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2 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]2 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]2 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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ExtremeBDS [4]

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3 years ago

A shopper walks westward 2.8 meters in the first 5 minutes and then eastward 9.2 meters in the next 10 minutes. What is the shop

Alecsey [184]

Answer:

12m

Explanation:

Given parameters:

Distance walked westward = 2.8m

Time of travel = 5min

Distance walked eastward = 9.2m

Time of travel = 10min

Unknown:

The total shopper's travel distance = ?

Solution:

Total distance traveled is the sum of the length of path covered by a body. It is a scalar quantity.

Total distance = distance walked westward + distance walked eastward

Total distance = 2.8m + 9.2m = 12m

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2 years ago

Please help<br> types of stimuli the sensory organs receive.

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2 years ago

A baseball pitcher throws a ball horizontally at a speed of 34.0 m/s. A catcher is 18.6 m away from the pitcher. Find the magnit

Sidana [21]

To develop this problem, it is necessary to apply the concepts related to the description of the movement through the kinematic trajectory equations, which include displacement, velocity and acceleration.

The trajectory equation from the motion kinematic equations is given by

$y = \frac{1}{2} at^2+v_0t+y_0$

Where,

a = acceleration

t = time

$v_0$ = Initial velocity

$y_0$ = initial position

In addition to this we know that speed, speed is the change of position in relation to time. So

$v = \frac{x}{t}$

x = Displacement

t = time

With the data we have we can find the time as well

$v = \frac{x}{t}$

$t = \frac{x}{v}$

$t = \frac{18.6}{34}$

$t = 0.547s$

With the equation of motion and considering that we have no initial position, that the initial velocity is also zero then and that the acceleration is gravity,

$y = \frac{1}{2} at^2+v_0t+y_0$

$y = \frac{1}{2} gt^2+0+0$