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3 years ago

5

When two hydrogen atoms bond, the positive nucleus of one atom attracts the a. negative nucleus of the other atom. c. negative e

lectron of the other atom. b. positive electron of the other atom. d. positive nucleus of the other atom.

1 answer:

Aleksandr-060686 [28]3 years ago

4 0

Answer:

c

Explanation:

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Electrons are accelerated in the picture tube of a television through potential difference of 8.00 * 10 ^ 3 V. (Use the values q

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Answer:

$\lambda=1.37\times 10^{-11}\ m$

Explanation:

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The potential difference of the electrons, $V=8\times 10^{3}\ V$

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Using the conservation of energy,

$2meV=\dfrac{h^2}{\lambda^2}\\\\\lambda=\sqrt{\dfrac{h^2}{2meV}}$

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2 years ago

A flutist assembles her flute in a room where the speed of sound is 342 m/s. When she plays the note A, it is in perfect tune wi

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Answer:

5.15348 Beats/s

4.55 mm

Explanation:

$v_1$ = Velocity of sound = 342 m/s

$v_2$ = Velocity of sound = 346 m/s

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Frequency is given by

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Beat frequency is given by

$|f_1-f_2|=|440-445.15348|=5.15348\ Beats/s$

Beat frequency is 5.15348 Hz

Wavelength is given by

$\lambda_1=\frac{v_1}{f}\\\Rightarrow \lambda_1=\frac{342}{440}\\\Rightarrow \lambda_1=0.77727\ m$

Relation between length of the flute and wavelength is

$\lambda_1=2L_1\\\Rightarrow L_1=\frac{\lambda_1}{2}\\\Rightarrow L_1=\frac{0.77727}{2}\\\Rightarrow L_1=0.38863\ m$

At v = 346 m/s

$\lambda_2=\frac{v_2}{f}\\\Rightarrow \lambda_2=\frac{346}{440}\\\Rightarrow \lambda_1=0.78636\ m$

$L_2=\frac{\lambda_2}{2}\\\Rightarrow L_2=\frac{0.78636}{2}\\\Rightarrow L_2=0.39318\ m$

Difference in length is

$\Delta L=L_2-L_1\\\Rightarrow \Delta L=0.39318-0.38863\\\Rightarrow \Delta L=0.00455\ m=4.55\ mm$

It extends to 4.55 mm

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