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Physics 64 Online
OpenStudy (anonymous):

An electron of mass 9.11×10^−31kg leaves one end of a TV picture tube with zero initial speed and travels in a straight line to the accelerating grid, which is a distance 2.65cm away. It reaches the grid with a speed of 3.20×10^6 m/s . The accelerating force is constant. How do i find the acceleration? An electron of mass 9.11×10^−31kg leaves one end of a TV picture tube with zero initial speed and travels in a straight line to the accelerating grid, which is a distance 2.65cm away. It reaches the grid with a speed of 3.20×10^6 m/s . The accelerating force is constant. How do i find the acceleration? @Physics

OpenStudy (anonymous):

I guess you can use the formula v^2=v'(initial speed)^2 + 2ad

OpenStudy (anonymous):

so if i solve for acceleration in that equation its a=v^2-v'^2/2d right?

OpenStudy (anonymous):

yeah you are right

OpenStudy (anonymous):

ok i am doing a = (3.2x10^6m/s)^2/(2x2.65cm) and i am getting 1.932075, but that answer is wrong in the homework software i am using...what am i doing wrong?

OpenStudy (anonymous):

2.65cm -> 0.0265m

OpenStudy (anonymous):

oh man i didnt even notice that, now i got the right answer:) thanks!

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