Three small but dense objects are located in the x-y plane as shown in the figure. The objects have the following masses: mA = 3
.19 kg, mB = 2.55 kg and mC = 1.41 kg.
Determine the x and the y coordinates of the center of the mass of this system. The objects are small in size, they can be treated as point masses.
x coordinate: ?
y coordinate: ?
Determine the x and the y coordinates of the center of the mass of this system. The objects are small in size, they can be treated as point masses.
x coordinate: ?
y coordinate: ?
1 answer:
The x and the y coordinates of the center of the mass of this system will be 43.1 m and 3.12 m respectively.
<h3>What is the center of mass?</h3>A location is established in relation to an object or set of objects in the center of mass. It is the system's average position across all of its components.
Given data;
There are three little objects that are densely spaced out in the x-y plane.;
The value of the masses
m₁=1.41 kg
m₂=2.55 kg
m₃=3.19 kg
Hence, the x and the y coordinates of the center of the mass of this system will be 4.31 m and 3.12 m respectively.
To learn more about the center of mass, refer;
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Answer:
The conservation of energy should be used to answer this question.
a)
At the position where the spring is unstretched, the elastic potential energy of the spring is zero.
since and
is equal to zero.
The roots of this quadratic equation can be solved by using discriminant.
We should use the positive root, so
x = 0.292 m.
b)
We should use energy conservation between the point where the spring is momentarily at rest, and the point where the spring is unstretched.
since the kinetic energy at point 2 and the potential energy at point 3 is equal to zero.
Explanation:
In questions with springs, the important thing is to figure out the points where kinetic or potential energy terms would be zero. When the spring is unstretched, the elastic potential energy is zero. And when the spring is at rest, naturally the kinetic energy is equal to zero.
In part b) the cookie slides back to its original position, so the distance traveled, x, is equal to the distance in part a). The frictional force is constant in the system, so it is quite simple to solve part b) after solving part a).
Answer:
A. xmax = 131.49 m
B. t = 8.74 s
C. ymax = 220.33 m
Explanation:
A. In order to find the horizontal distance which cannon travels you first calculate the flight time. The flight time can be calculated by using the following formula:
(1)
yo: height from the projectile is fired = 200m
vo: initial velocity of the projectile = 25m/s
g: gravitational acceleration = 9.8 m/s^2
θ: angle between the direction of the initial motion of the ball and the horizontal = 53°
t: time
You need the value of t when the projectile hits the ground. Then, in th equation (1) you make y = 0m.
When you replace the values of all parameters in the equation (1), you obtain the following quadratic formula:
(2)
You use the quadratic formula to obtain the value of t:
You use the positive value because it has physical meaning.
Now, you can calculate the horizontal range of the projectile by using the following formula:
The cannon ball travels a horizontal distance of 131.49 m
B. The cannon ball reaches the canon for t = 8.74s
C. The maximum height is obtained by using the following formula:
(3)
By replacing in the equation (3) the values of all parameters you obtain:
The maximum height reached by the cannon ball is 220.33m