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
Momentum, p = m.v
m of the girl = 60.0 kg
m of the boat = 180 kg
v of the girl = 4.0 m/s
A) Momentum of the girl as she is diving:
p = m.v = 60.0 kg * 4.0 m/s = 24.0 N/s
B) momentum of the raft = - momentum of the girl = -24.0 N/s
C) speed of the raft
p = m.v ; v = p/m = 24.0N/s / 180 kg = -0.13 m/s [i.e. in the opposite direction of the girl's velocity]