Ok ths is my last one i swear this is for u sheystar
1 answer:
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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:
a= 3.49 m/s^2
Explanation:
magnitude of total acceleration = sqrt{radial acceleration^2+tangential acceleration^2}.
we know that tangential acceleration a_t= change in velocity /time taken
now 90 km/h = 25 m/s
a_t = 25/17 = 1.47 m/s^2.
radial acceleration a_r = v^2/r
v= a_t×t = 1.47×13 = 19.11 m/s
a_r = 19.11^2/115= 3.175
now,
a= 3.49 m/s^2