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3 years ago

What is the basic formula for the harmonic of a standing wave

2 answers:

lisabon 2012 [21]3 years ago

6 0

The basic formula for the harmonic of a standing wave, describing the relation between the length and the wavelength is, $L=\left(\frac{n}{2}\right) \lambda$

<u>Explanation:</u>

Standing waves are produced if two waves of same frequencies interfere each other in such a way that it produce points that appears to be standing still in the medium.

On studying the wave patterns analytically, we see that there is a direct relationship between the length of the medium in which the wave pattern is seen and the wavelength of the waves.

With other experimental researches and studies, it has been concluded that the length of the medium that draws a string covered by a harmonic of a standing wave is equal to half of wavelength of the wave pattern i.e., the equation for the nth harmonic of a standing wave is,

$L=\left(\frac{n}{2}\right) \lambda$

Where,

L- Length of the string covered by the single loop of a complete wave

n- An integer representing the harmonic number

Pie3 years ago

4 0

fn=nf1 is the formula

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3 years ago

A solar cell, 3.0 cm square, has an output of 350 mA at 0.80 V when exposed to full sunlight. A solar panel that delivers close

Vesna [10]

Answer:

You will need 450 cells (3 cm each) to meet the voltage/current requirement.

The panel must be 3 cells in one side, by 150 cell in another side. 1350 cm^2 or 0.135 m^2. They must be connected 3 in row in parallel (to add current), then each of the former group must be connected in series to meet the voltage, so it would be 150 rows of connected in series.

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Explanation:

To meet the voltage:

120 [v] required voltage

0.8 [v] voltage of each cell

$\frac{120}{0.8} =150[v]\\$

So we need 150 cells in series for the voltage.

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3 years ago

A bicyclist of mass 60kg supplies 340W of power while riding into a 5 m/s head wind. The frontal area of the cyclist and bicycle

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Answer:

87.1 mph

Explanation:

We are given that

Mass,m=60 kg

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Drag coefficient, $C_d=0.88$

Coefficient of rolling resistance, $\mu_r=0.007$

Friction force, $f=\mu_rmg=0.007\times 60\times 9.8=4.1 N$

Where $g=9.8 m/s^2$

Let speed of cyclist=v'

Drag force, $F_d=\frac{1}{2}\rho_{air}AC_dv^2$

Density of air, $\rho_{air}=1.225 kg/m^3$

$F_d=\frac{1}{2}\times 1.225\times 0.344\times 0.88(5)^2=4.635N$

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3 years ago

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Katen [24]

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