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Alex Ar [27]
3 years ago
13

if two waves with equal amplitudes and wavelengths travel through a medium in such a way that a particular particle of the mediu

m is at the crest of one wave and at the trough of the other wave at the same time, what will happen to that particle?
Physics
2 answers:
Novosadov [1.4K]3 years ago
8 0
If two waves with equal amplitude and wavelength travel through a medium such that particular particle is at the crest of one wave and at the trough of another wave, then destructive interference will take place.Both waves will cancel each other out at that point such that a node will form in which the particle will remain at its rest position only and will not move. 
Irina-Kira [14]3 years ago
8 0

Answer:

As per the law of superposition we know that when two waves superimpose at a point on the medium then the resultant displacement of the medium particle is vector sum of the displacement due to each wave.

So here by superposition principle we can write

y_{net} = y_1 + y_2

now here we know that one of the wave reaches with crest while other wave reaches at trough position.

So we will have

y_1 = +A

y_2 = -A

so here by superposition we will have

y_{net} = A - A = 0

so after superposition the medium particle will come to its mean position.

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Answer: turns to gas

Explanation:

when a liquid gets heated up the chemicals start to boil/evaporate turning the liquid into a gas

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Which of the following would most likely utilize a particulate control device?
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Answer:

d

Explanation:

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What can be concluded about the atom from knowing that oxygen-18 has an atomic number of 8? A) An oxygen atom that includes 8 ba
anyanavicka [17]
B)

If it is known that the atomic number is 8, we know that the electrons are also 8. Since the atomic mass (O18) is 18, the neutrons are 18-8=10. Option B is the correct answer. 

Hope I helped :) 
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2 years ago
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Instead of moving back and forth, a conical pendulum moves in a circle at constant speed as its string traces out a cone (see fi
tigry1 [53]

Answer:

a

The  radial acceleration is  a_c  = 0.9574 m/s^2

b

The horizontal Tension is  T_x  = 0.3294 i  \ N

The vertical Tension is  T_y  =3.3712 j   \ N

Explanation:

The diagram illustrating this is shown on the first uploaded

From the question we are told that

   The length of the string is  L =  10.7 \ cm  =  0.107 \ m

     The mass of the bob is  m = 0.344 \  kg

     The angle made  by the string is  \theta  =  5.58^o

The centripetal force acting on the bob is mathematically represented as

         F  =  \frac{mv^2}{r}

Now From the diagram we see that this force is equivalent to

     F  =  Tsin \theta where T is the tension on the rope  and v is the linear velocity  

     So

          Tsin \theta  =   \frac{mv^2}{r}

Now the downward normal force acting on the bob is  mathematically represented as

          Tcos \theta = mg

So

       \frac{Tsin \ttheta }{Tcos \theta }  =  \frac{\frac{mv^2}{r} }{mg}

=>    tan \theta  =  \frac{v^2}{rg}

=>   g tan \theta  = \frac{v^2}{r}

The centripetal acceleration which the same as the radial acceleration  of the bob is mathematically represented as

      a_c  =  \frac{v^2}{r}

=>  a_c  = gtan \theta

substituting values

     a_c  =  9.8  *  tan (5.58)

     a_c  = 0.9574 m/s^2

The horizontal component is mathematically represented as

     T_x  = Tsin \theta = ma_c

substituting value

   T_x  = 0.344 *  0.9574

    T_x  = 0.3294 \ N

The vertical component of  tension is  

    T_y  =  T \ cos \theta  = mg

substituting value

     T_ y  =  0.344 * 9.8

      T_ y  = 3.2712 \ N

The vector representation of the T in term is of the tension on the horizontal and the tension on the vertical is  

         

       T  = T_x i  + T_y  j

substituting value  

      T  = [(0.3294) i  + (3.3712)j ] \  N

         

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Answer:

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