Answer:
Q = 5.06 x 10⁻⁸ m³/s
Explanation:
Given:
v=0.00062 m² /s and ρ= 850 kg/m³
diameter = 8 mm
length of horizontal pipe = 40 m
Dynamic viscosity =
μ = ρv
=850 x 0.00062
= 0.527 kg/m·s
The pressure at the bottom of the tank is:
P₁,gauge = ρ g h = 850 x 9.8 x 4 = 33.32 kN/m²
The laminar flow rate through a horizontal pipe is:
Q = 5.06 x 10⁻⁸ m³/s
Answer: MAP!!! it's always a good Idea to have a map nearby, cause you'll never know when you'll need it or like you said, lose reception
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:
The work furnished by the compressor is 
The minimum work required for the state to change is 
Explanation:
The explanation to these solution is on the first, second , third and fourth uploaded image respectively
