Question

In a TV set, an electron beam moves with horizontal velocity of 4.8 x 10^7 m/s across the cathode ray tube and strikes the screen, 50cm away. How far does the electron beam fall while traversing this distance?

Answer 1

`5.32* 10^(-16) m`

Step-by-step explanation:

`Speed=\frac {distance}{time}` hence making time the subject then

`Time=\frac {Distance}{Speed}`

Substituting 50 cm which is equivalent to 0.5 m for distance and velocity as given as
`4.8*10^(7)`we obtain

`Time=\frac {0.5}{4.8*10^(7)}=1.04167* 10^(-8) s`

From kinematic equation

`s=ut+0.5gt^(2)`and taking g as 9.81 then

`s=(0*1.04167* 10^(-8) s)+(0.5*9.81*(1.04167* 10^(-8) s)^(2))=5.32227* 10^(-16) m`

`s\approx 5.32* 10^(-16) m`