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An electron $\left(\mathrm{mass}=9.11 \times 10^{-31} \mathrm{kg}\right)$ leaves one end of a TV picture tube with zero initial speed and travels in a straight line to the accelerating grid, which is 1.80 $\mathrm{cm}$ away. It reaches the grid with a speed of $3.00 \times 10^{6} \mathrm{m} / \mathrm{s} .$ If the accelerating force is constant, compute (a) the acceleration of the electron, (b) the time it takes the electron to reach the grid, and (c) the net force that is accelerating the electron, in newtons. (You can ignore the gravitational force on the electron.)

$2.50 \times 10^{14} \mathrm{m} / \mathrm{s}^{2}$$1.2 \times 10^{-8} \mathrm{g}$$2.28 \times 10^{-16} \mathrm{N}$

Physics 101 Mechanics

Chapter 4

Newton's Laws of Motion

Physics Basics

Motion Along a Straight Line

Motion in 2d or 3d

University of Michigan - Ann Arbor

Hope College

McMaster University

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All right. So I had this problem and I found a typo. So I'm going to go through the solution I had partly correct. And then we'll add the other party different chance during this. Sorry. Wrote all this up So and what? We have years. We have an electron being protection and tube is going through. This tube hits this acceleration grid, which I've drawn is like a net. I don't know how to draw tube accurately. So this is my attempt. The yellow dot is the electron is a massive 9.11 times, 10th inning, 31 kilograms when it hits the net, which I've drawn a little gold X right there to illustrate that when it does, that is a final Boston green of three times 10 to 6 meters per second. The length of the two is one point in centimeters. First, we'll find out with the exploration of the particle is well that we know that the we know that we have the final velocity squared minus the initial velocity squared to celebration. Next times the the final distance next, minus the initial well, the neutral distance X zero and the initial velocity zero so 00 is there? Um, X was points 1.8 centimeters, which is 0.18 meters. God be careful that we're doing everything in meters stood consistency and then the final monsters three times to the six thought just gets squared. So you can see Here is I've just plug it all the numbers into this equation with a little bit of converting their for the cinema years. And then zeros are going to do anything. So I end up with nine times, tend to 12th. He acne and redo part of this year. Well, I talked to makes more sense do you do to do we have 90% 12 physical, 2.36 It was a necessary should mix, So he's divide both sides. But when I'm doing this Ah, you want calculator using concrete eventually. But I don't want to type in, you know, 12 decibels. So I'm gonna read, Write this and more convenient formas 90% of the 12 divided by 3.6 times 10 to the negative too. And in 95 with 3.6 is 2.5 and 10 of the 12 divided by 10:30 a.m. Tuesday 14th and that's what gets answered right here. So this is our exploration, Vex. Now we want all the time until impact. Well, if the exertion is constant, that means the velocity is changing. So you can't just divide the distance by the velocity. So organizes for really here the change next sequel to initial rosti time One after the expiration x squared. We know that the change in X was again 0.18 meters. Initial velocity was zero times time should recover. But look, it's a greater there tens times Transall for plus 1/2 times exploration, which we just found. Does he squared? Um, zero time. That's gonna be zero. Well, that is a bad zero. When is it all that stuff? Right? The head there, there, that it goes to zero. Okay, so and I'm gonna multiply for everything by to how I want. It is the best. I guess we could rewrite what we have first. So put 018 And again, let's stick to scientific notation. So 1.8 times 10 to the negative too is equal to 1.25 times 10 to the 14th T squared dividing both sides. I have 1.8 times 10 to the negative two divided by 1.25 times 10 to the 14th physical T squared. This becomes all right. I'm a cochlear for second because it's kind of mess to do 1.8 divided by 1.25 people to 1.44 times 10 to the negative. You have to minus 14. Negative 16. It's equal to he squared. Then I'm going to square root everything. So see the square root of 1.44 is 1.2 Nice. 1.2 times the square root tendency of 16th or less The same is divided by two sides. Just tend to the negative eight. And that is gonna be seconds. Is it going? T. All right, so that is done. And then all that's left was last question. Last question was net force acting a particle. We know the acceleration and we know that the masses 9 11 to 10 in a 31 kilograms forces able to masters acceleration. So I know that the force is equal to the mass times The acceleration I just plug doesn't here already, cause, like I said this earlier, didn't really be writing it. Number 11 times 2.5 is 20.775 Um, added extras. We could tend to the negative 17th. I want to shift this decimal over one place, so that's gonna make that 10 to the negative 16th. And that makes the 2.27775 which I could just round a 2.28 Because that it seems every day, uh, to dust bowl, please. It seems to be the preferred choice in the textbook, and that is the total four. So in Newton's include units. So Nunes for that seconds for time, meters per second, squared for acceleration. We're done.

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