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morpeh [17]
3 years ago
8

A uniform seesaw is balanced at its center of mass, as seen below. The smaller boy on the right has a mass of 40.0 kg. What is t

he mass of his friend is (A) 40 kg (B) 20 kg (C) 80 kg (D) Cannot be determined

Physics
1 answer:
Dominik [7]3 years ago
4 0

Answer:

<h2>C. 80kg</h2>

Explanation:

The question lacks the appropriate diagram. Find the diagram attached below:

Let the mass of his friend be M.

Using the principle of moment which states that the sum of clockwise moment is equal to the sum of its anticlockwise moment to solve the problem,.

Since Moment = Force * perpendicular distance.

The smaller boy will move in the clockwise direction and his friend will move in the anti clockwise direction.

Clockwise moment = 40 * 4 = 160kgm

anticlockwise moment = M * 2 = 2Mkgm

Equating both moments to get the mass M of his friend

160 = 2M

Divide both sides by 2

2M/2 = 160/2

M = 80kg

<em>Hence the mass of his friend that will keep the seesaw balanced horizontally is 80kg.</em>

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Answer:

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A car moves 50km in the first 30 minutes and 50km in the second 30 minutes and continues in this way which motion is this ------
egoroff_w [7]

Answer:

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Explanation:

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An 89 kg man drops from rest on a diving board −3.1 m above the surface of the water and comes to rest 0.5 s after reaching the
OLga [1]

To solve this problem we will use the linear motion kinematic equations, for which the change of speed squared with the acceleration and the change of position. The acceleration in this case will be the same given by gravity, so our values would be given as,

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Through the aforementioned formula we will have to

v_f^2-v_i^2 = 2ax

The particulate part of the rest, so the final speed would be

v_f^2 = 2gx

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Here,

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