How do very long radio waves reach stations beyond the curvature of the Earth?
2 answers:
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Answer:
Ф = 2.179 eV
Explanation:
This exercise has electrons ejected from a metal, which is why it is an exercise on the photoelectric effect, which is explained assuming the existence of energy quanta called photons that behave like particles.
E = K + Ф
the energy of the photons is given by the Planck relation
E = h f
we substitute
h f = K + Ф
Ф= hf - K
the speed of light is related to wavelength and frequency
c = λ f
f = c /λ
Φ =
let's reduce the energy to the SI system
K = 0.890 eV (1.6 10⁻¹⁹ J / 1eV) = 1.424 10⁻¹⁹ J
calculate
Ф = 6.63 10⁻³⁴ 3 10⁸/405 10⁻⁹ -1.424 10⁻¹⁹
Ф = 4.911 10⁻¹⁹ - 1.424 10⁻¹⁹
Ф = 3.4571 10⁻¹⁹ J
we reduce to eV
Ф = 3.4871 10⁻¹⁹ J (1 eV / 1.6 10⁻¹⁹ J)
Ф = 2.179 eV
Answer:
The density of the woman is 950.8 kg/m³
Explanation:
Given;
fraction of the woman's volume above the surface = 4.92%
then, fraction of the woman's volume below the surface = 100 - 4.92% = 95.08%
the specific gravity of the woman
The density of the woman is calculate as;
Density of fresh water = 1000 kg/m³
Density of the woman = 0.9508 x 1000 kg/m³
Density of the woman = 950.8 kg/m³
Therefore, the density of the woman is 950.8 kg/m³