W= F x dPower (formula)P= W/tForce (formula)F= m x anewton (N)the SI unit for force , equal to 1 kg*m/s2joule (J)<span>the SI unit of work, equal to 1 newton-meter</span>
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:
$3.51
Explanation:
The energy (E) delivered by a system is the product of the power (P) dissipated by the system and the time (t) taken for the dissipation. i.e
E = P x t -----------------------(i)
Where;
The power (P) is the product of the current (I) flowing through the system and the voltage (V) across the system. i.e
P = I x V
Substituting P = I x V into equation (i) gives;
E = (I x V) x t
=> E = IVt --------------------(ii)
<em>From the question;</em>
Current (I) = 0.3A
Voltage (V) = 1.3V
time (t) = 75 hours = 75h
<em>Substituting these values into equation (ii) gives;</em>
=> E = 0.3 x 1.3 x 75
=> E = 29.25 Kwh
The energy delivered is 29.25Kwh
But in the U.S;
$0.12 = 1Kwh
=> 29.25Kwh = 29.25 x $0.12
=> $3.51
Therefore, the cost of the energy delivered by this battery per kilowatt hour is $3.51
Answer:
a.8m/s is my ans it may help you
Nothing is shown, so I cannot answer correctly