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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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dexar [7]

Answer: b

Explanation:

8 0
3 years ago
Five kilograms of air at 427°C and 600 kPa are contained in a piston–cylinder device. The air expands adiabatically until the pr
son4ous [18]

Answer:

The entropy change of the air is 0.240kJ/kgK

Explanation:

T_{1} =427+273K,T_{1} =700K\\P_{1} =600kPa\\P_{2} =100kPa

T_{2}  is unknown

we can apply the following expression to find T_{2}

-w_{out} =mc_{v} (T_{2} -T_{1} )

T_{2} =T_{1} -\frac{w_{out } }{mc_{v} }

now substitute

T_{2} =700K-\frac{600kJ}{5kg*0.718kJ/kgK} \\T_{2}=533K

To find entropy change of the air we can apply the ideal gas relationship

Δs_{air}=c_{p} ln\frac{T_{2} }{T_{1} } -Rln\frac{P_{2} }{P_{1} }

Δs_{air} =1.005*ln(\frac{533}{700})-0.287* in(\frac{100}{600} )

Δs_{air} =0.240kJ/kgK

4 0
3 years ago
Radioactive wastes generating heat at a rate of 3 x 106 W/m3 are contained in a spherical shell of inner radius 0.25 m and outsi
MariettaO [177]

Answer:

Inner surface temperature= 783K.

Outer surface temperature= 873K

Explanation:

Parameters:

Heat,e= 3×10^6 W/m^3

Inner radius = 0.25 m

Outside radius= 0.30 m

Temperature at infinity, T(¶)= 10°c = 273. + 10 = 283K.

Convection coefficient,h = 500 W/m^2 . K

Temperature of the surface= T(s) = ?

Temperature of the inner= T(I) =?

STEP 1: Calculate for heat flux at the outer sphere.

q= r × e/3

This equation satisfy energy balance.

q= 1/3 ×3000000(W/m^3) × 0.30 m

= 3× 10^5 W/m^2.

STEP 2: calculus the temperature for the surface.

T(s) = T(¶) + q/h

T(s) = 283 + 300000( W/m^2)/500(W/m^2.K)

T(s) = 283+600

T(s)= 873K.

TEMPERATURE FOR THE OUTER SURFACE is 873 kelvin.

The same TWO STEPS are use for the calculation of inner temperature, T(I).

STEP 1: calculate for the heat flux.

q= r × e/3

q= 1/3 × 3000000(W/m^3) × 0.25 m

q= 250,000 W/m^2

STEP 2:

calculate the inner temperature

T(I) = T(¶) + q/h

T(I) = 283K + 250,000(W/m^2)/500(W/m^2)

T(I) = 283K + 500

T(I) = 783K

INNER TEMPERATURE IS 783 KELVIN

5 0
2 years ago
If anyone knows manufacturing plz help
sveta [45]

Answer:

I don't know ask my dad he would

Explanation:

but I can't ask him because he went to get milk and forgot to come back

8 0
2 years ago
Travel Time Problem: Compute the time of concentration using the Velocity, Sheet Flow Method for Non-Mountainous Orange County a
Vaselesa [24]

Answer:

Total time taken = 0.769 hour

Explanation:

using the velocity method

for sheet flow ;

Tt = \frac{0.007(nl)^{0.8} }{(Pl)^{5}s^{0.4}  }  

Tt = travel time

n = manning CaH

Pl = 25years

L = how length ( ft )

s = slope

For Location ( 1 )

s = 0.045

L = 1000 ft

n = 0.06 ( from manning's coefficient table )

Tt1 = 0.128 hour

For Location ( 2 )

s = 2.5 %

L= 750

n = 0.13

Tt2 = 0.239 hour

For Location ( 3 )

s = 1.5%

L = 500 ft

n = 0.15

Tt3 = 0.237  hour

For Location (4)

s = 0.5 %

L = 250 ft

n = 0.011

Tt4 = 0.165 hour

hence the Total time taken = Tt1 + Tt2 + Tt3 + Tt4

                                              = 0.128 + 0.239 + 0.237 + 0.165 = 0.769 hour

5 0
3 years ago
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