A meter stick is found to balance at the 49.7-cm mark when placed on a fulcrum. When a 40.0-gram mass is attached at the 14.0-cm mark, the fulcrum must be moved to the 39.2-cm mark for balance. What is the mass of the meter stick?

103 g

Explanation:

given that

Mass of the attached mass, m = 40 g

Initial distance of balance, x1 = 49.7 cm = 0.497 m

Final distance of balance, x2 = 39.2 cm = 0.392 m

Equilibrium point of attachment, x3 = 14 cm = 0.14 m

To get the mass of the meter stick, we use the relation

M = m.[(x2 - x3) / (x1 - x2)]

And on solving in full, we gave

M = 40.[(0.392 - 0.14) / [0.497 - 0.392)]

M = 40(0.258 / 0.105)

M = 40 * 2.457

M = 102.8 g

Therefore the needed work good for mass of the stick j

Is 103 g

find the sum of the two horizontal vectors and the sum of the two vertical vectors

isoceles right angled triangle