Answer:
Graph for object that is not moving: B
Graph for object that is speeding up: D
Explanation:
A.) In order to represent that an object is not moving, you must either show that there is no velocity (0 m/s) or show a position over time graph that is a horizontal line.
Because the position is the same as time increases, the graph shows that there the object must be at rest, as there is no change in position due to velocity. (Velocity must be 0m/s)
B.) In order to represent an object is speeding up, the position time graph must either be a positive exponential function, the velocity time graph must be a positive, linear line, or the acceleration over time graph must be a positive, horizontal line.
Why is D the correct answer? Because if an object is speeding up, you know that the value of its speed (velocity) is increasing at some rate. And since speeding up refers to positive change, the function of velocity over time graph must be a positive function.
Answer:
Explanation:
Its not a good practice to write all the functions including main() in the same class.
But as you opted for this, the code goes here like this:
public class CalcPyramidVolume {
/* Your solution goes here */
static double pyramidVolume(double baseLength, double baseWidth, double pyramidHeight)
{
double Volume,baseArea;
baseArea = baseLength * baseWidth;
Volume = baseArea * pyramidHeight * 1/3;
return Volume;
}
public static void main (String [] args) {
System.out.println("Volume for 1.0, 1.0, 1.0 is: " + pyramidVolume(1.0, 1.0, 1.0));
return;
}
}
And there is not caveat of integer division, as you are declaring all your variables of type double.
This is called the perihelion.
Answer: A
Explanation:
From the question, the given parameters are given.
Mass M = 30kg
Radius r = 2 m
Coefficient of static friction μ = 0.8
Coefficient of kinetic friction μ = 0.6
Kinetic friction Fk = μ × mg
Fk = 0.6 × 30 × 9.8
Fk = 176.4 N
The force acting on the merry go round is a centripetal force F.
F = MV^2/r
This force must be greater than or equal to the kinetic friction Fk. That is,
F = Fk
F = 176.4
Substitute F , M and r into the centripetal force formula above
176.4 = (30×V^2)/2
Cross multiply
352.8 = 30V^2
V^2 = 352.8/30
V = sqrt (11.76) m/s
V = 5.24 m/s
Therefore, the maximum speed of the merry go round before the child begins to slip is sqrt (12) m/s approximately
