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Olegator [25]
3 years ago
11

During fabrication processes, polymeric materials are generally subjected to which of the following conditions? The ambient temp

erature. Elevated temperatures. The ambient pressure. Elevated pressures.
Physics
1 answer:
geniusboy [140]3 years ago
3 0

Answer:

- Elevated temperatures.

- Elevated pressures.

Explanation:

In the manufacture of plastic products are used:

Raw material (pellets) which are the monomers that promote the chemical reaction.

To these are added the charges, in order to reduce the cost of the final product and improve some properties. These charges can be fiberglass, paper, metal structures.

Additives are also added whose mission is to improve or achieve certain properties, such as reducing friction, reducing chemical degradation, increasing electrical conductivity, coloring the product, and all this happens in the presence of a catalyst that is responsible for initiating and accelerating the chemical reaction process.

There are different methods of production of plastics, one of the most frequent is by injection.

Injection is a process that is carried out on machines similar to extrusion machines, in which the spindle, in addition to rotating, has an axial displacement.

In the injection, once the mold is filled, it is separated from the nozzle of the machine, breaking the feed channel. After a certain time, the piece already cooled is demoulded.

High pressures and temperatures are necessary, but parts of good finish and at high production speeds are obtained

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Each element has an atomic number state which is meant by atomic number
netineya [11]

Answer: The atomic number is the number of protons in an atom, and isotopes have the same atomic number but differ in the number of neutrons. The number of protons in an atom is called its atomic number. This number is very important because it is unique for atoms of a given element. All atoms of an element have the same number of protons, and every element has a different number of protons in its atoms.

Explanation:

3 0
3 years ago
A flat sheet of ice has a thickness of 2.20 cm. It is on top of a flat sheet of crystalline quartz that has a thickness of 1.50
kolezko [41]

Answer:

Distance_{vaccum}=5.19cm

Explanation:

The speed of light in these mediums shall be lower than that in vacuum thus the total time light needs to cross both the media are calculated as under

Total time = Time taken through ice + Time taken through quartz

Time taken through ice = Thickness of ice / (speed of light in ice)

T_{ice}=\frac{2.20\times 10^{-2} \times \mu _{ice}}{V_{vaccum}}

T_{quartz}=\frac{1.50\times 10^{-2} \times \mu _{quartz}}{V_{vaccum}}

Thus in the same time the it would had covered a distance of

Distance_{vaccum}=Totaltime\times V_{vaccum}\\\\Distance_{vaccum}=10^{-2}[2.20\mu _{ice+1.50\mu _{quartz}}]

we have

\mu _{ice}=1.309\\\\\mu _{quartz}=1.542

Applying values we have

Distance_{vaccum}=10^{-2}[2.20\times 1.309+1.50\times 1.542]

Distance_{vaccum}=5.19cm

6 0
3 years ago
A snail can move approximately 0.30 meters per minute.how many meters can the snail cover in 15 minutes?
ololo11 [35]

Answer:

4.5 meters is your Answer goood luck please give 5 star

Explanation:

3 0
3 years ago
Variations to the forehand stroke are not allowed. a. true b. false
timofeeve [1]
This is b. false

There are numerous ways of adjusting the forehand stroke and variations are allowed.
3 0
3 years ago
Read 2 more answers
An electron is released from rest at the negative plate of a parallel plate capacitor. The charge per unit area on each plate is
Andreas93 [3]

Answer:

v=1.08\times 10^7\ m/s

Explanation:

Initial speed of the electron, u = 0

The charge per unit area of each plate, \dfrac{Q}{A}=1.69\times 10^{-7}\ C/m^2

Separation between the plates, d=1.75\times 10^{-2}\ m

An electron is released from rest, u = 0

Using equation of kinematics,

v^2-u^2=2ad..........(1)

Acceleration of the electron in electric field, a=\dfrac{qE}{m}............(2)

Electric field, E=\dfrac{\sigma}{\epsilon_o}............(3)

From equation (1), (2) and (3) :

v=\sqrt{\dfrac{2q\sigma d}{m\epsilon_o}}

v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\times 1.69\times 10^{-7}\times 1.75\times 10^{-2}}{9.1\times 10^{-31}\times 8.85\times 10^{-12}}}

v = 10840393.1799 m/s

or

v=1.08\times 10^7\ m/s

So, the electron is moving with a speed of 1.08\times 10^7\ m/s before it reaches the positive plate. Hence, this is the required solution.

3 0
3 years ago
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