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Pachacha [2.7K]
4 years ago
8

Corrective maintenance _________________________.​

Engineering
1 answer:
andrew-mc [135]4 years ago
3 0

Answer:

b. ​diagnoses and corrects errors in an operational system

Explanation:

Corrective maintenance is one that is performed with the purpose of repairing faults or defects that occur in equipment and machinery.  As such, it is the most basic way of providing maintenance, as it simply involves repairing what has broken down. In this sense, corrective maintenance is a process that basically consists of locating and correcting faults or damages that are preventing the machine from performing its function in a normal way.

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What are the basic steps in the operation of a solar cell?
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Light is absorbed and knocks electrons loose. Loose electrons flow, creating a current. The current is captured and transferred to wires.

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What is the function of a fixed resistor?
agasfer [191]

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  add resistance to a circuit

Explanation:

It depends on the design in which it is incorporated. A fixed resistor has many uses, including, but not limited to ...

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8 0
4 years ago
Who was first credited with the discovery of the intensity of current within a circuit?
jeyben [28]

Answer:

Hans Christian Ørsted

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Explanation:

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8 0
3 years ago
A mercury thermometer has a cylindrical capillary tube with an internal diameter of 0.2 mm. If the volume of the thermometer and
vova2212 [387]

To solve this problem we will proceed to calculate the specific volume from the area of the cylinder and the sensitivity. Later we will calculate the volumetric coefficient of thermal expansion and finally we will be able to calculate the volume through the relation of the two terms mentioned above. Our values are

\text{Sensitivity}= 2mm/\°C

\text{Internal diameter } d= 0.2mm

\text{Differential expansion of Hg } \lambda_L = 1.82*10^{-4}/\°C

Let's start by calculating the specific volume which is given by

v = \pi (\frac{d}{2})^2 \gamma

Here,

d = Diameter

\gamma = Sensitivity

Replacing our values we have

v = (\frac{\pi}{4})(0.2mm)^2(2mm/\°C)

v = 0.0628mm^3 /\°C

Now we will obtain the value of the volumetric coefficient of thermal expansion of mercury through the differential expansion coefficient of Hg whic is three times, then

\lambda_V = 3\lambda_L

\lambda_V = 3(1.82*10^{-4}/\°C)

\lambda_V = 5.46*10^{-4}/\°C

Finally the relation to calculate the volume the bulb must is

\text{Specific volume} = \text{Bulb Volume} \times \text{Volumetric Coefficient}

v = v_B \times \lambda_V

v_B = \frac{v}{\lambda_V}

v_B = \frac{0.0628mm^3/\°C}{5.46*10^{-4}/\°C}

v_B = 115mm^3

Therefore the volume that the bulb must have is 115mm^3

5 0
3 years ago
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