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Ghella [55]
3 years ago
10

When do phase transitions occur in molecules?

Physics
1 answer:
jekas [21]3 years ago
5 0
It occurs when energy is supplied or withdrawn :)
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One bowling ball roll into an other bowling ball that is not moving the bowling ball bounces aft each other what must be true ab
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Answer:

Rebouncing

Explanation:

Since the two balls didn't move but bounced back, it's then said to be called rebouncing

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Two speakers placed 0.94 m apart produce pure tones in sync with each other at a frequency of 1630 Hz. A microphone can be moved
Alina [70]

Answer:

a. approximately 1.1\; \rm m (first minimum.)

b. approximately 2.2\; \rm m (first maximum.)

c. approximately 3.4\; \rm m (second minimum.)

d. approximately 4.7\; \rm m (second maximum.)

Explanation:

Let d represent the separation between the two speakers (the two "slits" based on the assumptions.)

Let \theta represent the angle between:

  • the line joining the microphone and the center of the two speakers, and
  • the line that goes through the center of the two speakers that is also normal to the line joining the two speakers.

The distance between the microphone and point P_0 would thus be 9.4\, \tan(\theta) meters.

Based on the assumptions and the equation from Young's double-slit experiment:

\displaystyle \sin(\theta) = \frac{\text{path difference}}{d}.

Hence:

\displaystyle \theta = \arcsin \left(\frac{\text{path difference}}{d}\right).

The "path difference" in these two equations refers to the difference between the distances between the microphone and each of the two speakers. Let \lambda denote the wavelength of this wave.

\displaystyle \begin{array}{c|c} & \text{Path difference} \\ \cline{1-2}\text{First Minimum} & \lambda / 2 \\ \cline{1-2} \text{First Maximum} & \lambda \\\cline{1-2} \text{Second Minimum} & 3\,\lambda / 2 \\ \cline{1-2} \text{Second Maximum} & 2\, \lambda\end{array}.

Calculate the wavelength of this wave based on its frequency and its velocity:

\displaystyle \lambda = \frac{v}{f} \approx 0.211\; \rm m.

Calculate \theta for each of these path differences:

\displaystyle \begin{array}{c|c|c} & \text{Path difference} & \text{approximate of $\theta$} \\ \cline{1-3}\text{First Minimum} & \lambda / 2 & 0.112 \\ \cline{1-3} \text{First Maximum} & \lambda & 0.226\\\cline{1-3} \text{Second Minimum} & 3\,\lambda / 2 & 0.343\\ \cline{1-3} \text{Second Maximum} & 2\, \lambda & 0.466\end{array}.

In each of these case, the distance between the microphone and P_0 would be 9.4\, \tan(\theta). Therefore:

  • At the first minimum, the distance from P_0 is approximately 1.1\; \rm m.
  • At the first maximum, the distance from P_0 is approximately 2.2\; \rm m.
  • At the second minimum, the distance from P_0 is approximately 3.4\; \rm m.
  • At the second maximum, the distance from P_0 is approximately 4.7\;\rm m.
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A scientist uses a camera to study the stars.
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B.

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