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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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Ok the velocity of an object in free fall is given by the equation :

v=v0-gt, where v0 is the original velocity, g is the gravitational constant (9.8 m/s^2) and t is the time. 

so, we substitute values into this equation. v=35.8-9.8*2.5; v=11.3 m/s
5 0
1 year ago
A hydrogen discharge lamp emits light with two prominent wavelengths: 656 nm (red) and 486 nm (blue). The light enters a flint-g
mezya [45]

Answer:

The angle between the red and blue light is 1.7°.

Explanation:

Given that,

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Wavelength of blue = 486 nm

Angle = 37°

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n_{r}=1.572

n_{b}=1.587

We need to calculate the angle for red wavelength

Using Snell's law,

n_{r}\sin\theta_{i}=n_{a}\sin\theta_{r}

Put the value into the formula

1.572\sin37=1\times\sin\theta_{r}

\theta_{r}=\sin^{-1}(\dfrac{1.572\sin37}{1})

\theta_{r}=71.0^{\circ}

We need to calculate the angle for blue wavelength

Using Snell's law,

n_{b}\sin\theta_{i}=n_{a}\sin\theta_{b}

Put the value into the formula

1.587\sin37=1\times\sin\theta_{b}

\theta_{b}=\sin^{-1}(\dfrac{1.587\sin37}{1})

\theta_{b}=72.7^{\circ}

We need to calculate the angle between the red and blue light

Using formula of angle

\Delta \theta=\theta_{b}-\theta_{r}

Put the value into the formula

\Delta \theta=72.7-71.0

\Delta \theta=1.7^{\circ}

Hence, The angle between the red and blue light is 1.7°.

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2 years ago
A stranded soldier shoots a signal flare into the air to attract the attention of a nearby plane. The flare has an initial verti
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Answer:

Explanation:

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1) v² = u² - 2 g h , v = 0 so

h = u² / 2g = 1500² / 2 x 32 = 35156.25 ft

2) v = u - gt

t = u / g = 1500 / 32 = 46.875 s

3) It will hit the ground after 2 x 46.875 = 93.75 s

4 ) time to reach 30000 ft height  t is given by

h = ut - 16 t²

30000 = 1500t - 16t²

16t²-1500t + 30000 = 0

t = 28.92 s  and 64.82 s

Time required to travel 50000 by plane

= 50000/880 = 56.82 . There is no match of timing so plane will not hit it.

8 0
2 years ago
A handbag weighing 162N is carried by two students each holding the handle of the bag next to him. If each handle is inclined at
Mashcka [7]

Answer:

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Download pdf
3 0
1 year ago
A 3.45-kg centrifuge takes 100 s to spin up from rest to its final angular speed with constant angular acceleration. A point loc
Dafna11 [192]

Answer:

(a) 18.75 rad/s²

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Explanation:

(a)

First we find the acceleration of the centrifuge using,

a = (v-u)/t......................... Equation 1

Where v = final velocity, u = initial velocity, t = time.

Given: v = 150 m/s,  u = 0 m/s ( from rest), t = 100 s

Substitute into equation 1

a = (150-0)/100

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Secondly we calculate for the angular acceleration using

α = a/r..................... Equation 2

Where α = angular acceleration, r = radius of the centrifuge

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substitute into equation 2

α = 1.5/0.08

α = 18.75 rad/s²

(b)

Using,

Ф = (ω'+ω).t/2........................... Equation 3

Where Ф = number of revolution of the centrifuge, ω' = initial angular velocity, ω = Final angular velocity.

But,

ω = v/r and ω' = u/r

therefore,

Ф = (u/r+v/r).t/2

where u = 0 m/s (at rest),  = 150 m/s, r = 0.08 m, t = 100 s

Ф = [(0/0.08)+(150/0.08)].100/2

Ф = 93750 rad

If,

1 rad = 0.159155 rev,

Ф = (93750×0.159155) rev

Ф = 14920.78 rev

6 0
2 years ago
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