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Arada [10]
3 years ago
10

In general, as the energy of a sound wave increases, A. the sound gets softer and then louder. B. the sound gets louder. C. the

sound always remains the same. D. the sound gets softer.
Physics
2 answers:
agasfer [191]3 years ago
8 0
<h3><u>Answer;</u></h3>

B. the sound gets louder

<h3><u>Explanation;</u></h3>
  • <em><u>Sound waves are examples of mechanical longitudinal waves.</u></em> Mechanical waves because they require a material medium for transmission. Longitudinal waves because the vibration of particles in sound waves is parallel to the direction of wave motion.
  • <em><u>As the energy of a sound wave increases, then the sound becomes louder.</u></em> The higher the energy of a wave the higher the amplitude of sound waves. Therefore, <em><u>greater amplitude of sound waves means the waves possess higher energy and thus greater intensity and therefore, the sound will be louder.</u></em>
Vadim26 [7]3 years ago
5 0
The sound gets louder, like hitting a metal pole on a thin metal beam, the energy increases as the sound does.
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3 years ago
What is the potential difference when the current in a circuit is 5mA and resistance is 30 Ohms
Mashcka [7]

<h2>\bf{ \underline{Given:- }}</h2>

\sf• \: The \:  current \:  in \:  a \:  circuit \:  is  \: 5 \: amps.  \: and  \: resistance \:  is \:  30 \:  Ohms.

\\

<h2>\bf{ \underline{To \:  Find :- }}</h2>

\sf{• \:  The  \: Potential  \: Difference. }

\\

\huge\bf{ \underline{ Solution:- }}

\sf According  \: to  \: the  \: question,

\sf•  \: Current \:  (I) = 5  \: Amps.

\sf• \:  Resistance  \: (R) = 30 \:  Ω

\sf{Potential \:  difference  \: means  \: Voltage \: ( V).}

\sf{We \:  know \: that, }

\bf \red{ \bigstar{ \: V = IR }}

\rightarrow \sf V =5 \times 30

\rightarrow \sf V =150

\\

\sf \purple{Therefore, \:  the \:  potential  \: difference  \: is  \: 150  \: v \: .}

4 0
3 years ago
The edge of a flying disc with a radius of 0.13 m spins with a tangential speed of 3.3 m/s. The centripetal acceleration of the
Marat540 [252]

Answer:

Centripetal acceleration = 83.77m/s²

Explanation:

<u>Given the following data;</u>

Radius, r = 0.13m

Velocity, v = 3.3m/s

To find centripetal acceleration;

Centripetal acceleration is given by the formula;

Acceleration, a = \frac {v^{2}}{r}

Substituting into the equation, we have;

Centripetal \; acceleration, a = \frac {3.3^{2}}{0.13}

Centripetal \; acceleration, a = \frac {10.89}{0.13}

<em>Centripetal acceleration = 83.77m/s²</em>

<em>Therefore, the centripetal acceleration of the edge of the disc is 83.77 m/s². </em>

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3 years ago
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Alina [70]
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