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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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A window‐mounted air‐conditioning unit (AC) removes energy by heat transfer from a room, and rejects energy by heat transfer to

Arada [10]

Solution :

Given :

The power of the air‐conditioning (AC) unit is , W = 0.434 kW

The coefficient of performance or the COP of the air‐conditioning (AC) unit is given by = 6.22

Therefore he heat removed is given by , $Q_H = 6.22 \times 0.434$

$Q_H = 2.7 \ kW$

Now if the electricity is valued at 0.10 dollar per kW hour, then the operating cost of the air conditioning unit in 24 hours is given by = 0.10 x 2.7 x 24

= 6.48

Therefore the operating cost = $ 6.48 for 24 hours.

3 0

3 years ago

Here you go!!!!!!!!!!!!!!!!!1

sweet [91]

Answer:

Im confused, what does this mean

Explanation:

i mean, thx lol

3 0

3 years ago

Given the potential field in cylindrical coordinates, V = 100/ (z2+1) rho cos φV, and point P at rho = 3m, φ = 60°, z = 2m, find

Margaret [11]

Answer:

V = 30 V

vector (E) = -10*a_p + 17.32*a_Q - 24*a_z

mag(E) = 31.24 V/m

dV / dN = 31.2 V /m

a_N = 0.32*a_p -0.55*a_Q + 0.77*a_z

p_v = 276 pC / m^3

Explanation:

Given:

- The Volt Potential in cylindrical coordinate system is given as:

- The point P is at p = 3 , Q = 60 , z = 2

Find:

values at P for:

a.) V

b.) E

c.) E

d.) dV/dN

e.) aN

f.) rhov in free space

Solution:

Evaluate the Volt potential function at the given point P, by simply plugging the values of position P. The following:

$V = \frac{100*3*cos(60)}{2^2 + 1}\\\\V = \frac{150}{5} = 30 V$

To compute the Electric field from Volt potential we have the following relation:

E = - ∀.V

Where, ∀ is a del function which denoted:

∀ = $\frac{d}{dp}.a_p + \frac{d}{p*dQ}*a_Q + \frac{d}{dz}*a_z$

Hence,

$E = \frac{100*cos(Q)}{z^2 + 1}.a_p+ \frac{100*sin(Q)}{z^2 + 1}.a_Q + \frac{-200*p*zcos(Q)}{(z^2 + 1)^2}.a_z$

Plug in the values for point P:

$E = \frac{100*cos(60)}{5}.a_p+ \frac{100*sin(60)}{5}.a_Q + \frac{-200*3*2*cos(60)}{(5)^2}.a_z\\\\E = - 10*a_p +17.32 *a_Q-24*a_z$

The magnitude of the Electric Field is given by:

E = √((-10)^2 + (17.32)^2 + (24)^2)

E = √975.9824

E = 31.241 N/C

The dV/dN is the Field Strength E in normal to the surface with vector N given by:

dV / dN = | - ∀.V |

dV / dN = | E | = 31.2 N/C

a_N is the unit vector in the direction of dV/dN or the electric field strength E, as follows:

$a_N = - vector(E) / (dV/dN)\\\\a_N = 10 /31.2 *a_p - 17.32/31.2 *a_Q + 24/31.2 *a_z\\\\a_N = 0.32 *a_p - 0.55 *a_Q + 0.77 *a_z$

The charge density in free space:

p_v = E.∈_o

Where, ∈_o is the permittivity of free space = 8.85*10^-12

p_v = 31.24.8.85*10^-12

p_v = 276 pC / m^3

4 0

3 years ago

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Kieran and Kurt spend most of their day performing physical labor, aligning materials, and inspecting projects for quality and s

xxMikexx [17]

Answer:

Kieran is a Carpenter, and Kurt is a Drafter

Explanation:

Among all the professions listed, Kurt can only possibly be a Drafter. A Drafter is an engineering technician who makes detailed technical drawings or plans for machinery, buildings, electronics, infrastructure, sections, etc. Drafters make use of computer software and manual sketches to convert the designs, plans, and layouts of engineers and architects into a set of technical drawings. All the other professions will at some point need for the professional to stand and work at an height above the ground except for a drafter, whose feet will remain on the ground at almost all times.

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3 years ago

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A solid shaft and a hollow shaft of the same material have same length and outer radius R. The inner radius of the hollow shaft

alexandr402 [8]

Answer with Explanation:

By the equation or Torque we have

$\frac{T}{I_{p}}=\frac{\tau }{r}=\frac{G\theta }{L}$

where

T is the torque applied on the shaft

$I_{p}$ is the polar moment of inertia of the shaft

$\tau$ is the shear stress developed at a distance 'r' from the center of the shaft

$\theta$ is the angle of twist of the shaft

'G' is the modulus of rigidity of the shaft

We know that for solid shaft $I_{p}=\frac{\pi R^4}{2}$

For a hollow shaft $I_{p}=\frac{\pi (R_o^4-R_i^4)}{2}$

Since the two shafts are subjected to same torque from the relation of Torque we have

1) For solid shaft

$\frac{2T}{\pi R^4}\times r=\tau _{solid}$

2) For hollow shaft we have

$\tau _{hollow}=\frac{2T}{\pi (R^4-0.7R^4)}\times r=\frac{2T}{\pi 0.76R^4}$

Comparing the above 2 relations we see

$\frac{\tau _{solid}}{\tau _{hollow}}=0.76$

Similarly for angle of twist we can see

$\frac{\theta _{solid}}{\theta _{hollow}}=\frac{\frac{LT}{I_{solid}}}{\frac{LT}{I_{hollow}}}=\frac{I_{hollow}}{I_{solid}}=1.316$

Part b)

Strength of solid shaft = $\tau _{max}=\frac{T\times R}{I_{solid}}$

Weight of solid shaft = $\rho \times \pi R^2\times L$

Strength per unit weight of solid shaft = $\frac{\tau _{max}}{W}=\frac{T\times R}{I_{solid}}\times \frac{1}{\rho \times \pi R^2\times L}=\frac{2T}{\rho \pi ^2R^5L}$

Strength of hollow shaft = $\tau '_{max}=\frac{T\times R}{I_{hollow}}$

Weight of hollow shaft = $\rho \times \pi (R^2-0.7R^2)\times L$

Strength per unit weight of hollow shaft = $\frac{\tau _{max}}{W}=\frac{T\times R}{I_{hollow}}\times \frac{1}{\rho \times \pi (R^2-0.7^2)\times L}=\frac{5.16T}{\rho \pi ^2R^5L}$

Thus $\frac{Strength/Weight _{hollow}}{Strength/Weight _{Solid}}=5.16$

3 0

3 years ago