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lubasha [3.4K]
3 years ago
6

Tech A says that bleeding an electronic brake control system is just like bleeding a non-electronic brake control system. Tech B

says that if air is allowed into the electronic brake control system, it can become trapped in the hydraulic control unit, requiring a scan tool to operate the solenoid valves to help bleed the air from the electronic brake control system. Who is correct
Engineering
1 answer:
PtichkaEL [24]3 years ago
3 0

Answer:

Tech A is correct.

Explanation:

An electric brake controller is a device that sends a signal to the trailer via vehicle's brakes. This reduces the wear and tear on the vehicle brakes. As a result, the vehicle stops.

Tech A says that bleeding an electronic brake control system is just like bleeding a non-electronic brake control system.

So,

Tech A is correct.

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An equal-tangent sag vertical curve (with a negative initial and a positive final grade) is designed for 55 mi/h. The PVI is at
Varvara68 [4.7K]

Answer:

The lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.

Explanation:

Length of curve is given as

L=2(PVT-PVI)\\L=2(242+30-240+00)\\L=2(230)\\L=460 ft

G_2 is given as

G_2=\frac{E_{PVT}-E_{PVI}}{0.5L}\\G_2=\frac{127.5-122}{0.5*460}\\G_2=0.025=2.5 \%

The K value is given from the table 3.3 for 55 mi/hr is 115. So the value of A is given as

A=\frac{L}{K}\\A=\frac{460}{115}\\A=4

A is given as

-G_1=A-G_2\\-G_1=4.0-2.5\\-G_1=1.5\\G_1=-1.5\%

With initial grade, the elevation of PVC is

E_{PVC}=E_{PVI}+G_1(L/2)\\E_{PVC}=122+1.5%(460/2)\\E_{PVC}=125.45 ft\\

The station is given as

St_{PVC}=St_{PVI}-(L/2)\\St_{PVC}=24000-(230)\\St_{PVC}=237+70\\

Low point is given as

x=K \times |G_1|\\x=115 \times 1.5\\x=172.5 ft

The station of low point is given as

St_{low}=St_{PVC}-(x)\\St_{low}=23770+(172.5)\\St_{low}=239+42.5 ft\\

The elevation is given as

E_{low}=\frac{G_2-G_1}{2L} x^2+G_1x+E_{PVC}\\E_{low}=\frac{2.5-(-1.5)}{2*460} (1.72)^2+(-1.5)*(1.72)+125.45\\E_{low}=124.16 ft

So the lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.

3 0
4 years ago
A 1000 kg turbine has a rotating unbalance of 0.1 kg.m. The turbine operates at a speed between 500 to 750 rpm. What is the maxi
raketka [301]

Answer:

maximum isolator stiffness k =1764 kN-m

Explanation:

mean speed of rotation =\frac{N_1 +N_2}{2}

Nm = \frac{500+750}{2} = 625 rpm

w =\frac{2\pi Nm}{60}

  =65.44 rad/sec

F_T = mw^2 e

F_T = mew^2

       = 0.1*(65.44)^2

F_T =428.36 N

Transmission ratio =\frac{300}{428.36} = 0.7

also

transmission ratio = \frac{1}{[\frac{w}{w_n}]^{2} -1}

0.7 =\frac{1}{[\frac{65.44}{w_n}]^2 -1}

SOLVING FOR Wn

Wn = 42 rad/sec

Wn = \sqrt {\frac{k}{m}

k = m*W^2_n

k = 1000*42^2 = 1764 kN-m

k =1764 kN-m

3 0
3 years ago
Much of the workd went to bed hungry
Marysya12 [62]
The workers went to bed hungry probably because they are hard workers and so didn’t want to eat because they didn’t want to take break┌(; ̄◇ ̄)┘
7 0
3 years ago
⦁ An ideal Otto cycle has a compression ratio of 8. At the beginning of the compression process, air is at 95 kPa and 27°C, and
Anon25 [30]

Answer:

<u><em>note:</em></u>

<u><em>solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment</em></u>

Download docx
8 0
4 years ago
A 3 m aluminum pole is kept at a residential site for construction
Aliun [14]

Answer:

I don't know sorry

Explanation:

5 0
3 years ago
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