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3 years ago

8

Technician A says that the enable criteria are the criteria that must be met before the PCM completes a monitor test. Technician

B says that a drive cycle includes operating the vehicle under specific conditions so that a monitor test can be completed. Who is correct?
a. Technician A
b. Technician B
c. Both A and B
d. Neither A nor B

1 answer:

DENIUS [597]3 years ago

5 0

Answer:

"A"

Explanation:

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A reservoir delivers water to a horizontal pipeline 39 long The first 15 m has a diameter of 50 mm, after which it suddenly beco

allsm [11]

Answer:

The difference of head in the level of reservoir is 0.23 m.

Explanation:

For pipe 1

$d_1=50 mm,f_1=0.0048$

For pipe 2

$d_2=75 mm,f_2=0.0058$

Q=2.8 l/s

$Q=2.8\times 10^{-3]$

We know that Q=AV

$Q=A_1V_1=A_2V_2$

$A_1=1.95\times 10^{-3}m^2$

$A_2=4.38\times 10^{-3} m^2$

$So V_2=0.63 m/s,V_1=1.43 m/s$

head loss (h)

$h=\dfrac{f_1L_1V_1^2}{2gd_1}+\dfrac{f_2L_2V_2^2}{2gd_2}+0.5\dfrac{V_1^2}{2g}$

Now putting the all values

$h=\dfrac{0.0048\times 15\times 1.43^2}{2\times 9.81\times 0.05}+\dfrac{0.0058\times 24\times 0.63^2}{2\times 9.81\times 0.075}+0.5\dfrac{1.43^2}{2\times 9.81}$

So h=0.23 m

So the difference of head in the level of reservoir is 0.23 m.

8 0

3 years ago

Consider two Carnot heat engines operating in series. The first engine receives heat from the reservoir at 1400 K and rejects th

Aleksandr-060686 [28]

Answer:

The temperature T= 648.07k

Explanation:

T1=input temperature of the first heat engine =1400k

T=output temperature of the first heat engine and input temperature of the second heat engine= unknown

T3=output temperature of the second heat engine=300k

but carnot efficiency of heat engine = $1 - \frac{Tl}{Th} \\$

where Th =temperature at which the heat enters the engine

Tl is the temperature of the environment

since both engines have the same thermal capacities <em> $n_{th}$ </em> therefore $n_{th} =n_{th1} =n_{th2}\\n_{th }=1-\frac{T1}{T}=1-\frac{T}{T3}\\ \\= 1-\frac{1400}{T}=1-\frac{T}{300}\\$

We have now that

$\frac{-1400}{T}+\frac{T}{300}=0\\$

multiplying through by T

$-1400 + \frac{T^{2} }{300}=0\\$

multiplying through by 300

- $420000+ T^{2} =0\\T^2 =420000\\\sqrt{T2}=\sqrt{420000} \\T=648.07k$

The temperature T= 648.07k

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3 years ago

How many robots does bailey nursery own

givi [52]

Answer:

The Bailey family has flourished during its business’ 110-year history. But Bailey Nurseries’ leaders still operate with the belief that the family doesn’t always know best. The company has grown from a one-man operation selling fruit trees and ornamental shrubs to one of the largest wholesale nurseries in the United States, thanks to insights from those who are family and those who aren’t.

“For a business to thrive, you have to ask for outside help,” says Terri McEnaney, president of the Newport-based company and a fourth-generation family member. “We get an outside perspective through family business programs, advisors and our board, because you can get a bit ingrained in your own way of thinking.”

When Bailey Nurseries chose its current leader in 2000, it brought in a facilitator who gathered insights from key employees, board members and owners. Third-generation leaders (and brothers) Gordie and Rod Bailey picked Rod’s daughter McEnaney, who had experience both inside and outside the company.

Explanation:

5 0

2 years ago

Explain the working and performance of a centrifugal clutch with a sketch

FinnZ [79.3K]

Answer:

A centrifugal clutch works, as the name suggests, through centrifugal force.

One of the most common methods is by mounting the clutch onto the parallel or taper crank shaft of the engine.

When the crank shaft rotates the shaft of the clutch rotates at the same speed as the engine

5 0

3 years ago

Read 2 more answers

A thick steel slab (rho= 7800 kg/m3 , cp= 480 J/kg K, k= 50 W/m K) is initially at 300 °C and is cooled by water jets impinging

dimaraw [331]

Answer:

t = 2244.3 sec

Explanation:

calculate the thermal diffusivity

$\alpha = \frac{k}{\rho c}$

$= \frac{50}{7800\times 480} = 1.34 \times 10^{-5} m^2/s$

Temperature at 28 mm distance after t time = = 50 degree C

we know that

$\frac[ T_{28} - T_s}{T_i -T_s} = erf(\frac{x}{2\sqrt{at}})$

$\frac{ 50 -25}{300-25} = erf [\frac{28\times 10^{-3}}{2\sqrt{1.34\times 10^{-5}\times t}}]$

$0.909 = erf{\frac{3.8245}{\sqrt{t}}}$

from gaussian error function table , similarity variable w calculated as

erf w = 0.909

it is lie between erf w = 0.9008 and erf w = 0.11246 so by interpolation we have

w = 0.08073

$erf 0.08073 = erf[\frac{3.8245}{\sqrt{t}}]$

$0.08073 = \frac{3.8245}{\sqrt{t}}$

solving fot t we get

t = 2244.3 sec

3 0

3 years ago