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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.
Answer:
Explanation:
The mass of one electron is
So the number of electrons contained in M=1.7 kg of mass is
The charge of one electron is
So, the total charge of these electrons is equal to the charge of one electron times the number of electrons: