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anonymous
 one year ago
A projectile has a mass m and, of course, it is charged (q). The tube, has an electric field E as shown which deflects the projectile as shown. The mass enters the tube with a velocity v (tube length is L) and, due to the electric field, it will be deflected in an upward direction. For a projectile mass of 1.0x10^10kg, an electric field strength of 1.0x10^6N/C, a plate length of 15mm, and a drop initial horizontal speed of 29m/s, calculate the drop charge necessary to result in a deflection of 1.3mm when it first emerges from the tube.
anonymous
 one year ago
A projectile has a mass m and, of course, it is charged (q). The tube, has an electric field E as shown which deflects the projectile as shown. The mass enters the tube with a velocity v (tube length is L) and, due to the electric field, it will be deflected in an upward direction. For a projectile mass of 1.0x10^10kg, an electric field strength of 1.0x10^6N/C, a plate length of 15mm, and a drop initial horizontal speed of 29m/s, calculate the drop charge necessary to result in a deflection of 1.3mm when it first emerges from the tube.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441757085723:dw Here is the diagram that I attempted to draw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Mertsj @texaschic101 @Nnesha @Luigi0210 @robtobey Would any of you be able to help me, please? :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@e.mccormick @UnkleRhaukus

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0F = qE is the force of the electric field on the charge. Are there any other forces acting?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0How about taking into consideration the difference in height when the charge moves across the field? \(\sf \Delta y = v_{i_x}t+\frac{1}{2}gt^2\)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2the equation of motions are: \[\Large \left\{ \begin{gathered} z\left( t \right) =  \frac{{\left q \rightE}}{{2m}}{t^2} + {h_0} \hfill \\ \hfill \\ x\left( t \right) = {v_0}t \hfill \\ \end{gathered} \right.\] dw:1441829580320:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2now the projectile emerges from tube at time \( \large \tau \), such that: \[\Large x\left( \tau \right) = d = {v_0}\tau \to \tau = \frac{d}{{{v_0}}}\] at that time the zcoordinate is: \[\Large z\left( \tau \right) =  \frac{{\left q \rightE}}{{2m}}{\tau ^2} + {h_0}\] dw:1441829944559:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2so, we can write this: \[\Large \begin{gathered} \delta = {h_0}  z\left( \tau \right) = \frac{{\left q \rightE}}{{2m}}{\tau ^2} \hfill \\ \hfill \\ \delta = \frac{{\left q \rightE}}{{2m}}\frac{{{d^2}}}{{v_0^2}} \hfill \\ \end{gathered} \] where \( \large \delta \) is the deflection

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Just wondering... but why are you using Tau?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2usually, when I write an equation, I use "t" for as variable, whereas I use "\tau" to denote a specific time @Jhannybean
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