In the photoelectric effect, the energy of the incoming photon (E=hf) is used in part to extract the photoelectron from the metal (work function) and the rest is converted into kinetic energy of the photoelectron:
where
h is the Planck constant
f is the frequency of the incident light
is the work function of the material
K is the kinetic energy of the photoelectron.
The photoelectron generally loses part of its kinetic energy inside the material; however, we are interested in its maximum kinetic energy, that is the one the electron has when it doesn't lose energy, so we can rewrite the previous equation as
The work function is (in Joule)
and using the data of the problem, we find the maximum kinetic energy of the photoelectrons
Answer:
4.24m/s,45 degrees in forward direction
Explanation:
Visceral epithelial cells
Answer:
1,85 m / s²
Explanation:
De la pregunta anterior, se obtuvieron los siguientes datos:
Velocidad inicial (u) = 40 km / h
Hora inicial (t₁) = 0
Tiempo final (t₂) = 6 s
Velocidad final (v) = 0
Aceleración (a) =?
A continuación, convertiremos 40 km / ha m / s. Esto se puede obtener de la siguiente manera:
1 km / h = 0,2778 m / s
Por lo tanto,
40 km / h = 40 km / h × 0,2778 m / s / 1 km / h
40 km / h = 11,11 m / s
Por tanto, 40 km / h equivalen a 11,11 m / s.
Finalmente, determinaremos la aceleración del móvil durante el período en el que desaceleró. Esto se puede obtener de la siguiente manera:
Velocidad inicial (u) = 11,11 m / s
Hora inicial (t₁) = 0
Tiempo final (t₂) = 6 s
Velocidad final (v) = 0
Aceleración (a) =?
a = (v - u) / (t₂ - t₁)
a = (0 - 11,11) / (6 - 0)
a = - 11,11 / 6
a = –1,85 m / s²
Por tanto, la aceleración del móvil durante el período en el que se ralentizó es de –1,85 m / s²