Answer:
pls put a question not random letters
Explanation:
Answer:
D
Explanation:
Masonry uses stone work, making a stone wall requires perfect masonry.
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:
The total tube surface area in m² required to achieve an air outlet temperature of 850 K is 192.3 m²
Explanation:
Here we have the heat Q given as follows;
Q = 15 × 1075 × (1100 - 
) = 10 × 1075 × (850 - 300) = 5912500 J
∴ 1100 - = 1100/3
= 733.33 K
Where
= Arithmetic mean temperature difference
= Inlet temperature of the gas = 1100 K
= Outlet temperature of the gas = 733.33 K
= Inlet temperature of the air = 300 K
= Outlet temperature of the air = 850 K
Hence, plugging in the values, we have;
Hence, from;
, we have
5912500 = 90 × A × 341.67
Hence, the total tube surface area in m² required to achieve an air outlet temperature of 850 K = 192.3 m².
Answer:
The specific weight of unknown liquid is found to be 15 KN/m³
Explanation:
The total pressure in tank is measured to be 65 KPa in the tank. But, the total pressure will be equal to the sum of pressures due to both oil and unknown liquid.
Total Pressure = Pressure of oil + Pressure of unknown liquid
65 KPa = (Specific Weight of oil)(depth of oil) + (Specific Weight of unknown liquid)(depth of unknown liquid)
65 KN/m² = (8.5 KN/m³)(5 m) + (Specific Weight of Unknown Liquid)(1.5 m)
(Specific Weight of Unknown Liquid)(1.5 m) = 65 KN/m² - 42.5 KN/m²
(Specific Weight of Unknown Liquid) = (22.5 KN/m²)/1.5 m
<u>Specific Weight of Unknown Liquid = 15 KN/m³</u>
