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Chapter 2: Problem 40

In a cathode ray tube in an old TV, electrons are accelerated from rest with a constant acceleration of magnitude $7.03 \times 10^{13} \mathrm{m} / \mathrm{s}^{2}\( during the first \)2.0 \mathrm{cm}$ of the tube's length; then they move at essentially constant velocity another \(45 \mathrm{cm}\) before hitting the screen. (a) Find the speed of the electrons when they hit the screen. (b) How long does it take them to travel the length of the tube?

Short Answer

Expert verified
Answer: The final speed of the electrons when they hit the screen is \(1.675 \times 10^6 \mathrm{m} / \mathrm{s}\), and the total time it takes to travel the length of the tube is approximately \(2.926 \times 10^{-7} \mathrm{s}\).

Step by step solution

01

Identify given values and convert units

In the given problem, the initial velocity (u) is 0, the acceleration (a) is \(7.03 \times 10^{13} \) m/s², and the distance (s) is 2.0 cm which we need to convert into meters. \(s = 2.0 \times 10^{-2}\) m
02

Use the equation of motion to find the final velocity (v) after acceleration

We will use the second equation of motion \(v^{2} = u^{2} + 2as\), where u is the initial velocity (0) and s is the distance during acceleration. \(v^{2} = u^{2} + 2as\) \(v^{2} = 0 + (2)(7.03 \times 10^{13} \mathrm{m} / \mathrm{s}^{2})(2 \times 10^{-2} \mathrm{m})\) \(v^{2} = 2.806 \times 10^{12} \mathrm{m^2} / \mathrm{s^2}\) \(v = \sqrt{2.806 \times 10^{12} \mathrm{m^2} / \mathrm{s^2}}\) \(v = 1.675 \times 10^6 \mathrm{m} / \mathrm{s}\)
03

Calculate time to travel the first 2.0 cm of the tube

We will use the first equation of motion: \(v = u + at\). Since initial velocity is 0, we can rewrite the equation as follows: \(v = at\) \(t_1 = \frac{v}{a}\) \(t_1 = \frac{1.675 \times 10^6 \mathrm{m} / \mathrm{s}}{7.03 \times 10^{13} \mathrm{m} / \mathrm{s}^{2}}\) \(t_1 \approx 2.384 \times 10^{-8} \mathrm{s}\)
04

Find the distance remaining in the tube

The total length of the tube is given as 47 cm, and 2.0 cm is already traveled. The remaining distance is 45 cm, which we should convert to meters. \(45 \mathrm{cm} = 0.45 \mathrm{m}\)
05

Calculate time to travel the remaining 45 cm

In this part of the tube, the electrons travel at a constant velocity (v from Step 2). We can use the following equation: \(t_2 = \frac{s}{v}\) \(t_2 = \frac{0.45 \mathrm{m}}{1.675 \times 10^6 \mathrm{m} / \mathrm{s}}\) \(t_2 \approx 2.687 \times 10^{-7} \mathrm{s}\)
06

Find the total time to travel the entire tube

Now, we'll simply add the time it takes to travel each part of the tube (from Steps 3 and 5). \(t = t_1 + t_2\) \(t = (2.384 \times 10^{-8} \mathrm{s}) + (2.687 \times 10^{-7} \mathrm{s})\) \(t \approx 2.926 \times 10^{-7} \mathrm{s}\) We have now found everything we need: (a) The final speed of the electrons when they hit the screen is \(1.675 \times 10^6 \mathrm{m} / \mathrm{s}\). (b) The total time it takes to travel the length of the tube is about \(2.926 \times 10^{-7} \mathrm{s}\).

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