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3 years ago

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What happens when you pass a current through an insulated wire?

1 answer:

ad-work [718]3 years ago

8 0

As electricity passes through a wire, some of the electrical energy is converted to heat. The more current that flows through a wire, the more heat is generated. If you double the current passing through a wire, the heat generated will increase 4 times!

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Technician A says that the enable criteria are the criteria that must be met before the PCM completes a monitor test. Technician

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Answer:

"A"

Explanation:

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3 years ago

Three identical fatigue specimens (denoted A, B, and C) are fabricated from a nonferrous alloy. Each is subjected to one of the

Law Incorporation [45]

Answer:

B A and C

Explanation:

Given:

Specimen σ $_{max}$ σ $_{min}$

A +450 -150

B +300 -300

C +500 -200

Solution:

Compute the mean stress

σ $_{m}$ = (σ $_{max}$ + σ $_{min}$ )/2

σ $_{mA}$ = (450 + (-150)) / 2

= (450 - 150) / 2

= 300/2

σ $_{mA}$ = 150 MPa

σ $_{mB}$ = (300 + (-300))/2

= (300 - 300) / 2

= 0/2

σ $_{mB}$ = 0 MPa

σ $_{mC}$ = (500 + (-200))/2

= (500 - 200) / 2

= 300/2

σ $_{mC}$ = 150 MPa

Compute stress amplitude:

σ $_{a}$ = (σ $_{max}$ - σ $_{min}$ )/2

σ $_{aA}$ = (450 - (-150)) / 2

= (450 + 150) / 2

= 600/2

σ $_{aA}$ = 300 MPa

σ $_{aB}$ = (300- (-300)) / 2

= (300 + 300) / 2

= 600/2

σ $_{aB}$ = 300 MPa

σ $_{aC}$ = (500 - (-200))/2

= (500 + 200) / 2

= 700 / 2

σ $_{aC}$ = 350 MPa

From the above results it is concluded that the longest fatigue lifetime is of specimen B because it has the minimum mean stress.

Next, the specimen A has the fatigue lifetime which is shorter than B but longer than specimen C.

In the last comes specimen C which has the shortest fatigue lifetime because it has the higher mean stress and highest stress amplitude.

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A direct contact heat exchanger (where the fluid mixes completely) has three inlets and one outlet. The mass flow rates of the i

lara31 [8.8K]

Answer:

Enthalpy at outlet=284.44 KJ

Explanation:

$m_1=1 Kg/s,m_2=1.5 Kg/s,m_3=22 Kg/s$

$h_1=100 KJ/Kg,h_2=120 KJ/Kg,h_3=500 KJ/Kg$

We need to Find enthalpy of outlet.

Lets take the outlet mass m and outlet enthalpy h.

So from mass conservation

$m_1+m_2+m_3=m$

m=1+1.5+2 Kg/s

m=4.5 Kg/s

Now from energy conservation

$m_1h_1+m_2h_2+m_3h_3=mh$

By putting the values

$1\times 100+1.5\times 120+2\times 500=4.5\times h$

So h=284.44 KJ

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