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3 years ago

14

You have an empty 20 oz. soda bottle and you blow air over the opening to excite a fundamental standing wave. Now, you slice off

the bottom of the bottle (it’s plastic) without changing its length very much. You blow over the opening and excite a fundamental standing wave in the bottle with its bottom end open. The frequency of the standing wave in the second case:______________________________.

1 answer:

Tatiana [17]3 years ago

3 0

Answer:

The frequency of the standing wave in the second case is higher than that in the first case

Explanation:

The frequency and wavelength of a wave are related.

The moment you sliced the bottle, you've reduced the wavelength of the bottle.

When wavelength decreases, frequency increases and vice versa.

So, When frequency increases in the second case, more wave crests pass a fixed point each second. That means the wavelength shortens. So, as frequency increases, wavelength decreases. The opposite is also true—as frequency decreases, wavelength increases.

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What's the mass show the work?

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2 years ago

A charge of -3.30 nC is placed at the origin of an xy-coordinate system, and a charge of 2.05 nC is placed on the y axis at y =

Elis [28]

Answer:

$F_{3h}=39065.298\times 10^9\ N$ attractive toward +x axis is the net horizontal force

$F_v=80062.47\times 10^9$ attractive toward +y axis is the net vertical force

Explanation:

Given:

charge at origin, $Q_0=-3.35\times 10^{-6}\ C$

magnitude of second charge, $Q_2=2.05\times 10^{-6}\ C$

magnitude of third charge, $Q_3=5\times 10^{-6}\ C$

position of second charge, $(x_2,y_2)\equiv(0,4.35)\ cm$

position of third charge, $(x_3,y_3)\equiv(3.1,3.8)\ cm$

<u>Now the distance between the charge at at origin and the second charge:</u>

$d_2=\sqrt{(x_2-0)^2+(y_2-0)^2}$

$d_2=\sqrt{(0-0)^2+(4.35-0)^2}$

$d_2=0.0435\ m$

<u>Now the distance between the charge at at origin and the third charge:</u>

$d_3=\sqrt{(x_3-0)^2+(y_3-0)^2}$

$d_3=\sqrt{(3.1-0)^2+(3.8-0)^2}$

$d_3=0.04904\ m$

<u>Now the force due to second charge:</u>

$F_2=\frac{1}{4\pi.\epsilon_0} \times \frac{Q_0.Q_2}{d_2^2}$

$F_2=9\times 10^9\times \frac{3.3\times 2.05}{0.0435^2}$

$F_2=32175.98\times 10^9\ N$ attractive towards +y

<u>Now the force due to third charge:</u>

$F_3=\frac{1}{4\pi.\epsilon_0} \times \frac{Q_0.Q_3}{d_3^2}$

$F_3=9\times 10^9\times \frac{3.3\times 5}{0.04904^2}$

$F_3=61748.38\times 10^9\ N$ attractive

<u>Now the its horizontal component:</u>

$F_{3h}=\frac{3.1}{4.9} \times 61748.38\times 10^9$

$F_{3h}=39065.298\times 10^9\ N$ attractive toward +x axis

<u>Now the its vertical component:</u>

$F_{3v}=\frac{3.8}{4.9} \times 61748.38\times 10^9$

$F_{3v}=47886.49\times 10^9\ N$ upwards attractive

Now the net vertical force:

$F_v=F_{3v}+F_2$

$F_v=47886.49\times 10^9+32175.98\times 10^9$

$F_v=80062.47\times 10^9$

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