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Lelu [443]
4 years ago
11

Sonar is used to determine the speed of an object. A 38.0-kHz signal is sent out, and a 40.0-kHz signal is returned. If the spee

d of sound is 341 m/s, how fast is the object moving?
Physics
1 answer:
ANEK [815]4 years ago
7 0

Answer:

The velocity is  v  =  8.743 \ m/s

Explanation:

From the question we are told that

    The frequency of the signal sent out  is  f_s  =  38.0 \ kHz  =  38.0 *10^{3} \ Hz

    The frequency of the signal received is  f_r  =  40.0 \ kHz  =  40.0 *10^{3} \ Hz

     The  speed of sound is  v_s  =  341 \ m/s

Generally the frequency of the sound received is  mathematically represented as

         f_r =  f_s  [\frac{v_s  + v}{v_s  - v} ]

where v is the velocity of the object

       =>      40 *10^{3} =  38 *10^{3} *   [\frac{341  + v}{341  - v} ]

       =>      1.05263 = \frac{341+v }{341-v}

       =>   358.94 - 1.05263v  =  341 + v

      =>    17.947 =  2.05263 v

      =>    v  =  8.743 \ m/s

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A car moves with the speed of 120m/s for 4 minutes ,calculate the distance covered by the car​
Vsevolod [243]

Answer:

960 m

Explanation:

Given that,

  • Speed = 120 m/s
  • Time taken = 4 minutes

We have to find the distance covered.

Firstly, let's convert time in seconds.

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→ 4 minutes = (4 × 60) seconds

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3 years ago
three girls were pushing the same car with a net force of 450 N [N48°E]. Two of the girls were pushing with forces of 310 N [N25
ElenaW [278]

The net force is the vector

∑ F = (450 N) (cos(42°) i + sin(42°) j)

and two of the forces provided by the girls are

F₁ = (310 N) (cos(115°) i + sin(115°) j)

F₂ = (250 N) (cos(285°) i + sin(285°) j)

Then the force provided by the third girl is the vector

F₃ = ∑ F - F₁ - F₂

F₃ = ((450 N) cos(42°) - (310 N) cos(115°) - (250 N) cos(285°)) i

… … … + ((450 N) sin(42°) - (310 N) sin(115°) - (250 N) sin(285°)) j

F₃ ≈ (400.722 N) i + (261.635 N) j

So, the third girl provided a force of magnitude

||F₃|| = √((400.722 N)² + (261.635 N)²) ≈ 478.572 N ≈ 480 N

pointing in a direction

arctan((261.635 N)/(400.722 N)) ≈ 33.1409° ≈ 33°

relative to East which refers to 0°; that is, 33° N of E or E33°N. Since the other forces are given relative to North or South, we can write this direction as N57°E.

So, the third girl pushed with force 480 N [N57°E].

5 0
3 years ago
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