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
1 answer:
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:
(a) 25 m
(b) 75 m
Explanation:
Given that the jogger runs at a constant rate of 10.0 m every 2.0 seconds.
So, the speed of the jogger,
Let d be the distance covered by him in time, t s.
As distance=(speed) x (time)
So,
From equation (i)
As the jogger starts from origin, so, the distance, , also represents the position of the jogger at the time
s.
The position-time graph has been shown.
(a) From equation (ii), for t=5.0 s
So, the jogger is at a distance of 25 m from the origin.
(b) Similarly, for t=15.0 s
So, the jogger is at a distance of 75 m from the origin.
