Physics: Principles with Applications
Physics: Principles with Applications
6th Edition
ISBN: 9780130606204
Author: Douglas C. Giancoli
Publisher: Prentice Hall
bartleby

Concept explainers

bartleby

Videos

Textbook Question
Book Icon
Chapter 3, Problem 10P

(a) given the vectors A and B shown in Fig. 3-35, determine B A . (b) Determine A B without using your answer in (a) then compare your results and see if they are opposite.

Expert Solution & Answer
Check Mark
Solution

To Determine:

  1. BA
  2. AB without using results obtained in (a)comparingresults.

Solution:

  1. The vector BA has a magnitude of 53.6 with a direction of 178.5o counterclockwise from +x-axis.
  2. The vector AB has a magnitude of 53.6 with a direction of 358.5o counterclockwise from +x-axis. Comparing vector BA with vector AB we can observe that both vectors have the same magnitude both points in opposite directions.

Explanation:

Given: The magnitude |A| is 44.0 with a direction θ'A of 28.0˚ counterclockwise from the positive x -axis. For |B| the magnitude is 26.5with a direction θ'B of 56.0˚ clockwise from the negative x -axis

Formula used:

The x -component Vx of vector V is defined as:

  Vx=|V|cosθ

The y-component of Vy vector V is defined as:

  Vy=|V|sinθ

Where:

  • |V| is the magnitude of the vector V
  • θ is the direction of the vector  V measured respect to the positive x -axis counter-clockwise.

The magnitude of vector can be obtained by:

  |V|=(Vx2)+( V y )2

And, the direction θ is defined as:

  θ=tan1(VyVx)

Calculation:Section 1: x -component of A

  Ax=44cos28o=38.8

y -component of A

  Ay=44sin28o=20.6

x -component of B

  Bx=26.5cos( 180o 56o)=14.8

y -component of B

  By=26.5sin( 180o 56o)=22

Therefore, the x-component of BA is:

  BxAx=14.838.8=53.6

Similarly, the y-component of BA is:

  ByAy=2220.6=1.4

Therefore, the magnitude of BA is:

  |BA|= ( 53.6 )2+ ( 1.4 )2=53.6

  θ=tan1( 1.4 53.6)=1.5o=180o1.5o=178.5o

Therefore, the vector BA has a magnitude of 53.6 with a direction of 178.5˚ counterclockwise from + x -axis.

Section 2: The x-component of AB is:

  AxBx=38.8(14.8)=53.6

Similarly, the y-component of AB is:

  AyBy=20.622=1.4

Therefore, the magnitude of AB is:

  |AB|= ( 53.6 )2+ ( 1.4 )2=53.6

  θ=tan1( 1.4 53.6)=1.5o=360o1.5o=358.5o

Conclusion:

The vector BA has a magnitude of 53.6 with a direction of 178.5o counterclockwise from +x-axis.The vector AB has a magnitude of 53.6 with a direction of 358.5o counterclockwise from +x-axis. Comparing vector BA with vector AB we can observe that both vectors have the same magnitude but point in opposite directions.

