The center of mass of a 0.30 kg (non-uniform) meter stick is located at its 48-cm mark. what is the magnitude of the torque (in n⋅m) due to gravity if it is supported at the 25-cm mark? (use g = 9.79 m/s2). (never include units with the answer to any numerical question.)

Explanation:

It is given that,

Mass of the meter stick, m = 0.3 kg

Meter stick is located at its 48 cm mark, l = 48 cm

We need to find the magnitude of the torque if it is supported at the 25-cm mark, l' = 25 cm

It is given by :

So, the magnitude of the torque is 0.675 N-m. Hence, this is the required solution.

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.