Question

An electron in a cathode ray tube (CRT) accelerates from 2.00 x10 m/s to 6.00 x 10" m/s over 1.50 cm. (a) How long does the electron take to travel this 1.50 cm? (c) What is its acceleration?​

Answer 1

Answer:

Solution



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We have V[i]=2.00∗104m/s,Vf=6.00∗106m/s and Xf−Xi=1.50∗10−2m

(a) Xf−Xi=21(Vi+Vf)t

t=Vi+Vf2(Xf−Xi)=2.00∗104m/s+6.00∗106m/s2(1.50∗10−2m)

=4.98∗10−9s

(b) Vf2=Vi2+2ax(Xf−Xi