Answer:
a) True. The number of photoelectrons is proportional to the amount (intensity) of the incident beam. From the expression above we see that threshold frequency cannot emit electrons.
b) λ = c / f
Therefore, as the wavelength increases, the frequency decreases and therefore the energy of the photoelectrons emitted,
c) threshold energy
h f =Ф
Explanation:
It's photoelectric effect was fully explained by Einstein by the expression
Knox = h f - fi
Where K is the kinetic energy of the photoelectrons, f the frequency of the incident radiation and fi the work function of the metal
a) True. The number of photoelectrons is proportional to the amount (intensity) of the incident beam. From the expression above we see that threshold frequency cannot emit electrons.
b) wavelength is related to frequency
λ = c / f
Therefore, as the wavelength increases, the frequency decreases and therefore the energy of the photoelectrons emitted, so there is a wavelength from which electrons cannot be removed from the metal.
c) As the work increases, more frequency radiation is needed to remove the electrons, because there is a threshold energy
h f =Ф
It would depend on how she jumped off but based on it sounds it would be a curving motion
Answer:
D) 11 m/s
Explanation:
The problem asks us to calculate the velocity of the hot dog with respect to the observer stationary outside the train. This velocity is given by:
where
is the velocity of the train (towards right)
is the velocity of the man (towards right)
is the velocity of the hot-dog (towards left, so we put a negative sign)
Substituting the numbers into the equation, we find
and the positive sign means the velocity is toward right.
The solution would be like
this for this specific problem:
<span>5.5 g = g + v^2/r </span><span>
<span>4.5 g =
v^2/r </span>
<span>v^2 = 4.5
g * r </span>
<span>v = sqrt
( 4.5 *9.81m/s^2 * 350 m) </span>
v = 124
m/s</span>
So the pilot will black out for this dive at 124
m/s. I am hoping that these answers have satisfied your query and it
will be able to help you in your endeavors, and if you would like, feel free to
ask another question.
