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anygoal [31]
3 years ago
11

A submarine can detect an approaching enemy submarine using sonar. If a stationary submerged submarine emits a 100kHz tone and r

eceives a tone Doppler shifted by 200Hz, calculate the speed of the enemy submarine if the speed of sound in salt water is 1,482m/s. Note: Because the echo is off of a moving object, the moving object amplifies the Doppler effect by a factor of 2! Therefore, the speed must be cut in half. A. 1.0m/s B. 1.5m/s C.2.2m/s OD.3.3m/s E. 4.4m/s
Physics
1 answer:
kotegsom [21]3 years ago
7 0

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

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kvasek [131]

Answer:

a) a= \frac{4 \pi^2 (0.02 m)^2}{0.00162 s}=9.74 \frac{m}{s^2}

b) k = \frac{9.8}{9.74}=1.006

Explanation:

Part a

For this case we can begin finding the period like this:

T= \frac{1}{w} =\frac{1}{617 rad/s}=0.00162 s

Then we know that the centripetal acceleration is given by:

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If we replace this into the acceleration we got:

a = \frac{(\frac{2\pi r}{T})^2}{r}= \frac{4 \pi^2 r}{T^2}

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a= \frac{4 \pi^2 (0.02 m)^2}{0.00162 s}=9.74 \frac{m}{s^2}

Part b

For this case we want to find a value of k such that:

a= k 9.8

Where a = 9.74, so then we can solve for k like this:

k = \frac{9.8}{9.74}=1.006

8 0
3 years ago
What are the units of acceleration?
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