11:44

Chapter 3 Solutions

Physics: Principles with Applications

Ch. 3 - Prob. 11QCh. 3 - Prob. 12QCh. 3 - Prob. 13QCh. 3 - A projectile is launched at an upward angle of 300...Ch. 3 - Prob. 15QCh. 3 - Two cannonballs, A and B, are fired from the...Ch. 3 - 18. A person sitting in an enclosed train car,...Ch. 3 - Prob. 18QCh. 3 - Prob. 19QCh. 3 - Prob. 20QCh. 3 - A car is driven 225 km west and then 98 km...Ch. 3 - A delivery truck travels 21 blocks north, 16...Ch. 3 - If Vx=9.80 units and Vy=6.40 units, determine the...Ch. 3 - Graphically determine the resultant of the...Ch. 3 - V is a vector 24.8 units in magnitude and points...Ch. 3 - Vector V is 6.6 using long and points along the...Ch. 3 - Figure 3-33 shows two vectors, A and B , whose...Ch. 3 - Prob. 8PCh. 3 - Three vectors are shown in Fig. 3-35 Q. Their...Ch. 3 - (a) given the vectors A and B shown in Fig. 3-35,...Ch. 3 - Determine the vector AC , given the vectors A and...Ch. 3 - For the vectors shown in Fig. 3—35, determine (a)...Ch. 3 - Prob. 13PCh. 3 - Prob. 14PCh. 3 - Prob. 15PCh. 3 - Prob. 16PCh. 3 - Prob. 17PCh. 3 - 17. (l) A tiger leaps horizontally from a...Ch. 3 - 18. (l) A diver running 2.5 m/s dives out...Ch. 3 - Prob. 20PCh. 3 - Prob. 21PCh. 3 - Prob. 22PCh. 3 - Prob. 23PCh. 3 - Prob. 24PCh. 3 - Prob. 25PCh. 3 - Prob. 26PCh. 3 - Prob. 27PCh. 3 - Prob. 28PCh. 3 - Prob. 29PCh. 3 - Prob. 30PCh. 3 - Prob. 31PCh. 3 - Prob. 32PCh. 3 - Prob. 33PCh. 3 - Prob. 34PCh. 3 - Prob. 35PCh. 3 - Prob. 36PCh. 3 - Prob. 37PCh. 3 - Prob. 38PCh. 3 - Prob. 39PCh. 3 - Prob. 40PCh. 3 - Prob. 41PCh. 3 - Prob. 42PCh. 3 - Prob. 43PCh. 3 - Prob. 44PCh. 3 - Prob. 45PCh. 3 - Prob. 46PCh. 3 - Prob. 47PCh. 3 - Prob. 48PCh. 3 - Prob. 49PCh. 3 - Prob. 50PCh. 3 - Prob. 51PCh. 3 - Prob. 52PCh. 3 - Prob. 53GPCh. 3 - Prob. 54GPCh. 3 - Prob. 55GPCh. 3 - Prob. 56GPCh. 3 - Prob. 57GPCh. 3 - Prob. 58GPCh. 3 - Prob. 59GPCh. 3 - Prob. 60GPCh. 3 - Prob. 61GPCh. 3 - Prob. 62GPCh. 3 - Prob. 63GPCh. 3 - Prob. 64GPCh. 3 - Prob. 65GPCh. 3 - Prob. 66GPCh. 3 - Prob. 67GPCh. 3 - Prob. 68GPCh. 3 - Prob. 69GPCh. 3 - Prob. 70GPCh. 3 - Prob. 71GPCh. 3 - Prob. 72GPCh. 3 - Prob. 73GPCh. 3 - Prob. 74GPCh. 3 - Prob. 75GP
Knowledge Booster
Background pattern image
Physics
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Text book image
College Physics
Physics
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Cengage Learning
Text book image
University Physics (14th Edition)
Physics
ISBN:9780133969290
Author:Hugh D. Young, Roger A. Freedman
Publisher:PEARSON
Text book image
Introduction To Quantum Mechanics
Physics
ISBN:9781107189638
Author:Griffiths, David J., Schroeter, Darrell F.
Publisher:Cambridge University Press
Text book image
Physics for Scientists and Engineers
Physics
ISBN:9781337553278
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
Lecture- Tutorials for Introductory Astronomy
Physics
ISBN:9780321820464
Author:Edward E. Prather, Tim P. Slater, Jeff P. Adams, Gina Brissenden
Publisher:Addison-Wesley
Text book image
College Physics: A Strategic Approach (4th Editio...
Physics
ISBN:9780134609034
Author:Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field
Publisher:PEARSON
Position/Velocity/Acceleration Part 1: Definitions; Author: Professor Dave explains;https://www.youtube.com/watch?v=4dCrkp8qgLU;License: Standard YouTube License, CC-BY