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scZoUnD [109]
3 years ago
13

The frequency of the second harmonic of a certain musical instrument is 100 Hz. What is the fundamental frequency of the instrum

ent? 100 Hz 50 Hz 150 Hz 200 Hz
Physics
1 answer:
ruslelena [56]3 years ago
4 0
The harmonic frequency of a musical instrument is the minimum frequency at which a string that is fixed at both ends in the instrument may vibrate. The harmonic frequency is known as the first harmonic. Each subsequent harmonic has a frequency equal to:
n*f, where n is the number of the harmonic and f is the harmonic frequency. Therefore, the harmonic frequency may be calculated using:
f = 100 / 2
f = 50 Hz
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Explanation:

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A chimpanzee sitting against his favorite tree gets up and walks 70.9 m due east and 31.9 m due south to reach a termite mound,
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A right triangle is formed by the 70.9 m walked east and 31.9 m walked south.
The legs of this right triangle are 70.9 and 31.9.
The shortest distance between two points is a straaight line. Therefore the hypotenuse of this triangle is going to be that shortest distance.
We can use the Pythagorean Theorem to find the hypotenuse of this triangle.
a^2+b^2=c^2\\70.9^2+31.9^2=c^2\\5026.81+1017.61=c^2\\6044.42=c^2\\c=\sqrt{6044.42}

c=\sqrt{6044.42}\approx\boxed{77.7458681}\ (decimal\ form)\\\\c=\sqrt{6044.42}=\sqrt{\frac{604442}{100}}=\boxed{\frac{\sqrt{604442}}{10}}\ (exact\ form)

As for the second part of the question, we want to find the angle formed by the hypotenuse and the 31.9 walked east.
We could use any of the three trigonometric ratios here since we know all 3 sides.
sine = opposite / hypotenuse
cosine = adjacent / hypotenuse
tangent = opposite / adjacent
I am going to use tangent, because then I won't have to deal with the hypotenuse and so the answer will be more accurate.

If you haven't already drawn yourself a diagram, now is a good time to.
The side opposite our angle is the 31.9, and the adjacent is 70.9.
Therefore, \tan(m\angle)=\frac{31.9}{70.9}.

We can use inverse trig ratios here to find the measure of our angle.
\tan^{-1}(\tan(m\angle))=\tan^{-1}(\frac{31.9}{70.9})\\\\m\angle=\tan^{-1}(\frac{31.9}{70.9})\approx\boxed{24.2243851\°\ or\ 0.422795279\ rad}
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What is the simplest way to build a graph of kinetic and potential energy?
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Explanation:

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The kinetic energy of a proton and that of an a-particle are 4 eV and 1 eV,
madam [21]

Answer:

  • <u><em>(b) 1:1</em></u>

Explanation:

<u></u>

<u>1. Formulae:</u>

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  • λ = h/(mv)
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Where:

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  • v = speed of the particle
  • h = Planck constant

<u><em>2. Reasoning</em></u>

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Thus, the relative masses of an alpha particle and a proton are:

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For the kinetic energies you find:

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Thus:

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          m_\alpha v_\alpha=m_pv_p

From de-Broglie equation, λ = h/(mv)  

       \dfrac{\lambda_p}{\lambda_\alpha}=\dfrac{m_\lambda v_\lambda}{m_pv_p}=\dfrac{1}{1}=1:1

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