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Afina-wow [57]
3 years ago
14

A spring stretches 2.6 cm when a 7 g object is hung from it. The object is replaced with a block of mass 28 g which oscillates i

n simple harmonic motion. Calculate the period of motion. The acceleration of gravity is 9.8 m/s 2 . Answer in units of s.
Physics
1 answer:
PSYCHO15rus [73]3 years ago
4 0

Answer:

period = 0.65 sec

Explanation:

from the question we are given the following

extension (x) = 2.6 cm = 0.026 m

mass of object (Mo) = 7 g = 0.007 kg

mass of block (Mb) = 28 g = 0.028 g

acceleration due to gravity (g) = 9.8 m/s^{2}

period = 2π x \sqrt{\frac{Mb}{k}}

where k is the spring constant of the spring

and k = \frac{Mo x g}{x}

k =  \frac{0.007 x 9.8}{0.026}

k = 2.64 N/m

now period = 2π x \sqrt{\frac{0.028}{2.64}}

period = 0.65 sec

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

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

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E = φ  + K.E

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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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