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

Answer:

$5.32\times 10^{-16} m$

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

$Time=\frac {0.5}{4.8*10^{7}}=1.04167\times 10^{-8} s$

From kinematic equation

$s=ut+0.5gt^{2}$and taking g as 9.81 then

$s=(0*1.04167\times 10^{-8} s)+(0.5*9.81*(1.04167\times 10^{-8} s)^{2})=5.32227\times 10^{-16} m$

$s\approx 5.32\times 10^{-16} m$