Answer:
When the air pressure in the throat and outside the body is less than the air pressure in the middle ear, barotrauma occurs.
Explanation:
Ear barotrauma is a medical condition that describes discomfort in the ear which is caused by pressure differences in the inner and outer ear drum.
Usually, the air pressure in the middle ear is the same as the air pressure in the throat and outside the body.
When we swallow, the eustachian tube opens up and air flows out of and into the middle ear, this balances the pressure. But if the eustachian tube is blocked, the air pressure in the throat and outer body become different from the air pressure in the middle ear.
Calm, sunny days with wind moving away from the center.
Answer:
a) 
For this case we know the following values:




So then if we replace we got:

b) 
With 
And replacing we have:

And then the scattered wavelength is given by:

And the energy of the scattered photon is given by:

c) 
Explanation
Part a
For this case we can use the Compton shift equation given by:
For this case we know the following values:
So then if we replace we got:
Part b
For this cas we can calculate the wavelength of the phton with this formula:
With
And replacing we have:
And then the scattered wavelength is given by:
And the energy of the scattered photon is given by:
Part c
For this case we know that all the neergy lost by the photon neds to go into the recoiling electron so then we have this:
Answer:
H = 6.93 m
Explanation:
given data
velocity v = 35 m/s
horizontal component Vx = 33 m/s
solution
we get here maximum height so first we get vertical component here that is express as
Vy =
.........................1
put here value
Vy =
Vy = 11.66 m/s
and
now we get height
H =
.............................2
put here value
H = 
H = 6.93 m
Answer:

3257806.62409 m/s
Explanation:
G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²
M = Mass of Sun = 
r = Radius of Star = 20 km
u = Initial velocity = 0
v = Final velocity
s = Displacement = 16 m
a = Acceleration
Gravitational acceleration is given by

The gravitational acceleration at the surface of such a star is 

The velocity of the object would be 3257806.62409 m/s