If F2 has a mass of 400 grams and is placed 20 cm from the fulcrum, where does F1 that has a mass of 200 grams need to be placed in order to have the board balance? A. 40 cm from the fulcrum B. 40 cm from the end of the meter stick C. Directly on the fulcrum D. 20 cm from the fulcrum

A. 40 cm from the fulcrum

Explanation:

In order to have a balanced board, the masses should be placed such that their moments are equal in magnitude but opposite in direction. Therefore,

F₁ d₁ = F₂ d₂

where,

F₁ = Weight of Mass 1 = m₁g = (0.2 kg)(9.8 m/s²) = 1.96 N

d₁ = Distance of Mass 1 from Fulcrum = ?

F₂ = Weight of Mass 2 = m₂g = (0.4 kg)(9.8 m/s²) = 3.92 N

d₂ = Distance of Mass 2 from Fulcrum = 20 cm

Therefore,

(1.96 N) d₁ = (3.92 N)(20 cm)

d₁ = 78.4 N.cm/1.96 N

d₁ = 40 cm

Hence, the correct answer is:

A. 40 cm from the fulcrum

i think they are all correct

explanation:

they seem good

resonant frequency