The figure(Figure 1) shows the angular-velocity-versus-time graph for a particle moving in a circle, starting from θ0=0 rad at t
=0s. Draw the angular-position-versus-time graph. Include an appropriate scale on both axes.
1 answer:
Answer:
As position is the product of velocity and time, the area under the velocity-time curve will be the position.
Between 2 and 4 s, the area increases linearly to 20(4) = 80
Between 4 and 6 s, the additional area is 0(2) = 0
Between 6 and 8 s the area increases linearly by -10(2) = -20
which is a decrease.
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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