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mr Goodwill [35]
3 years ago
12

A wave hits a wall as shown. As the wave interacts with a wall, which kind of wave interaction is shown? absorption diffraction

refraction reflection

Physics
2 answers:
dangina [55]3 years ago
8 0
The answer is reflection.

The drawing is simple but illustrates the concept beautifully.
lorasvet [3.4K]3 years ago
8 0

The correct answer to the question is reflection.

EXPLANATION:

Before coming into any conclusion, first we have to understand reflection.

Reflection is the type of phenomenon in which a wave is reflected back to the same medium when it  incidents on a obstacle. The obstacle may be any opaque body, mirror etc.

As per the diagram, the wave incidents obliquely on the wall.

After incident on the wall, the wave is reflected back to the same medium.Here, the wall acts as a obstacle. This property of wave is called as reflection.

Diffraction and reflection can not be the suitable answer for it . It is so because diffraction refers to the bending nature of light at the edge of an obstacle while refraction is the optical phenomenon in which the light ray will be bent towards or away from the normal at the interface of the refracting surface.

Hence, the correct answer to the question is reflection.

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Leni [432]

Answer:

(a)0.531m/s

(b)0.00169

Explanation:

We are given that

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Speed of bullet, v=357m/s

Mass of block 1,m_1=1177g=1.177kg

Mass of block 2,m_2=1626 g=1.626 kg

Velocity of block 1,v_1=0.681m/s

(a)

Let velocity of the second block  after the bullet imbeds itself=v2

Using conservation of momentum

Initial momentum=Final momentum

mv=m_1v_1+(m+m_2)v_2

4.67\times 10^{-3}\times 357+1.177(0)+1.626(0)=1.177\times 0.681+(4.67\times 10^{-3}+1.626)v_2

1.66719=0.801537+1.63067v_2

1.66719-0.801537=1.63067v_2

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v_2=\frac{0.865653}{1.63067}

v_2=0.531m/s

Hence, the  velocity of the second block after the bullet imbeds itself=0.531m/s

(b)Initial kinetic energy before collision

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Final kinetic energy after collision

K_f=\frac{1}{2}m_1v^2_1+\frac{1}{2}(m+m_2)v^2_2

K_f=\frac{1}{2}(1.177)(0.681)^2+\frac{1}{2}(4.67\times 10^{-3}+1.626)(0.531)^2

K_f=0.5028 J

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