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:
A) Upper bound modulus of elasticity; E = 165.6 GPa
B) Lower bound modulus of elasticity; E = 83.09 GPa
Explanation:
A) Formula for upper bound modulus is given as;
E = E_m(1 - V_f) + E_f•V_f
We are given;
E_m = 60 GPa
E_f = 380 GPa
V_f = 33% = 0.33
Thus,
E = 60(1 - 0.33) + 380(0.33)
E = (60 x 0.67) + 125.4
E = 165.6 GPa
B) Formula for lower bound modulus is given as;
E = 1/[(V_f/E_f) + ((1 – V_f)/E_m)]
E = 1/[(0.33/380) + ((1 – 0.33)/60)]
E = 1/(0.0008684 + 0.01116667)
E = 1/0.01203507
E = 83.09 GPa
Answer:
Statement 1: All balls hit the ground at the same time
Explanation:
When there is no resistance of air, the acceleration due to gravity experienced by all the bodies are same. So for falling bodies, neglecting the air resistance, the falling object will be weightless and therefore all the objects will hit the ground at the same time when there is nor air resistance and the objects are considered to be falling in vacuum.
Answer:
(a) Current density at P is
.
(b) Total current I is 3.257 A
Explanation:
Because question includes symbols and formulas it can be misunderstood. In the question current density is given as below;

where
and
unit vectors.
(a) In order to find the current density at a specific point <em>(P)</em>, we can simply replace the coordinates in the current density equation. Therefore

(b) Total current flowing outward can be calculated by using the relation,

where integral is calculated through the circular band given in the question. We can write the integral as below,

due to unit vector multiplication. Then,

where
. Therefore
