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love history [14]
3 years ago
5

In a swimming meet, the swimmers swim a total of 8 laps of a 50-meter-long swimming pool. What is the distance traveled by a swi

mmer in this meet? What is swimmer’s displacement?
Physics
1 answer:
N76 [4]3 years ago
7 0

Answer:

The swimmer has a distance traveled of 800 meters.

The final displacement of the swimmer is 0 meters.

Explanation:

A lap is a round trip made by a swimmer in the pool, so that the distance traveled by swimmer is sixteen times the length of the swimming pool. That is:

s = \left(2\,\frac{travels}{lap} \right)\cdot \left(8\,laps \right)\cdot \left(50\,\frac{m}{travel}\right)

s = 800\,m

A swimmer has a distance traveled of 800 meters.

The displacement is the distance between swimmer and a reference point, let suppose that reference point is located at the beginning of the first lap. Hence, the final displacement of the swimmer is 0 meters.

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Answer:

a

 \theta _2  = 13^o

b

 \theta _1  =32.94^o

c

 \theta_c  =  53.05^o    

Explanation:

From the question we are told that

    The angle of incidence is  \theta_1 =  10^o

    The refractive index of water is  n_1 = 1.3

  Generally Snell's law is mathematically represented as

          n_1 sin(\theta_1) =  n_2 sin(\theta_ 2)

Here n_2 is the refractive index of air with value  n_2 =  1

         \theta_2  is the angle of refraction

So  

        \theta _2  =  sin^{-1}[\frac{n_1 * sin(\theta _1)}{n_2} ]

=>     \theta _2  =  sin^{-1}[\frac{1.3 * sin(10)}{1} ]

=>     \theta _2  = 13^o

Given that the angle should not be greater than \theta _2 =45^o  then the angle of incidence will be

       \theta _1  =  sin^{-1}[\frac{n_2 * sin(\theta _2)}{n_1} ]

=>     \theta _1  =  sin^{-1}[\frac{1 * sin(45)}{1.3} ]

=>     \theta _1  =32.94^o

Generally for critical angle is mathematically represented as

        \theta_c  =  sin^{-1}[\frac{n_2}{n_1} ]

=>     \theta_c  =  sin^{-1}[\frac{1}{1.3} ]  

=>     \theta_c  =  53.05^o            

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