A meter stick is balanced on a stand at the 50 cm mark. A 42 -g mass is placed on one side at the 29 cm mark (i.e., not at 29 cm from the 50, but at the point marked as 29 cm). At what cm mark would you place a 47-g mass so that the system balances? Write your answer in terms of the position in cm on the meter stick (to one place of decimal ) .

Let’s denote the position where the 47-g mass should be placed as x (in cm from the 0 cm mark). The balance point is at the 50 cm mark, so the distance from the balance point to the 47-g mass is |50 - x|. Setting the sum of moments on each side of the balance point equal gives us: (50 - 29) cm * 42 g = |50 - x| cm * 47 g Solving this equation will give us the value of x. Note that there will be two solutions, one for x < 50 and one for x > 50, but since we already have a mass at the 29 cm mark, we are interested in the solution where x > 50. Let’s solve this equation: Simplify the left side: 21 cm * 42 g = 882 g*cm Divide both sides by 47 g: 882 g*cm / 47 g = 18.77 cm Add 50 cm to get x: 50 cm + 18.77 cm = 68.77 cm So, you would place the 47-g mass at the 68.8 cm mark (rounded to one decimal place) on the meter stick to balance the system.