/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 16 An electron \(\left(\mathrm{mass... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 16

An electron \(\left(\mathrm{mass}=911 \times 10^{-31} \mathrm{kg}\right)\) 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; (b) the time to reach the grid; (c) the net force, in newtons. (You can ignore the gravitational force on the electron.)

Short Answer

Expert verified
(a) Acceleration is \(2.50 \times 10^{14} \, \text{m/s}^2\), (b) Time is \(1.20 \times 10^{-8} \, \text{s}\), (c) Force is \(2.28 \times 10^{-16} \, \text{N}\).

Step by step solution

01

Identify the known values

We start by identifying the given values:- Initial speed of the electron, \( u = 0 \, \text{m/s} \).- Final speed of the electron, \( v = 3.00 \times 10^6 \, \text{m/s} \).- Distance to the grid, \( s = 1.80 \, \text{cm} = 0.018 \, \text{m} \).- Mass of the electron, \( m = 9.11 \times 10^{-31} \, \text{kg} \).We also know that the gravitational force on the electron is negligible.
02

Calculate the acceleration

Use the kinematic equation to solve for acceleration \( a \):\[v^2 = u^2 + 2as\]Substitute the known values:\[(3.00 \times 10^6)^2 = 0^2 + 2a(0.018)\]\[9.00 \times 10^{12} = 0.036a\]Solve for \( a \):\[a = \frac{9.00 \times 10^{12}}{0.036} = 2.50 \times 10^{14} \, \text{m/s}^2\]
03

Calculate the time to reach the grid

Use the kinematic equation:\[v = u + at\]Solve for time \( t \):\[3.00 \times 10^6 = 0 + (2.50 \times 10^{14})t\]\[t = \frac{3.00 \times 10^6}{2.50 \times 10^{14}} = 1.20 \times 10^{-8} \, \text{s}\]
04

Calculate the net force acting on the electron

Use Newton's second law of motion:\[F = ma\]Substitute the known values:\[F = (9.11 \times 10^{-31} \, \text{kg})(2.50 \times 10^{14} \, \text{m/s}^2)\]\[F = 2.28 \times 10^{-16} \, \text{N}\]

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Key Concepts

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

acceleration calculation
In physics, acceleration is a measure of how fast an object changes its velocity. In this exercise, we need to find the acceleration of the electron as it travels towards the accelerating grid. Since we were provided with the initial velocity (\(u = 0 \text{ m/s}\), the final velocity (\(v = 3.00 \times 10^6 \text{ m/s}\), and the distance (\(s = 0.018 \text{ m}\), we can use a kinematic equation to find the electron's acceleration.
  • The kinematic formula connecting these quantities is \[v^2 = u^2 + 2as\]. This equation helps us determine the acceleration when the initial velocity, final velocity, and distance are known.
  • Given that the initial speed is zero, simplifying our calculation: \[v^2 = 2as\].
  • Next, substitute the given values: \[(3.00 \times 10^6)^2 = 2a(0.018)\].
  • Solving this equation gives us the acceleration: \[a = 2.50 \times 10^{14} \text{ m/s}^2\].
This calculation of acceleration is crucial in understanding how forces influence velocity changes, especially in particle physics.
Newton's second law
Newton's second law of motion provides the relationship between an object's mass, the force applied to it, and its acceleration. This law is represented by the equation \(F = ma\). In this exercise, the same law helps us ascertain the net force acting on the electron.
  • We already calculated the acceleration of the electron as \(2.50 \times 10^{14} \text{ m/s}^2\).
  • The mass of the electron is a constant, known to be \(9.11 \times 10^{-31} \text{ kg}\).
  • By substituting these values into the equation \(F = ma\), we find the force: \[F = (9.11 \times 10^{-31} \text{ kg})(2.50 \times 10^{14} \text{ m/s}^2) = 2.28 \times 10^{-16} \text{ N}\].
This force calculation reveals how even tiny forces can lead to significant accelerations due to the small mass of electrons, an essential principle in understanding electron dynamics.
electron dynamics
Electron dynamics refers to the study of the movement and behavior of electrons, especially when subject to forces like electric fields. In a television tube, this principle is crucial as electrons are accelerated to create images on the screen.
  • Initially, the electron is at rest, but once a force is applied, it accelerates. This acceleration is found to be \(2.50 \times 10^{14} \text{ m/s}^2\), showcasing how electrons can reach high speeds swiftly.
  • Using the dynamics principles, the time it takes for the electron to reach the grid can be calculated. This involves rearranging the kinematic equation: \[v = u + at\]. With known values, it becomes \[t = \frac{3.00 \times 10^6}{2.50 \times 10^{14}} = 1.20 \times 10^{-8} \text{ s}\].
  • Observing such dynamics explains not only the electron's journey in displays but also their fundamental behaviors in circuits and other technologies.
Understanding electron dynamics is crucial in many fields, from developing better electronic devices to gaining insights into quantum mechanics.

