# An electron moves with a constant horizontal velocity of 3.0 × 10^6 m/s and no initial vertical velocity as it enters a deflector inside a TV tube. The electron strikes the screen after traveling 38.0 cm horizontally and 57.0 cm vertically upward with no horizontal acceleration. What is the constant vertical acceleration provided by the deflector? (The effects of gravity can beignored.) Show work please. The electron has no initial vertical speed

`V_(0y) =0 m/s`

`V_(0x) =3*10^6 m/s`

On the horizontal the motion is uniform. Time to travel horizontally from deflector to the screen is

`t = x/V_(0x) =0.38/(3*10^6)=1.267*10^-7 seconds`

On the vertical the motion is uniform accelerated over the time t. We suppose the...

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The electron has no initial vertical speed

`V_(0y) =0 m/s`

`V_(0x) =3*10^6 m/s`

On the horizontal the motion is uniform. Time to travel horizontally from deflector to the screen is

`t = x/V_(0x) =0.38/(3*10^6)=1.267*10^-7 seconds`

On the vertical the motion is uniform accelerated over the time t. We suppose the vertical deflection takes place over all distance from start to the screen.

`y =V_(0y)*t +(a*t^2)/2`

`V_(0y) =0 m/s`

`a =2*y/t^2 =2*0.57/(1.267*10^-7)^2 =7.1*10^13 m/s^2`

Observation: In real life, the deflection takes place only on a short distance at the beginning of the TV tube. To compute the vertical acceleration, one needs to know the horizontal and vertical dimensions of the deflector in the neck of the tube.

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