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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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tatyana61 [14]

Answer:

A. Increase in temperature is 0.0176 degree Celsius. b. the remaining energy will be lost.

Explanation:

The mass of copper block = 7kg

Initial speed = 4.0 m/s

Specific heat of copper = 0.385 j/g degree Celcius.

a. The increase in temperature is calculated below:

\text{Change in kinetic energy} = \frac{1}{2}mv^{2} \\=  \frac{1}{2} \times 2.7 \times 4^{2} = 21.6 J \\

85% of energy is converted into internal energy.

mc\DeltaT = 21.6 \times 0.85 \\2.7 \times 385 \times \Delta T = 21.6 \times 0.85 \\\Delta T = 0.0176 degree \ celsius

b. The remaining  15 per cent of kinetic energy will be lost and it will be changed into other forms.

7 0
2 years ago
The speed of sound in seawater is 1470 m/s. A dolphin sends out a click that reflects off of an
Nitella [24]

Answer: 0.204 s

Explanation:

The speed of sound V is defined as the distance traveled d in a especific time t:  

V=\frac{d}{t}  

Where:  

V=1470 m/s is the speed of sound  in seawater

t is the time the sound wave travels from the dolphin and then returns after the reflection

d=2(150 m) is twice the distance between the dolphin and the object to which the sound waves are reflected

Finding t:

t=\frac{d}{V}  

t=\frac{2(150 m)}{1470 m/s}

<u>Finally:</u>

t=0.204 s

3 0
2 years ago
Two horizontal rods are each held up by vertical strings tied to their ends. Rod 1 has length L and mass M; rod 2 has length 2L
antiseptic1488 [7]

Answer:

Rod 1 has greater initial angular acceleration; The initial angular acceleration for rod 1 is greater than for rod 2.

Explanation:

For the rod 1 the angular acceleration is

\tau_1 = I_1\alpha _1 \\\\\alpha_1 = \dfrac{\tau_1}{I_1}

Similarly, for rod 2

\alpha_2 = \dfrac{\tau_2}{I_2}.

Now, the moment of inertia for rod 1 is

I_1 = \dfrac{1}{3}ML^2,

and the torque acting on it is (about the center of mass)

\tau_1 = Mg\dfrac{L}{2};

therefore, the angular acceleration of rod 1 is  

\alpha_1 = \dfrac{Mg\dfrac{L}{2}}{\dfrac{1}{3}ML^2},

\boxed{\alpha_1 = \dfrac{3g}{2L} }

Now, for rod 2 the moment of inertia is

I_2 = \dfrac{1}{3}(2M)(2L)^2

I_2 = \dfrac{8}{3} ML^2,

and the torque acting is (about the center of mass)

\tau _2 = (2M)g \dfrac{(2L)}{2}

\tau _2 = 2MgL;

therefore, the angular acceleration \alpha_2 is

\alpha_2 = \dfrac{2MgL;}{\dfrac{8}{3} ML^2,}.

\boxed{\alpha_2 = \dfrac{3g}{4L}}

We see here that

\dfrac{3g}{2L} > \dfrac{3g}{4L}

therefore

\boxed{\alpha_1 > \alpha_2.}

In other words , the initial angular acceleration for rod 1 is greater than for rod 2.

7 0
1 year ago
Consider a spring that does not obey Hooke’s law very faithfully. One end of the spring is fixed. To keep the spring stretched o
IRINA_888 [86]

Answer:

a) W=-0.0103125\ J

b) W=0.0059375\ J

c) Compressing is easier

Explanation:

Given:

Expression of force:

F=kx-bx^2+cx^3

where:

k=100\ N.m^{-1}

b=700\ N.m^{-2}

c=12000\ N.m^{-3}

x when the spring is stretched

x when the spring is compressed

hence,

F=100x-700x^2+12000x^3

a)

From the work energy equivalence the work done is equal to the spring potential energy:

here the spring is stretched so, x=-0.05\ m

Now,

The spring constant at this instant:

j=\frac{F}{x}

j=\frac{100\times (-0.05)-700\times (-0.05)^2+12000\times (-0.05)^3}{-0.05}

j=-8.25\ N.m^{-1}

Now work done:

W=\frac{1}{2} j.x^2

W=0.5\times -8.25\times (-0.05)^2

W=-0.0103125\ J

b)

When compressing the spring by 0.05 m

we have, x=0.05\ m

<u>The spring constant at this instant:</u>

j=\frac{F}{x}

j=\frac{100\times (0.05)-700\times (0.05)^2+12000\times (0.05)^3}{0.05}

j=4.75\ N.m^{-1}

Now work done:

W=\frac{1}{2} j.x^2

W=0.5\times 4.75\times (0.05)^2

W=0.0059375\ J

c)

Since the work done in case of stretching the spring is greater in magnitude than the work done in compressing the spring through the same deflection. So, the compression of the spring is easier than its stretching.

8 0
2 years ago
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horsena [70]
The brick, even though the brick would end up traveling faster, it most likely has a larger surface area therefore it would have more air resistance.
6 0
2 years ago
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