Answer:
a) 6.4 mg/l
b) 5.6 mg/l
Explanation:
Given data:
effluent Discharge Q_w = 1.0 m^3.s
Ultimate BOD L_w = 40 mg/l
Discharge of stream Q_r = 10 m^3.s
Stream ultimate BOD L_r = 3 mg/l
a) Ultimate BOD of mixture
b) utlimate BOD at 10,000 m downstream
putting 
t = 0.578 days
we know
Answer:
View Image
Explanation:
You didn't provide me a picture of the opamp.
I'm gonna assume that this is an ideal opamp, therefore the input impedance can be assumed to be ∞ . This basically implies that...
- no current will go in the inverting(-) and noninverting(+) side of the opamp
- V₊ = V₋ , so whatever voltage is at the noninverting side will also be the voltage at the inverting side
Since no current is going into the + and - side of the opamp, then
i₁ = i₂
Since V₊ is connected to ground (0V) then V₋ must also be 0V.
V₊ = V₋ = 0
Use whatever method you want to solve for v_out and v_in then divide them. There's so many different ways of solving this circuit.
You didn't give me what the input voltage was so I can't give you the entire answer. I'll just give you the equations needed to plug in your values to get your answers.
Answer:
Check the 2nd, 3rd and 4th statements.
Explanation:
Answer:While heat cycles cause oil to darken, soot causes oil to turn black
Explanation:
Most people associate soot with diesel engines, but gasoline engines can produce soot as well, particularly modern gasoline-direct-injection engines. ... Any finer filtration and the filter could catch dissolved additives in the motor oil
Answer:
- public class Main {
- public static void main(String[] args) {
- String testString = "abscacd";
-
- String evenStr = "";
- String oddStr = "";
-
- for(int i=testString.length() - 1; i >= 0; i--){
-
- if(i % 2 == 0){
- evenStr += testString.charAt(i);
- }
- else{
- oddStr += testString.charAt(i);
- }
- }
-
- System.out.println(evenStr + oddStr);
- }
- }
Explanation:
Firstly, let declare a variable testString to hold an input string "abscacd" (Line 1).
Next create another two String variable, evenStr and oddStr and initialize them with empty string (Line 5-6). These two variables will be used to hold the string at even index and odd index, respectively.
Next, we create a for loop that traverse the characters of the input string from the back by setting initial position index i to testString.length() - 1 (Line 8). Within the for-loop, create if and else block to check if the current index, i is divisible by 2, (i % 2 == 0), use the current i to get the character of the testString and join it with evenStr. Otherwise, join it with oddStr (Line 10 -14).
At last, we print the concatenated evenStr and oddStr (Line 18).
