Answer:
a = 6, b = 0
Explanation:
The loop ran 3 times before b == 0. The statement "while ((b != 0)" is essentially saying: 'While b is not equal to 0, do what's in my loop'. Same general thing with "&& ((a / b) >= 0)". The "&&" is specifying that there should be another loop condition. The final part of the while loop states: 'as long as a ÷ b is greater than 0, do what's in my loop'. If all of these conditions are met, the loop will run. It will continue to run until at least one of the conditions are not met.
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<em>Side note: I can't help but notice you posted the same question a while ago, so I just copied and pasted my previous response with some tweaking here and there. Hope this helps you! :)</em>
Answer:
sorry,but I can't understand your question.
Answer:
Following are the code in the java language
numBers = scnr.nextInt ( ) ;
Here scnr is an instance of scanner class .
Explanation:
In this code, we take the input by using the object of Scanner class i.e "scnr". The scanner class in the java programming language is used for taking the input by the user. The scnr.nextInt ( ) is taking the input which is stored in the "numbers" variable.
So the whole program is looking like that
import java.util.*; // import package
public class Main
{
public static void main(String[] args) // main function
{
int numBers; // variable declaration
Scanner scnr=new Scanner(System.in); // create the instance of scanner class
numBers=scnr.nextInt( ) ; // taking input
System.out.println(numBers); // display the value of numBers
}
}
Output:
78
78
the second answer is correct
Answer:
h = 8.588 m
Explanation:
Given:
Mass of hockey puck: m = 200 g = 0.2 K
g
Angle of incline: θ = 30º (with respect to horizontal)
Coefficient of static and kinetic friction between the puck and ramp:
μ
s = 0.4 and μk = 0.3
initial speed:
v = 3.5 m
/s
Let h be the vertical height reached by the puck, above the ground. This corresponds to a distance of
d = h
*sin 30º = 2
*h
along the incline. As the puck is moving, only kinetic friction comes into play as it climbs up the incline. Let N be the normal reaction exerted by the metal ramp on the ball. Then,
N = m*g
*cos θ = 0.2
*9.8
1*cos 30
º = 1.6991 N
Therefore, kinetic friction acting on the puck is:
F
k = μ
k
*N = 0.3
*1.6974 = 0.50974 N
From work energy theorem, the change in kinetic energy should equal the work done by friction and gravity. Therefore,
0.5* 0.2
*16² = 0.50974
*2h + 0.2*9.81*h
⇒ h = 8.588 m
