Answer:
Java program explained below
Explanation:
FindSpecialNumber.java
import java.util.Scanner;
public class FindSpecialNumber {
public static void main(String[] args) {
//Declaring variable
int number;
/*
* Creating an Scanner class object which is used to get the inputs
* entered by the user
*/
Scanner sc = new Scanner(System.in);
//getting the input entered by the user
System.out.print("Enter a number :");
number = sc.nextInt();
/* Based on user entered number
* check whether it is special number or not
*/
if (number == -99 || number == 0 || number == 44) {
System.out.println("Special Number");
} else {
System.out.println("Not Special Number");
}
}
}
_______________
Output#1:
Enter a number :-99
Special Number
Output#2:
Enter a number :49
Not Special Number
Answer:
C. 1 ft long with a weight of 1 lb
Answer:
Fa = 57.32 N
Explanation:
given data
mass = 5 kg
acceleration = 4 m/s²
angular velocity ω = 2 rad/s
solution
first we take here moment about point A that is
∑Ma = Iα + ∑Mad ...............1
put here value and we get
so here I = (
) × m × L² ................2
I = (
) × 5 × 0.8²
I = 0.267 kg-m²
and
a is = r × α
a = 0.4 α
so now put here value in equation is 1
0 = 0.267 α + m r α (0.4) - m A (0.4)
0 = 0.267 α + 5 (0.4α) (0.4 ) - 5 (4) 0.4
so angular acceleration α = 7.5 rad/s²
so here force acting on x axis will be
∑ F(x) = m a(x) ..............3
a(x) = m a - m rα
put here value
a(x) = 5 × 4 - 5 × 0.4 × 7.5
a(x) = 5 N
and
force acting on y axis will be
∑ F(y) = m a(y) .............. 4
a(y) - mg = mrω²
a(y) - 5 × 9.81 = 5 × 0.4 × 2²
a(y) = 57.1 N
so
total force at A will be
Fa =
...............5
Fa =
Fa = 57.32 N