/*! 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 1 A baseball has a mass of \(0.20 ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 1

A baseball has a mass of \(0.20 \mathrm{~kg}\). What is the kinetic, energy relative to home plate of a \(160 \mathrm{~km} / \mathrm{hr}\) fastball, in \(\mathrm{kJ}\) ?

Short Answer

Expert verified
The kinetic energy is approximately 0.198 kJ.

Step by step solution

01

- Convert Speed to Meters per Second

Firstly, the speed needs to be converted from \( \text{km/hr} \) to \( \text{m/s} \). Use the conversion factor: \( 1 \text{ km/hr} = \frac{1}{3.6} \text{ m/s} \). Therefore, \( 160 \text{ km/hr} = 160 \times \frac{1}{3.6} \text{ m/s} = 44.44 \text{ m/s} \).
02

- Write the Formula for Kinetic Energy

The kinetic energy (\text{KE}) can be calculated using the formula: \[ KE = \frac{1}{2}mv^2 \], where \( m \) is the mass and \( v \) is the velocity.
03

- Substitute the Values and Perform the Calculation

Substituting the given values into the formula: \[ KE = \frac{1}{2} \times 0.20 \times (44.44)^2 = \frac{1}{2} \times 0.20 \times 1975.14 = 197.514 \text{ J} \].
04

- Convert Joules to KiloJoules

Finally, convert the energy from joules to kilojoules by dividing by 1000: \[ 197.514 \text{ J} = \frac{197.514}{1000} \text{ kJ} = 0.197514 \text{ kJ} \].

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Unit Conversion
Unit conversion is a fundamental skill in physics that allows us to express measurements in different units. This is often necessary when dealing with different physical quantities or even within the same quantity but different unit systems.
In our problem, we need to convert speed from kilometers per hour (\text{km/hr}) to meters per second (\text{m/s}).
To do this, we use the conversion factor:
1 \text{ km/hr} = \frac{1}{3.6} \text{ m/s}.
Thus, converting 160 \text{ km/hr} results in:
\[\begin{equation} 160 \times \frac{1}{3.6} = 44.44 \text{ m/s} \end{equation}\]
This conversion is important to ensure we use consistent units in the kinetic energy formula.
Kinetic Energy Formula
The kinetic energy (\text{KE}) of an object is the energy it possesses due to its motion. The formula to calculate kinetic energy is:
\[\begin{equation} KE = \frac{1}{2}mv^2 \end{equation}\]
Here,
\begin{itemize}
  • | \textbf{m} | is the mass of the object.

  • | \textbf{v} | is the velocity of the object.
    • This formula shows that kinetic energy depends on both the mass and the square of the velocity of the object.
    Remember to have the mass in kilograms (kg) and velocity in meters per second (\text{m/s}) to get the kinetic energy in joules (J).
    Velocity
    Velocity is the speed of an object in a specified direction.
    When calculating kinetic energy, it is crucial that velocity is measured in meters per second (\text{m/s}) to maintain consistency within the SI (International System of Units).
    From the problem, the given velocity was 160 \text{km/hr}.
    Using the conversion factor, we convert this to:
    \[\begin{equation} 44.44 \text{ m/s} \end{equation}\]
    This ensures accurate calculation when applying the kinetic energy formula.
    Mass
    Mass is a measure of the amount of matter in an object and is usually measured in kilograms (kg) for physics calculations.
    For kinetic energy calculations, it's essential to keep the mass in kilolibrary of SI unit.
    In our exercise, the baseball has a mass of 0.20 kg.
    We use this value directly in our kinetic energy formula:
    \[\begin{equation} KE = \frac{1}{2} \times 0.20 \times (44.44)^2 \end{equation}\]
    This contributes to the final kinetic energy result.

    One App. One Place for Learning.

    All the tools & learning materials you need for study success - in one app.

