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GuDViN [60]
2 years ago
11

A 5.2 kg cat and a 2.5 kg bowl of tuna fish are at opposite ends of the 4.0-m-long seesaw. How far to the left of the pivot must

a 4.0 kg cat stand to keep the seesaw balanced
Physics
1 answer:
algol [13]2 years ago
6 0

Answer:

The cat must stand 0.96 m from the left side of the pivot to keep the seesaw balanced.

Explanation:

Given;

mass of the cat, m₁ = 5.2 kg

mass of the tuna, m₂ = 2.5 kg

length of the seesaw, L = 4.0 m

Apply the following approach;

                                                                   

                  x m                      <-------------2m ---------->              

        0------------------------------Δ------------------------------4m

                  ↓                         2m                                 ↓

                 5.2 kg                                                      2.5 kg

Apply the principle of moment; anticlockwise moment = clockwise moment.

5.2(x ) = 2.5(2)

5.2x = 5

x = 5/5.2

x = 0.96 m

Therefore, the cat must stand 0.96 m from the left side of the pivot to keep the seesaw balanced.

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2 years ago
You find it takes 200 N of horizontal force tomove an unloaded pickup truck along a level road at a speed of2.4 m/s. You then lo
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F(h)= 230 N is the horizontal force you will need to move the pickup along the same road at the same speed.

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Since the velocity is constant so acceleration is zero; a=0

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