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Ulleksa [173]
2 years ago
13

In an experiment to illustrate the propagation of sound through fluids, identical sound sources emitting at 440 Hz are used to p

ropagate sound waves through mercury and through ethanol.
Medium Sound Speed
Ethanol 1,160 m/s
Mercury 1,450 m/s

Which of the following statements is correct, assuming that the dampening effect in each medium is negligible?

A. In ethanol, the frequency of the wave is greater to compensate for the smaller wavelength
B. In mercury, the frequency of the wave is the same as in ethanol, but the wavelength is greater.
C. In ethanol and mercury, the wavelength of the wave is the same, even if the speed of sound is different.
Physics
1 answer:
earnstyle [38]2 years ago
3 0

Answer:

B. In mercury, the frequency of the wave is the same as in ethanol, but the wavelength is greater.

Explanation:

To solve this easily, we can just calculate the wavelength of the sound in Ethanol and in Mercury.

In Ethanol, the wavelength will be:

λ = c/f

λ = 1160/440

λ = 2.63 m

In Mercury, the wavelength will be:

λ = c/f

λ = 1450/440

λ = 3.3 m

The wavelength of sound is greater in Mercury than in Ethanol but the frequency is the same.

Frequency of sound is not dependent on medium, but velocity and wavelength change depending on the medium.

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Answer: The final volume V₂ of the container is  0.039 m³.

Explanation:

Since the temperature is constant, the gas would expand isothermally.

For isothermal expansion,

P₁V₁=P₂V₂

Where, P₁ and P₂ are the initial and final pressure and V₁ and V₂ are initial and final volume.

It is given that:

V₁ = 0.0250 m³

P₁ = 1.5 × 10⁶ Pa

P₂ = 0.950 × 10⁶ Pa

V₂ = ?

⇒ 1.5 × 10⁶ Pa × 0.0250 m³ = 0.950 × 10⁶ Pa × V₂

⇒V₂ = 0.039 m³

Hence, the final volume V₂ of the container is  0.039 m³.

4 0
1 year ago
*An inductor is capable of dissipating 50W of heat energy when a current 0.8A flows through it at a certain frequency. Calculate
ale4655 [162]
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8 0
2 years ago
Walt ran 5 kilometers in 25 minutes, going eastward. What was his average velocity?
storchak [24]
1km per 5 mins
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7 0
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A circuit is made of 0.4 ohm wire, 150 ohm bulb and a 120ohm rheo stat connected inseries. Determine the total resistance of the
UNO [17]
Answer:
Total resistance of the circuit is 270.4 ohm

Explanation:
We are given that:
resistance of wire = 0.4 ohm
resistance of bulb = 150 ohm
resistance of rheostat = 120 ohm

We are also given that these components are connected in series. This means that the total resistance is summation of all the series components.

Therefore
Total resistance = 0.4 + 150 + 120 = 270.4 ohm

Hope this helps :)
8 0
2 years ago
A nucleus whose mass is 3.499612×10^(−25) kg undergoes spontaneous alpha decay. The original nucleus disappears and there appear
Elanso [62]

Answer:

The sum of the kinetic energies of the alpha particle and the new nucleus = (6.5898 × 10⁻¹³) J

Explanation:

Old nucleus ---> New nucleus + alpha particle.

We will use the conservation of energy theorem for extremely small particles,

Total energy before split = total energy after split

That is,

Total energy of the original nucleus = (total energy of the new nucleus) + (total energy of the alpha particle)

Total energy of these subatomic particles is given as equal to (rest energy) + (kinetic energy)

Rest energy = mc² (Einstein)

Let Kinetic energy be k

Kinetic energy of original nucleus = k₀ = 0 J

Kinetic energy of new nucleus = kₙ

Kinetic energy of alpha particle = kₐ

Mass of original nucleus = m₀ = (3.499612 × 10⁻²⁵) kg

Mass of new nucleus = mₙ = (3.433132 × 10⁻²⁵) kg

Mass of alpha particle = mₐ = (6.640678 × 10⁻²⁷) kg

Speed of light = c = (3.0 × 10⁸) m/s

Total energy of the original nucleus = m₀c² (kinetic energy = 0, since it was originally at rest)

Total energy of new nucleus = (mₙc²) + kₙ

Total energy of the alpha particle = (mₐc²) + kₐ

(m₀c²) = (mₙc²) + kₙ + (mₐc²) + kₐ

kₙ + kₐ = (m₀c²) - [(mₙc²) + (mₐc²)

(kₙ + kₐ) = c² (m₀ - mₙ - mₐ)

(kₙ + kₐ) = (3.0 × 10⁸)² [(3.499612 × 10⁻²⁵) - (3.433132 × 10⁻²⁵) - (6.640678 × 10⁻²⁷)]

(kₙ + kₐ) = (9.0 × 10¹⁶)(0.00007322 × 10⁻²⁵) = (6.5898 × 10⁻¹³) J

5 0
2 years ago
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