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3 years ago

Seperate real and imaginary parts tan(2x+i3y)

Engineering

1 answer:

ahrayia [7]3 years ago

3 0

Answer:

First you have to separate real and imaginary parts of Tan(x+iy)=Tan(z)=sin(z)/cos(z)

sinz=sin(x+iy)=sinxcos(iy)+cosxsin(iy)=sinxcoshy-icosx sinhy

cosz=cos(x+iy)=cosxcos(iy)-sinxsin(iy)=cosxcoshy−isinxsinhy

Now if you plug in Tan(z) and simplify (it is easy!) you get

Tan(z)=(sin(2x)+isinh(2y))/(cos(2x)+cosh(2y))= A+iB.

This means that

A=sin(2x)/(cos(2x)+cosh(2y)) and B= sinh(2y)/(cos(2x)+cosh(2y))

Now,

A/B=sin(2x)/sinh(2y)

If any questions, let me know.

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3 years ago

Read 2 more answers

The specific grounding and bonding requirements for an information technology room are located in which article of the nec?

Oksi-84 [34.3K]

The specific grounding and bonding requirements for an information technology room are located in Article 645.

<h3>What does NEC stand for?</h3>

The National Electrical Code (NEC) is a collection of requirements for the secure installation of electrical wiring in the US that are routinely updated.

Bonding is the joining of conductors that do not convey current, such as enclosures and buildings. Bonded systems are attached to the earth by grounding. To protect persons and property from electric risks, both are required.

Article 645 compliance is optional. The two main pardons provided by Article 645 are. First, underneath a raised floor, non-plenum-rated cables are acceptable. Second, cables, boxes, and similar items for the mentioned IT equipment are exempt from the need for security. sted IT equipment cables, boxes, and the like are not required to be secured in place.

To learn more about NEC refer to:

brainly.com/question/20348027

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2 years ago

A three-point bending test was performed on an aluminum oxide specimen having a circular cross section of radius 3.5 mm (0.14 in

RideAnS [48]

To resolve this problem we have,

$R=3.5mm\\F_f1=950N\\L_1=50mm\\b=12mm\\L_2=40mm$

$F_{f2}$ is unknown.

With these dates we can calculate the Flexural strenght of the specimen,

$\sigma{fs}=\frac{F_{f1}L}{\pi R^3}\\\sigma{fs}=\frac{(950)(50*10^{-3})}{\pi 3.5*10^{-3}}\\\sigma{fs}=352.65Mpa$

After that, we can calculate the flexural strenght for the square cross section using the previously value.

$\sigma{fs}=\frac{F_{f2}L}{\pi R^3}\\(352.65*10^6)=\frac{3Ff(40*10^{-3})}{2(12*10^{-10})}\\F_{f2}=\frac{352.65*10^6}{34722.22}\\F_{f2}=10156.32N\\F_{f2}=10.2kN$

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4 years ago