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jenyasd209 [6]
2 years ago
11

Tech B says that long-term fuel trims that are positive means that the PCM is leaning out the fuel mixture from the base pulse-w

idth setting. Who is correct
Engineering
2 answers:
Masteriza [31]2 years ago
4 0

Answer:

include a pictrue and mark me brainliest

Explanation:

konstantin123 [22]2 years ago
4 0
Include a picture and mark
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The four strokes in a four stroke cycle engine in proper order.
Morgarella [4.7K]

Answer:

A four-stroke cycle engine is an internal combustion engine that utilizes four distinct piston strokes (intake, compression, power, and exhaust) to complete one operating cycle. The piston make two complete passes in the cylinder to complete one operating cycle.

Explanation:

6 0
3 years ago
⚠️Answer needed quick!!⚠️
Allisa [31]

Answer:

A

Explanation:

3 gifts

gift cost = x

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5 0
3 years ago
The lattice constant of a simple cubic primitive cell is 5.28 Å. Determine the distancebetween the nearest parallel ( a ) (100),
skelet666 [1.2K]

Answer:

a)5.28 Å , b)3.73 Å , c)3.048 Å

Explanation:

the atoms are situated only at the corners of cube.Each and every atom in simple cubic primitive at the corner is shared with 8 adjacent unit cells.

Therefore, a particular unit cell consist only 1/8th part of an atom.

The lattice constant of a simple cubic primitive cell is 5.28 Å

We know formula of distance,

d = \frac{a}{\sqrt{h^{2}+k^{2}+l^{2}}}

a)(100)

a=5.28 Å

Distance = \frac{5.28 Å}{\sqrt{1^{2}+0^{2}+0^{2}}}=5.28 Å

b)(110)

Distance = \frac{5.28}{\sqrt{1^{2}+1^{2}+0^{2}}} = 3.73 Å

c)(111)

Distance= \frac{5.28}{\sqrt{1^{2}+1^{2}+1^{2}}}= 3.048 Å

6 0
3 years ago
For some metal alloy, the following engineering stresses produce the corresponding engineering plastic strains prior to necking.
kirza4 [7]

Answer:

203.0160

Explanation:

Because you add then subtract then multiply buy 7 the subtract then divide then you add that to the other numbers you got than boom

7 0
2 years ago
The hot and cold inlet temperatures to a concentric tube heat exchanger are Th,i = 200°C, Tc,i = 100°C, respectively. The outlet
alexgriva [62]

Answer:Counter,

0.799,

1.921

Explanation:

Given data

T_{h_i}=200^{\circ}C

T_{h_o}=120^{\circ}C

T_{c_i}=100^{\circ}C

T_{c_o}=125^{\circ}C

Since outlet temperature of cold liquid is greater than hot fluid outlet temperature therefore it is counter flow heat exchanger

Equating Heat exchange

m_hc_{ph}\left [ T_{h_i}-T_{h_o}\right ]=m_cc_{pc}\left [ T_{c_o}-T_{c_i}\right ]

\frac{m_hc_{ph}}{m_cc_{pc}}=\frac{125-100}{200-120}=\frac{25}{80}=C\left ( capacity rate ratio\right )

we can see that heat capacity of hot fluid is minimum

Also from energy balance

Q=UA\Delta T_m=\left ( mc_p\right )_{h}\left ( T_{h_i}-T_{h_o}\right )

NTU=\frac{UA}{\left ( mc_p\right )_{h}}=\frac{\left ( T_{h_i}-T_{h_o}\right )}{T_m}

T_m=\frac{\left ( 200-125\right )-\left ( 120-100\right )}{\ln \frac{75}{20}}

T_m=41.63^{\circ}C

NTU=1.921

And\ effectiveness \epsilon =\frac{1-exp\left ( -NTU\left ( 1-c\right )\right )}{1-c\left ( -NTU\left ( 1-c\right )\right )}

\epsilon =\frac{1-exp\left ( -1.921\left ( 1-0.3125\right )\right )}{1-0.3125exp\left ( -1.921\left ( 1-0.3125\right )\right )}

\epsilon =\frac{1-exp\left ( -1.32068\right )}{1-0.3125exp\left ( -1.32068\right )}

\epsilon =\frac{1-0.2669}{1-0.0834}

\epsilon =0.799

5 0
3 years ago
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