x direction. And it’s going come pass with this electrical field the is between these fee plates, and it will exit the plates going at part velocity v1 and it’ll have actually a vertical component to its velocity and also a horizontal component. Therefore in component a, we number out what is this upright displacement and in component b, we figure out what is the upright component the its velocity after the passes through the plates and also then component c is what is this angle. So part a, an initial of every we need to figure out exactly how long does the spend between the plates and also we use that to ultimately calculate the vertical displacement due to the fact that it’s walking to it is in one half times the upright acceleration times time squared. So let’s very first figure the end the time. We know that the horizontal ingredient of that velocity will certainly be consistent because this electric field will certainly impart a vertical velocity and it’ll have actually no effect on that is horizontal velocity that it had actually initially therefore the horizontal ingredient of velocity stays constant. And so we know the x displacement i beg your pardon we’re offered is 4 centimeters, is walk to equal the horizontal component of velocity main point by time, and also so we’ll settle for t by dividing both sides by vx here. And also so t is dx separated by vx, and the horizontal displacement is 4 centimeters written as time ten come the minus two meters, and divided by the x component of that is velocity which is three times ten to the six meters per second. And we get 1.3333 times ten come the minus eight seconds, maintaining lots of digits in there due to the fact that this is an intermediary calculation. Now the upright component the its displacement will certainly be the initial y component of its velocity times time, add to one fifty percent times vertical acceleration times time squared, yet there is no initial ingredient to that velocity vertically, and also so this hatchet is zero. Therefore this functions out to one half ay t squared. Now the y ingredient of the network force equals mass time the y ingredient of acceleration and this needs to equal the force exerted by the electrical field top top the charge. And also so we’ll fix for the upright acceleration by separating both sides by m here, and so vertical acceleration is the electric field strength times the charge separated by that is mass, and also then us substitute that into the upright acceleration, we carry out that here. Therefore the upright component that its displacement ~ passing with the plate is one fifty percent Eq over m time time squared. Therefore it’s one fifty percent times 100 newtons every coulomb, electric field strength, times 1.6 time ten come the minus 19 coulombs, elementary school charge, split by the massive of one electron 9.11 time ten come the minus 31 kilograms, times the moment squared, and also we get 1.56 millimeters. Then in part b, we desire to recognize what is the vertical component that its velocity after it passes in between the plates. Well that’s walking to same the initial ingredient of that velocity the is vertical which is zero, to add the upright acceleration time time. And also the acceleration we currently found is Eq over m from up here, therefore we deserve to substitute the again down right here so we’re taking that multiply it by time. And also so electrical field strength times elementary charge divided by massive of electron multiply by time, offering us 2.34 time ten come the 5 meters per second will be the upright component of its velocity. And also the angle of that velocity will certainly be the train station tangent of the y component divided by the x component, so that’s inverse tangent the this answer for part b, split by the x component the we’re given, and this is 4.46 degrees over the confident horizontal.">

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This is college Physics Answers with Shaun Dychko. We have an electron with some initial rate of 3 times ten to the six meters per 2nd and the initial rate is completely in the x direction. And it’s going to pass v this electric field the is in between these fee plates, and also it will leave the plates going at some velocity v1 and also it’ll have a vertical component come its velocity as well as a horizontal component. For this reason in part a, we figure out what is this upright displacement and also in component b, we figure out what is the upright component the its velocity after the passes v the plates and then component c is what is this angle. So part a, an initial of all we need to number out how long does that spend in between the plates and also we use that to eventually calculate the upright displacement since it’s walk to it is in one fifty percent times the vertical acceleration times time squared. So let’s very first figure out the time. We know that the horizontal ingredient of that velocity will be constant because this electric field will impart a vertical velocity and it’ll have no effect on that horizontal velocity that it had initially therefore the horizontal component of velocity stays constant. And also so we understand the x displacement which we’re provided is 4 centimeters, is walking to same the horizontal ingredient of velocity main point by time, and also so we’ll deal with for t by splitting both sides by vx here. And so t is dx split by vx, and the horizontal displacement is four centimeters written as time ten come the minus two meters, and also divided by the x ingredient of the velocity i m sorry is three times ten to the six meters per second. And we gain 1.3333 times ten to the minus eight seconds, maintaining lots of digits in there because this is an intermediate calculation. Now the upright component of its displacement will certainly be the early stage y component of the velocity time time, to add one half times vertical acceleration times time squared, yet there is no initial component to that is velocity vertically, and so this term is zero. So this functions out come one fifty percent ay t squared. Now the y component of the network force equals mass time the y ingredient of acceleration and also this needs to equal the pressure exerted by the electric field on the charge. And also so we’ll solve for the upright acceleration by separating both political parties by m here, and also so upright acceleration is the electric field strength times the charge split by that mass, and then we substitute that into the upright acceleration, we carry out that here. So the upright component that its displacement after passing v the bowl is one fifty percent Eq over m time time squared. Therefore it’s one fifty percent times 100 newtons every coulomb, electric field strength, times 1.6 time ten to the minus 19 coulombs, primary school charge, split by the mass of an electron 9.11 time ten to the minus 31 kilograms, times the time squared, and we gain 1.56 millimeters. Climate in part b, we desire to know what is the vertical component the its velocity after that passes in between the plates. Well that’s walking to same the initial component of the velocity the is vertical which is zero, to add the vertical acceleration time time. And the acceleration we already found is Eq over m from increase here, therefore we deserve to substitute the again down below so we’re acquisition that multiplying it by time. And also so electric field stamin times primary school charge split by massive of electron main point by time, offering us 2.34 times ten to the 5 meters per 2nd will be the vertical component that its velocity. And also the angle of its velocity will be the inverse tangent that the y component separated by the x component, so that’s inverse tangent of this answer for component b, divided by the x component the we’re given, and this is 4.46 degrees above the hopeful horizontal.
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