Relatively hot objects emit visible light.
Some examples:
==> the wire coils in the toaster;
==> the spoon that you stuck in the flame on the stove;
==> the fine wire in the lightbulb when current goes through it.
VERY radioactive objects also do that. But if you're actually
standing there watching an object that's THAT radioactive,
then you're in big trouble.
The answer: Solanum tuberosum
Answer:
The formula that links energy and power is: Energy = Power x Time. The unit of energy is the joule, the unit of power is the watt, and the unit of time is the second.
Explanation:
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