A satellite orbiting the moon very near the surface has a period of110 min. What is free-fall acceleration on the surface of the
1 answer:
Answer:
1.54 m/s²
Explanation:
The free-fall acceleration is calculated as
g = w²r
Where w is the angular velocity of the satellite and r is the radius of the moon.
The angular velocity can be calculated as
Where T is the period, so
T = 110 min = 110 x 60 s = 6600 s
Then,
Finally, the radius of the moon is r = 1.7 x 10⁶ m, so the free-fall acceleration is
Therefore, the answer is 1.54 m/s²
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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