Answer: Trajectory=51m
Displacement=41m
Explanation:
Let's begin by stating clear that <u>movement is the change of position of a body at a certain time.</u> So, during this movement, the body will have a trajectory and a displacement, being both different:
The trajectory is the path followed by the body (is a scalar magnitude).
The displacement is the distance in a straight line between the initial and final position (is a vector magnitude).
According to this, in the description of the object placed at x= -7m on a number line and moving some 12m to the left and then to the right, stopping at x=34m; we are talking about the path followed by the object, hence its <u>trajectory</u>. So, 51 m is its trajectory.
But, if we talk about displacement, we have to draw a straight line between the initial position of the object (x=-7m) to its final position (x=+34m).
Now, being this an unidimensional problem, the displacement vector for this object is 41m.
Answer:
Explanation:
We can use the conservation of the angular momentum.
Now the Inertia is I(professor_stool) plus mR², that is the momentum inertia of a hoop about central axis.
So we will have:
Now, we just need to solve it for ω.
I hope it helps you!
Answer:
The work done by non-conservative forces on the car from the top of the first hill to the top of the second hill is 6574.75 joules.
Explanation:
By Principle of Energy Conservation and Work-Energy Theorem we present the equations that describe the situation of the roller coaster car on each top of the hill. Let consider that bottom has a height of zero meters.
From top of the first hill to the bottom
(1)
From the bottom to the top of the second hill
(2)
Where:
- Mass of the roller coaster car, in kilograms.
- Speed of the roller coaster car at the bottom between the two hills, in meters per second.
- Gravitational acceleration, in meters per square second.
- Height of the first top of the hill with respect to the bottom, in meters.
- Work done by non-conservative forces on the car between the top of the first hill and the bottom, in joules.
- Speed of the roller coaster car at the top of the second hill, in meters per seconds.
- Height of the second top of the hill with respect to the bottom, in meters.
- Work done by non-conservative forces on the car bewteen the bottom between the two hills and the top of the second hill, in joules.
By using (1) and (2), we reduce the system of equation into a sole expression:
(3)
Where is the work done by non-conservative forces on the car from the top of the first hill to the top of the second hill, in joules.
If we know that ,
,
,
and
, then the work done by non-conservative force is:
The work done by non-conservative forces on the car from the top of the first hill to the top of the second hill is 6574.75 joules.