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
Answer:
W = -1844.513 J
Explanation:
GIVEN DATA:
mass of spider man is m 74 kg
vertical displacement if spider is 11 m
final displacement = 11 cos 60.6 = - 6.753 m
change in displacement is = -6.753 - (-11) = 4.25 m
gravity force act on spiderman is f = mg = 74 × 9.8 = 725.2 N
work done by gravity is 
where 180 is the angle between spiderman weight and displacement
W = -1844.513 J
Answer: 56.72 ft/s
Explanation:
Ok, initially we only have potential energy, that is equal to:
U =m*g*h
where g is the gravitational acceleration, m the mass and h the height.
h = 50ft and g = 32.17 ft/s^2
when the watermelon is near the ground, all the potential energy is transformed into kinetic energy, and the kinetic energy can be written as:
K = (1/2)*m*v^2
where v is the velocity.
Then we have:
K = U
m*g*h = (m/2)*v^2
we solve it for v.
v = √(2g*h) = √(2*32.17*50) ft/s = 56.72 ft/s
Answer:
600Hz
Explanation:
In electrical systems of alternating current, the harmonics are, as in acoustics, frequencies multiples of the fundamental working frequency of the system and whose amplitude decreases as the multiple increases. For example, if we have systems fed by the 50 Hz network, harmonics of 100, 150, 200, etc. may appear.
In our case having a fundamental wave of 100Hz, I can have harmonics of 200,300,400, ..., 600Hz
Answer:
67
Explanation:
- The atomic number (Z) of an atom is equal to the number of protons in the nucleus
- The mass number (A) of an atom is equal to the sum of protons and neutrons in the nucleus
Therefore, calling p the number of protons and n the number of neutrons, for element X we have:
Z = p = 23
A = p + n = 90
Substituting p=23 into the second equation, we find the number of neutrons:
n = 90 - p = 90 - 23 = 67
