Question

The electrons that produce the picture in a TV set are acelerated by a very large electric force as they pass through a small region in the neck of the picture tube. This region is 1.6 cm in length and the electrons enter with a speed of 1x10^5 m/s and leave with a speed of 2.5 x10^6 m/s. What is their acceleration over this 1.6 cm length? Answer in units of m/s2.

Answer 1

Average electron speed in the 1.6 cm long region is given by:
\[\large \frac{25\times10^5+1\times10^5}{2}=1.3\times10^6 m/s\]
The time t to travel the 1.6 cm distance is found from:
\[\large t=\frac{distance}{speed}=\frac{1.6\times10^{-2}}{1.3\times10^6}=12\times10^{-9}\ seconds\]
The acceleration is given by:
\[\large a=\frac{v-v _{0}}{t}=\frac{25\times10^{5}-1\times10^{5}}{12\times10^{-9}}=2\times10^{14}\ ms ^{-2}\]

Answer 2

thank you. we are working problems with our high schooler tonight. This one was a doozie. I was working the problem as follows:
2.5EE6x²-1EE5x²÷2x0.016=2.4999999999999e+36

I was using the following formula:

a = (v^2 - u^2)/2s

Answer 3

If you use the formula:
\[\large a=\frac{v ^{2}-v _{0}^{2}}{2s}\]
The calculation is as follows:
\[\large a=\frac{(2.5\times10^{6})^{2}-(1\times10^{5})^{2}}{2\times1.6\times10^{-2}}\]
which simplifies to:
\[\large a=\frac{6.24\times10^{12}}{2\times1.6\times10^{-2}}=1.95\times10^{14}\ ms^{-2}\]