Answer:
The photoelectric effect occurs only for frequencies above the cutoff frequency, regardless of the intensity.
Explanation:
The photoelectric effect occurs when light is shined on metals such as zinc beyond a certain frequency (the threshold frequency), which causes electrons to escape from the zinc. The electrons which are fleeing are called photo electrons.
Therefore photo electric effect is
The photoelectric effect occurs only for frequencies above the cutoff frequency, regardless of the intensity.
Given required solution
M=10kg W=? W=Fd
v=5.0m/s F=mg
t=2.40s =10*10=100N
S=VT
=5m/s*2.4s
=12m
so W=12*100
W=1200J
<em>Answer:</em>
<em>well..</em>
<em>Explana</em><em>tion</em><em>:</em>
<em>L</em><em>iquid</em><em> can flow but solid cannot because of differences in their properties</em>
<em>property of liquid which lets it flow:</em>
- <em>i</em><em>nter-particular</em><em> space is large</em>
- <em>inter-particular attraction is small</em><em> </em><em>t</em><em>hese</em><em> properties tend to make the molecules of liquid free to flow</em><em> </em>
<em>property</em><em> </em><em>of</em><em> </em><em>solid</em><em> </em><em>which</em><em> </em><em>tends</em><em> </em><em>to</em><em> </em><em>obstruct</em><em> </em><em>flow</em><em>:</em>
- <em>inter-particular</em><em> </em><em>spa</em><em>c</em><em>e</em><em> </em><em>is</em><em> </em><em>small</em><em> </em><em>and</em><em> </em><em>so</em><em> </em><em>it's </em><em>compac</em><em>t</em>
- <em>inter-molecular</em><em> </em><em>attra</em><em>ction</em><em> </em><em>is</em><em> </em><em>strong</em><em> </em><em>hence</em><em> </em><em>no</em><em> </em><em>tenden</em><em>cy</em><em> </em><em>to</em><em> </em><em>flow</em>
<em>H</em><em>o</em><em>p</em><em>e</em><em> </em><em>this</em><em> </em><em>helps</em><em>!</em>
Answer:
θ = 1.591 10⁻² rad
Explanation:
For this exercise we must suppose a criterion when two light sources are considered separated, we use the most common criterion the Rayleigh criterion that establishes that two light sources are separated census the central maximum of one of them coincides with the first minimum of the other source
Let's write the diffraction equation for a slit
a sin θ = m λ
The first minimum occurs for m = 1, also field in these we experience the angles are very small, we can approximate the sin θ = θ
θ = λ / a
In our case, the pupil is circular, so the system must be solved in polar coordinates, so a numerical constant is introduced.
θ = 1.22 λ / D
Where D is the diameter of the pupil
Let's apply this equation to our case
θ = 1.22 600 10⁻⁹ / 0.460 10⁻²
θ = 1.591 10⁻² rad
This is the angle separation to solve the two light sources
here we know that the speed of the signal is same as the speed of light
so here we will have
the altitude of the airplane is given as
now we know that time taken by the signal to reach the control tower is given as
now it is given as
so above is the time taken by the signal to reach the control tower
