Velocity of submarine A is vs = 11.0m/s
frequency emitted by submarine A. F = 55.273 × 10∧3HZ
Velocity of submarine B = vO = 3.00m/s
The given equation is
f' = ((V + vO) ((v - vS)) × f
The observer on submarine detects the frequency f'.
The sign of vO should be positive as the observer of submarine B is moving away from the source of submarine A.
The speed of the sound used in seawater is 1533m/s
The frequency which is detected by submarine B is
fo = fs (V -vO/ v +vs)
= 53.273 × 10∧3hz) ((1533 m/s - 4.5 m/s)/ (1533 m/s +11 m/s)
fo = 5408 HZ
Now really !
If there were some kind of treatment of the results that made them
more difficult to prove by experiment, or made them more confusing,
or made them less appealing to the public, then there would be no
reason on Earth to treat them that way.
Choice ' B ' is the only positive effect, so it's the only reason you'd
WANT to handle your findings that way.
I believe it is A. positive acceleration since it is speeding up
Explanation:
We need to apply the conservation law of linear momentum to two dimensions:
Let 
= momentum of the 1st ball
= momentum of the 2nd ball
In the x-axis, the conservation law can be written as
or
Since we are dealing with identical balls, all the m terms cancel out so we are left with
Putting in the numbers, we get
In the y-axis, there is no initial y-component of the momentum before the collision so we can write
or
Taking the ratio of the sine equation to the cosine equation, we get
or
Solving now for 
,
