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Andre45 [30]
3 years ago
11

Wave A has an amplitude of 2 and wave B has an amplitude of 2 as shown below. What will happen when the crest of wave A meets th

e trough of wave B? They will interfere to create a crest with an amplitude of 4. They will interfere to create a crest with an amplitude of 2. They will interfere to create a crest with an amplitude of 0. They will bounce off each another.
Physics
2 answers:
puteri [66]3 years ago
6 0
Since the two waves have equal amplitudes, if the crest of one wave
meets the trough of the other one, they'll add to produce a level of zero
at that location.
Ugo [173]3 years ago
3 0

The correct answer to the question is- They will interfere to create a crest with an amplitude of 0.

EXPLANATION:

Before going to answer this question, first we have to understand destructive interference.

The two super imposing waves are said to be doing destructive interference if the crest of one wave falls on the trough of another wave.

Let the amplitudes of  two waves are A and B.

The phases difference (\theta) will be 180 degree for destructive interference.

Hence,the amplitude of the resultant wave is calculated as -

                                          R =  \sqrt{A^2+B^2+2ABcos\theta}

                                             = \sqrt{A^2+B^2+2ABcos180}

                                             = \sqrt{A^2+B^2-2AB}

                                             = A-B

Here, the amplitude of first wave A = 2

  The amplitude of second wave B = 2

Hence, amplitude of resultant wave R = 2-2

                                                                = 0

Hence, the correct answer will be - They will interfere to create a crest with an amplitude of 0.

                               

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92 protons

Explanation:

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g You drop a 3.6-kg ball from a height of 3.5 m above one end of a uniform bar that pivots at its center. The bar has mass 9.9 k
Salsk061 [2.6K]

Answer:

h = 3.5 m

Explanation:

First, we will calculate the final speed of the ball when it collides with a seesaw. Using the third equation of motion:

2gh = v_f^2 - v_i^2\\

where,

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vi = initial speed = 0 m/s

Therefore,

(2)(9.81\ m/s^2)(3.5\ m) = v_f^2 - (0\ m/s)^2\\v_f = \sqrt{68.67\ m^2/s^2}\\v_f = 8.3\ m/s

Now, we will apply the law of conservation of momentum:

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where,

m₁ = mass of colliding ball = 3.6 kg

m₂ = mass of ball on the other end = 3.6 kg

v₁ = vf = final velocity of ball while collision = 8.3 m/s

v₂ = vi = initial velocity of other end ball = ?

Therefore,

(3.6\ kg)(8.3\ m/s)=(3.6\ kg)(v_i)\\v_i = 8.3\ m/s

Now, we again use the third equation of motion for the upward motion of the ball:

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h = height = ?

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Answer:

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