Answer:
B) 1.5 m/s
Explanation:
The apparent frequency will be enhanced due to Doppler effect
If f be the apparent frequency , F be the real frequency , V be the velocity of sound and v be the velocity of approaching submarine then f is given by
f = F \frac{V+v}{V-v}\\
\frac{f}{F} =\frac{V+v}{V-v}\\
\frac{f}{F}-1 =\frac{V+v}{V-v}-1\\
\Delta f = \frac{2vf}{V-v}\\
200=\frac{2\times v\times 100\times 1000}{1482-v}\\
v=1.48 m/s
Answer:
Explanation:
according to third equation of motion
2as=vf²-vi²
vf²=2as+vi²
vf=√2as+vi²
vf=√2as+vi
vf=√2*2*4+3
vf=√16+3
vf=4+3=7
so final velocity is 7 m/s
The beat frequency is the concept required to develop this process. The phenomenon is generated when you have two waves but their frequencies do not differ greatly. So the beat frequency will be the difference between those two waves. For this case we have the difference and one of the waves, therefore,

Our values are given as,


Using the formula of the frequency I will have,

Replacing we have,



The possible values are two: 1045Hz and 1055Hz
Answer:
The distance between the ships changing at 6PM is 21.29Km/h
Explanation:
Ship A is sailing east at 35Km/h and ship B is sailing West at 25Km/h
Given
dx/dt= 35
dy/dt= 25
dv/dt= ???? at t= 6PM - 2PM= 4
Therefore t=4
We know ship A travels at 150km in the x-direction and Ship A at t=4 travels at 4.35 Which is 140 also in x-direction
So, we use:
;

But ship B travels at t=4, at 4.25 =100 in the y-direction
so, let's use the equation:


Lets use 2DD' = 2xx' + 2yy'
Differentiating with respect to t we have:
D•d(D)/dt = -(10)•dx/dt + 100•dy/dt
=100.5 d(D)/dt = (-10)•35 + (100)•25
When t=4, we have x=(140-150) =10 and y=100

=100.5
= 100.5 dD/dt = 10.35 +100.25
= dD/dt = 21.29km/h