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.
Answer:
vapor fraction = 0.4 and 0.08
Explanation:
At reasonably high temperatures, a mixture will exist in the form of a sub cooled liquid. Between these extremes, the mixture exists in a two phrase region where it is a vapor liquid equilibrium. From a vapor-liquid phase diagram, a mixture of 40% A, 39% B, and 21% C separates to give the vapor compositions of 0.4 and 0.08.
Answer:
a) the power consumption of the LEDs is 0.25 watt
b) the LEDs drew 0.0555 Amp current
Explanation:
Given the data in the question;
Three AAA Batteries;
<---- 1000mAh [ + -] 1.5 v ------1000mAh [ + -] 1.5 v --------1000mAh [ + -] 1.5 v------
so V_total = 3 × 1.5 = 4.5V
a) the power consumption of the LEDs
I_battery = 1000 mAh / 18hrs { for 18 hrs}
I_battery = 1/18 Amp { delivery by battery}
so consumption by led = I × V_total
we substitute
⇒ 1/18 × 4.5
P = 0.25 watt
Therefore the power consumption of the LEDs is 0.25 watt
b) How much current do the LEDs draw
I_Draw = I_battery = 1/18 Amp = 0.0555 Amp
Therefore the LEDs drew 0.0555 Amp current
Answer:
im sorry but i cant find any studies about this and im 3 days late
Answer:
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