Question

An electron in a cathode-ray tube accelerates uniformly from 2.00 3 104 m/s to 6.00 3 106 m/s over 1.50 cm. (a) In what time interval does the electron travel this 1.50 cm? (b) What is its acceleration?

Answer 1

(a)
`t=5.02* 10^(-9)\ s`

(b)
`a=1.19* 10^(15)\ m/s^2`

Step-by-step explanation:

It is given that,

Initial speed of an electron,
`u=2* 10^4\ m/s`

Final speed of an electron,
`v=6* 10^6\ m/s`

Distance, d = 1.5 cm = 0.015 m

(b) If a is the acceleration of the electron. It is given by the rate of change of its velocity. It can be calculated using third equation of motion as :

`a=(v^2-u^2)/(2d)`

`a=((6* 10^6)^2-(2* 10^4)^2)/(2* 0.015)`

`a=1.19* 10^(15)\ m/s^2`

(a) Let t is the time interval in which the electron travel 1.50 cm. It is given by :

`t=(v-u)/(a)`

`t=(6* 10^6-2* 10^4)/(1.19* 10^(15))`

`t=5.02* 10^(-9)\ s`

So, the time interval in which the electron travel 1.50 cm is
`5.02* 10^(-9)\ s`.

Hence, this is the required solution.

Answer 2

(a). The time interval is
`4.98*10^(-9)\ sec`

(b). The acceleration is
`1.2*10^(15)\ m/s^2`

Step-by-step explanation:

Given that,

Initial velocity
`u= 2.003*10^(4)\ m/s`

Final velocity
`v=6.003*10^(6)\ m/s`

Distance = 1.50 cm

(b) We need to calculate the acceleration

Using equation of motion

`v^2=u^2+2as`

Where, v = final velocity

u = initial velocity

s = distance

Put the value into the formula

`(6.003*10^(6))^2=(2.003*10^(4))^2+2* a*1.50*10^(-2)`

`a=((6.003*10^(6))^2-(2.003*10^(4))^2)/(2*1.50*10^(-2))`

`a=1.2*10^(15)\ m/s^2`

(a). We need to calculate the time interval

Using equation of motion

`v=u+at`

`t=(v-u)/(a)`

Put the value into the formula

`t=(6.003*10^(6)-2.003*10^(4))/(1.2*10^(15))`

`t=4.98*10^(-9)\ sec`

Hence, (a). The time interval is
`4.98*10^(-9)\ sec`

(b). The acceleration is
`1.2*10^(15)\ m/s^2`