    Get started for free

    Most popular questions from this chapter

    \(.\) A block of mass \(10 \mathrm{~kg}\) moves along a surface inclined \(30^{\circ}\) relative to the horizontal. The center of gravity of the block. is elevated by \(3.0 \mathrm{~m}\) and the kinetic energy of the block decreases by \(50 \mathrm{~J}\). The block is acted upon by a constant force \(\mathbf{R}\) parallel to the incline, and by the force of gravity. Assume frictionless surfaces and let \(g=9.81 \mathrm{~m} / \mathrm{s}^{2}\). Determine the magnitude and direction of the constant force \(\mathbf{R}\), in \(\mathrm{N}\).

    Experimental molecular motors are reported to exhibit movement upon the absorption of light, thereby achieving a conversion of electromagnetic radiation into motion. Should the incident light be considered work or heat transfer?

    Steam in a piston-cylinder assembly undergoes a polytropic process, with \(n=2\), from an initial state where \(p_{1}=3.45 \mathrm{MPa}, v_{1}=0.106 \mathrm{~m}^{3} / \mathrm{kg}, u_{1}=3,171.1 \mathrm{~kJ} / \mathrm{kg}\), to a final state where \(u_{2}=2,303.9 \mathrm{~kJ} / \mathrm{kg}\). During the process, there is a heat transfer from the steam of magnitude \(361.76 \mathrm{~kJ}\). The mass of steam is \(0.54 \mathrm{~kg}\). Neglecting changes in kinetic and potential energy, determine the work, in \(\mathrm{kJ}\), and the final specific volume, in \(\mathrm{m}^{3} / \mathrm{kg}\).

    An object whose mass is \(2 \mathrm{~kg}\) is accelerated from a velocity of \(200 \mathrm{~m} / \mathrm{s}\) to a final velocity of \(500 \mathrm{~m} / \mathrm{s}\) by the action of a resultant force \(\mathbf{F}\). Determine the work done by the resultant force, in \(\mathrm{kJ}\), if there are no other interactions between the object and its surroundings.

    The two major forces opposing the motion of a vehicle moving on a level road are the rolling resistance of the tires, \(F_{t}\), and the aerodynamic drag force of the air flowing around the vehicle, \(F_{\mathrm{d}}\), given respectively by $$ F_{r}=f m g, \quad F_{d}=C_{d} \mathrm{~A}_{2} \rho \mathrm{V}^{2} $$ where \(f\) and \(C_{\mathrm{d}}\) are constants known as the rolling resistance coefficient and drag coefficient, respectively, \(\mathrm{m}\) and \(\mathrm{A}\) are the vehicle mass and projected frontal area, respectively, \(\mathrm{V}\) is the vehicle velocity, and \(\rho\) is the air density. For a passenger car with \(\mathrm{m}=1,610 \mathrm{~kg}, \mathrm{~A}=2.2 \mathrm{~m}^{2}\), and \(C_{\mathrm{d}}=0.34\), and when \(f=0.02\) and \(\rho=1.28 \mathrm{~kg} / \mathrm{m}^{3}\) (a) determine the power required, in \(\mathrm{kW}\), to overcome rolling resistance and aerodynamic drag when \(\mathrm{V}\) is \(88.5\) \(\mathrm{km} / \mathrm{hr}\)\ (b) plot versus vehicle velocity ranging from 0 to \(120.7 \mathrm{~km} / \mathrm{hr}\) (i) the power to overcome rolling resistance, (ii) the power to overcome aerodynamic drag, and (iii) the total power, all in \(\mathrm{kW}\). What implication for vehicle fuel economy can be deduced from the results of part (b)?
    See all solutions

    Recommended explanations on Physics Textbooks

    Work Energy and Power

    Read Explanation

    Mechanics and Materials

    Read Explanation

    Fluids

    Read Explanation

    Further Mechanics and Thermal Physics

    Read Explanation

    Kinematics Physics

    Read Explanation

    Circular Motion and Gravitation

    Read Explanation
    View all explanations

    What do you think about this solution?

    We value your feedback to improve our textbook solutions.

    Study anywhere. Anytime. Across all devices.