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3 years ago

Describe electroplating as it would be used to batch process small components.

Engineering

1 answer:

denpristay [2]3 years ago

4 0

Answer:

Electroplating is the method of hydrolysis placing one layer of metal on another.

Explanation:

Electroplating is the method of hydrolysis placing one layer of metal on another, most frequently for ornamental reasons or to avoid metal deterioration due to corrosion. some precise electroplating are copper, silver and chromium plating are also available.Electroplating enables producers to utilize cheap metals like steel or zinc for the bulk of the item and then apply differently.

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The author claims that engineers should always have the final say and last word when it comes to setting safety standards. Selec

Lisa [10]

Answer:

I select false because engineers are not the only thing we have, we have scientists doctors mathematicians and much more to give safety standards

4 0

4 years ago

Read 2 more answers

A condenser accepts steam from the turbine in problem 2 at a pressure of 2.34 kPa. Saturated water at the same pressure leaves t

vaieri [72.5K]

Answer:

The answer is "83.98, 1889.195, and 1889.195"

Explanation:

Given value:

$\bold{P_{4}=2.34 \ kPa}$

In point a:

The value of $h_{f4}=83.915 \ \ \frac{Kj}{kg}\\$

$V_4=0.001002 \ \ \frac{Kj}{kg}\\\\U_4= 83.98 \ \ \frac{Kj}{kg}\\\\$

In point b:

calculating heat leaves formula= $h_3-h_{f4}$

$= 1973.11-83.915\\\\= 1889.195 \ \ \frac{KJ}{kg}$

In point c:

calculating Heat transfer rate formula $=m(h_3-h_4)$

$= 1(1889.195)\\\\ = 1889.19 \ \ kw.$

7 0

3 years ago

Do you think for security reasons everything that happens on the internet should be analyzed by the public security services ?

givi [52]

Answer: yes

Explanation: People post bad things that i think should get taken down i was on an app the other day and people were posting bad things

8 0

3 years ago

0 - 1" -20 -15 -10 5 0 1 2 3 0

faust18 [17]

Answer:

#WeirdestQuestionOfAllTime

Explanation:

8 0

3 years ago

Air enters the combustor of a jet engine at p1=10 atm, T1=1000°R, and M1=0.2. Fuel is injected and burned, with a fuel/air mass

snow_lady [41]

Answer:

M2 = 0.06404

P2 = 2.273

T2 = 5806.45°R

Explanation:

Given that p1 = 10atm, T1 = 1000R, M1 = 0.2.

Therefore from Steam Table, Po1 = (1.028)*(10) = 10.28 atm,

To1 = (1.008)*(1000) = 1008 ºR

R = 1716 ft-lb/slug-ºR cp= 6006 ft-lb/slug-ºR fuel-air ratio (by mass)

F/A =???? = FA slugf/slugaq = 4.5 x 108ft-lb/slugfx FA slugf/sluga = (4.5 x 108)FA ft-lb/sluga

For the air q = cp(To2– To1)

(Exit flow – inlet flow) – choked flow is assumed For M1= 0.2

Table A.3 of steam table gives P/P* = 2.273,

T/T* = 0.2066,

To/To* = 0.1736 To* = To2= To/0.1736 = 1008/0.1736 = 5806.45 ºR Gives q = cp(To* - To) = (6006 ft-lb/sluga-ºR)*(5806.45 – 1008)ºR = 28819500 ft-lb/slugaSetting equal to equation 1 above gives 28819500 ft-lb/sluga= FA*(4.5 x 108) ft-lb/slugaFA =

F/A = 0.06404 slugf/slugaor less to prevent choked flow at the exit

5 0

3 years ago