An electron moving with a velocity of 5×10^3 m/s enters into a uniform electric field and acquiers a uniform acceleration of 10^3 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
Answers
Answer:
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 m
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Answer:
37.5×10^4 m
Explanation:
Given initial velocity, u = 5 × 10^4 m/s and acceleration, a = 10^4ms-2
(i) final velocity = v = 2 u = 2 × 5 ×10^4 m/s =10 × 10^4 m/s
To find t, use v = at or t = u – u / a = (5 × 10^4)/10^4
=5s
(ii) Using s = ut + 12at 2 = (5 ×10^4) × 5 + 12 (10 ) × (5) 2
= 25 ×10^4 + 25 /2 ×10^4
= 37.5×10^4 m