All categories

2 years ago

What are the constraints of the problem for thermoforming

Engineering

1 answer:

Inga [223]2 years ago

6 0

Answer:

here are four options so you can choose the ones of your choice.

Explanation:

1. warpage

2 Dimensional Inconsistencies

3 Part Thickness Inconsistencies

4 Lack of Detail in Part Geometry and Aesthetics

please rate brainliest if helps and follow

You might be interested in

A converging nozzle has an exit area of 0.001 m2. Air enters the nozzle with negligible velocity at a pressure of 1 MPa and a te

Artyom0805 [142]

Explanation:

a converging nozzle has an exit area of 0.001 m2. Air enters the nozzle with negligible velocity at a pressure of 1 MPa and a temperature of 360 K. For isentropic flow of an ideal gas with k = 1.4 and the gas constant R = Ru/MW = 287 J/kg-K, determine the mass flow rate in kg/s and the exit Mach number for back pressures

100% (3 ratings)

A_2 = 0.001 m^2 P_1 = 1 MPa, T_1 = 360 k P_2 = 500 kpa p^gamma - 1/gamma proportional T (1000/500)^1.4 - 1/1.4 = (360/T_2) 2^4/14 = 360/T_2 T_2

3 0

3 years ago

Ext

Galina-37 [17]

Engineering is the technical

8 0

2 years ago

A 10-mm-diameter Brinell hardness indenter produced an indentation 1.55 mm in diameter in a steel alloy when a load of 500 kg wa

BigorU [14]

Answer:

HB = 3.22

Explanation:

The formula to calculate the Brinell Hardness is given as follows:

$HB = \frac{2P}{\pi D\sqrt{D^{2}- d^{2} } }$

where,

HB = Brinell Hardness = ?

P = Applied Load in kg = 500 kg

D = Diameter of Indenter in mm = 10 mm

d = Diameter of the indentation in mm = 1.55 mm

Therefore, using these values, we get:

$HB = \frac{(2)(500)}{\pi (10)\sqrt{10^{2}- 1.55^{2} } }$

4 0

3 years ago

Label the following statements as either T (true) or F (false).

IgorLugansk [536]

Answer:

1 its 1 bc you have to do it step by step

Explanation:

step by step

6 0

2 years ago

6msection of 150lossless line is driven by a source with vg(t) = 5 cos(8π × 107t − 30◦ ) (V) and Zg = 150 . If the line, which h

Rainbow [258]

Answer:

a. 5m

b. r = 0.16 e^-80.5◦

c. Zpn = (115.7 + j27.4) ohms

d. Vi = 2.2e^-j22.56◦ volts

e. Vi(t) = 2.2 cos (8π × 107t − 22.56◦ ) Volts

Explanation:

In this question, we are tasked with calculating a series of terms.

Please check attachment for complete solution and step by step explanation

7 0

3 years ago