Question

An electron moving with a velocity of 5 ×10 power 4 metre per second enter into a uniform electrical field and equation uniform acceleration of 10 power 4 metre per second square in the direction of its initial motion
a)Calculate the time in which the electron with secure velocity double of its initial velocity
b)How much distance the electron with cover in this time?​

Answer 1

Answer:

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Answer 2

Given data:Initial velocity of the electron is,

u = 5 × 10^4 m/s

Acceleration of the electron is,

a = 10^4 m/s2

(i)

Using, v = u + at

=> 2u = u + at. [since v is double of Initial velocity (u)]

=> t = u/a = (5 × 10^4)/( 10^4) = 5 s

(ii)

Distance travelled in this time is,

S = ut + ½ at2

=> S = (5 × 10^4) × 5 + ½ × 10^4 × 5^2

=> S = 3,75,000 metre