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Juli2301 [7.4K]
3 years ago
11

What is the recommended processing technique for manufacturing plastic bottles?

Engineering
1 answer:
charle [14.2K]3 years ago
6 0

Answer Explanation:

THERE ARE DIFFERENT PROCESS TECHNIQUES FOR MANUFACTURING PLASTIC BOTTLES:

  1. EXTRUSION BLOW MOLDING PROCESS: In this process a mold cavity composed of two halves, closes around the parison and pinches off one end. compressed air expands the parison to conform against the cold mold cavity walls
  2. INJECTION BLOW MOLDING: The process of injection blow modeling is used for the production of hollow plastic bottles
  3. STRECH BLOW MOLDING: Stretch  blow molding uses either the injection or extrusion blow molding process as a basic foundation

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In the construction of a large reactor pressure vessel, a new steel alloy with a plane strain fracture toughness of 55 MPa-m1/2
sp2606 [1]

Answer:

l=24mm

Explanation:

From the question we are told that:

Plane strain fracture toughness of T=55 MPa-m1/2

Y value Y=1.0

Stress level of\sigma =200 MPa

Generally the equation for length of a surface crack is mathematically given by

l=\frac{1}{\pi}(\frac{T}{Y*\sigma})^2

l=\frac{1}{3.142}(\frac{55}{1*200})^2

l=0.024m

Therefore

in mm

l=24mm

6 0
3 years ago
1. A flywheel is suspended by resting the inside of the rim on a horizontal knife edge so that the wheel can swing in a vertical
sammy [17]

Answer: A fly wheel having a mass of 30kg was allowed to swing as pendulum about a knife edge at inner side of the rim as shown in figure.

Explanation:

8 0
3 years ago
With 64 KB of memory and 8 bits in each memory location, how wide should the address bus be to access all 64 KB of memory? (k =
marishachu [46]

Answer:

16-bit wide

Explanation:

In order to find the width of the address bus, we need first to know how many memory cells it is needed to address.

If the size memory is 64 KB, this means that the memory size, in bytes, is equal to the following quantity:

64 KB = 2⁶ * 2¹⁰ bytes = 2¹⁶ bytes.

In order to address this quantity of cell positions, the address bus must be able to address 2¹⁶ bytes, so it must have 16-bit wide.

3 0
3 years ago
6.48 programming project 1: encode/decode tic -tac-toe
natita [175]

Answer:

What do you need help with?

Explanation:

5 0
3 years ago
Using a forked rod, a 0.5-kg smooth peg P is forced to move along the vertical slotted path r = (0.5 θ) m, whereθ is in radians.
-BARSIC- [3]

Answer:

N_c = 3.03 N

F = 1.81 N

Explanation:

Given:

- The attachment missing from the question is given:

- The given expressions for the radial and θ direction of motion:

                                       r = 0.5*θ

                                       θ = 0.5*t^2              ...... (correction for the question)

- Mass of peg m = 0.5 kg

Find:

a) Determine the magnitude of the force of the rod on the peg at the instant t = 2 s.

b) Determine the magnitude of the normal force of the slot on the peg.

Solution:

- Determine the expressions for radial kinematics:

                                        dr/dt = 0.5*dθ/dt

                                        d^2r/dt^2 = 0.5*d^2θ/dt^2

- Similarly the expressions for θ direction kinematics:

                                        dθ/dt = t

                                        d^2θ/dt^2 = 1

- Evaluate each at time t = 2 s.

                                        θ = 0.5*t^2 = 0.5*2^2 = 2 rad -----> 114.59°

                                        r = 1 m , dr / dt = 1 m/s , d^2 r / dt^2 = 0.5 m/s^2

- Evaluate the angle ψ between radial and horizontal direction:

                                        tan Ψ = r / (dr/dθ) = 1 / 0.5

                                        Ψ = 63.43°

- Develop a free body diagram (attached) and the compute the radial and θ acceleration:

                                        a_r = d^2r / dt^2 - r * dθ/dt

                                        a_r = 0.5 - 1*(2)^2 = -3.5 m/s^2

                                        a_θ =  r * (d^2θ/dt^2) + 2 * (dr/dt) * (dθ/dt)

                                        a_θ = 1(1) + 2*(1)*(2) = 5 m/s^2

- Using Newton's Second Law of motion to construct equations in both radial and θ directions as follows:

Radial direction:              N_c * cos(26.57) - W*cos(24.59) = m*a_r

θ direction:                      F  - N_c * sin(26.57) + W*sin(24.59) = m*a_θ

Where, F is the force on the peg by rod and N_c is the normal force on peg by the slot. W is the weight of the peg. Using radial equation:

                                       N_c * cos(26.57) - 4.905*cos(24.59) = 0.5*-3.5

                                       N_c = 3.03 N

                                       F  - 3.03 * sin(26.57) + 4.905*sin(24.59) = 0.5*5

                                       F = 1.81 N

4 0
3 years ago
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