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enyata [817]
2 years ago
8

An open-pipe resonator has a length of 2.39 m. Calculate the frequency of its third harmonic if the speed of sound is 343 m/s.

Physics
1 answer:
Verdich [7]2 years ago
8 0

Answer:

f = 215.27 Hz

Explanation:

We have,

An open-pipe resonator has a length of 2.39 m.

It is required to find the frequency of its third harmonic if the speed of sound is 343 m/s.

For third harmonic, its length is equal to, l=\dfrac{3\lambda}{2}

So,

\dfrac{3v}{2f}=2.39

f is frequency in its third harmonics

f=\dfrac{3v}{2\times 2.39}\\\\f=\dfrac{3\times 343}{2\times 2.39}\\\\f=215.27\ Hz

So, the frequency of its third harmonic is 215.27 Hz.

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Sandra takes a class field trip to a geology museum. She sees an interesting sample of igneous rock. She wonders if the material
Aleonysh [2.5K]

Answer:

Yes

Explanation:

It is possible for sedimentary rocks to be converted to igneous rocks. Under conditions of high temperature and pressure, sedimentary rocks can be broken down into igneous rock by melting this rock type.

When the rock is broken down, it forms melt which when cooled and solidifies will form igneous rocks.

Sedimentary rocks are formed from the breaking down of pre-existing rocks through the action of weathering, erosion and sediment transportation. Within a basin, the sediments are compacted and lithified.

When this is subjected to intense pressure and temperature, the rock hardens and might further break down to melt.

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2 years ago
The movement of electrons through a wire is called.....
Elenna [48]

Answer:

Electric Current

Explanation:

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Blue light of wavelength λ passes through a single slit of width d and forms a diffraction pattern on a screen. If we replace th
ololo11 [35]

Answer:

We can retain the original diffraction pattern if we change the slit width to d) 2d.

Explanation:

The diffraction pattern of a single slit has a bright central maximum and dimmer maxima on either side. We will retain the original diffraction pattern on a screen if the relative spacing of the minimum or maximum of intensity remains the same when changing the wavelength and the slit width simultaneously.

Using the following parameters: <em>y</em> for the distance from the center of the bright maximum to a place of minimum intensity, <em>m</em> for the order of the minimum, <em>λ </em>for the wavelength, <em>D </em>for the distance from the slit to the screen where we see the pattern and <em>d </em>for the slit width. The distance from the center to a minimum of intensity can be calculated with:

                                                    y\approx\frac{m\lambda D}{d}

From the above expression we see that if we replace the blue light of wavelength λ by red light of wavelength 2λ in order to retain the original diffraction pattern we need to change the slit width to 2d:

<em>                                                 </em>y\approx\frac{m\lambda D}{d} =\frac{m2\lambda D}{2d}

7 0
3 years ago
The law of conservation of momentum states that the total momentum of interacting objects does not change . This means the total
pickupchik [31]

Answer:

The momentum of an object is equal to the product of its mass and its velocity.

Explanation:

Consider an object of mass m travelling at a velocity \vec{v}. The momentum \vec{p} of this object would be:

\vec{p} = m \cdot \vec{v}.

For the law of conservation of momentum, consider two objects: object \rm a and object \rm b. Assume that these two objects collided with each other.

  • Let m_{\rm a} and m_{\rm b} denote the mass of the two objects.
  • Let \vec{v}_{\rm a}(\text{initial}) and \vec{v}_{\rm b}(\text{initial}) denote the velocity of the two object right before the interaction.
  • Let \vec{v}_{\rm a}(\text{final}) and \vec{v}_{\rm b}(\text{final}) denote the velocity of the two objects right after the interaction.
  • The momentum of the two objects right before the collision would be m_{\rm a}\cdot \vec{v}_{\rm a}(\text{initial}) and m_{\rm b}\cdot \vec{v}_{\rm b}(\text{initial}), respectively.
  • The momentum of the two objects right after the collision would be m_{\rm a}\cdot \vec{v}_{\rm a}(\text{final}) and m_{\rm b}\cdot \vec{v}_{\rm b}(\text{final}), respectively.

The sum of the momentum of the two objects would be:

  • m_{\rm a}\cdot \vec{v}_{\rm a}(\text{initial}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{initial}) right before the collision, and
  • m_{\rm a}\cdot \vec{v}_{\rm a}(\text{final}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{final}) right after the collision.

Assume that the system of these two objects is isolated. By the law of conservation of momentum, the sum of the momentum of these two objects should be the same before and after the collision. That is:

m_{\rm a}\cdot \vec{v}_{\rm a}(\text{initial}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{initial}) = m_{\rm a}\cdot \vec{v}_{\rm a}(\text{final}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{final}).

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Answer:

Weight of the car, normal force, drag force

Explanation:

The forces acting on the car are:

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Friction is ignored, so the force due to friction is assumed negligible

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