Answer:
0.000025s
Explanation:
Period it’s. : T(s)= 1/f(Hz)=1/40000Hz=0.000025s
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:
= 54,000 Joules or 54 kJ
Explanation:
Electrical energy is given by the formula;
E = VIt; where V is the potential difference in volts, I is the current and t is the time in seconds.
Therefore;
Electrical energy = 120 V × 0.50 A × 15 ×60 seconds
= 54,000 Joules
Thus; the electrical energy is 54,000 joules or 54 kJ