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3 years ago

A uniform meter rule with a mass of 200g is suspended at zero mark pivotes at 22.0cm mark. calculate the mass of the rule.

Physics

1 answer:

denpristay [2]3 years ago

3 0

Answer:

The mass of the rule is 56.41 g

Explanation:

Given;

mass of the object suspended at zero mark, m₁ = 200 g

pivot of the uniform meter rule = 22 cm

Total length of meter rule = 100 cm

0 22cm 100cm

-------------------------Δ------------------------------------

↓ ↓

200g m₂

Apply principle of moment

(200 g)(22 cm - 0) = m₂(100 cm - 22 cm)

(200 g)(22 cm) = m₂(78 cm)

m₂ = (200 g)(22 cm) / (78 cm)

m₂ = 56.41 g

Therefore, the mass of the rule is 56.41 g

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