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3 years ago

We can block light by placing obstacles in its path, but it’s much more difficult to block sound. Why?

Physics

2 answers:

Ivanshal [37]3 years ago

8 0

Answer:

Because light propagate in a straight path, while sound moves in all directions.

Explanation:

The rectilinear propagation of light proves that light travels on a straight path. With this, any opaque obstacle in its path would absorb the major fraction and reflect little fraction of the light. Light is an electromagnetic wave and transverse in nature.

Sound travels in all directions and it requires material medium (majorly air) for its propagation. Thus, it is not easy to obstruct the propagation of sound. It is a mechanical wave and longitudinal in nature.

Olin [163]3 years ago

7 0

Sound spreads through some kind of medium. Most of the time the medium is usually air. Sound can go through walls, like when you are playing really loud music and the neighbors tell you its to loud, you can't block it because it can travel through objects. Light does not need a medium to travel with. if you put a light in a room with no windows and no cracks, and you seal off the door the light can't escape that room, but if you did the same thing with sound it would travel trough the walls.

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Firemen are shooting a stream of water at a burning building using a high-pressure hose that shoots out the water with a speed o

Mrac [35]

Answer:

a)α= 53.13°

b)The velocity at the highest point = 15 m/s

The acceleration at the highest point = 9.8 $m/s^2$

c)h=15 m

V=18.02 m/s

Explanation:

Speed of water ,u= 25 m/s

So the horizontal component of speed u = u cos α

Given that horizontal distance cover by water in 3 s is 45 m.

So We know that in projectile motion horizontal acceleration is zero.

In horizontal direction

Distance = Velocity x time

45 = u cos α x 3

u cos α = 45

45 = 25 cos α x 3

cos α = 45/75

α= 53.13°

So the velocity at the highest point = u cos α

The velocity at the highest point = 15 m/s

The acceleration at the highest point = 9.8 $m/s^2$

Now the velocity along vertical direction(Vo) = u sin α

Vo= 25 sin 53.13°

Vo =20 m/s

$h=V_o.t-\dfrac{1}{2}gt^2$

$h=20\times 3-\dfrac{1}{2}\times 10\times 3^2$

h=15 m

So at 15 m above the ground water will strike .

The y-component of velocity after 3 sec

Vy= Vo - g t

Vy = 20 - 10 x 3

Vy= -10 m/s

The horizontal component of velocity will remain 15 m/s.

The resultant velocity

$V=\sqrt{10^2+15^2}\ m/s$

V=18.02 m/s

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4 years ago

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saul85 [17]

Answer:

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love history [14]

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iris [78.8K]

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