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2 years ago

10

While servicing a vehicle, you retrieve the diagnostic code P0401, "EGR (exhaust gas recirculatior flow insufficient." Which of

these is the most likely cause?

1 answer:

n200080 [17]2 years ago

3 0

Answer: Clogged EGR ports or passages

Explanation: The clogging causes insufficient flow.

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A device is needed to accelerate a 3000 lb vehicle into a barrier with constant velocity to test its 5 mph bumpers. The vehicle

Arte-miy333 [17]

Answer:

The device acceleration is

a = f/m = f/(w/g)=fg/w

The velocity v is

v= u +at

When u =o

v = at

Substitute the value of a

v = fgt/w

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3 years ago

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the voltage across a 5mH inductor is 5[1-exp(-0.5t)]V. Calculate the current through the inductor and the energy stored in the i

Serggg [28]

Given Information:

Inductance = L = 5 mH = 0.005 H

Time = t = 2 seconds

Required Information:

Current at t = 2 seconds = i(t) = ?

Energy at t = 2 seconds = W = ?

Answer:

Current at t = 2 seconds = i(t) = 735.75 A

Energy at t = 2 seconds = W = 1353.32 J

Explanation:

The voltage across an inductor is given as

$V(t) = 5(1-e^{-0.5t})$

The current flowing through the inductor is given by

$i(t) = \frac{1}{L} \int_0^t \mathrm{V(t)}\,\mathrm{d}t \,+ i(0)$

Where L is the inductance and i(0) is the initial current in the inductor which we will assume to be zero since it is not given.

$i(t) = \frac{1}{0.005} \int_0^t \mathrm{5(1-e^{-0.5t}}) \,\mathrm{d}t \,+ 0\\\\i(t) = 200 \int_0^t \mathrm{5(1-e^{-0.5t}}) \,\mathrm{d}t \\\\i(t) = 200 \: [ {5\: (t + \frac{e^{-0.5t}}{0.5})]_0^t \\i(t) = 200\times5\: \: [ { (t + 2e^{-0.5t} + 2 )] \\$

$i(t) = 1000t +2000e^{-0.5t} -2000\\$

So the current at t = 2 seconds is

$i(t) = 1000(2) +2000e^{-0.5(2)} -2000\\\\i(t) = 735.75 \: A$

The energy stored in the inductor at t = 2 seconds is

$W = \frac{1}{2}Li(t)^{2}\\\\W = \frac{1}{2}0.005(735.75)^{2}\\\\W = 1353.32 \:J$

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3 years ago

The annual storage in Broad River watershed is 0 cm/y. Annual precipitation is 100 cm/y and evapotranspiration is 50 cm/y. The s

REY [17]

Answer:

0.34232

Explanation:

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Who can work on a fixed ladder that extends more than 24 feet?

baherus [9]

Answer:

Explanation:

If the fixed ladder will reach more than twenty-four (24) feet above a lower level, you as the employer are required to incorporate a personal fall arrest or ladder safety system into the installation of the ladder. For reference, this requirement is cited in OSHA Section 1910.28(b)(9)(i).

If the total length of the climb on a fixed ladder equals or exceeds 24 feet (7.3 m), the ladder must be equipped with ladder safety devices; or self-retracting lifelines and rest platforms at intervals not to exceed 150 feet (45.7 m); or a cage or well and multiple ladder sections with each ladder section not

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Storing parts outside doesn't cause any environmental risks as long as the items are covered.

telo118 [61]

Answer:

False

Explanation:

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