Answered

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An electron emitted from a filament is travelling at 1.5 x 105 m/s when it enters an acceleration of an electron gun in a television tube. It is constantly accelerated while travelling 0.01 m, and leaves the gun at 5.4 x 106 m/s. What was the acceleration of the electron

Sagot :

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Answer:

The acceleration of the electron is 1.457 x 10Ā¹āµ m/sĀ².

Explanation:

Given;

initial velocity of the emitted electron, u = 1.5 x 10āµ m/s

distance traveled by the electron, d = 0.01 m

final velocity of the electron, v = 5.4 x 10ā¶ m/s

The acceleration of the electron is calculated as;

vĀ² = uĀ² + 2ad

(5.4 x 10ā¶)Ā² = (1.5 x 10āµ)Ā² + (2 x 0.01)a

(2 x 0.01)a = (5.4 x 10ā¶)Ā² - (1.5 x 10āµ)Ā²

(2 x 0.01)a = 2.91375 x 10Ā¹Ā³

[tex]a = \frac{2.91375 \ \times \ 10^{13}}{2 \ \times \ 0.01} \\\\a = 1.457 \ \times \ 10^{15} \ m/s^2[/tex]

Therefore, the acceleration of the electron is 1.457 x 10Ā¹āµ m/sĀ².

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