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**please look at attached photo for proper understanding* Inside a cathode ray tube, an electron is in the presence of a uniform electric field with a magnitude of 295 N/C.


What is the magnitude of the acceleration of the electron (in m/s2)?

_____m/s2


The electron is initially at rest. What is its speed (in m/s) after 8.50 ✕ 10−9 s?

________m/s

Please Look At Attached Photo For Proper Understanding Inside A Cathode Ray Tube An Electron Is In The Presence Of A Uniform Electric Field With A Magnitude Of class=

Sagot :

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The magnitude of the acceleration of the electron is 5.187 x 10¹³ m/s².

The speed of the electron at the given time is 4.41 x 10⁵ m/s.

Acceleration of the electron

The magnitude of the acceleration of the electron is calculated as follows;

Force on the electron, F = ma = Eq

ma = Eq

a = Eq/m

where;

  • E is electric field
  • q is charge of electron
  • m is mass of electron

a = (295 x 1.6 x 10⁻¹⁹) / (9.1 x 10⁻³¹)

a = 5.187 x 10¹³ m/s²

Speed of the electron

The speed of the electron at the given time is calculated as follows;

v = at

v = ( 5.187 x 10¹³)(8.5 x 10⁻⁹)

v = 4.41 x 10⁵ m/s

Thus, the magnitude of the acceleration of the electron is 5.187 x 10¹³ m/s².

The speed of the electron at the given time is 4.41 x 10⁵ m/s.

Learn more about acceleration here: https://brainly.com/question/605631

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