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Nadya [2.5K]
3 years ago
13

On a nice winter day at the South Pole, the temperature rises to −54°F. What is the approximate temperature in degrees Celsius?

Physics
2 answers:
hram777 [196]3 years ago
5 0
<h2>Answer:</h2><h3>-47.7778</h3><h2>Explanation:</h2><h3>To translate degrees Fahrenheit to degrees Celsius, deduct 30 from the temperature in degrees Fahrenheit and divide by 2 to get the temperature in degrees Celsius. = 27 °C (In fact, 84 °F is equal to 28.89 °C—again, a close approximation!).</h3>
Evgesh-ka [11]3 years ago
3 0

Answer:

Its going to be about -47.78°C

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When two capacitors are connected in parallel and then connected to a battery, the total stored energy is 6.9 times greater than
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Ultraviolet light of wavelength 270 nm strikes a metal whose work function is 2.3 eV.What is the shortest de Broglie wavelength
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Answer:

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

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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

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3 years ago
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