Answer:
The wavelength of the emitted radiation is inversely proportional to its frequency, or λ = c/ν. The value of Planck's constant is defined as 6.62607015 × 10−34 joule∙second.
Explanation:
Planck's quantum theory. According to Planck's quantum theory, Different atoms and molecules can emit or absorb energy in discrete quantities only. The smallest amount of energy that can be emitted or absorbed in the form of electromagnetic radiation is known as quantum.
Hope this helps!
Answer:
![\vec{E} = \frac{\lambda}{2\pi\epsilon_0}[\frac{1}{y}(\^y) - \frac{1}{x}(\^x)]](https://tex.z-dn.net/?f=%5Cvec%7BE%7D%20%3D%20%5Cfrac%7B%5Clambda%7D%7B2%5Cpi%5Cepsilon_0%7D%5B%5Cfrac%7B1%7D%7By%7D%28%5C%5Ey%29%20-%20%5Cfrac%7B1%7D%7Bx%7D%28%5C%5Ex%29%5D)
Explanation:
The electric field created by an infinitely long wire can be found by Gauss' Law.
For the electric field at point (x,y), the superposition of electric fields created by both lines should be calculated. The distance 'r' for the first wire is equal to 'y', and equal to 'x' for the second wire.
![\vec{E} = \vec{E}_1 + \vec{E}_2 = \frac{\lambda}{2\pi\epsilon_0 y}(\^y) + \frac{-\lambda}{2\pi\epsilon_0 x}(\^x)\\\vec{E} = \frac{\lambda}{2\pi\epsilon_0 y}(\^y) - \frac{\lambda}{2\pi\epsilon_0 x}(\^x)\\\vec{E} = \frac{\lambda}{2\pi\epsilon_0}[\frac{1}{y}(\^y) - \frac{1}{x}(\^x)]](https://tex.z-dn.net/?f=%5Cvec%7BE%7D%20%3D%20%5Cvec%7BE%7D_1%20%2B%20%5Cvec%7BE%7D_2%20%3D%20%5Cfrac%7B%5Clambda%7D%7B2%5Cpi%5Cepsilon_0%20y%7D%28%5C%5Ey%29%20%2B%20%5Cfrac%7B-%5Clambda%7D%7B2%5Cpi%5Cepsilon_0%20x%7D%28%5C%5Ex%29%5C%5C%5Cvec%7BE%7D%20%3D%20%5Cfrac%7B%5Clambda%7D%7B2%5Cpi%5Cepsilon_0%20y%7D%28%5C%5Ey%29%20-%20%5Cfrac%7B%5Clambda%7D%7B2%5Cpi%5Cepsilon_0%20x%7D%28%5C%5Ex%29%5C%5C%5Cvec%7BE%7D%20%3D%20%5Cfrac%7B%5Clambda%7D%7B2%5Cpi%5Cepsilon_0%7D%5B%5Cfrac%7B1%7D%7By%7D%28%5C%5Ey%29%20-%20%5Cfrac%7B1%7D%7Bx%7D%28%5C%5Ex%29%5D)
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:
Normalization is a systematic approach of decomposing tables to eliminate data redundancy. It is a multi-step process that puts data into tabular form, removing duplicated data from the relation tables.
Explanation:
Ask ya mother .. she birthed you so she’ll know .
