Physics
Physics
5th Edition
ISBN: 9781260486919
Author: GIAMBATTISTA
Publisher: MCG
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 3, Problem 30P

(a)

To determine

The x and y component of B.

(a)

Expert Solution
Check Mark

Answer to Problem 30P

The x component of B is 6.9 and y component of B is 1.7.

Explanation of Solution

Since angle is below x axis, take it as negative.

Write the expression for x component of B

Bx=Bcosθ (I)

Here, Bx is the x component of B , B is the magnitude of B and θ is the angle that B makes below the x axis .

Write the expression for y x component of B

By=Bsinθ (II)

Here, By is the y component of B and θ is the angle that B makes below the x axis.

Conclusion:

Substitute 14° for θ and 7.1 for B in equation (I) to get Bx.

Bx=7.1cos(14°)=6.9

Substitute 14° for θ and 7.1 for B in equation (II) to get By.

By=7.1sin(14°)=1.7

Negative sign for By indicate that it lies along negative y axis.

Therefore, the x component of B is 6.9 and y component of B is 1.7.

(b)

To determine

The magnitude and direction of C.

(b)

Expert Solution
Check Mark

Answer to Problem 30P

The magnitude of C is 6.9 and it is 225° counterclockwise from the +x axis.

Explanation of Solution

Write the expression for the magnitude of C.

C=Cx2+Cy2 (III)

Here, C is the magnitude of vector C , Cx is the x component of C and Cy is the y component of C.

Write the expression for direction of C.

θc=tan1(CyCx) (IV)

Here, θc is the angle than C makes with x axis.

Conclusion:

Substitute 6.7 for Cy and 1.8 for Cx in equation (III) to get C.

C=(1.8)2+(6.7)2=6.9

Substitute 6.7 for Cy and 1.8 for Cx in equation (IV) to get θc.

θc=tan1(6.71.8)=75°

The x component and y component are negative. Therefore, C lies at 180°+75°=255° counterclockwise from the x axis.

Therefore, the magnitude of C is 6.9 and it is 225° counterclockwise from the +x axis.

(c)

To determine

The magnitude and direction of C+B.

(c)

Expert Solution
Check Mark

Answer to Problem 30P

The magnitude of C+B is 9.8 and it is directed 59° clockwise from the +x axis.

Explanation of Solution

Write the expression for the magnitude of C+B.

|C+B|=(Cx+Bx)2+(Cy+By)2 (V)

Here, |C+B| is the magnitude of vector C+B , Cx is the x component of C and Cy is the y component of C.

Write the expression for direction of C+B.

θC+B=tan1(Cy+ByCx+Bx) (VI)

Here, θC+B is the angle than C+B makes with +x axis.

Conclusion:

Substitute 6.9 for Bx, 1.7 for By, 6.7 for Cy and 1.8 for Cx in equation (V) to get |C+B|

|C+B|=(1.8+6.9)2+(6.7+1.7)2=9.8

Substitute 6.9 for Bx, 1.7 for By,6.7 for Cy and 1.8 for Cx in equation (VI) to get θC+B.

θC+B=tan1(6.7+1.71.8+6.9)=59°

Therefore, the magnitude of C+B is 9.8 and it is directed 59° clockwise from the +x axis.

(d)

To determine

The magnitude and direction of CB.

(d)

Expert Solution
Check Mark

Answer to Problem 30P

The magnitude of CB is 10 and it is directed 210° counterclockwise from the +x axis.

Explanation of Solution

Write the expression for the magnitude of CB.

|CB|=(CxBx)2+(CyBy)2 (V)

Here, |CB| is the magnitude of vector CB , Cx is the x component of C and Cy is the y component of C.

Write the expression for direction of CB.

θC-B=tan1(CyByCxBx) (VI)

Here, θCB is the angle than CB makes with +x axis.

Conclusion:

Substitute 6.9 for Bx, 1.7 for By, 6.7 for Cy and 1.8 for Cx in equation (V) to get |C+B|

|C+B|=(1.86.9)2+(6.7(1.7))2=10

Substitute 6.9 for Bx, 1.7 for By,6.7 for Cy and 1.8 for Cx in equation (VI) to get θC+B.

θC+B=tan1(6.7(1.7)1.86.9)=30°

Since x component of CB is negative , it lies at 180°+30°=210° counterclockwise from the +x axis.

