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Olegator [25]
2 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]2 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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The position function x(t) of a particle moving along an x axis is x = 4.00 - 6.00t2, with x in meters and t in seconds. (a) at
elena-14-01-66 [18.8K]

The position function x(t) of a particle moving along an x axis is x=4.00 - 6.00t^2

a) The point at which particle stop, it's velocity = 0 m/s

  So dx/dt = 0

        0 = 0- 12t = -12t

  So when time t= 0, velocity = 0 m/s

    So the particle is starting from rest.

At t = 0 the particle is (momentarily) stop

b) When t = 0

 x=4.00 - 6.00*0^2 = 4m

SO at x = 4m the particle is (momentarily) stop

c) We have x=4.00 - 6.00t^2

   At origin x = 0

  Substituting

         0 = 4.00 - 6.00t^2\\ \\ t^2 = \frac{2}{3}

         t = 0.816 seconds or t = - 0.816 seconds

So when  t = 0.816 seconds and t = - 0.816 seconds, particle pass through the origin.

5 0
2 years ago
Two identical horizontal sheets of glass have a thin film of air of thickness t between them. The glass has refractive index 1.4
Gre4nikov [31]

Answer:

the wavelength λ of the light when it is traveling in air = 560 nm

the smallest thickness t of the air film = 140 nm

Explanation:

From the question; the path difference is Δx = 2t  (since the condition of the phase difference in the maxima and minima gets interchanged)

Now for constructive interference;

Δx= (m+ \frac{1}{2} \lambda)

replacing ;

Δx = 2t   ; we have:

2t = (m+ \frac{1}{2} \lambda)

Given that thickness t = 700 nm

Then

2× 700 = (m+ \frac{1}{2} \lambda)     --- equation (1)

For thickness t = 980 nm that is next to constructive interference

2× 980 = (m+ \frac{1}{2} \lambda)     ----- equation (2)

Equating the difference of equation (2) and equation (1); we have:'

λ = (2 × 980) - ( 2× 700 )

λ = 1960 - 1400

λ = 560 nm

Thus;  the wavelength λ of the light when it is traveling in air = 560 nm

b)  

For the smallest thickness t_{min} ; \ \ \ m =0

∴ 2t_{min} =\frac{\lambda}{2}

t_{min} =\frac{\lambda}{4}

t_{min} =\frac{560}{4}

t_{min} =140 \ \  nm

Thus, the smallest thickness t of the air film = 140 nm

7 0
2 years ago
Read 2 more answers
For a given initial projectile speed Vo, calculate what launch angle A gives the longest range R. Show your work, don't just quo
pickupchik [31]
The optimal angle of 45° for maximum horizontal range is only valid when initial height is the same as final height. 

<span>In that particular situation, you can prove it like this: </span>

<span>initial velocity is Vo </span>
<span>launch angle is α </span>

<span>initial vertical velocity is </span>
<span>Vv = Vo×sin(α) </span>

<span>horizontal velocity is </span>
<span>Vh = Vo×cos(α) </span>

<span>total time in the air is the the time it needs to fall back to a height of 0 m, so </span>
<span>d = v×t + a×t²/2 </span>
<span>where </span>
<span>d = distance = 0 m </span>
<span>v = initial vertical velocity = Vv = Vo×sin(α) </span>
<span>t = time = ? </span>
<span>a = acceleration by gravity = g (= -9.8 m/s²) </span>
<span>so </span>
<span>0 = Vo×sin(α)×t + g×t²/2 </span>
<span>0 = (Vo×sin(α) + g×t/2)×t </span>
<span>t = 0 (obviously, the projectile is at height 0 m at time = 0s) </span>
<span>or </span>
<span>Vo×sin(α) + g×t/2 = 0 </span>
<span>t = -2×Vo×sin(α)/g </span>

<span>Now look at the horizontal range. </span>
<span>r = v × t </span>
<span>where </span>
<span>r = horizontal range = ? </span>
<span>v = horizontal velocity = Vh = Vo×cos(α) </span>
<span>t = time = -2×Vo×sin(α)/g </span>
<span>so </span>
<span>r = (Vo×cos(α)) × (-2×Vo×sin(α)/g) </span>
<span>r = -(Vo)²×sin(2α)/g </span>

<span>To find the extreme values of r (minimum or maximum) with variable α, you must find the first derivative of r with respect to α, and set it equal to 0. </span>

<span>dr/dα = d[-(Vo)²×sin(2α)/g] / dα </span>
<span>dr/dα = -(Vo)²/g × d[sin(2α)] / dα </span>
<span>dr/dα = -(Vo)²/g × cos(2α) × d(2α) / dα </span>
<span>dr/dα = -2 × (Vo)² × cos(2α) / g </span>

<span>Vo and g are constants ≠ 0, so the only way for dr/dα to become 0 is when </span>
<span>cos(2α) = 0 </span>
<span>2α = 90° </span>
<span>α = 45° </span>
4 0
1 year ago
After an eye examination, you put some eyedrops on your sensitive eyes. The cornea (the front part of the eye) has an index of r
mina [271]

Answer:

t = 103.45 n m

Explanation:

given,

refractive index of cornea = 1.38

refractive index of eye drop = 1.45

wavelength of refractive index = 600 nm

refractive index of eye drop is greater than refractive index of cornea and the air.

Formula used in this case

for constructive interference

2 n t = (m + \dfrac{1}{2})\lambda

At m = 0 for the minimum thickness, so

2\times 1.45 \times t = (0 + 0.5)\times 600

2.9 \times t =300

t =\dfrac{300}{2.9}

t = 103.45 n m

the minimum thickness of the film of eyedrops t = 103.45 n m

6 0
2 years ago
A truck collides with a car on horizontal ground. At one moment during the collision, the magnitude of the acceleration of the t
Mice21 [21]

Answer:

The magnitude of the acceleration of the car is 35.53 m/s²

Explanation:

Given;

acceleration of the truck, a_t = 12.7 m/s²

mass of the truck, m_t = 2490 kg

mass of the car, m_c = 890 kg

let the acceleration of the car at the moment they collided = a_c

Apply Newton's third law of motion;

Magnitude of force exerted by the truck = Magnitude of force exerted by the car.

The force exerted by the car occurs in the opposite direction.

F_c = -F_t\\\\m_ca_c = -m_t a_t\\\\a_c =- \frac{m_ta_t}{m_c} \\\\a_c = -\frac{2490 \times 12.7}{890} \\\\a_c = - 35.53 \ m/s^2

Therefore, the magnitude of the acceleration of the car is 35.53 m/s²

3 0
2 years ago
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