Answer:
In my opinion the unstoppable object will hit the unmovable object and stop but the wheels will still be rolling and trying to move but can't.
<h3>Hope this helps.</h3><h3>Good luck ✅.</h3>
The free-body diagram of the forces acting on the flag is in the picture in attachment.
We have: the weight, downward, with magnitude

the force of the wind F, acting horizontally, with intensity

and the tension T of the rope. To write the conditions of equilibrium, we must decompose T on both x- and y-axis (x-axis is taken horizontally whil y-axis is taken vertically):


By dividing the second equation by the first one, we get

From which we find

which is the angle of the rope with respect to the horizontal.
By replacing this value into the first equation, we can also find the tension of the rope:
The answer to this would inFact be A