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NemiM [27]
3 years ago
9

While in a stream 39 cm deep, they look down into the water and see a craw fish at the bottom. How deep does the stream appear t

o the student? (nwater = 1.33)
Physics
1 answer:
Helga [31]3 years ago
7 0

Answer:

The  depth of stream to the student is  d_1  =  0.2932 \  m

Explanation:

From the question we are told that

   The actual  depth of the stream is d =  39 \ cm  =  0.39 \ m  

    The  refractive index of the water is  n =  1.33

Generally the apparent depth of the stream is mathematically represented as

         d_1  =  \frac{d}{1.33}

substituting values

        d_1  =  \frac{ 0.39}{1.33}

        d_1  =  0.2932 \  m

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There is one mistake in the question.The Correct question is here

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Answer:

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Explanation:

Given data

time=1/2 sec to 1 sec

v(t)=-9.8t m/s

To find

Distance

Solution

As the acceleration as first derivative of velocity with respect to time  

So

acceleration(-g)=  dv/dt

Solve it

dv  =  a dt

dv =  -g dt

v - v₀  =  -gt

v=  dy/dt

dy  =  v dt

dy =  ( v₀ - gt ) dt

y(1s) - y(1/2s)  =  ( v₀ ) ( 1 - 1/2 ) - ( g/2 )[ ( t1)² -( t1/2s )² ]

y(1s) - y(1/2s)  = ( - 9.8/2 ) [ ( 1 )² - ( 1/2 )² ]

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