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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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Write short notes on: (any four) a) Suspended ground floor b) Soil exploration c) Baulking of sand d) Bearing capacity of soil e
vredina [299]

Answer:

a) A suspended floor is a ground floor with a void underneath the structure. The floor can be formed in various ways, using timber joists, precast concrete panels, block and beam system or cast in-situ with reinforced concrete. However, the floor structure is supported by external and internal walls.

b) Soil exploration consists of determining the profile of the natural soil deposits at the site, taking the soil samples and determining the engineering properties of soils using laboratory tests as well as in-situ testing methods

c) Bulking in sand Occurs When dry sand interacts with the atmospheric moisture. Presence of moisture content forms a thin layer around sand particles. This layer generates the force which makes particles to move aside to each other. This results in the increase of the volume of sand.

d) In a nutshell, bearing capacity is the capacity of soil to support the loads that are applied to the ground above. It depends primarily on the type of soil, its shear strength and its density. It also depends on the depth of embedment of the load – the deeper it is founded, the greater the bearing capacity.

Explanation:

<h2>please follow me</h2>
6 0
3 years ago
What type of engineer would be most likely to develop a design for cars? chemical civil materials mechanical
Andreyy89
I don’t know but good luck
4 0
3 years ago
Design process 8 steps with definition​
Troyanec [42]

Answer:

Step 1: Define the Problem.  

Step 2: Do Background Research. .

Step 3: Specify Requirements. .

Step 4: Brainstorm, Evaluate and Choose Solution.  

Step 5: Develop and Prototype Solution.  

Step 6: Test Solution.

Step 7: Does Your Solution Meet the Requirements?  

Step 8: Communicate Results.

can u tell me the definition tho?

palled correctly as “though” which is an alternate form of “although”) at the end is informal usage. It's better placed before “she seems better today

8 0
3 years ago
For unrestrained cube made from linear, isotropic, homogeneous material the temperature increase causes strain in_____ direction
LenKa [72]

Answer: The answer is four; four

Explanation: This is because of the mixture of material used and the number of directions it causes strain I directly proportional to the number of times it causes stress.

7 0
3 years ago
A pipe of 0.3 m outer diameter at a temperature of 160°C is insulated with a material having a thermal conductivity of k = 0.055
Alekssandra [29.7K]

Answer:

Q=0.95 W/m

Explanation:

Given that

Outer diameter = 0.3 m

Thermal conductivity of material

K= 0.055(1+2.8\times 10^{-3}T)\frac{W}{mK}

So the mean conductivity

K_m=0.055\left ( 1+2.8\times 10^{-3}T_m \right )

T_m=\dfrac{160+273+40+273}{2}

T_m=373 K

K_m=0.055\left ( 1+2.8\times 10^{-3}\times 373 \right )

K_m=0.112 \frac{W}{mK}

So heat conduction through cylinder

Q=kA\dfrac{\Delta T}{L}

Q=0.112\times \pi \times 0.15^2\times 120

Q=0.95 W/m

4 0
3 years ago
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