A meter stick is supported by a pivot at its center of mass. Assume that the meter stick is uniform and that the center of mass is at the 50 cm mark.a)If a massm1= 80 g is suspended at the 30 cm mark, at which cm mark would a mass m2= 110 g need to be suspended for the system to be in equilibrium?

Answer :

Chenk13

Answer:

64.5 cm

Explanation:

30 * 80 + x * 110 = 50* 190 => x = 64.5

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