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# 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 $to one place of decimal$ .

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## Answer to a math question 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 $to one place of decimal$ .

Timmothy
4.8
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.
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