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Answer:
The extension of the wire is 0.362 mm.
Explanation:
Given;
mass of the object, m = 4.0 kg
length of the aluminum wire, L = 2.0 m
diameter of the wire, d = 2.0 mm
radius of the wire, r = d/2 = 1.0 mm = 0.001 m
The area of the wire is given by;
A = πr²
A = π(0.001)² = 3.142 x 10⁻⁶ m²
The downward force of the object on the wire is given by;
F = mg
F = 4 x 9.8 = 39.2 N
The Young's modulus of aluminum is given by;

Where;
Young's modulus of elasticity of aluminum = 69 x 10⁹ N/m²

Therefore, the extension of the wire is 0.362 mm.
The total number of revolutions made by the wheel
is closest to is 28.2 revolutions. I am hoping that this
answer has satisfied your query and it will be able to help you in your
endeavor, and if you would like, feel free to ask another question.
Weight = mass * gravity = 60 kg * 3.75 m/s² = 225 N
<span>Option D.</span>
The total momentum should come out to be <span>2.0 x 10^4 kilogram meters/second </span>