Suppose that a meter stick is balanced at its center. A 0.11 kg mass is then positioned at the 17-cm mark. At what cm mark must a 0.38 kg mass be placed to balance the 0.11 kg mass

59.55 cm

Explanation:

Note: A meter stick has a length of 100 cm, and it is balanced at 50 cm.

From the principle of moment,

Sum of clockwise moment = sum of anti clockwise moment

Taking moment about the center

mg(50-x) = m'(y-50)g Equation 1

Where m = first mass, m' = second mass, x = position of the first mass, y = position of the second mass, g = acceleration

make y the subject of the equation

y = (m(50-x)/m')+50 Equation 2

y = (0.11(50-17)/0.38)+50

y = (0.11(33)/0.38)+50

y = 9.55+50

y = 59.55 cm

the answer would be a