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2 years ago

Which of the following tells the computer wha to do

Engineering

1 answer:

Sever21 [200]2 years ago

8 0

That’s a very hard question! But I believe it’s the operating system, hope I helped!

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What is the maximum number of 12-2 with ground nonmetallic-sheathed cables permitted in an 18-cubic-inch device box if two singl

Novosadov [1.4K]

Answer:i think it is 35

Explanation:

i just guessed sorry im only in 5th grade

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3 years ago

Read 2 more answers

14. An engine is brought into the shop with a

Lostsunrise [7]

Answer:

B. To accurately measure spark advance, use a timing light that incorporates an

ignition advance meter. The spark advance cannot be determined by listening to the way the engine sounds.

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2 years ago

An operating gear box (transmission) has 350 hp at its input shaft while 250. hp are delivered to the output shaft. The gear box

True [87]

Answer:

Rate of Entropy =210.14 J/K-s

Explanation:

given data:

power delivered to input = 350 hp

power delivered to output = 250 hp

temperature of surface = 180°F

rate of entropy is given as

$Rate\ of\ entropy = \frac{Rate\ of \ heat\ released}{Temperature}$

T = 180°F = 82°C = 355 K

Rate of heat = (350 - 250) hp = 100 hp = 74600 W

Rate of Entropy $= \frac{74600}{355} = 210.14 J/K-s$

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3 years ago

You don't know which insert you have, and the inserts are different sizes, meaning the amount needed for a 1:3 ratio is differen

VashaNatasha [74]

Answer:

Explanation:

For ligation process the 1:3 vector to insert ratio is the good to utilize . By considering that we can take 1 ratio of vector and 3 ratio of insert ( consider different insert size ) and take 10 different vials of ligation ( each calculated using different insert size from low to high ) and plot a graph for transformation efficiency and using optimum transformation efficiency we can find out the insert size.

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3 years ago

(a) A non-cold-worked 1040 steel cylindrical rod has an initial length of 100 mm and initial diameter of 7.50 mm. is to be defor

serg [7]

Answer:

A) 1040 steel is not a possible candidate for this application

B) 35.94%

Explanation:

Initial length = 100 mm = 0.1 m

Initial diameter ( d ) = 7.5 mm = 0.0075 m

Tensile load ( p ) = 18,000 N

Condition : The 1040 steel must not experience plastic deformation or a diameter reduction of more than 1.5 * 10^-5 m

A) would the 1040 steel be a possible candidate for this application

Yield strength of 1040 steel < stress ( in order to be a possible candidate )

stress = p / A0 = ( 18000 ) / ( $\frac{\pi }{4}$ ) * 0.0075^2

= 18,000 / (4.418 * 10^-5 ) = 407.424 MPa

Yield strength of 1040 steel = 450 MPa

stress = 407.424 MPa

∴ Yield strength ( 450 MPa ) > stress ( 407.424 MPa )

Therefore 1040 steel is not a possible candidate for this application

B) Determine How much cold work would be required to reduce the diameter of the steel to 6.0 mm

Area1 = ( $\frac{\pi }{4}$ ) ( 0.006 )^2 = 2.83 * 10^-5 m^2

therefore % of cold work done = ( A0 - A1 ) / A0 * 100 = 35.94%

6 0

2 years ago