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!!!
Answer:
Im guessing this is for CEA for PLTW, if so look up the exact assignment number and look at online examples of the exact same assignment.
Explanation:
Answer:
The required diameter of the fuse wire should be 0.0383 cm to limit the current to 0.53 A with current density of 459 A/cm²
.
Explanation:
We are given current density of 459 A/cm² and we want to limit the current to 0.53 A in a fuse wire. We are asked to find the corresponding diameter of the fuse wire.
Recall that current density is given by
j = I/A
where I is the current flowing through the wire and A is the area of the wire
A = πr²
but r = d/2 so
A = π(d/2)²
A = πd²/4
so the equation of current density becomes
j = I/πd²/4
j = 4I/πd²
Re-arrange the equation for d
d² = 4I/jπ
d = √4I/jπ
d = √(4*0.53)/(459π)
d = 0.0383 cm
Therefore, the required diameter of the fuse wire should be 0.0383 cm to limit the current to 0.53 A with current density of 459 A/cm²
.
