If the load weighs 350 N, the effort that is required to lift the weight is 70 N
To answer the question, we need to know what mechanical advantage is.
<h3>What is mechanical advantage?</h3>
The mechanical advantage of the hydraulic lift M.A = L/E where
<h3>Effort required to lift the weight</h3>
Making E subject of the formula, we have
E = L/M.A
Given that
substituting the values of the variables into the equation, we have
E = L/M.A
E = 350 N/5
E = 70 N
So, the effort that is required to lift the weight is 70 N
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Answer:
1. The data will not be tallying up - there will great variations and deviance.
2. There will be not enough data points
Explanation:
The management style, though it uses one manager, is not devoid of problems in the system. For example, there will be great difficulty in the collection of accurate data to properly define the system. This presents a great misinformation on the outcome of the data process. This is likely to be caused by the first factor that is the data having a lot of discrepancies. The discrepancy causes the data to be meaningless in terms of the trends displayed. In addition, the lack of enough points posses a challenge to the data collection and processing procedure. The data base management system can be used to concentrate data on a central point of reference.
Answer:
- public class Main {
-
- public static void main(String[] args) {
- int myArray[] = {3, 7, 2, 5, 9,11, 24, 6, 10, 12};
- int myArray2 [] = {1, 2, 3, 4 ,5};
-
- displayValue(myArray);
- reverseDisplay(myArray);
- displaySum(myArray);
- displayLess(myArray, 10);
- displayHighAvg(myArray);
- displayBoth(myArray, myArray2);
- }
-
- public static void displayValue(int arr[]){
- for(int i = 0; i < arr.length; i++){
- System.out.print(arr[i] + " ");
- }
-
- System.out.println();
- }
-
- public static void reverseDisplay(int arr[]){
- for(int i = arr.length-1; i >= 0; i--){
- System.out.print(arr[i] + " ");
- }
-
- System.out.println();
- }
-
- public static void displaySum(int arr[]){
- int sum = 0;
- for(int i = 0; i < arr.length; i++){
- sum += arr[i];
- }
- System.out.println(sum);
- }
-
- public static void displayLess(int arr[], int limit){
-
- for(int i = 0; i < arr.length; i++){
- if(arr[i] < limit){
- System.out.print(arr[i] + " ");
- }
- }
- System.out.println();
- }
-
-
- public static void displayHighAvg(int arr[]){
- int sum = 0;
- for(int i = 0; i < arr.length; i++){
- sum += arr[i];
- }
-
- double avg = sum / arr.length;
-
- for(int i = 0; i < arr.length; i++){
- if(arr[i] > avg){
- System.out.print(arr[i] + " ");
- }
- }
- System.out.println();
- }
-
- public static void displayBoth(int arr1[], int arr2 []){
- for(int i = 0; i < arr2.length; i++){
- for(int j = 0; j < arr1.length; j++){
- if(arr1[j] == arr2[i]){
- System.out.print(arr1[j] + " ");
- }
- }
- }
- System.out.println();
- }
- }
Explanation:
There are five methods written to solve all the problems stated in the question.
Method 1 : displayValue (Line 15 - 21)
This is the method that take one input array and use the print() method to display the all the elements in the array.
Method 2: reverseDisplay (Line 23 - 26)
This method will take one input array and print the value in the reverse order. We just need to start with the last index when running the for-loop to print the value.
Method 3: displaySum (Line 31 - 37)
This method will take one input array and use a for-loop to calculate the total of the values in the array.
Method 4: displayLess (Line 39 - 47)
This method will take one two inputs, a array and a limit. We use the limit as the condition to check if any value less than the limit, then the value will only be printed.
Method 5: displayHighAvg (Line 50 - 64)
This method will take one input array and calculate the average. The average will be used to check if any element in the array higher than it, then the value will only be printed.
Method 6: displayBoth (Line 66 - 75)
This method will take two input arrays and compare both of them to find out if any value appears in both input arrays and print it out.