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Brums [2.3K]
3 years ago
13

Your brother is insisting that you’ll never hear a sound produced behind a barrier wall at the end of your yard you notice that

the door within the wall is open and quickly explain to him that the sound will be able to reach you due to
A.Reverberation
B.Diffusion
C.Focusing
D.Diffraction
Physics
2 answers:
Tresset [83]3 years ago
8 0

Answer

D.Diffraction

Explanation

Diffraction is a property that is experienced by waves when they come across a barrier when they are in motion.

The ways tends to curve behind the barrier. This is called diffraction of waves.

Now, sound is a wave and it also experience diffraction. . So the brother will be able to hear the sound due to diffraction

oksian1 [2.3K]3 years ago
7 0

Answer: D. Diffraction

Explanation:

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If an object is not accelerating, it can exist in what 2 other states of motion?
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Friction is a force that acts in a direction opposite to the motion of the moving object.

Friction will cause a moving object to slow down and finally stop.

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When an object is placed 110 cm from a diverging thin lens, its image is found to be 55 cm from the lens. The lens is removed, a
Kazeer [188]

Explanation:

Given that,

Object distance u= -110 cm

Image distance v= 55 cm

We need to calculate the focal length for diverging lens

Using formula of lens

\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}

Put the value into the formula

\dfrac{1}{-f}=\dfrac{1}{55}-\dfrac{1}{-110}

\dfrac{1}{f}=-\dfrac{3}{110}

f=-36.6\ cm

The focal length of the diverging lens is 36.6 cm.

Now given a thin lens with same magnitude of focal length 36.6 cm is replaced.    

Here, The object distance is again the same.

We need to calculate the image distance for converging lens

Using formula of lens

\dfrac{1}{36.6}=\dfrac{1}{v}-\dfrac{1}{-110}

Here, focal length is positive for converging lens

\dfrac{1}{v}=\dfrac{1}{36.6}-\dfrac{1}{110}

\dfrac{1}{v}=\dfrac{367}{20130}

v=54.85\ cm

The distance of the image is 54.85 cm from converging lens.

Hence, This is the required solution.

5 0
3 years ago
A 4 kg toy car moves horizontally on a rough road with coefficient of kinetic friction 0.2. It accelerates from rest to 20 m/s i
attashe74 [19]

The total work done on the car is 784Joule.

<h3>What's the acceleration of the car?</h3>
  • As per Newton's equation of motion, V= U+at
  • U= initial velocity= 0 m/s

V= vinal velocity= 20m/s

t= time = 10s

a= acceleration

  • So, 20= 0+ 10a

=> a= 20/10= 2m/s²

<h3>What's the distance covered by the car in 10 seconds?</h3>
  • As per Newton's equation of motion,

V²-U² = 2aS

  • S= distance covered by the car
  • So, 20²-0=2×2×S=4S

=> 400= 4S

=> S= 400/4= 100m

<h3>What's the work done on the car due to frictional force?</h3>

Work done by frictional force= frictional force × distance

= (0.2×4×9.8)×100

= 784Joule

Thus, we can conclude that the work done on the car is 784Joule.

Learn more about the work done here:

brainly.com/question/25573309

#SPJ1

4 0
2 years ago
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