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An electron in a cathode-ray tube is traveling horizontally at 2.10×10^9 cm/s when deflection plates give it an upward acceleration of 5.30×10^17 cm/s^2 . B.) What is its vertical displacement during this time? My work http://i.imgur.com/AqKlTuX.jpg I am asking this because the answer my teacher gave is 2.3cm Thanks in advance |
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$y = y_0 + ut + 0.5at^2$ Since $y_0$ and $u$ are $0$, we have $y = 0.5~at^2$. In your calculation $t = 2.95 \times 10^{-9}$, which is correct. So, $y = 0.5\times5.30×10^{17}\times (2.95 \times 10^{-9})^2$ Therefore, $y = 2.3~\text{cm}$ |
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