E=274J
h=140cm=1,4m
g≈9,8m/s²
m=?
E=mgh
m=E/gh=274J/9,8m/s²*1,4m≈20kg
"Non nobis Domine, non nobis, sed Nomini tuo da gloriam."
Regards M.Y.
Amplitude, is the answer to the question
Answer:
Explanation:
My speed after the interaction will depend upon the impulse the ball will make on me . Now impulse can be expressed as follows
Impulse = change in momentum
change in momentum in the ball will be maximum when the ball bounces back with the same velocity which can be shown as follows
change in momentum = mv - ( - mv ) = 2mv
So when ball is bounced back with same velocity , it suffers greatest impulse from my hand . In return , it reacts with the same impulse on my hand pushing me with greatest impulse according to third law of motion. this maximizes my speed after the interaction.
We Know, F = m*a
F = 2200 * 3.4
F = 7480 Kg m/s²
So, your final answer is 7480
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
