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wolverine [178]
3 years ago
7

Tech A says that in some cases, the electronic brake control module can be programmed with a new tire size to restore proper ele

ctronic brake control performance. Tech B says that there could be an electronic brake control module available for installation that matches a new tire size. Who is correct?A. Tech AB. Tech BC. Both Techs A and BD. Neither Tech A nor B
Engineering
1 answer:
Vlad [161]3 years ago
3 0

Answer:

Both Techs A and B

Explanation:

Electronic braking systems are controlled by the electronic brake control module. It is a microprocessor that processes information from wheel-speed sensors and the hydraulic brake system to determine when to release braking pressure at a wheel that's about to lock up and start skidding  and activates the anti lock braking system or traction system when it detects it is necessary.

Some electronic brake control modules can be programmed to the size of the vehicle's new tires to restore proper electronic brake control performance. While others may require replacing the module to match the module's programming to the installed tire size. So, both technicians A and B are correct.

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If you are a mechanical engineer answer these questions:
Natasha_Volkova [10]

Answer:

1. Yes, they are all necessary.

2. Both written and verbal communication skills are of the utmost importance in business, especially in engineering. Communication skills boost you or your teams' performance because they provide clear information and expectations to help manage and deliver excellent work.

3 0
3 years ago
For some transformation having kinetics that obey the Avrami equation (Equation 10.17), the parameter n is known to have a value
OleMash [197]

Answer:

t = 25.10 sec

Explanation:

we know that Avrami equation

Y = 1 - e^{-kt^n}

here Y is percentage of completion  of reaction = 50%

t  is duration of reaction = 146 sec

so,

0.50 = 1 - e^{-k^146^2.1}

0.50 = e^{-k306.6}

taking natural log on both side

ln(0.5) = -k(306.6)

k = 2.26\times 10^{-3}

for 86 % completion

0.86 = 1 - e^{-2.26\times 10^{-3} \times t^{2.1}}

e^{-2.26\times 10^{-3} \times t^{2.1}} = 0.14

-2.26\times 10^{-3} \times t^{2.1} = ln(0.14)

t^{2.1} = 869.96

t = 25.10 sec

5 0
3 years ago
A heat pump operates on a vapor-compression refrigeration cycle with R-134a as the working fluid. The refrigerant enters the com
Rudiy27

Answer:

Hello your question has some missing information below are the missing information

The refrigerant enters the compressor as saturated vapor at 140kPa Determine The coefficient of performance of this heat pump

answer : 2.49

Explanation:

For  vapor-compression refrigeration cycle

P1 = P4  ; P1 = 140 kPa

P2( pressure at inlet ) = P3 ( pressure at outlet ) ; P2 = 800 kPa

<u>From pressure table of R 134a refrigerant</u>

h1 ( enthalpy of saturated vapor at 140kPa ) = 239.16 kJ/kg

h2 ( enthalpy of saturated liquid at P2 = 800 kPa and t = 60°C )

= 296.8kJ/kg

h3 ( enthalpy of saturated liquid at P3 = 800 kPa ) = 95.47 kJ/kg

also h4 = 95.47 kJ/kg

To determine the coefficient of performance  

Cop = ( h1 - h4 ) / ( h2 - h1 )

∴ Cop = 2.49

3 0
3 years ago
In poor weather, you should __ your following distance
Ratling [72]

Answer:

I think reduce your following distance

5 0
4 years ago
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How should you mark a component listed on the VCU form that
Anna [14]
I’m not that shareholding about it but I think the answer is C mark it green I hope I don’t get it rong
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