It is to be noted that it is impossible to find the Maclaurin Expansion for F(x) = cotx.
<h3>What is
Maclaurin Expansion?</h3>
The Maclaurin Expansion is a Taylor series that has been expanded around the reference point zero and has the formula f(x)=f(0)+f′. (0) 1! x+f″ (0) 2! x2+⋯+f[n](0)n!
<h3>
What is the explanation for the above?</h3>
as indicated above, the Maclaurin infinite series expansion is given as:
F(x)=f(0)+f′. (0) 1! x+f″ (0) 2! x2+⋯+f[n](0)n!
If F(0) = Cot 0
F(0) = ∝ = 1/0
This is not definitive,
Hence, it is impossible to find the Maclaurin infinite series expansion for F(x) = cotx.
Learn more about Maclaurin Expansion at;
brainly.com/question/7846182
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Answer:
15625 moles of methane is present in this gas deposit
Explanation:
As we know,
PV = nRT
P = Pressure = 230 psia = 1585.79 kPA
V = Volume = 980 cuft = 27750.5 Liters
n = number of moles
R = ideal gas constant = 8.315
T = Temperature = 150°F = 338.706 Kelvin
Substituting the given values, we get -
1585.79 kPA * 27750.5 Liters = n * 8.315 * 338.706 Kelvin
n = (1585.79*27750.5)/(8.315 * 338.706) = 15625
Answer:
Assumption:
1. The kinetic and potential energy changes are negligible
2. The cylinder is well insulated and thus heat transfer is negligible.
3. The thermal energy stored in the cylinder itself is negligible.
4. The process is stated to be reversible
Analysis:
a. This is reversible adiabatic(i.e isentropic) process and thus 
From the refrigerant table A11-A13
sat vapor
m=
b.) We take the content of the cylinder as the sysytem.
This is a closed system since no mass leaves or enters.
Hence, the energy balance for adiabatic closed system can be expressed as:
ΔE
ΔU
![w_{b, out} =m([tex]u_{1} -u_{2](https://tex.z-dn.net/?f=w_%7Bb%2C%20out%7D%20%3Dm%28%5Btex%5Du_%7B1%7D%20-u_%7B2)
)
workdone during the isentropic process
=5.8491(246.82-219.9)
=5.8491(26.91)
=157.3993
=157.4kJ
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Answer:
V₂ = 20 V
Vt = 20 V
V₁ = 20 V
V₃ = 20 V
I₁ = 10 mA
I₃ = 3.33 mA
It = 18.33 mA
Rt = 1090.91 Ω
Pt = 0.367 W
P₁ = 0.2 W
P₂ = 0.1 W
P₃ = 0.067 W
Explanation:
Part of the picture is cut off. I assume there is a voltage source Vt there?
First, use Ohm's law to find V₂.
V = IR
V₂ = (0.005 A) (4000 Ω)
V₂ = 20 V
R₁ and R₃ are in parallel with R₂ and the voltage source Vt. That means V₁ = V₂ = V₃ = Vt.
V₁ = 20 V
V₃ = 20 V
Vt = 20 V
Now we can use Ohm's law again to find I₁ and I₃.
V = IR
I = V/R
I₁ = (20 V) / (2000 Ω)
I₁ = 0.01 A = 10 mA
I₃ = (20 V) / (6000 Ω)
I₃ = 0.00333 A = 3.33 mA
The current It passing through Vt is the sum of the currents in each branch.
It = I₁ + I₂ + I₃
It = 10 mA + 5 mA + 3.33 mA
It = 18.33 mA
The total resistance is the resistance of the parallel resistors:
1/Rt = 1/R₁ + 1/R₂ + 1/R₃
1/Rt = 1/2000 + 1/4000 + 1/6000
Rt = 1090.91 Ω
Finally, the power is simply each voltage times the corresponding current.
P = IV
Pt = (0.01833 A) (20 V)
Pt = 0.367 W
P₁ = (0.010 A) (20 V)
P₁ = 0.2 W
P₂ = (0.005 A) (20 V)
P₂ = 0.1 W
P₃ = (0.00333 A) (20 V)
P₃ = 0.067 W
