Answer:
AN electric current is required for an electric charge
Explanation:
Answer:
a) E₀ = 2.125 eV, b) # photon2 = 9.2 10¹⁵ photons / mm²
Explanation:
a) To calculate the energy of a photon we use Planck's education
E = h f
And the ratio of the speed of light
c = λ f
We replace
E = h c /λ
Let's calculate
E₀ = 6.63 10⁻³⁴ 3 10⁸/585 10⁻⁹
E₀ = 3.40 10⁻¹⁹ J
Let's reduce
E₀ = 3.4 10⁻¹⁹ J (1 eV / 1.6 10⁻¹⁹ J)
E₀ = 2.125 eV
b) Let's look for the energy in each pulse
P = E / t
E = P t
E = 20.0 0.45 10⁻³
E = 9 10⁻³ J
let's use a ratio of proportions (rule of three) if we have the energy of a photon (E₀), to have the energy of 9 10⁻³ J
# photon = 9 10⁻³ /3.40 10⁻¹⁹
# photon = 2.65 10¹⁶ photons
Let's calculate the areas
Focus area
A₁ = π r²
A₁ = π (3.4/2)²
A₁ = 9,079 mm²2
Area requested for calculation r = 1 mm
A₂ = π 1²
A₂ = 3.1459 mm²
Let's use another rule of three. If we have 2.65 106 photons in an area A1 how many photons in an area A2
# photon2 = 2.65 10¹⁶ 3.1459 / 9.079
# photon2 = 9.2 10¹⁵ photons / mm²
Answer: 0.43 V
Explanation:
L = [μ(0) * N² * A] / l
Where
L = Inductance of the solenoid
N = the number of turns in the solenoid
A = cross sectional area of the solenoid
l = length of the solenoid
7.3*10^-3 = [4π*10^-7 * 450² * A] / 0.24
1.752*10^-3 = 4π*10^-7 * 202500 * A
1.752*10^-3 = 0.255 * A
A = 1.752*10^-3 / 0.255
A = 0.00687 m²
A = 6.87*10^-3 m²
emf = -N(ΔΦ/Δt).........1
L = N(ΔΦ/ΔI) so that,
N*ΔΦ = ΔI*L
Substituting this in eqn 1, we have
emf = - ΔI*L / Δt
emf = - [(0 - 3.2) * 7.3*10^-3] / 55*10^-3
emf = 0.0234 / 0.055
emf = 0.43 V
Answer:
23376 days
Explanation:
The problem can be solved using Kepler's third law of planetary motion which states that the square of the period T of a planet round the sun is directly proportional to the cube of its mean distance R from the sun.
where k is a constant.
From equation (1) we can deduce that the ratio of the square of the period of a planet to the cube of its mean distance from the sun is a constant.
Let the orbital period of the earth be 
and its mean distance of from the sun be 
.
Also let the orbital period of the planet be 
and its mean distance from the sun be 
.
Equation (2) therefore implies the following;
We make the period of the planet the subject of formula as follows;
But recall that from the problem stated, the mean distance of the planet from the sun is 16 times that of the earth, so therefore
Substituting equation (5) into (4), we obtain the following;
cancels out and we are left with the following;
Recall that the orbital period of the earth is about 365.25 days, hence;
True because it has "falling" ability
