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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To develop the problem it is necessary to apply the equations related to the moment of inertia.
The given values can be defined as,
According to the definition of the moment of inertia applied to the exercise we can arrive at the equation that,
Where n is the number of spokes necessary to construct the wheel.
Replacing the values at the general equation we have,
Solving for n,
Therefore the number of spokes necessary to construct the wheel is 36
PART B) The mass of the wheel is given by the sum of all masses and the total spokes, then
Therefore the mass of the wheel must be of 1.36Kg
Answer:
M = 222 fringes
Explanation:
given
λ = 559 n m = 559 × 10⁻⁹ m
radius = 0.026 mm = 0.026 ×10⁻³ m
length of the glass plate = 22.1 ×10⁻² m
using relation
= 221.79
= 221 (approx.)
hence no of bright fringe
M = m + 1
= 221 +1
M = 222 fringes