Question

n electron moving with a velocity of 5 * 10 power 4 m/s enters with a uniform acceleration of 10 power 4m/s-² in the direction of initial motion.

Calculate :
i)Calculate the time in which the electron would acquires a velocity double to its initial velocity.
ii)How much distance electron would cover in this time.

Answer 1

Explanation:

Given initial velocity, u = 5 × 104 m s-1 and acceleration, a = 104 m s–2 (i) final velocity = v = 2 u = 2 × 5 ×104 m s–1 =10 × 104 m s–1 To find t, use v = at or t = u - u / a (ii) Using s = ut + 1/ 2 at 2 = (5 ×104) × 5 + 1/ 2 (10 ) × (5) 2 = 25 ×104 + 25 /2 ×104 = 37.5×104 mRead more on Sarthaks.com - https://www.sarthaks.com/10357/an-electron-moving-with-velocity-of-10-s-1-enters-into-uniform-electric-field-and-acquires

Answer 2

Answer:

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Given :

initial velocity, u = 5 × 104 m s-^1

acceleration, a = 104 m s–^2

(i) final velocity = v = 2 u = 2 × 5 ×104 m s–1 =10 × 104 m s–1

To find t ,

use v = at

or t = u - u / a

$(\frac{10 \times {10}^{2} - 5 \times {10}^{4} }{ {10}^{4} } ) = \frac{5 \times {10}^{4} }{ {10}^{4} } = 5s$

(ii) Using s = ut + 1/ 2 at 2

= (5 ×10^4) × 5 + 1/ 2 (10 ) × (5)^ 2

= 25 ×10^4 + 25 /2 ×10^4

= 37.5×10^4 m

Explanation:

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