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:
no ma'am ill help you look
Explanation:
The answer to your question is the Origin
Answer:
a)0.22 m/s².
Explanation:
Given that
Net force ,F= 6.8 N
mass ,m = 31 kg
From the second law of Newton's
F = m a ---------------1
Where
F=Net force ,m=mass
a=Acceleration
Now by putting the values in the equation 1
F = m a
6.8 = 31 x a
Therefore the acceleration of the scooter will be 0.22 m/s².
The answer will be "a".
a)0.22 m/s².
