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IRISSAK [1]
1 year ago
6

Nec ________ covers selection of time-delay fuses for motor- overload protection.

Engineering
1 answer:
Murljashka [212]1 year ago
6 0

Nec Article 430 covers selection of time-delay fuses for motor- overload protection.

<h3>What article in the NEC covers motor overloads?</h3>

Article 430 that is found in  National Electrical Code (NEC) is known to be state as “Motors, Motor Circuits and Controllers.” .

Note that the article tells that it covers areas such as motors, motor branch-circuit as well as feeder conductors, motor branch-circuit and others.

Therefore, Nec Article 430 covers selection of time-delay fuses for motor- overload protection.

Learn more about motor- overload from

brainly.com/question/20738481

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The housing for a certain machinery product is made of two components, both aluminum castings. The larger component has the shap
Dmitry_Shevchenko [17]

Answer:

The pouring of the molten metal during casting is done very slowly hence the molten metal froze before reaching all parts of the mould cavity. also the early freezing can occur if the temperature of the molten metal was lower than the required temperature for casting

Explanation:

since the same components are being casted at other foundries and they don't have the defects ,

Hence the reason for the defects experienced by these components can be caused when the pouring of the molten metal during casting is done very slowly hence the molten metal froze before reaching all parts of the mould cavity. also the early freezing can occur if the temperature of the molten metal was lower than the required temperature for casting .

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2 years ago
Complete the following sentence. The skills and content of several subject areas were combined to form a new field known as a me
Svetlanka [38]

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Explanation:

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3 years ago
Read 2 more answers
How many meters per second is 100 meters and 10 seconds
Elden [556K]

Answer:

the velocity = 10 m / sec if an object moves 100 m in 10s

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2 years ago
The collar A, having a mass of 0.75 kg is attached to a spring having a stiffness of k = 200 N/m . When rod BC rotates about the
gladu [14]

Answer:

Speed=1.633 m/s

Force= 20 N

Explanation:

Ideally, v^{2}=\frac {ks^{2}}{m} hence v=s\sqrt {\frac {k}{m}} where v is the speed of collar, m is the mass of collar, k is spring constant and s is the displacement.

In this case, s=100-0=100mm=0.1m since 1 m is equivalent to 1000mm

k is given as 200 N/m and mass is 0.75 Kg

Substituting the given values

v=0.1 m\sqrt \frac {200 N/m}{0.75 Kg}=1.632993162 m/s\approx 1.633 m/s

Therefore, <u>the speed is 1.633 m/s</u>

The sum of vertical forces is given by mg where g is acceleration due to gravity and it's value taken as 9.81 m/s^{2}

Therefore, F_y=0.75\times 9.81=7.3575 N\approx 7.36 N

The sum of forces in normal direction is given by Ma_n=Ks therefore

Ma_n=200*0.1=20 N

Therefore, <u>normal force on the rod is 20 N</u>

5 0
3 years ago
A flywheel made of Grade 30 cast iron (UTS = 217 MPa, UCS = 763 MPa, E = 100 GPa, density = 7100 Kg/m, Poisson's ratio = 0.26) h
hram777 [196]

Answer:

N = 38546.82 rpm

Explanation:

D_{1} = 150 mm

A_{1}= \frac{\pi }{4}\times 150^{2}

              = 17671.45 mm^{2}

D_{2} = 250 mm

A_{2}= \frac{\pi }{4}\times 250^{2}

              = 49087.78 mm^{2}

The centrifugal force acting on the flywheel is fiven by

F = M ( R_{2} - R_{1} ) x w^{2} ------------(1)

Here F = ( -UTS x A_{1} + UCS x A_{2} )

Since density, \rho = \frac{M}{V}

                        \rho = \frac{M}{A\times t}

                        M = \rho \times A\times tM = 7100 \times \frac{\pi }{4}\left ( D_{2}^{2}-D_{1}^{2} \right )\times t

                        M = 7100 \times \frac{\pi }{4}\left ( 250^{2}-150^{2} \right )\times 37

                        M = 8252963901

∴ R_{2} - R_{1} = 50 mm

∴ F = 763\times \frac{\pi }{4}\times 250^{2}-217\times \frac{\pi }{4}\times 150^{2}

  F = 33618968.38 N --------(2)

Now comparing (1) and (2)

33618968.38 = 8252963901\times 50\times \omega ^{2}

∴ ω = 4036.61

We know

\omega = \frac{2\pi N}{60}

4036.61 = \frac{2\pi N}{60}

∴ N = 38546.82 rpm

7 0
3 years ago
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