<span>If two wheels are exactly the same but spin at different speeds, wheel b is twice te speed of wheel a, it is possible to find the ratio of the magnitude of radial acceleration at a singular point of the rim on wheel be to the spot is four.</span>
Answer:
I have no clue sorry I wish I could help
Answer:
0.231 N
Explanation:
To get from rest to angular speed of 6.37 rad/s within 9.87s, the angular acceleration of the rod must be

If the rod is rotating about a perpendicular axis at one of its end, then it's momentum inertia must be:

According to Newton 2nd law, the torque required to exert on this rod to achieve such angular acceleration is

So the force acting on the other end to generate this torque mush be:

Answer:
0.3858 Nm
Explanation:
The torque of the couple is the dot product of the force vector and the couple vector from 1 end of the ruler to the center. This equals to the product of their magnitude times the cosine() of the angle made by their direction:

It would either break or stop depends on the density