Answer:
Steps:
1. Create a text file that contains blade diameter (in feet), wind velocity (in mph) and the approximate electricity generated for the year
2. load the data file for example, in matlab, use ('fileame.txt') to load the file
3. create variables from each column of your data
for example, in matlab,
x=t{1}
y=t{2}
4. plot the wind velocity and electricity generated.
plot(x, y)
5. Label the individual axis and name the graph title.
title('Graph of wind velocity vs approximate electricity generated for the year')
xlabel('wind velocity')
ylabel('approximate electricity generated for the year')
Answer:
1.0MG
Explanation:
to solve this problem we use this formula
S₀-S/t = ksx --- (1)
the values have been given as
concentration = S₀ = 250mg
effluent concentration = S= 10mg
value of K = 0.04L/day
x = 3000 mg
when we put these values into this equation,
250-10/t = 0.04x10x3000
240/t = 1200
we cross multiply from this stage
240 = 1200t
t = 240/1200
t = 0.2
remember the question says that 5MGD is required to be treated
so the volume would be
v = 0.2x5
= 1.0 MG
Answer: MAP!!! it's always a good Idea to have a map nearby, cause you'll never know when you'll need it or like you said, lose reception
Answer:
for 1st question the answer is 5th option.
for 2nd question the answer is 2nd option
hope it helps you mate
please mark me as brainliast
Answer:
Option ‘a’ is the cheapest for this house.
Explanation:
Cheapest method of heating must have least cost per kj of energy. So, convert all the energy in the same unit (say kj) and take select the cheapest method to heat the house.
Given:
Three methods are given to heat a particular house are as follows:
Method (a)
Through Gas, this gives energy of amount $1.33/therm.
Method (b)
Through electric resistance, this gives energy of amount $0.12/KWh.
Method (c)
Through oil, this gives energy of amount $2.30/gallon.
Calculation:
Step1
Change therm to kj in method ‘a’ as follows:
$/kj.
Step2
Change kWh to kj in method ‘b’ as follows:
$/kj.
Step3
Change kWh to kj in method ‘c’ as follows:
$/kj.
Thus, the method ‘a’ has least cost as compare to method b and c.
So, option ‘a’ is the cheapest for this house.
