Answer:
# Program is written in python
# 22.1 Using the count method, find the number of occurrences of the character 's' in the string 'mississippi'.
# initializing string
Stringtocheck = "mississippi"
# using count() to get count of s
counter = Stringtocheck.count('s')
# printing result
print ("Count of s is : " + str(counter))
# 2.2 In the string 'mississippi', replace all occurrences of the substring 'iss' with 'ox
# Here, we'll make use of replace() method
# Prints the string by replacing iss by ox
print(Stringtocheck.replace("iss", "ox"))
#2.3 Find the index of the first occurrence of 'p' in 'mississippi'
# declare substring
substring = 'p'
# Find index
index = Stringtocheck.find(substring)
# Print index
print(index)
# End of program
Answer: um wuh anyways thxs for the points!
Explanation: ....:/
Algorithm of the Nios II assembly program.
- Attain data for simulation from the SW11-0, on the DE2-115 Simulator
- The data will be read from the switches in loop.
- The decimal output is displayed using the seven-segment displays and done using the loop.
- The program is ended by the user operating the SW1 switch
and
The decimal equivalent on the seven-segment displays HEX3-0 is
- DE2-115
- DE2-115_SW11
- DE2-115_HEX3
- DE2-115_HEX4
- DE2-115_HEX5
- DE2-115_HEX6
- DE2-115_HEX7
<h3>The Algorithm and
decimal equivalent on the
seven-segment displays HEX3-0</h3>
Generally, the program will be written using a cpulator simulator in order to attain best result.
We are to
- Attain data for simulation from the SW11-0, on the DE2-115 Simulator
- The data will be read from the switches in loop.
- The decimal output is displayed using the seven-segment displays and done using the loop.
- The program is ended by the user operating the SW1 switch
This will be the Algorithm of the Nios II assembly program .
Hence, the decimal equivalent on the seven-segment displays HEX3-0 is
- DE2-115
- DE2-115_SW11
- DE2-115_HEX3
- DE2-115_HEX4
- DE2-115_HEX5
- DE2-115_HEX6
- DE2-115_HEX7
For more information on Algorithm
brainly.com/question/11623795
Answer:
471 days
Explanation:
Capacity of Carvins Cove water reservoir = 3.2 billion gallons i.e. 3.2 x 10˄9 gallons
As,
1 gallon = 0.133 cubic feet (cf)
Therefore,
Capacity of Carvins Cove water reservoir in cf = 3.2 x 10˄9 x 0.133
= 4.28 x 10˄8
Applying Mass balance i.e
Accumulation = Mass In - Mass out (Eq. 01)
Here
Mass In = 0.5 cfs
Mass out = 11 cfs
Putting values in (Eq. 01)
Accumulation = 0.5 - 11
= - 10.5 cfs
Negative accumulation shows that reservoir is depleting i.e. at a rate of 10.5 cubic feet per second.
Converting depletion of reservoir in cubic feet per hour = 10.5 x 3600
= 37,800
Converting depletion of reservoir in cubic feet per day = 37, 800 x 24
= 907,200
i.e. 907,200 cubic feet volume is being depleted in days = 1 day
1 cubic feet volume is being depleted in days = 1/907,200 day
4.28 x 10˄8 cubic feet volume will deplete in days = (4.28 x 10˄8) x 1/907,200
= 471 Days.
Hence in case of continuous drought reservoir will last for 471 days before dry-up.
Answer:
power developed by the turbine = 6927.415 kW
Explanation:
given data
pressure = 4 MPa
specific enthalpy h1 = 3015.4 kJ/kg
velocity v1 = 10 m/s
pressure = 0.07 MPa
specific enthalpy h2 = 2431.7 kJ/kg
velocity v2 = 90 m/s
mass flow rate = 11.95 kg/s
solution
we apply here thermodynamic equation that
energy equation that is
put here value with
turbine is insulated so q = 0
so here
solve we get
w = 579700 J/kg = 579.7 kJ/kg
and
W = mass flow rate × w
W = 11.95 × 579.7
W = 6927.415 kW
power developed by the turbine = 6927.415 kW
