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DaniilM [7]
3 years ago
9

He goes 12km north 6km east 12km south 3km west what is the distance and displacement

Physics
1 answer:
Hunter-Best [27]3 years ago
6 0

Distance is 33 km, displacement is 3 km east

Explanation:

- Distance is a scalar quantity equal to the total length of the path covered by an object during its motion, regardless of its direction.

- Displacement is a vector connecting the starting position of motion of an object to its final position. It is a vector quantity, therefore it has both magnitude and direction.

In this problem, the motion of the person is:

12 km north

6 km east

12 km south

3 km west

Therefore, the distance is:

d=12+6+12+3=33 km

The displacement instead can be calculated by comparing the final position with the initial position. We notice that:

- the person goes 12 km north and 12 km south, so he didn't move in the north-direction

- The person goes 6 km east and 3 km west, so his final position is 3 km east

Therefore, the displacement is 3 km east.

Learn more about distance and displacement:

brainly.com/question/3969582

#LearnwithBrainly

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Two coherent sources of radio waves, A and B, are 5.00 meters apart. Each source emits waves with wavelength 6.00 meters. Consid
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Answer:

a

    z= 2.5 \ m

b

   z =  (1 \ m ,  4 \ m )

Explanation:

From the question we are told that

     Their distance apart is  d =  5.00 \ m

      The  wavelength of each source wave \lambda =  6.0 \ m

Let the distance from source A  where the construct interference occurred be z

Generally the path difference for constructive interference is

              z - (d-z) =  m \lambda

Now given that we are considering just the straight line (i.e  points along the line connecting the two sources ) then the order of the maxima m =  0

  so

        z - (5-z) =  0

=>     2 z - 5 =  0

=>     z= 2.5 \ m

Generally the path difference for destructive  interference is

           |z-(d-z)| = (2m + 1)\frac{\lambda}{2}

=>         |2z - d |= (0 + 1)\frac{\lambda}{2}

=>        |2z - d| =\frac{\lambda}{2}

substituting values

          |2z - 5| =\frac{6}{2}

=>      z  =  \frac{5 \pm 3}{2}

So  

      z =  \frac{5 + 3}{2}

      z =  4\ m

and

      z =  \frac{ 5 -3 }{2}

=>   z =  1 \ m

=>    z =  (1 \ m ,  4 \ m )

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3 years ago
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