Question

An electron which is moving with a velocity of 5×10^6 m/sec emerges from a sheet of paper of thickness 2.1×10^-4 cm, with a velocity of 2×10^6m/sec. Calculate the time taken by the electron to pass through the sheet of the paper.

Answer 1

I hope this answer helpful

Answer 2

Answer:

12×10^-13

Explanation:

Given,

u=5×10^6 m/sec

v=2×10^6 m/sec

s=2.1×10^-4 cm

= 2.1×10^-4/100 m, (:change cm into m)

=2.1×10^-4/10^2 m

s=2.1×10^-6 m

v^2-u^2=2as.

a=v^2-u^2/2s...............e^n(1)

a=v-u/t...............e^n(2)

equation (1) and equation (2) ,we get,

= v^2-u^2/2s=v-u/t

= (v+u)(v-u)/2s=v-u/t. ,{;a^2-b^2=(a+b)(a-b)}

= (v+u)/2s=1/t

= t=2s/v+u. ,by cross multiple

= t=2×(2.1×10^-6)/2×10^6+5×10^6

= t=4.2×2×10^-6/7×10^6

= t=0.6×2×10^-6/10^6

= t=6×2×10^-6/10^7

= t=12×10^-13 Ans

if you like my solving style and then plz guys comments and follow me.i hope this ans is help for you.