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Katyanochek1 [597]
3 years ago
5

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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Answer:

b) using an indicator to measure the hydrogen ion concentration of a solution

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The Kinetic energy, K, of an object with mass m moving with velocity v can be found using the formula - E_{\text{k}}={\tfrac {1}
tester [92]

Answer:

The ratio of kinetic energies of 5 kg object to 20 kg object is 1:1.

Explanation:

Kinetic energy is defined as energy possessed by an object due to its motion.It is calculated by:

K.E=\frac{1}{2}mv^2

Kinetic energy of the 5 kg object.

Mass of object,m = 5 kg

Velocity of an object = v

K.E=\frac{1}{2}mv^2=\frac{1}{2}\times 5kg\times v^2

Kinetic energy of the 20 kg object.

Mass of object,m' = 20 kg

Velocity of an object = v'

K.E=\frac{1}{2}mv^2=\frac{1}{2}\times 20kg\times v'^2

The ratio of the kinetic energy of the 5 kilogram object to the kinetic energy of the 20-kilogram object:

\frac{K.E}{K.E'}=\frac{\frac{1}{2}\times 5kg\times v^2}{\frac{1}{2}\times 20kg\times v'^2}

Given that, v = 2v'

\frac{K.E}{K.E'}=\frac{1}{1}

The ratio of kinetic energies of 5 kg object to 20 kg object is 1:1.

3 0
3 years ago
A soccer player kicks a ball horizontally off a 40.0-m-high cliff into a pool of water. If the player hears the sound of the spl
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the speed of the ball before it reaches the pool of water would be 9.91 m/s.

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A girl pushes a 1.04 kg book across a table with a horizontal applied force 10 points
mr Goodwill [35]

Answer:

Approximately 11.0\; \rm m \cdot s^{-1}. (Assuming that g = 9.81 \; \rm N \cdot kg^{-1}, and that the tabletop is level.)

Explanation:

Weight of the book:

W = m \cdot g = 1.04 \; \rm kg \times 9.81\; \rm N \cdot kg^{-1} \approx 10.202\; \rm N.

If the tabletop is level, the normal force on the book will be equal (in magnitude) to weight of the book. Hence, F(\text{normal force}) \approx 10.202\; \rm N.

As a side note, the F_N and W on this book are not equal- these two forces are equal in size but point in the opposite directions.

When the book is moving, the friction F(\text{kinetic friction}) on it will be equal to

  • \mu_{\rm k}, the coefficient of kinetic friction, times
  • F(\text{normal force}), the normal force that's acting on it.

That is:

\begin{aligned}& F(\text{kinetic friction}) \\ &= \mu_{\rm k}\cdot F(\text{normal force})\\ &\approx 0.35 \times 10.202\; \rm N \approx 3.5708\; \rm N\end{aligned}.

Friction acts in the opposite direction of the object's motion. The friction here should act in the opposite direction of that 15.0\; \rm N applied force. The net force on the book shall be:

\begin{aligned}& F(\text{net force})  \\ &= 15.0 \; \rm N - F(\text{kinetic friction}) \\& \approx 15.0 - 3.5708\; \rm N \approx 11.429\; \rm N\end{aligned}.

Apply Newton's Second Law to find the acceleration of this book:

\displaystyle a = \frac{F(\text{net force})}{m} \approx \frac{11.429\; \rm N}{1.04\; \rm kg} \approx 11.0\; \rm m \cdot s^{-2}.

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