a meter stick balances horizontally on a knife-edge at the 50.0 cm mark. with two 5.00 g coins stacked over the 12.0 cm mark, the stick is found to balance at the 45.5 cm mark. what is the



Answer :

Qwdog

The mass of the knife which is kept on the meter stick is 66 grams.

The mass of the two coins = 2 x 5 g

                                            = 10 g

The position of the coins is at the 0.12 m mark

The position of the knife is at the 0.45 m mark

The length of the meter stick is 0.5 m

The mass of the knife can be found using the formula,

                           M = (x₂ - x₁)m / (x₃ - x₁)

where M is the mass of the knife

           m is the mass of the coins

           x₁ is the position of the coins

           x₂ is the position of the knife

           x₃ is the length of the knife.

Let us substitute the known values in the above equation, we get

          M = (0.45 - 0.12)(10) / (0.5 - 0.45)

              = (0.33)(10) / 0.05

              = 3.3 / 0.05

              = 66 g

Therefore, the mass of the knife is 66 gram

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