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Most popular questions from this chapter

A hockey puck with mass 0.160 \(\mathrm{kg}\) is at rest at the origin \((x=0)\) on the horizontal, frictionless surface of the rink. At time \(t=0\) a player applies a force of 0.250 \(\mathrm{N}\) to the puck, parallel to the \(x\) -axis; he continues to apply this force until \(t=2.00 \mathrm{s}\) . (a) What are the position and speed of the puck at \(t=2.00 \mathrm{s}\) ? (b) If the same force is again applied at \(t=5.00 \mathrm{s},\) what are the position and speed of the puck at \(t=7.00 \mathrm{s} ?\)

A small 8.00 -kg rocket burns fuel that exerts a time-varying upward force on the rocket as the rocket moves upward from the launch pad. This force obeys the equation \(F=A+B t^{2}\) . Measurements show that at \(t=0,\) the force is \(100.0 \mathrm{N},\) and at the end of the first 2.00 s, it is 150.0 \(\mathrm{N}\) . (a) Find the constants \(A\) and \(B\) , including their SI units. (b) Find the net force on this rocket and its acceleration (i) the instant after the fuel ignites and (ii) 3.00 s after fuel ignition. (c) Suppose you were using this rocket in outer space, far from all gravity. What would its acceleration be 3.00 s after fuel ignition?

At the surface of Jupiter's moon Io, the acceleration due to gravity is \(g=1.81 \mathrm{m} / \mathrm{s}^{2} .\) A watermelon weighs 44.0 \(\mathrm{N}\) at the surface of the earth. (a) What is the watermelon's mass on the earth's surface? (b) What are its mass and weight on the surface of Io?

A 6.50 -kg instrument is hanging by a vertical wire inside a space ship that is blasting off at the surface of the earth. This ship starts from rest and reaches an altitude of 276 \(\mathrm{m}\) in 15.0 \(\mathrm{s}\) with constant acceleration. (a) Draw a free-body diagram for the instrument during this time. Indicate which force is greater. (b) Find the force that the wire exerts on the instrument.

An astronaut is tethered by a strong cable to a spacecraft. The astronaut and her spacesuit have a total mass of 105 \(\mathrm{kg}\) , while the mass of the cable is negligible. The mass of the spacecraft is \(9.05 \times 10^{4} \mathrm{kg} .\) The spacecraft is far from any large astronomical bodies, so we can ignore the gravitational forces on it and the astronaut. We also assume that both the spacecraft and the astronaut are initially at rest in an inertial reference frame. The astronaut then pulls on the cable with a force of 80.0 \(\mathrm{N}\) (a) What force does the cable exert on the astronaut? (b) since \(\Sigma \vec{\boldsymbol{F}}=m \vec{\boldsymbol{a}}\) , how can a "massless" \((m=0)\) cable exert a force? (c) What is the astronaut's acceleration? (d) What force does the cable exert on the spacecraft? (e) What is the acceleration of the spacecraft?
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