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  • An electron moving with a velocity of 5 × 10^4 ms^-1 enters into a uniform electric field and acquires a uniform acceleration of 104 ms^–2 in the direction of its initial motion.

An electron moving with a velocity of 5 \times 10^4 ms^{-1 }enters into a uniform electric field and acquires a uniform acceleration of 10^4 ms^{-2 } in the direction of its initial motion.

(i) Calculate the time in which the electron would acquire a velocity double of its initial velocity.

(ii) How much distance the electron would cover in this time?

Answers (1)

(i) Given that, Initial velocity of the electron is u =5 \times 10^4 ms^{-1 }

Acceleration of the electron is a = 10^4 ms^{-2 }

We know that first equation of motion is,

v = u + at

2u = u + at

t = \frac{u}{a} =\frac{ (5 \times 10^4) }{ (10^4) }= 5 seconds

(ii) Now we know that,

s = ut + \frac{1}{2}at^{2}

s =( 5 \times 10^{4})5+ \frac{1}{2}(10)^{4}(5)^{2}

s = 37.5 \times 10^{4}m

 

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infoexpert21

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