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Answer to Question #22618 in Electric Circuits for Yvette Heath

Question #22618
In a TV set an electron beam moves horizontal velocity of 4.3 x 10 to the 7th m/s across the cathode ray tube and strikes the screen, 45 cm. away. The acceleration of gravity is 9.8 m/s to the 2nd. How far does the electron beam fall while traversing this distance? Answer in units of m
Expert's answer
the electron beam fall while traversing this distance on:
d = g*t^2/2
g - Theacceleration of gravity
t - time of moving
t = l/v
l - distance to thescreen
v - horizontalvelocity
d = g* (l/v)^2/2 = 9.8m/s^2 (0.45 m/ (4.3*10^7 m/s))^2/2 = 9.8*(0.45/(4.3*10^7))^2/2 m
= 5.37*10^-16 m
Answer: d = 5.37*10^-16m

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