Answer:
if i were you i would try to do the work because if you let someone else you wont be able to understand the question
The frequency of the
scattered photon decreases or it will be lower compare to the frequency of
incident photon. An x-ray photon scatters in one direction after a collision
and some energy is transferred to the electron as it recoils in another
direction resulting to have less energy in the scattered photon. In addition, the
frequencies will also depend on the differences of the angle at which the
scattered photon leaves the collision and this incident is called Compton Effect.
Answer: option 1 : the electric potential will decrease with an increase in y
Explanation: The electric potential (V) is related to distance (in this case y) by the formulae below
V = kq/y
Where k = 1/4πε0
Where V = electric potential,
k = electric constant = 9×10^9,
y = distance of potential relative to a reference point, ε0 = permittivity of free space
q = magnitude of electronic charge = 1.609×10^-19 c
From the formulae, we can see that q and k are constants, only potential (V) and distance (y) are variables.
We have that
V = k/y
We see the potential(V) is inversely proportional to distance (y).
This implies that an increase in distance results to a decreasing potential and a decrease in distance results to an increase in potential.
This fact makes option 1 the correct answer
Answer:
The angle it subtend on the retina is 
Explanation:
From the question we are told that
The length of the warbler is 
The distance from the binoculars is 
The magnification of the binoculars is 
Without the 8 X binoculars the angle made with the angular size of the object is mathematically represented as
Now magnification can be represented mathematically as
Where 
is the angle the image of the warbler subtend on your retina when the binoculars i.e the binoculars zoom.
So
=> 
Generally the conversion to degrees can be mathematically evaluated as
