Create a daily schedule that includes focused time for you to work toward your goals.

Find the length (in m) of an organ pipe closed at one end that produces a fundamental frequency of 494 Hz when air temperature i

elena-14-01-66 [18.8K]

Answer:

0.173 m.

Explanation:

The fundamental frequency of a closed pipe is given as

fc = v/4l .................. Equation 1

Where fc = fundamental frequency of a closed pipe, v = speed of sound l = length of the pipe.

Making l the subject of the equation,

l = v/4fc ................ Equation 2

also

v = 331.5×0.6T ................. Equation 3

Where T = temperature in °C, T = 18.0 °c

Substitute into equation 3

v = 331.5+0.6(18)

v = 331.5+10.8

v = 342.3 m/s.

Also given: fc = 494 Hz,

Substitute into equation 2

l = 342.3/(4×494)

l = 342.3/1976

l =0.173 m.

Hence the length of the organ pipe = 0.173 m.

7 0

3 years ago

Ultraviolet light of wavelength 270 nm strikes a metal whose work function is 2.3 eV.What is the shortest de Broglie wavelength

soldier1979 [14.2K]

Answer:

The shortest de Broglie wavelength for the electrons that are produced as photoelectrons is 0.81 nm

Explanation:

Given;

wavelength of ultraviolet light, λ = 270 nm

work function of the metal, φ = 2.3 eV = 2.3 x 1.602 x 10⁻¹⁹ J = 3.685 x 10⁻¹⁹ J

The energy of the ultraviolet light is given by;

$E = \frac{hc}{\lambda}\\\\E = \frac{(6.626*10^{-34} )(3*10^{8}) }{270*10^{-9} }\\\\E = 7.362 * 10^{-19} \ J$

The energy of the incident light is related to kinetic energy of the electron and work function of the metal by the following equation;

E = φ + K.E

K.E = E - φ

K.E = (7.362 x 10⁻¹⁹ J) - (3.685 x 10⁻¹⁹ J )

K.E = 3.677 x 10⁻¹⁹ J

K.E = ¹/₂mv²

mv² = 2K.E

velocity of the electron is given by;

$V = \sqrt{\frac{2K.E}{m} }\\\\V = \sqrt{\frac{2(3.677*10^{-19}) }{9.1*10^{-31} } }\\\\V = 8.99*10^{5} \ m/s$

the shortest de Broglie wavelength for the electrons is given by;

$\lambda = \frac{h}{mv}\\ \\\lambda = \frac{6.626*10^{-34} }{(9.1*10^{-31})( 8.99*10^{5} )}\\\\\lambda = 8.10*10^{-10} \ m\\\\\lambda = 0.81 \ nm$

Therefore, the shortest de Broglie wavelength for the electrons that are produced as photoelectrons is 0.81 nm

7 0

4 years ago

A planet has a surface temperature of 95.0 K and an atmospheric pressure of 1.60 atm. If 4.87 L of atmosphere has a mass of 28.6

mr Goodwill [35]

Answer:

M=28.88 gm/mol

Explanation:

Given that

T= 95 K

P= 1.6 atm

V= 4.87 L

m = 28.6 g

R=0.08206L atm .mol .K

We know that gas equation for ideal gas

P V = n R T

P=Pressure , V=Volume ,n=Moles,T= Temperature ,R=gas constant

Now by putting the values

P V = n R T

1.6 x 4.87 = n x 0.08206 x 95

n=0.99 moles

We know that number of moles given as

$n=\dfrac{m}{M}$

M=Molar mass

$n=\dfrac{m}{M}$

$0.99=\dfrac{28.6}{M}$

M=28.88 gm/mol

3 0

3 years ago

A particle with a charge of 3.00 elementary charges moves through a potential difference of 4.50 volts. What is the change in el

GuDViN [60]

Answer:

$7.2\cdot 10^{-19} J$

Explanation:

The change in electrical potential energy of a charged particle moving through a potential difference is given by

$\Delta U = q \Delta V$

where

q is the magnitude of the charge of the particle

$\Delta V$ is the potential difference

In this problem:

- the charge of the particle is 3.00 elementary charges, so

$q=3e=3\cdot 1.6\cdot 10^{-19} J=4.8\cdot 10^{-19}J$

- the potential difference is

$\Delta V=4.50 V$

So, the change in electrical potential energy is

$\Delta U=(1.6\cdot 10^{-19}C)(4.50 V)=7.2\cdot 10^{-19} J$

7 0

4 years ago

Consider electrons of kinetic energy 6.0 eV and 600 keV. For each electron, find the de Broglie wavelength, particle speed, phas

irinina [24]

Answer:

For 6.0 eV

0.5 nm, 1.45*10^6 m/s, 6.17*10^10 m/s, 1.45*10^6 m/s

For 600 eV

1.26*10^-3 nm, 2.66*10^8 m/s, 3.37*10^8 m/s, 2.66*10^8 m/s

Explanation:

See attachment for calculation

5 0

3 years ago