Answer: the half-angle "alpha" of the Mach cone = 30⁰
Explanation:
To calculate the half-angle "alpha" of the Mach cone.
we say ;
Sin∝ = 1 / Ma
given that Ma = 2
now we substitute
Sin∝ = 1 / 2
Sin∝ = 0.5
∝ = Sin⁻¹ 0.5
∝ = 30⁰
Therefore, the half-angle "alpha" of the Mach cone is 30⁰
Death benefit from a Life insurance policy
Answer:
The radius of a wind turbine is 691.1 ft
The power generation potential (PGP) scales with speed at the rate of 7.73 kW.s/m
Explanation:
Given;
power generation potential (PGP) = 1000 kW
Wind speed = 5 mph = 2.2352 m/s
Density of air = 0.0796 lbm/ft³ = 1.275 kg/m³
Radius of the wind turbine r = ?
Wind energy per unit mass of air, e = E/m = 0.5 v² = (0.5)(2.2352)²
Wind energy per unit mass of air = 2.517 J/kg
PGP = mass flow rate * energy per unit mass
PGP = ρ*A*V*e

r = 210.64 m = 691.1 ft
Thus, the radius of a wind turbine is 691.1 ft
PGP = CVᵃ
For best design of wind turbine Betz limit (c) is taken between (0.35 - 0.45)
Let C = 0.4
PGP = Cvᵃ
take log of both sides
ln(PGP) = a*ln(CV)
a = ln(PGP)/ln(CV)
a = ln(1000)/ln(0.4 *2.2352) = 7.73
The power generation potential (PGP) scales with speed at the rate of 7.73 kW.s/m
Answer:
Explanation:
The python code to generate this is quite simple to run.
i hope you understand everything written here, you can as well try out other problems to understand better.
First to begin, we import the package;
Code:
import pandas as pd
import matplotlib.pyplot as plt
name = input('Enter name of the file: ')
op = input('Enter name of output file: ')
df = pd.read_csv(name)
df['Date'] = pd.to_datetime(df["Date"].apply(str))
plt.plot(df['Date'],df['Absent']/(df['Present']+df['Absent']+df['Released']),label="% Absent")
plt.legend(loc="upper right")
plt.xticks(rotation=20)
plt.savefig(op)
plt.show()
This should generate the data(plot) as seen in the uploaded screenshot.
thanks i hope this helps!!!
Solution:
Given that :
Volume flow is, 
So, 
Therefore, the equation of a single straight vessel is given by
......................(i)
So there are 100 similar parallel pipes of the same cross section. Therefore, the equation for the area is

or 
Now for parallel pipes
...........(ii)
Solving the equations (i) and (ii),




Therefore,

or 
Thus the answer is option A). 10