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Marrrta [24]
2 years ago
7

If you're swimming underwater and knock two rocks together, you will hear a very loud noise. But if your friend above the water

knocks two rocks together, you'll barely hear the sound.
Match the words.

The air-water interface is an example of boundary. The( )portion of the initial wave energy is way smaller than the( )portion. This makes the( ) wave hard to hear.

When both the source of the sound and your ears are located underwater, the sound is louder because the sound waves can( ) .

1. reflect more efficiently
2. transmitted
3. travel directly to your ears
4. boundary
5. reflected
6. discontinuity
Physics
1 answer:
Svetradugi [14.3K]2 years ago
4 0

Answer:

The air-water interface is an example of<em> </em>boundary. The <u><em>transmitted</em></u><em> </em> portion of the initial wave energy is way smaller than the <u><em>reflected</em></u><em> </em> portion. This makes the <u><em>boundary</em></u>  wave hard to hear.

When both the source of the sound and your ears are located underwater, the sound is louder because the sound waves can <u><em>travel directly to your ear</em></u>.

Explanation:

The air-to-water sound wave transmission is inhibited because more of reflection than transmission of the wave occurs at the boundary. In the end, only about 30% of the sound wave eventually reaches underwater. For sound generated underwater, all the wave energy is transmitted directly to the observer. Sound wave travel faster in water than in air because, the molecules of water are more densely packed together, and hence can easily transmit their vibration to their neighboring molecules, when compared to air.

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krok68 [10]

Answer:

r = 4.21 10⁷ m

Explanation:

Kepler's third law It is an application of Newton's second law where the forces of the gravitational force, obtaining

            T² = (\frac{4\pi }{G M_s} ) r³             (1)

           

in this case the period of the season is

            T₁ = 93 min (60 s / 1 min) = 5580 s

            r₁ = 410 + 6370 = 6780 km

            r₁ = 6.780 10⁶ m

for the satellite

           T₂ = 24 h (3600 s / 1h) = 86 400 s

if we substitute in equation 1

            T² = K r³

            K = T₁²/r₁³

            K = \frac{ 5580^2}{ (6.780 10^6)^2}

            K = 9.99 10⁻¹⁴ s² / m³

we can replace the satellite values

            r³ = T² / K

            r³ = 86400² / 9.99 10⁻¹⁴

            r = ∛(7.4724 10²²)

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this distance is from the center of the earth

7 0
2 years ago
A 2 000-kg sailboat experiences an eastward force of 3 000 N by the ocean tide and a wind force against its sails with a magnitu
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Answer:

The magnitude of the resultant acceleration is 2.2 m/s^2

Explanation:

Mass (m) of the sailboat =  2000 kg

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Eastwards means takes place along the positive x direction

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Then  

F_{2x} = -(6000N) cos 45 degree = -4242.6 N

F_{2y}  = (6000N) cos 45 degree = 4242.6 N

Hence  , the net force acting on the sailboat in x direction is  

F_x = F_{1x}+ F_{2x}

=  - 3000 N + 4242.6 N

=  - 3000 N +4242.6 N

= 1242.6N

Net Force acting on the sailboat in y direction is  

F_y = F_{1y}+ F_{2y}

= 0+ 4242.6N

= 4242.6N

The magnitude of the resultant force =

Using pythagorean theorm of 1243 N and 4243 N

\sqrt{(1242.6)^2 + (4242.6)^2

\sqrt{(1544054.76) + (17999654.8)}

\sqrt{(19543709.5)^2}

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F = ma

a = \frac{F}{m}

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4 0
2 years ago
This problem explores the behavior of charge on conductors. We take as an example a long conducting rod suspended by insulating
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Answer:

rod end A is strongly attracted towards the balls

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Answer with Explanation:

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8 0
2 years ago
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Or

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15 + 3 = 18

15 x 2 = 30

3 x 2 = 6

30 - 6 = <u>24km/h</u>

8 0
2 years ago
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