Answer:
10042.6 ohm
Explanation:
f = 10 kHz = 10000 Hz, L = 36 mH = 0.036 H, R = 10 kilo Ohm = 10000 ohm
C = 5 nF = 5 x 10^-9 F
XL = 2 x π x f x L
XL = 2 x 3.14 x 10000 x 0.036 = 2260.8 ohm
Xc = 1 / ( 2 x π x f x C) = 1 / ( 2 x 3.14 x 10000 x 5 x 10^-9)
Xc = 3184.7 ohm
Total impedance is Z.
Z^2 = R^2 + (XL - Xc)^2
Z^2 = 10000^2 + ( 2260.8 - 3184.7 )^2
Z = 10042.6 ohm
Answer:
It is a measure of the electric force per unit charge on a test charge.
Explanation:
The magnitude of the electric field is defined as the force per charge on the test charge.
Since we define electric field as the force per charge, it will have the units of force divided by the unit of charge. This implies that the SI unit of electric field is given as Newton/Coulomb (N/C).
Answer:
#_photon = 5 10²⁰ photons / s
Explanation:
For this exercise let's calculate the energy of a single quantum of energy, use Planck's law
E = h f
c= λ f
E = h c / λ
λ= 1000 nm (1 m / 109 nm) = 1000 10⁻⁹ m
Let's calculate
E₀ = 6.6310⁻³⁴ 3 10⁸/1000 10⁻⁹
E₀ = 19.89 10⁻²⁰ J
This is the energy emitted by a photon let's use a proportions rule to find the number emitted in P = 100 w
#_photon = P / E₀
#_photon = 100 / 19.89 10⁻²⁰
#_photon = 5 10²⁰ photons / s
