Answer:
import pandas pd
def read_prices(tickers):
price_dict = {}
# Read ingthe ticker data for all the tickers
for ticker in tickers:
# Read data for one ticker using pandas.read_csv
# We assume no column names in csv file
ticker_data = pd.read_csv("./" + ticker + ".csv", names=['date', 'price', 'volume'])
# ticker_data is now a panda data frame
# Creating dictionary
# for the ticker
price_dict[ticker] = {}
for i in range(len(ticker_data)):
# Use pandas.iloc to access data
date = ticker_data.iloc[i]['date']
price = ticker_data.iloc[i]['price']
price_dict[ticker][date] = price
return price_dict
Answer:
18 kJ
Explanation:
Given:
Initial volume of air = 0.05 m³
Initial pressure = 60 kPa
Final volume = 0.2 m³
Final pressure = 180 kPa
Now,
the Work done by air will be calculated as:
Work Done = Average pressure × Change in volume
thus,
Average pressure =
= 120 kPa
and,
Change in volume = Final volume - Initial Volume = 0.2 - 0.05 = 0.15 m³
Therefore,
the work done = 120 × 0.15 = 18 kJ
Answer:
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Explanation:
Answer:
Both model building codes and NFPA 220 can be used to determine the type of construction used in a building.
Answer:
T1 = 625.54 K
Explanation:
We are given;
T_α = Tsur = 25°C = 298K
h = 20 W/m².K,
L = 0.15 m
K = 1.2 W/m.K
ε = 0.8
Ts = T2 = 100°C = 373K
T1 = ?
Assumption:
-Steady- state condition
-One- dimensional conduction
-No uniform heat generation
-Constant properties
From Energy balance equation;
E°in - E°out = 0
Thus,
q"cond – q"conv – q"rad = 0
K[(T1 - T2)/L] - h(Ts-T_α) - εσ (Ts⁴ – Tsur⁴)
Where σ is Stephan-Boltzmann constant and has a value of 5.67 x 10^(-8)
Thus;
K[(T1 - T2)/L] - h(Ts-T_α) - εσ (Ts⁴ – Tsur⁴) = 1.2[(T1 - 373)/0.15] - 20(373 - 298] - 0.8x5.67x10^(-8)[373⁴ - 298⁴] = 0
This gives;
(8T1 - 2984) - (1500) - 520.31 = 0
8T1 = 2984 + 1500 + 520.31
8T1 = 5004.31
T1 = 5004.31/8
T1 = 625.54 K