A 5.00 A current runs through a 12 gauge copper wire (diameter 2.05 mm) and through a light bulb. Copper has 8.5*10^28 free electrons per cubic metre.
a) How many electrons pass through the light bulb each second?
b) What is the current density in the wire? (answer in A/m^2)
<span>c) At what speed does a typical electron pass by any given point in the wire? (answer in m/s)
</span>a) 5.0 A = 5.0 C/s
. Number of electrons in 5.0C = 5.0 / 1.60^-19 = 3.125^19
. 5.0 A ►= 3.125^19 electrons/s
b) A/m² = 5.0 / π(1.025^-3 m)² .. .. ►= 1.52^6 A/m²
c) Charge density (q/m³) = 8.50^28 e/m³ x 1.60^-19 = 1.36^10 C/m³
(q/m³)(m²)(m/s) = q/s (current i in C/s [A])
(m²) = Area
(m/s) = mean drift speed
(q/m³)(A)(v) = i
v = i.[(q/m³)A]ˉ¹
<span>v = 5.0 [1.36^10 * π(1.025^-3 m)²]ˉ¹.. .. ►v = 1.10^-4 m/s</span>
Answer: 3.7×10¹²watts
Explanation:
Radiation is one of the mode of heat transfer and modes differs from each other based on their medium of heat transfer. Radiation is a process of transferring heat energy from one point to another without heating the intervening medium (no material medium is required).
According to Stefan's law of radiation, the rate of emission of radiant energy is directly proportional to the fourth power of its absolute temperature.
Mathematically, R = eAT⁴
e is constant of proportionality called emissivity. Emissivity varies depending on the type of body being considered.
For the question, we are considering black body and emissivity of black body is 1 being a perfect body.
A is the area of the body
T is the absolute temperature
e = 1
A = 0.5cm²
T = 1650°C
Rate of radiation = 1×0.5×1650⁴
= 3.7×10¹²watts.
The hole will therefore radiate 3.7×10¹²watts
Answer: 170km/hr to the South
Explanation:
The resultant velocity of the plane ilwill be gotten by adding the value of the velocities of the plane with the tallwind. It should be noted that the direction of the wind is.takwn into consideration.
Therefore, the resultant velocity of the plane will be:
= 20km/hr + 150km/hr
= 170km/hr to the South
