A ball is thrown at a 60.0° angle above the horizontal across level ground. It is released from a
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<h2>Answer: Toward the center of the circle.</h2>
This situation is characteristic of the uniform circular motion , in which the movement of a body describes a circumference of a given radius with constant speed.
However, in this movement the velocity has a constant magnitude, but its direction varies continuously.
Let's say is the velocity vector, whose direction is perpendicular to the radius
of the trajectory, therefore
the acceleration
is directed toward the center of the circumference.
The block moves with constant velocity: for Newton's second law, this means that the resultant of the forces acting on the block is zero, because the acceleration is zero.
We are only concerned about the horizontal direction, and there are only two forces acting along this direction: the force F pushing the block and the frictional force
acting against the motion. Since their resultant must be zero, we have:
The frictional force is
where
is the coefficient of kinetic friction
is the weight of the block.
Substituting these values, we find the magnitude of the force F:
We are only concerned about the horizontal direction, and there are only two forces acting along this direction: the force F pushing the block and the frictional force
The frictional force is
where
Substituting these values, we find the magnitude of the force F:
To solve this problem we will use the concepts related to gravitational acceleration and centripetal acceleration. The equality between these two forces that maintains the balance will allow to determine how the rigid body is consistent with a spherically symmetric mass distribution of constant density. Let's start with the gravitational acceleration of the Star, which is
Here
Mass inside the orbit in terms of Volume and Density is
Where,
V = Volume
Density
Now considering the volume of the star as a Sphere we have
Replacing at the previous equation we have,
Now replacing the mass at the gravitational acceleration formula we have that
For a rotating star, the centripetal acceleration is caused by this gravitational acceleration. So centripetal acceleration of the star is
At the same time the general expression for the centripetal acceleration is
Where is the orbital velocity
Using this expression in the left hand side of the equation we have that
Considering the constant values we have that
As the orbital velocity is proportional to the orbital radius, it shows the rigid body rotation of stars near the galactic center.
So the rigid-body rotation near the galactic center is consistent with a spherically symmetric mass distribution of constant density