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enyata [817]
3 years ago
8

An open-pipe resonator has a length of 2.39 m. Calculate the frequency of its third harmonic if the speed of sound is 343 m/s.

Physics
1 answer:
Verdich [7]3 years ago
8 0

Answer:

f = 215.27 Hz

Explanation:

We have,

An open-pipe resonator has a length of 2.39 m.

It is required to find the frequency of its third harmonic if the speed of sound is 343 m/s.

For third harmonic, its length is equal to, l=\dfrac{3\lambda}{2}

So,

\dfrac{3v}{2f}=2.39

f is frequency in its third harmonics

f=\dfrac{3v}{2\times 2.39}\\\\f=\dfrac{3\times 343}{2\times 2.39}\\\\f=215.27\ Hz

So, the frequency of its third harmonic is 215.27 Hz.

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A binary star system consists of two equal mass stars that revolve in circular orbits about their center of mass. The period of
Vladimir79 [104]

Answer:

m = 2.23 \times 10^{-32} kg

Explanation:

Given data:

PERIOD OF MOTION IS T = 25.5 days

orbital speeds = 220 km/s

we know that

acceleration due to centripetal force isa =   \frac{F}{m} = \frac{V^2}{r}

Gravitational forceF= \frac{Gm m}{d^2}

we know that

v = \frac{2\pi R}{T}

solving for

R = \frac{vT}{2\pi}

F = \frac{Gm^2}{4(\frac{vT}{2\pi})^2}

F = G\times \frac{\pi m}{(vT)^2}

a = \frac{v^2}{\frac{vT}{2\pi}}

a = \frac{2\pi v}{T}

we know that

f =ma

G\times \frac{\pi m}{(vT)^2} = a = \frac{2\pi m v}{T}

solving for m

m = \frac{2Tv^3}{\pi G}

m = \frac{2\times 25.5 \times 86400 \times 220000^3\ m/s}{\pi \times 6.67\times 10^{-11}}

m = 2.23 \times 10^{-32} kg

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3 years ago
describes how the color and texture of a surface affect absorption and reflection of solar radiation?
Wewaii [24]

Most of the radiation, however, is absorbed by the earth's surface. ... Every surface on earth absorbs and reflects energy at varying degrees, based on its color and texture. Dark-colored objects absorb more visible radiation; light-colored objects reflect more visible radiation.

3 0
3 years ago
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padilas [110]
11.86 years.  Usually memorized as "12 years".
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Answer:

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Explanation:

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E= \frac{kQ}{r^2} r^

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