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Answer :
Final answer:
The moment of inertia for a door with a width of 0.822 m and a mass of 35.9 kg, assuming it is thin and rotates about an axis at one edge (y-axis), is approximately 8.113 kgm^2.
Explanation:
To calculate the moment of inertia for a door, we can use the formula for a rectangular object rotating around an edge perpendicular to the width. In this case, let us consider the width of the door, 0.822 m, to be distance a from the axis of rotation, and the mass of the door to be 35.9 kg. The moment of inertia I for a rectangular object is given by I = 1/3 * mass * (width)^2. Therefore, plugging in the values, we get:
I = 1/3 * 35.9 kg * (0.822 m)^2
This calculation yields
I = 1/3 * 35.9 kg * 0.676 m^2
= 8.113 kgm^2 approximately.
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