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jek_recluse [69]
3 years ago
13

Provide main reasons for the short shot during the injection molding.

Engineering
1 answer:
Mariulka [41]3 years ago
5 0

Answer:

some cause of short shot is

1) due to the restriction in the flow

2) air pockets

3) high viscosity.

Explanation:

short shot is a word defined for major defect, it is actually occur when molten material does not  fully occupy the cavities in a mold. Due to which mold remained incomplete after cooling. short shot may be because of restriction in the flow of molten material through the cavities and other main cause is present of large percentage of entrapped air.

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Why does the the diffusion capacitance fall off at high frequencies?
Vitek1552 [10]

Answer:

The answer 1 is correct.

Explanation:

When a junction called the p-n is forward biased, the capacitance of diffusion will form or structured across depletion layers.

When at a higher frequency or frequencies, the dopants, are vibrated across the metallurgical junction and the doping profiles smooths out.

4 0
3 years ago
The moisture content in air (humidity) is measured by weight and expressed in pounds or ____________________.
VikaD [51]

Moisture content is measured in terms of pounds of water per pound of air (lb water/lb air) or grains of water per pound of air (gr. of water/lb air).

Hope this helps❤

3 0
2 years ago
Find the error in the following pseudo code
IRISSAK [1]

Pseudocodes are used as a prototype of an actual program.

The error in the pseudocode is that, the while loop in the pseudocode will run endlessly.

From the pseudocode, the first line is:

<em>Declare Boolean finished = false</em>

The while loop is created to keep running as long as <em>finished = false.</em>

So, for the while loop to end, the finished variable must be updated to true.

This action is not implemented in the pseudocode.

Hence, the error in the pseudocode is that, the while loop is an endless loop

Read more about pseudocodes at:

brainly.com/question/17442954

8 0
2 years ago
An inventor claims to have developed a refrigerator that at steady state requires a net power input of 1.1 horsepower to remove
Lynna [10]

Answer:

The inventor's claim is false in the sense that no thermal machine can violate the first thermodynamic law.

Explanation:

The inventor's claim could not be possible as no thermal machine can transfer more heat than the input work consumed. If we expose the thermal efficiency:

n=Q out / W in

Where Q and W both must be in the same power unit, so we will convert the remove heat from BTU/hr to hp:

12000 BTU/hr = 4.72 hp

Therefore by comparing, we notice that the removing heat of 4.75 hp is large than the delivered work of 1.11 hp. By evaluating the efficiency:

[tex]n=4.75 hp / 1.1 hp  = 4.3 > 1[/tex]

6 0
3 years ago
Consider a Carnot refrigeration cycle executed in a closed system in the saturated liquid–vapor mixture region using 1.06 kg of
Alexxandr [17]

Answer:

P_m_i_n = 442KPA

Explanation:

We are given:

m = 1.06Kg

T_H = 1.2T_L

T = 22kj

Therefore we need to find coefficient performance or the cycle

COP_R = \frac {1}{(T_R/T_l) -1}

= \frac {1 }{1.2-1}

= 5

For the amount of heat absorbed:

Q_l = COP_R Wm

= 5 × 22 = 110KJ

For the amount of heat rejected:

Q_H = Q_L + W_m

= 110 + 22 = 132KJ

[tex[ q_H = \frac{Q_L}{m} [/tex];

= = \frac{132}{1.06}

= 124.5KJ

Using refrigerant table at hfg = 124.5KJ/Kg we have 69.5°c

Convert 69.5°c to K we have 342.5K

To find the minimum temperature:

T_L = \frac{T_H}{1.2};

T_L = \frac{342.5}{1.2}

= 285.4K

Convert to °C we have 12.4°C

From the refrigerant R -134a table at T_L = 12.4°c we have 442KPa

6 0
3 years ago
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