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:
Mechanical Advantage Formula
The efficiency of a machine is equal to the ratio of its output to its input. It is also equal to the ratio of the actual and theoretical MAs. But, it does not mean that low-efficiency machines are of limited use. An automobile jack, for example, have to overcome a great deal of friction and therefore it has low efficiency. But still, it is extremely valuable because small effort can be applied to lift a great weight.
Also, in another way the mechanical advantage is the force generated by a machine to the force applied to it which is applied in assessing the performance of the machine.
The mechanical advantage formula is:
MA = FBFA
Explanation:
MAmechanical advantageFBthe force of the object
FAthe effort to overcome the force
True strain and engineering strain? True stress is defined as the load divided by the cross-sectional area of the specimen at that instant and is a true indication of the internal pressures. ... Engineering stress is defined as the load divided by the initial cross-sectional area of the specimenAnswer:
Explanation:
Answer:
y ≈ 2.5
Explanation:
Given data:
bottom width is 3 m
side slope is 1:2
discharge is 10 m^3/s
slope is 0.004
manning roughness coefficient is 0.015
manning equation is written as
where R is hydraulic radius
S = bed slope
P is perimeter 
![Q = (2+2y) y) \times 1/0.015 [\frac{(3+2y) y}{(3+2\sqrt{5} y)}]^{2/3} 0.004^{1/2}](https://tex.z-dn.net/?f=Q%20%3D%20%282%2B2y%29%20y%29%20%5Ctimes%201%2F0.015%20%5B%5Cfrac%7B%283%2B2y%29%20y%7D%7B%283%2B2%5Csqrt%7B5%7D%20y%29%7D%5D%5E%7B2%2F3%7D%200.004%5E%7B1%2F2%7D)
solving for y![100 =(2+2y) y) \times (1/0.015) [\frac{(3+2y) y}{(3+2\sqrt{5} y)}]^{2/3} \times 0.004^{1/2}](https://tex.z-dn.net/?f=100%20%3D%282%2B2y%29%20y%29%20%5Ctimes%20%281%2F0.015%29%20%5B%5Cfrac%7B%283%2B2y%29%20y%7D%7B%283%2B2%5Csqrt%7B5%7D%20y%29%7D%5D%5E%7B2%2F3%7D%20%5Ctimes%200.004%5E%7B1%2F2%7D)
solving for y value by using iteration method ,we get
y ≈ 2.5
Answer:
k = 4.21 * 10⁻³(L/(mol.s))
Explanation:
We know that
k = Ae
------------------- euqation (1)
K= rate constant;
A = frequency factor = 4.36 10^11 M⁻¹s⁻¹;
E = activation energy = 93.1kJ/mol;
R= ideal gas constant = 8.314 J/mol.K;
T= temperature = 332 K;
Put values in equation 1.
k = 4.36*10¹¹(M⁻¹s⁻¹)e![^{[(-93.1*10^3)(J/mol)]/[(8.314)(J/mol.K)(332K)}](https://tex.z-dn.net/?f=%5E%7B%5B%28-93.1%2A10%5E3%29%28J%2Fmol%29%5D%2F%5B%288.314%29%28J%2Fmol.K%29%28332K%29%7D)
k = 4.2154 * 10⁻³(M⁻¹s⁻¹)
here M =mol/L
k = 4.21 * 10⁻³((mol/L)⁻¹s⁻¹)
or
k = 4.21 * 10⁻³((L/mol)s⁻¹)
or
k = 4.21 * 10⁻³(L/(mol.s))
