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Svetllana [295]
3 years ago
11

The frequency of the middle c note on a piano is 261.63 hz. What is the wavelength of this note in centimeters? The speed of sou

nd in air is 343.06 m/s.
Physics
1 answer:
Sholpan [36]3 years ago
8 0

Answer:

1.31 m

Explanation:

The relationship between frequency and wavelength of a sound wave is

c=f \lambda

where

c is the speed of the wave

f is the frequency

\lambda is the wavelenfth

In this problem, we have

c = 343.06 m/s

f = 261.63 Hz

So we can solve the formula for the wavelength:

\lambda=\frac{c}{f}=\frac{343.06 m/s}{261.63 Hz}=1.31 m

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Talja [164]

Given:

Horizontal distance between two boats = x = 14 m

One boat is at trough, the other is at crest.

As there is no crests between them meaning the boat are next to each other.

Wavelength is the distance between two consecutive crests/troughs = w

The distance between a crest and a trough next to it = w/2

Complete cycles = c = 5

Time taken for c cycles = t = 15 s

Vertical distance between two boats = y = 2.4 m

To find:

wavelength = w = 2x = 28 m

Amplitude =  A = Displacement from mean to extreme position = y/2 = 1.2 m

Time period for one cycle = T = t/c = 15/5 = 3 s/cycle

frequency = 1/T = 1/3 = 0.33 hertz

speed = wavelength/Period = w/T = 28/3 = 9.33 m/s

4 0
3 years ago
Calculate the number of free electrons per cubic meter for some hypothetical metal, assuming that there are 1.3 free electrons p
boyakko [2]

Answer:

The number of free electrons per cubic meter is 7.61\times 10^{28}\ m^{-3}

Explanation:

It is given that,

The number of free electrons per cubic meter is, 1.3

Electrical conductivity of metal, \sigma=6.8\times 10^7\ \Omega^{-1}m^{-1}

Density of metal, \rho=10.5\ g/cm^3

Atomic weight, A = 107.87 g/mol

Let n is the number of  free electrons per cubic meter such that,

n=1.3\ N

n=1.3(\dfrac{\rho N_A}{A})

Where

\rho is the density of silver atom

N_A is the Avogadro number

A is the atomic weight of silver

n=1.3\times (\dfrac{10.5\ g/cm^3\times 6.02\times 10^{23}\ atoms/mol}{107.87\ g/mol})

n=7.61\times 10^{22}\ cm^{-3}

or

n=7.61\times 10^{28}\ m^{-3}

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Answer:

Energy can flow into and out of the system but matter cannot.

Explanation:

In a closed system, energy can flow in and out of the system but matter cannot.

  • A closed system prevents double way flow of matter.
  • A closed system conserves matter.

For an isolated system, energy and matter cannot flow out of the system.

For open systems, energy and matter can flow out of the system.

Such systems are used for certain thermodynamics experiment.

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