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The electrons that produce the picture in a TV set are accelerated by a very large electric force as they pass through a small region in the neck of the picture tube. This region is 1.9 cm in length, and the electrons enter with a speed of 1 × 105 m/s and leave with a speed of 2.5 × 106 m/s.

physics
 Jun 6, 2015

Best Answer 

 #2
avatar+33603 
+5

Is it the average acceleration that you are after?

 

If so, use v2 = u2 + 2as where v = final velocity, u = initial velocity a = acceleration and s = distance

 

Rearrange as:  a = (v2 - u2)/(2s)

 

$${\mathtt{a}} = {\frac{\left({\left({\mathtt{2.5}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{6}}}\right)}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\left({\mathtt{1}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{5}}}\right)}^{{\mathtt{2}}}\right)}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{0.019}}\right)}} \Rightarrow {\mathtt{a}} = {\mathtt{164\,210\,526\,315\,789.473\: \!684\: \!210\: \!526\: \!315\: \!8}}$$

This is in m/sec2

.

 Jun 7, 2015
 #2
avatar+33603 
+5
Best Answer

Is it the average acceleration that you are after?

 

If so, use v2 = u2 + 2as where v = final velocity, u = initial velocity a = acceleration and s = distance

 

Rearrange as:  a = (v2 - u2)/(2s)

 

$${\mathtt{a}} = {\frac{\left({\left({\mathtt{2.5}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{6}}}\right)}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\left({\mathtt{1}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{5}}}\right)}^{{\mathtt{2}}}\right)}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{0.019}}\right)}} \Rightarrow {\mathtt{a}} = {\mathtt{164\,210\,526\,315\,789.473\: \!684\: \!210\: \!526\: \!315\: \!8}}$$

This is in m/sec2

.

Alan Jun 7, 2015

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