Answer:
Q = 125.538 W
Explanation:
Given data:
D = 30 cm
Temperature 
degree celcius
Heat coefficient = 12 W/m^2 K
Efficiency 80% = 0.8
Q = 125.538 W
Answer:
M_c = 61.6 Nm
Explanation:
Given:
F_a = 2.5 KN
Find:
Determine the moment of this force about C for the two casesshown.
Solution:
- Draw horizontal and vertical vectors at point A.
- Take moments about point C as follows:
M_c = F_a*( 42 / 150 ) *88
M_c = 2.5*( 42 / 150 ) *88
M_c = 61.6 Nm
- We see that the vertical component of force at point A passes through C.
Hence, its moment about C is zero.
Answer: N has to be lesser than or equal to 1666.
Explanation:
Cost of parts N in FPGA = $15N
Cost of parts N in gate array = $3N + $20000
Cost of parts N in standard cell = $1N + $100000
So,
15N < 3N + 20000 lets say this is equation 1
(cost of FPGA lesser than that of gate array)
Also. 15N < 1N + 100000 lets say this is equation 2
(cost of FPGA lesser than that of standardcell)
Now
From equation 1
12N < 20000
N < 1666.67
From equation 2
14N < 100000
N < 7142.85
AT the same time, Both conditions must hold true
So N <= 1666 (Since N has to be an integer)
N has to be lesser than or equal to 1666.
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).
