It you balance a weight of 20 N at the 15 cm MARK and want to balance a weight of on the other side of the meter stick at 70 cm MARK, what weight should you place there

The weight that will balance the meter stick at 70 cm mark is 35N

Explanation:

Given;

first weight at 15 cm mark, W₁ = 20 N

second weight at 70 cm mark, W₂ = ?

A sketch of the question in diagram form;

The center of gravity of the meter stick is 50 cm

                     15cm               50cm         70cm

0Δ100cm

                     ↓                                           ↓

                     20N                                      W₂

                     <>|<>

                                35cm               20cm

Take moment about the pivot;

Clockwise moment = anticlockwise moment

W₂(20cm) = 20N(35 cm)

W₂ = (20 x 35) / (20)

W₂ = 35 N

Therefore, the weight that will balance the meter stick at 70 cm mark is 35N

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more height = more potential energy.

greatest increase in height = greatest increase in potential energy

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has to be c-e or d-e .

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this was a big portion of territory and expanded new france which before then only consisted of quebec province in northeastern canada