A meter stick of uniform density is pivoted at the 75 cm mark and a 250 g mass is placed at the 90.0 cm mark. If it balances the meter stick, what is the mass of the meter stick

The mass of the meter stick is 50 g.

Explanation:

make a sketch of the problem statement;

075cm90cm100cm

↓                                                          Δ              ↓

W                                                                       250 g

where;

W is the mass of the meter stick

Apply the principle of moment; take moment about the pivot.

sum of the anticlockwise moment = sum of the clockwise moment

W(75 - 0) = 250(90 - 75)

W(75) = 250(15)

W(75) = 3750

W = 3750 / 75

W = 50 g

Therefore, the mass of the meter rule is 50 g.

spare some change and give it to the poor

explanation:

plz : ((

the components of a vector are the parts of the vectors that are to each other.

b perpendicular