friction between the tires and the road surface, g is the acceleration due to gravity (32.2 ft/s²), and a is the grade.

Since the initial grade is -3.0%, we can assume that the entire vertical curve has a grade of -3.0%. Therefore, a = -3.0%.

The coefficient of friction depends on several factors such as the condition of the road surface, the weather conditions, and the type of tires. A typical value for the coefficient of friction for dry pavement is 0.7.

Braking Distance = (60² / (30 x 0.7)) + (60/2 x 32.2)(-0.03 + 0.7)

Braking Distance = 410.96 + 180.18 = 591.14 ft

Therefore, the stopping sight distance is:

Stopping Sight Distance = Perception Distance + Reaction Distance + Braking Distance

Stopping Sight Distance = 220.5 + 90 + 591.14 = 901.64 ft

To ensure sufficient stopping sight distance, the driver must be able to see at least 901.64 ft ahead of their vehicle. This means that the lowest possible elevation of the bottom of the overpass should be such that the vertical distance from the driver's eye height (5 ft) to the overpass is greater than or equal to 901.64 ft.

Let h be the height of the overpass above the PVI. Then the vertical distance from the driver's eye height to the overpass is h - 5 ft.

h - 5 ft ≥ 901.64 ft

h ≥ 906.64 ft

Therefore, the lowest possible elevation of the bottom of the overpass should be 906.64 ft above the PVI.

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Answer:

english

Explanation:

Explanation:

(1) 174 inch

(2)1 foot 3.6 inches

(3) 39.649 centimetre

(4)175 decimeter

(5)26.25 feet

The quartiles divide a set of observations into four portions, each representing 25% of the observations, together with the minimum and maximum values of the data set. The interquartile range, a measurement of variation around the median, is calculated using quartiles.

Statistical file: {2, -6, 5}

Quartile Q1: -6

Quartile Q2: 2

Quartile Q3: 5.

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Answer:

Gauge Pressure = 408.3 KPa

Explanation:

The pressure inside the pipe can be given in terms of the elevation, by the following formula:

P = ρgΔz

where,

P = Absolute Pressure = ?

ρ = Density of Water = 1000 kg/m³

g = acceleration due to gravity = 9.8 m/s²

Δz = elevation = 52 m

Therefore,

P = (1000 kg/m³)(9.8 m/s²)(52 m)

P = 509.6 KPa

Now, for gauge pressure:

Gauge Pressure = P - Atmospheric Pressure

Gauge Pressure = 509.6 KPa - 101.3 KPa

Gauge Pressure = 408.3 KPa

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.