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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:

The object accelerates downward at 4 m/s² since the tension on the rope is less than weight of the object.

Explanation:

Given;

mass of the object, m = 2 kg

weigh of the object, W = 20 N

tension on the rope, T = 12 N

The acceleration of the object is calculated by applying Newton's second law of motion as follows;

T = F + W

T = ma + W

ma = T - W

a = \frac{T-W}{m} \\\\a = \frac{12 - 20}{2} \\\\a = -4 \ m/s^2 (the negative sign indicates deceleration of the object)

The object accelerates downward at 4 m/s² since the tension on the rope is less than weight of the object.

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Explanation:

The solid in the mixture to be separate would be found at the bottom of the distillation flask.

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The gravitational potential energy is calculated as -

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3 0
3 years ago
Read 2 more answers
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