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liubo4ka [24]
3 years ago
9

Which of the following statements does not correctly describe a harmonic plane wave traveling in some medium.

Physics
1 answer:
Nostrana [21]3 years ago
3 0

Answer:

Incorrect statement is (b).

Explanation:

Option (a) : There is an inverse relation between the time and the frequency. The time taken depends on the frequency of the number of oscillations. Statement 1 is correct i.e. the time taken by any point of the wave to make one complete oscillation does not depend on the amplitude.

Option (b) : Speed of a wave is given by the product of its frequency and wavelength. It is not necessary that doubling the wavelength of the wave will halve its frequency as speed depends on the medium.

Option (c) : Doubling the amplitude has no effect on on the wavelength as amplitude does not depends on its wavelength.

Option (d) : Since, v=f\times \lambda

Speed is directly proportional to the frequency and wavelength. So, doubling the frequency of the wave will double its speed. So, the incorrect statement is (b).

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An electron is released from rest at a distance of 6.00 cm from a proton. If the proton is held in place, how fast will the elec
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We know that only only electric field due to proton causes to move from a distance of 6 cm from proton to 3 cm distance from it. This means the electric force force does work on the electron to move it from one initial position to the final position which is equal to the change in potential energy of the electron due to proton.

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\therefore \textrm{Kinetic energy change}= \textrm{Change in potential energy}\\\Rightarrow \dfrac{1}{2}m(v^2-u^2)= \dfrac{kpe}{y}-\dfrac{kpe}{x}\\\Rightarrow \dfrac{1}{2}m(v^2-(0)^2)= \dfrac{kpe}{0.03}-\dfrac{kpe}{0.06}\\\Rightarrow \dfrac{1}{2}mv^2= \dfrac{100kpe}{3}-\dfrac{100kpe}{6}\\\Rightarrow \dfrac{1}{2}mv^2= \dfrac{100kpe}{6}\\

\Rightarrow v^2= \dfrac{100kpe\times 2}{6m}\\\Rightarrow v^2= \dfrac{100kpe}{3m}\\\Rightarrow v^2= \dfrac{100\times 9\times 10^9\times 1.6\times 10^{-19}\times 1.6\times 10^{-19}}{3\times 9.1\times 10^{-31}}\\\Rightarrow v^2=8.44\times 10^3\\\Rightarrow v=91.87\ m/s\\

Hence, when the electron is at a distance of c cm from the proton, it moves with a velocity of 91.87 m/s.

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