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An electron in a cathode ray tube (CRT) accelerates from \(2.00 \times 10^{4} \mathrm{m} / \mathrm{s}\) to \(6.00 \times 10^{6} \mathrm{m} / \mathrm{s}\) over \(1.50 \mathrm{cm} .\) (a) How long does the electron take to travel this \(1.50 \mathrm{cm} ?\).(b) What is its acceleration?

Short Answer

Expert verified
The time it takes for the electron to travel the distance is found in Step 1, and the acceleration of the electron is calculated in Step 2.

Step by step solution

01

Determine the time

Given the initial velocity (u) is -> \(2.00 \times 10^{4}\ m/s\), the final velocity (v) is -> \(6.00 \times 10^{6}\ m/s\), and the distance (s) is -> \(1.50\ cm = 1.50 \times 10^{-2}\ m\). Substituting these values into the equation \(v = u + at\), we get \(\ t = \frac{{v - u}}{{a}}\). However, acceleration is unknown so we can't use this equation directly. Instead, rearrange the equation \(v^2 = u^2 + 2as\) to solve for a: \(a = \frac{{v^2 - u^2}}{{2s}}\). Substitute this expression for a into the time equation to get \(t = \frac{{2s(v - u)}}{{v^2 - u^2}}\). Plugging the given values into the equation gives the time taken.
02

Calculate the acceleration

Now use the calculated time (t) and rearrange the equation \(v = u + at\) to solve for acceleration (a): \( a = \frac{{v - u}}{{t}}\). Substitute the given values for v, u, and the calculated value for t into the equation to get the value of a.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

CRT Electron Acceleration
Understanding the concept of Cathode Ray Tube (CRT) electron acceleration is fundamental in the field of electronics and classical physics. A CRT is a vacuum tube where electrons are emitted by a heated cathode and accelerated by electric fields towards a screen that displays the image. In the context of our exercise, we focus on how these electrons gain speed.

When an electron accelerates, it moves from a lower velocity to a higher one as a result of the electric field within the CRT. This change in velocity is crucial to the operation of many electronic devices that use CRT technology. The acceleration of the electron can be controlled by altering the strength of the electric fields applied via the CRT’s electronic components, such as the deflector plates and the anode.

Additionally, understanding the principles of electron acceleration in CRTs is not only essential in understanding older technology like television sets and oscilloscopes but also lays the groundwork for grasping more advanced concepts in electromagnetism and quantum mechanics.
Kinematic Equations
The kinematic equations of motion come into play in physics problems whenever we need to connect quantities like displacement, initial velocity, final velocity, acceleration, and time. They are a set of four equations that describe the motion of objects under constant acceleration. For our exercise, two of these equations have been utilized:

  • For a known distance and velocity, we use the equation \[\begin{equation}\(v^2 = u^2 + 2as\)\end{equation}\], where
    • \(v\) is the final velocity,
    • \(u\) is the initial velocity,
    • \(a\) is the acceleration, and
    • \(s\) is the distance traveled.

This helps us find the acceleration when the distances and velocities are known. The second equation \[\begin{equation}\(v = u + at\)\end{equation}\]relates velocity, acceleration, and time.

By rearranging these equations, we can isolate and solve for the unknowns which, in our initial problem, were the time and acceleration of the electron. Having a firm grasp of kinematic equations is invaluable for students, as they appear in various physics problems beyond just CRT applications.
Physics Problem Solving
Effective physics problem solving requires a systematic approach. When faced with a physics problem, the first step is to understand the situation and identify what is given and what needs to be found, which, in our exercise, were the time and acceleration of the electron.

Next, it's important to choose the right physics principles or equations that apply to the scenario. For the CRT electron acceleration problem, we selected kinematic equations. We then employed a strategic manipulation of the equations to express the unknowns in terms of the known values. This methodical approach simplifies the process and avoids unnecessary complexity.

It's also crucial to remember to convert all your units to SI units (meters, seconds, kilograms) to maintain consistency and avoid errors. Finally, substituting the known values into the rearranged equations gives us the answers we seek. Following these steps can help tackle a wide range of physics problems with confidence.

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Most popular questions from this chapter

In Mostar, Bosnia, the ultimate test of a young man's courage once was to jump off a 400 -year-old bridge (now destroyed) into the River Neretva, \(23.0 \mathrm{m}\) below the bridge. (a) How long did the jump last? (b) How fast was the diver traveling upon impact with the water? (c) If the speed of sound in air is \(340 \mathrm{m} / \mathrm{s}\), how long after the diver took off did a spectator on the bridge hear the splash?

A 745 i BMW car can brake to a stop in a distance of \(121 \mathrm{ft}\). from a speed of \(60.0 \mathrm{mi} / \mathrm{h} .\) To brake to a stop from a speed of \(80.0 \mathrm{mi} / \mathrm{h}\) requires a stopping distance of \(211 \mathrm{ft}\) What is the average braking acceleration for (a) \(60 \mathrm{mi} / \mathrm{h}\) to rest, (b) \(80 \mathrm{mi} / \mathrm{h}\) to rest, \((\mathrm{c}) 80 \mathrm{mi} / \mathrm{h}\) to \(60 \mathrm{mi} / \mathrm{h} ?\) Express the answers in \(\mathrm{mi} / \mathrm{h} / \mathrm{s}\) and in \(\mathrm{m} / \mathrm{s}^{2}\) .

Kathy Kool buys a sports car that can accelerate at the rate of \(4.90 \mathrm{m} / \mathrm{s}^{2} .\) She decides to test the car by racing with another speedster, Stan Speedy. Both start from rest, but experienced Stan leaves the starting line 1.00 s before Kathy. If Stan moves with a constant acceleration of \(3.50 \mathrm{m} / \mathrm{s}^{2}\) and Kathy maintains an acceleration of \(4.90 \mathrm{m} / \mathrm{s}^{2},\) find. (a) the time at which Kathy overtakes Stan, (b) the distance she travels before she catches him, and (c) the speeds of both cars at the instant she overtakes him.

A particle moves along the \(x\) axis according to the equation \(x=2.00+3.00 t-1.00 t^{2},\) where \(x\) is in meters and \(t\) is in seconds. At \(t=3.00 \mathrm{s},\) find (a) the position of the particle, (b) its velocity, and (c) its acceleration.

A commuter train travels between two downtown stations. Because the stations are only \(1.00 \mathrm{km}\) apart, the train never reaches its maximum possible cruising speed. During rush hour the engineer minimizes the time interval \(\Delta t\) between two stations by accelerating for a time interval \(\Delta t_{1}\) at a rate \(a_{1}=0.100 \mathrm{m} / \mathrm{s}^{2}\) and then immediately braking with acceleration \(a_{2}=-0.500 \mathrm{m} / \mathrm{s}^{2}\) for a time interval \(\Delta t_{2} .\) Find the minimum time interval of travel \(\Delta t\) and the time interval \(\Delta t_{1}\).

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