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3 years ago

14

Sound will travel slowest through which medium?

2 answers:

SSSSS [86.1K]3 years ago

8 0

The speed of sound depends on the medium in which it is transported.

Sound travels fastest through solids, slower through liquids and slowest through gases.

So it will travel slowest through water at 55 degrees

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disa [49]3 years ago

5 0

Answer:

water at 55°C

Explanation:

Sound is a mechanical wave, which travels through a medium by the vibrations of the particles in the medium, which collide with each other, allowing the sound wave to propagate through the medium.

Based on this explanation, we can understand two facts:

1) sound waves travel faster in solids than in liquids. This is because the particles inside a solid are much closer together than in a liquid, so the wave can be transferred faster because there are more collisions between the particles of the solid

2) The speed of a sound wave increases as the temperature increases. This is because at higher temperatures, particles move faster, so the wave can be transmitted faster.

Therefore, #1 tells us that sound travels slower in water than in ice, while #2 tells us that sound travels slower at 55°C than at 75°C . Therefore, sound travels slowest in water at 55°C .

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Two charged particles separated by a distance of = 3 and experienced electrostatic forces of = 60 . What would be this force if

klemol [59]

Answer: 539.4 N

Explanation:

Let's begin by explaining that Coulomb's Law establishes the following:

"The electrostatic force $F_{E}$ between two point charges $q_{1}$ and $q_{2}$ is proportional to the product of the charges and inversely proportional to the square of the distance $d$ that separates them, and has the direction of the line that joins them"

What is written above is expressed mathematically as follows:

$F_{E}= K\frac{q_{1}.q_{2}}{d^{2}}$ (1)

Where:

$F_{E}=60 N$ is the electrostatic force

$K=8.99(10)^{9} Nm^{2}/C^{2}$ is the Coulomb's constant

$q_{1}$ and $q_{2}$ are the electric charges

$d=3 m$ is the separation distance between the charges

Then:

$60 N= 8.99(10)^{9} Nm^{2}/C^{2}\frac{q_{1}.q_{2}}{(3 m)^{2}}$ (2)

Isolating $q_{1}$ and $q_{2}$ :

$q_{1}q_{2}=6(10)^{-8} C^{2}$ (3)

Now, if we keep the same charges but we decrease the distance to $d_{1}=1 m$ , (1) is rewritten as:

$F_{E}=8.99(10)^{9} Nm^{2}/C^{2}\frac{6(10)^{-8} C^{2}}{(1 m)^{2}}$ (4)

Then, the new electrostatic force will be:

$F_{E}= 539.4 N$ (5) As we can see, the electrostatic force is increased when we decrease the distance between the charges.

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