All categories

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.

You might be interested in

Consider the combustion of ethanol C2H5OH with air. Assume the air is dry and comprised of 21% oxygen and 79% nitrogen on a mola

babunello [35]

Answer:

a) 14.285

b) 8.956

Explanation:

Given :

The combustion of the ethanol is with air

Air is 21% oxygen

and, 79% nitrogen

thus, for 1 O₂ we have (79/21)N₂

thus,

the stochiometric equation for the combustion is as:

C₂H₅OH + 3[O₂ + (79/21)N₂] ⇒ 2CO₂ + 3H₂O + 3 × (79/21)N₂

Now,

the molecular weight of the fuel (C₂H₅OH) = (2× 12) + (5 × 1) + 16 + 1 = 46 g/mol

Molecular weight of the air = (2 × 16) + ((79/21) × 28) = 137.33 g/mol

a) air/fuel ratio on a molar basis

we have

air-fuel ratio = moles of air / moles of fuel

air-fuel ratio = [3 × 1 + 3 × (79/21)] / 1 = 14.285

b) air/fuel ratio on a mass basis = Mass of air / mass of fuel

air/fuel ratio on a mass basis = (number of moles of air × molar mass of air) / (number of moles of fuel × molar mass of fuel)

on substituting the values, we have

air/fuel ratio on a mass basis = (3 × 137.33) / (1 × 46) = 8.956

8 0

3 years ago

PLEASE HURRY TIMED TEST

ladessa [460]

Answer:

Explanation:

5 0

3 years ago

Read 2 more answers

Three capillary tubes with different radius (r1=1.0mm, r2=0.1mm, r3=0.01mm) are inserted into the same cup of water. The surface

kobusy [5.1K]

Answer:

you fgykyiihohuoououu

Explanation:

jjj

4 0

3 years ago

Using the Breguet range and endurance equation, estimate the amount of kerosene

love history [14]

The amount of kerosene fuel needed for an aircraft weighing 10 metric tons to fly from Boston to Los Angeles, assuming a distance of 5,000 [km] is 10000 kg.

<h3>What is Breguet equation?</h3>

Breguet equation is used to determine the range of airplane fly in some specified set of parameters.

$R=\dfrac{h_f}{g}\dfrac{L}{D}\times \eta \ln \left(\dfrac{W_{initial}}{W_{final}}\right)$

Here, (L/D) is lift to drag ratio,<em> g</em> is gravitation acceleration,<em> h </em>is height and <em>W</em> is weight.

The weight of aircraft is 10 metric tons (this is the dry weight that includes passengers and cargo) The distance of 5,000 [km], with velocity at 300 [m/s] from Boston to Los Angeles has to be cover by the airplane.

The initial weight is 10 metric tons or 10000 kg. Thus, the final weight with total fuel burn is,

$W_{final}=W_{initial}-W_{fuel}\\W_{final}=1-\dfrac{W_{fule}}{W_{initial}}$

The lift-to-drag ratio is 15 and the overall efficiency is 0.3, The standard air density at an altitude of approximately 6,000 [m] is half.

Thus, put the values, in above formula,

$5000=\dfrac{6000}{9.81}(15)\times(0.3) \ln \left(\dfrac{10000}{1-\dfrac{W_{fuel}}{10000}}}\right)\\$

$\ln \left(\dfrac{10000}{1-\dfrac{W_{fuel}}{10000}}}\right)=1.8167\\\ln \left(\dfrac{1}{10000-W_{fuel}}}\right)=1.8167$

Solving this equation, we get,

$W_{fuel}=10000\rm\; kg\\$

Thus, the amount of kerosene fuel needed for an aircraft weighing 10 metric tons to fly from Boston to Los Angeles, assuming a distance of 5,000 [km] is 10000 kg.

Learn more about the Breguet equation here:

brainly.com/question/15218094

#SPJ1

7 0

2 years ago

True/False

Temka [501]

Answer:

because it maybe have been broken or can cause minor accident.

8 0

3 years ago