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mel-nik [20]
1 year ago
12

Technician a s ays both an ohmmeter and a self-powered test light may be used to test for continuity. technician b says both may

be used to test fuses. who is correct?
Engineering
1 answer:
amm18121 year ago
5 0

Both A and B technicians are correct because both might be used to test fuses, according to technician B.

<h3>What is continuity?</h3>

The behavior of a function at a certain point or section is described by continuity. The limit can be used to determine continuity.

From the question:

We can conclude:

The technician claims that you may check for continuity using both an ohmmeter and a self-powered test light. Both might be used to test fuses, according to technician B.

Thus, both A and B technicians are correct because both might be used to test fuses, according to technician B.

Technician A says both an ohmmeter and a self-powered test light may be used to test for continuity. Technician B says both may be used to test fuses. Who is correct?

Learn more about the continuity here:

brainly.com/question/15025692

#SPJ1

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Explanation:

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3 years ago
When the compression process is non-quasi-equilibrium, the molecules before the piston face cannot escape fast enough, forming a
muminat

Answer:

a. true

Explanation:

Firstly, we need to understand what takes places during the compression process in a quasi-equilibrium process. A quasi-equilibrium process is a process in during which the system remains very close to a state of equilibrium at all times.  When a compression process is quasi-equilibrium, the work done during the compression is returned to the surroundings during expansion, no exchange of heat, and then the system and the surroundings return to their initial states. Thus a reversible process.

While for a non-quasi equilibrium process, it takes more work to move the piston against this high-pressure region.

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3 years ago
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Hydraulic fluid is primarily used to
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8 0
3 years ago
A gas enters a compressor that provides a pressure ratio (exit pressure to inlet pressure) equal to 8. If a gage indicates the g
olga55 [171]

Answer:

P_2_{abs}=160\ psia (absolute).

Explanation:

Given that

Pressure ratio r

r=8

r=\dfrac{P_2_{abs}}{P_1_{abs}}

  8=\dfrac{P_2_{abs}}{P_1_{abs}}                                  -----1

P₁(gauge) = 5.5 psig

We know that

Absolute pressure = Atmospheric pressure  + Gauge  pressure

Given that

Atmospheric pressure = 14.5 lbf/in²

P₁(abs) = 14.5 + 5.5  psia

P₁(abs) =20 psia

Now by putting the values in the above equation 1

8=\dfrac{P_2_{abs}}{20}

P_2_{abs}=8\times 20\ psia

P_2_{abs}=160\ psia

Therefore the exit gas pressure will be 160 psia (absolute).

7 0
3 years ago
Calculate the convective heat-transfer coefficient for water flowing in a round pipe with an inner diameter of 3.0 cm. The water
olasank [31]

Answer:

h = 10,349.06 W/m^2 K

Explanation:

Given data:

Inner diameter = 3.0 cm

flow rate  = 2 L/s

water temperature 30 degree celcius

Q = A\times V

2\times 10^{-3} m^3 = \frac{\pi}{4} \times (3\times 10^{-2})^2 \times velocity

V = \frac{20\times 4}{9\times \pi} = 2.83 m/s

Re = \frac{\rho\times V\times D}{\mu}

at 30 degree celcius = \mu = 0.798\times 10^{-3}Pa-s , K  = 0.6154

Re = \frac{10^3\times 2.83\times 3\times 10^{-2}}{0.798\times 10^{-3}}

Re = 106390

So ,this is turbulent flow

Nu = \frac{hL}{k} = 0.0029\times Re^{0.8}\times Pr^{0.3}

Pr= \frac{\mu Cp}{K} = \frac{0.798\times 10^{-3} \times 4180}{0.615} = 5.419

\frac{h\times 0.03}{0.615}  = 0.0029\times (1.061\times 10^5)^{0.8}\times 5.419^{0.3}

SOLVING FOR H

WE GET

h = 10,349.06 W/m^2 K

6 0
3 years ago
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