Answer:
Hello your question is incomplete attached below is the complete question
Answer : Factor of safety for point A :
i) using MSS
(Fos)MSS = 3.22
ii) using DE
(Fos)DE = 3.27
Factor of safety for point B
i) using MSS
(Fos)MSS = 3.04
ii) using DE
(Fos)DE = 3.604
Explanation:
Factor of safety for point A :
i) using MSS
(Fos)MSS = 3.22
ii) using DE
(Fos)DE = 3.27
Factor of safety for point B
i) using MSS
(Fos)MSS = 3.04
ii) using DE
(Fos)DE = 3.604
Attached below is the detailed solution
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).
Answer:
938.7 milliseconds
Explanation:
Since the transmission rate is in bits, we will need to convert the packet size to Bits.
1 bytes = 8 bits
1 MiB = 2^20 bytes = 8 × 2^20 bits
5 MiB = 5 × 8 × 2^20 bits.
The formula for queueing delay of <em>n-th</em> packet is : (n - 1) × L/R
where L : packet size = 5 × 8 × 2^20 bits, n: packet number = 48 and R : transmission rate = 2.1 Gbps = 2.1 × 10^9 bits per second.
Therefore queueing delay for 48th packet = ( (48-1) ×5 × 8 × 2^20)/2.1 × 10^9
queueing delay for 48th packet = (47 ×40× 2^20)/2.1 × 10^9
queueing delay for 48th packet = 0.938725181 seconds
queueing delay for 48th packet = 938.725181 milliseconds = 938.7 milliseconds
Answer:
Only Technician B is right.
Explanation:
The cylindrical braking system for a car works through the mode of pressure transmission, that is, the pressure applied to the brake pedals, is transmitted to the brake pad through the cylindrical piston.
Pressure applied on the pedal, P(pedal) = P(pad)
And the Pressure is the applied force/area for either pad or pedal. That is, P(pad) = Force(pad)/A(pad) & P(pedal) = F(pedal)/A(pedal)
If the area of piston increases, A(pad) increases and the P(pad) drops, Meaning, the pressure transmitted to the pad reduces. And for most cars, there's a pressure limit for the braking system to work.
If the A(pad) increases, P(pad) decreases and the braking force applied has to increase, to counter balance the dropping pressure and raise it.
This whole setup does not depend on the length of the braking lines; it only depends on the applied force and cross sectional Area (size) of the piston.
