Given Information:
Current = I = 20 A
Diameter = d = 0.205 cm = 0.00205 m
Length of wire = L = 1 m
Required Information:
Energy produced = P = ?
Answer:
P = 2.03 J/s
Explanation:
We know that power required in a wire is
P = I²R
and R = ρL/A
Where ρ is the resistivity of the copper wire 1.68x10⁻⁸ Ω.m
L is the length of the wire and A is the area of the cross-section and is given by
A = πr²
A = π(d/2)²
A = π(0.00205/2)²
A = 3.3x10⁻⁶ m²
R = ρL/A
R = 1.68x10⁻⁸*(1)/3.3x10⁻⁶
R = 5.09x10⁻³ Ω
P = I²R
P = (20)²*5.09x10⁻³
P = 2.03 Watts or P = 2.03 J/s
Therefore, 2.03 J/s of energy is produced in 1.00 m of 12-gauge copper wire carrying a current of 20 A
Answer:
moving a magnet into a coil of wire in a closed circuit.
Ed 2020
Answer:
P = 5880 J
Explanation:
Given that,
The mass of the block, m = 30 kg
The block is sitting at a height of 20 m.
The block will have gravitational potential energy. The formula for gravitational potential energy is given by :
So, the required potential energy is equal to 5880 J.
To reach a vertical height of 13.8 ft against gravity, which has an acceleration of 32 ft/s^2, the required vertical speed can be calculated from the equation:
vi^2 - vf^2 = 2*g*h
Given that it has vf = 0 (it is not moving vertically at its maximum height), g = 32, and h = 13.8, we can solve for vi:
vi^2 = 29.72 ft/s
This is only its vertical speed, so this is equivalent to its original speed multiplied by the sine of the angle:
29.72 ft/s = (v_original)*(sin 42.2<span>°</span>)
v_original = 44.24 ft/s
Converting to m/s, this can be divided by 3.28 to get 13.49 m/s.
Well i had the same question on my test, and when the graph in 2.5 seconds goes up from the equilibrium it reaches a positive maximum, so that would be your answer.
b. positive maximum
