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3 years ago

if the speed of sound is 34,500 cm/s and a note has a frequency of 256 hz, what is the wavelength in meters

Physics

1 answer:

zloy xaker [14]3 years ago

5 0

Answer:

Wavelength of sound in meters = 1.348 m

Explanation:

Given

speed of sound = 34500 cm $s^{-1}$ = 345 m $s^{-1}$ (since 1 cm = 0.01m)

frequency of sound = 256 Hz = 256 $s^{-1}$

Wavelength of sound in meters = ?

We know that all waves have same relationship among speed s, frequency f and wavelength λ, which is given by the equation

v = fλ

Wavelength λ = v/f

= 345 m $s^{-1}$ / 256 $s^{-1}$

=1.348 m

Hence wavelength of sound in meters = 1.348 m

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A mixture of nitrogen and xenon gases, at a total pressure of 836 mm Hg, contains 2.80 grams of nitrogen and 24.9 grams of xenon

larisa86 [58]

Answer: Partial pressure of nitrogen and xenon are 288mmHg and 548 mmHg respectively.

Explanation:

The partial pressure of a gas is given by Raoult's law, which is:

$p_A=p_T\times \chi_A$

where,

$p_A$ = partial pressure of substance A

$p_T$ = total pressure

$\chi_A$ = mole fraction of substance A

We are given:

$m_{N_2}=2.80g$

$m_{Xe}=24.9g$

Mole fraction of a substance is given by:

$\chi_A=\frac{n_A}{n_A+n_B}$

And,

$n_A=\frac{m_A}{M_A}$

Mole fraction of nitrogen is given as:

$\chi_{N_2}=\frac{\frac{m_{N_2}}{M_{N_2}}}{(\frac{m_{N_2}}{M_{N_2}}+\frac{m_{Xe}}{M_{Xe}})}$

Molar mass of $N_2$ = 28 g/mol

Molar mass of $Xe$ = g/mol

Putting values in above equation, we get:

$\chi_{N_2}=\frac{\frac{2.80}{28}}{\frac{2.80}{28}+\frac{24.9}{131}}$

$\chi_{N_2}=\frac{0.100}{0.100+0.190}=0.345$

To calculate the mole fraction of xenon, we use the equation:

$\chi_{Xe}+\chi_{N_2}=1\\\\\chi_{Xe}=1-0.345=0.655$

$p_{N_2}=836mmHg\times 0.345=288mmHg$

$p_{Xe}=836mmHg\times 0.655=548mmHg$

Thus partial pressure of nitrogen and xenon are 288mmHg and 548 mmHg respectively.

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3 years ago

When a source of dim orange light shines on a photosensitive metal, no photoelectrons are ejected from its surface. what could b

padilas [110]

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Explanation:

To expel electrons from a piece of metal, the incoming light must have a minimum frequency to cause a photoelectric effect, i.e., the ejection of photoelectrons from a metal surface, which is also known as the metal's threshold frequency.

If v = frequency of incident photon and vth= threshold frequency, then,

For v < vth, there will be no ejection of photoelectron.

For v = vth, photoelectrons are just ejected from the metal surface, in this case, the kinetic energy of the electron ejected is zero

For v > vth, then photoelectrons will come out of the surface along with kinetic energy

Therefore, we would have to increase the frequency of incident light so that it becomes greater than the threshold frequency of that surface, and consequently a photoelectric process takes place.

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1 year ago