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3 years ago

Plz help electrical technology

Engineering

2 answers:

oksano4ka [1.4K]3 years ago

8 0

Answer:

OPTION A,Larger

HOPE IT HELPS

ozzi3 years ago

3 0

The answer should be option A. Larger

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Consider a solid circular shaft subjected to bending and torsion so that the state of stress of interest involves only a normal

Alex17521 [72]

Answer:

The detailed explanation of answer is given in attached file.

Explanation:

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4 years ago

Explain orthographic<br>and multi-view<br>projection

Elina [12.6K]

Answer:

Orthographic projection is a projection method to present three-dimensional shapes in two-dimensional format in which the projection lines are drawn to be orthogonal to the plane of projection, such that each view of the three-dimensional object is translated to a view of the orthographic projection and orthogonal to the view

A multiview projection is the representation of a three-dimensional projection by two or more two-dimensional views

Explanation:

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4 years ago

How deep does electrical conduit need to be buried?

GenaCL600 [577]

In general, bury metal conduits at least 6 inches below the soil surface. You may also run them at a depth of 4 inches under a 4-inch concrete slab. Under your driveway, the conduits must be below a depth of 18 inches, and under a public road or alleyway, they must be buried below 24 inches.

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3 years ago

A car hits a tree at an estimated speed of 10 mi/hr on a 2% downgrade. If skid marks of 100 ft. are observed on dry pavement (F=

Dafna11 [192]

Answer:

$v_{o} = 22.703\,\frac{m}{s}$ $\left(50.795\,\frac{m}{s}\right)$

Explanation:

The deceleration of the car on the dry pavement is found by the Newton's Law:

$\Sigma F = -\mu_{k,1}\cdot m\cdot g \cdot \cos \theta + m\cdot g \cdot \sin \theta = m\cdot a_{1}$

Where:

$a_{1} = (-\mu_{k,1}\cdot \cos \theta + \sin \theta)\cdot g$

$a_{1} = (-0.33\cdot \cos 1.146^{\textdegree}+\sin 1.146^{\textdegree})\cdot \left(9.807\,\frac{m}{s^{2}} \right)$

$a_{1} = -3.040\,\frac{m}{s^{2}}$

Likewise, the deceleration of the car on the unpaved shoulder is:

$a_{2} = (-\mu_{k,2}\cdot \cos \theta + \sin \theta)\cdot g$

$a_{2} = (-0.28\cdot \cos 1.146^{\textdegree}+\sin 1.146^{\textdegree})\cdot \left(9.807\,\frac{m}{s^{2}} \right)$

$a_{2} = -2.549\,\frac{m}{s^{2}}$

The speed just before the car entered the unpaved shoulder is:

$v_{o} = \sqrt{\left(4.469\,\frac{m}{s} \right)^{2}-2\cdot \left(-2.549\,\frac{m}{s^{2}} \right)\cdot (60.88\,m)}$

$v_{o} = 18.175\,\frac{m}{s}$

And, the speed just before the pavement skid was begun is:

$v_{o} = \sqrt{\left(18.175\,\frac{m}{s} \right)^{2}-2\cdot \left(-3.040\,\frac{m}{s^{2}} \right)\cdot (30.44\,m)}$

$v_{o} = 22.703\,\frac{m}{s}$ $\left(50.795\,\frac{m}{s}\right)$

8 0

3 years ago

Which stroke in the four stroke cycle begins as the piston reaches BDC after combustion takes place

Svet_ta [14]

Explanation:

Which stroke in the four stroke cycle begins as the piston reaches BDC after combustion takes place?

Answer :Exhaust stroke is the waste gasses discharging process into the atmosphere after the combustion stroke. This step takes place after the piston reaches the BDC at the end of the combustion stroke. When the piston reaches BDC, the piston will definitely move back to TDC.

This is the Answer for your question :3

I hoped I helped you :3

I hope you are having a great day ❤️❤️❤️

8 0

3 years ago