Question

An electron moving with the velocity of 5 x 10⁴m/s enters into uniform electric field and acquires a uniform acceleration of 10⁴m/s² in the direction of its initial motion
(i)calculate the time in which the electron would acquire velocity double of its initial velocity.
(ii)how much distance the electron would cover in this time.

Answer 1

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QUESTION:- An electron moving with a velocity of 5×10^4 m/s enters into a uniform electric field and acquiers a uniform acceleration of 10^4 m/s^2 in the direction of its initial motion
1. Calculate the time in which the electron would acquire a velocity doubke of its initial velociry and
2) how much distance the electron would cover in this time


ANSWER:-

First of all we can write the given information

An electron moving with velocity of 5X10^4m/s

U= 5X10 ^4 m/s

And here the acceleration of the electron is
Given by

a= 10^4m/s

(1):- V=U+at

2U=U+at

=> t=u/a

So, => (5x10^4)/ (10^4)

By cutting 10^4 and 10^4

We will get the answer as 5sec

(2). So the second part is the distance traveled by the electron


S=U+1/2 at

=> s= (5x10^4)+5x 1/2 x 10^4X 5^2

=> s= 3,57,000 m

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Hope it help

Answer 2

Explanation:

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