Therefore, the magnitude of CB is 10 and it is directed 210° counterclockwise from the +x axis.

(e)

To determine

The x and y components of CB.

(e)

Expert Solution
Check Mark

Answer to Problem 30P

The x components of CB is 8.7 and y component of CB is 5.0.

Explanation of Solution

Write the expression for the x component of CB.

|CB|x=CxBx (VII)

Here, |CB|x is the x component of vector CB.

Write the expression for the y component of CB.

|CB|y=CyBy (VII)

Here, |CB|y is the y component of vector CB.

Conclusion:

Substitute 6.9 for Bx and 1.8 for Cx in equation (VII) to get |C+B|x.

|CB|x=1.86.9=8.7

Substitute 1.7 for By and 6.7 for Cy in equation (VI) to get θC+B.

|CB|y=6.7(1.7)=5.0

Therefore, the x components of CB is 8.7 and y component of CB is 5.0.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
What is the error determined by the 2/3 rule?
Your colleague gives you a sample that are supposed to consist of Pt-Ni nanoparticles, TiO2 nanorod arrays, and SiO2 monolith plates (see right panel schematic). The bimetallic Pt-Ni nanoparticles are expected to decorate on the side surfaces of the aligned TiO2 nanorod arrays. These aligned TiO2 nanoarrays grew on the flat SiO2 monolith. Let's assume that the sizes of the Pt-Ni nanoparticles are > 10 nm. We further assume that you have access to a modern SEM that can produce a probe size as small as 1 nm with a current as high as 1 nA. You are not expected to damage/destroy the sample. Hint: keep your answers concise and to the point. TiO₂ Nanorods SiO, monolith a) What do you plan to do if your colleague wants to know if the Pt and Ni formed uniform alloy nanoparticles? (5 points) b) If your colleague wants to know the spatial distribution of the PtNi nanoparticles with respect to the TiO2 nanoarrays, how do you accomplish such a goal? (5 points) c) Based on the experimental results…
Find the current in 5.00 and 7.00 Ω resistors. Please explain all reasoning

Chapter 3 Solutions

Physics

Ch. 3.5 - Prob. 3.6PPCh. 3.5 - Prob. 3.5ACPCh. 3.5 - Prob. 3.7PPCh. 3.5 - Prob. 3.5BCPCh. 3.6 - Prob. 3.6CPCh. 3.6 - Prob. 3.8PPCh. 3.6 - Prob. 3.9PPCh. 3 - Prob. 1CQCh. 3 - Prob. 2CQCh. 3 - Prob. 3CQCh. 3 - Prob. 4CQCh. 3 - Prob. 5CQCh. 3 - Prob. 6CQCh. 3 - Prob. 7CQCh. 3 - Prob. 8CQCh. 3 - Prob. 9CQCh. 3 - Prob. 10CQCh. 3 - Prob. 11CQCh. 3 - Prob. 12CQCh. 3 - Prob. 13CQCh. 3 - Prob. 14CQCh. 3 - Prob. 15CQCh. 3 - Prob. 1MCQCh. 3 - Prob. 2MCQCh. 3 - 4. A runner moves along a circular track at a...Ch. 3 - Prob. 4MCQCh. 3 - Prob. 5MCQCh. 3 - Prob. 6MCQCh. 3 - Prob. 7MCQCh. 3 - Prob. 8MCQCh. 3 - Prob. 9MCQCh. 3 - Prob. 10MCQCh. 3 - Prob. 11MCQCh. 3 - Prob. 12MCQCh. 3 - Prob. 13MCQCh. 3 - Prob. 14MCQCh. 3 - Prob. 15MCQCh. 3 - Prob. 16MCQCh. 3 - Prob. 1PCh. 3 - Prob. 2PCh. 3 - Prob. 3PCh. 3 - Prob. 4PCh. 3 - Prob. 5PCh. 3 - Prob. 6PCh. 3 - Prob. 7PCh. 3 - Prob. 8PCh. 3 - Prob. 9PCh. 3 - Prob. 10PCh. 3 - Prob. 11PCh. 3 - 12. Michaela is planning a trip in Ireland from...Ch. 3 - Prob. 13PCh. 3 - Prob. 14PCh. 3 - Prob. 15PCh. 3 - Prob. 16PCh. 3 - Prob. 17PCh. 3 - Prob. 18PCh. 3 - Prob. 19PCh. 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. 53PCh. 3 - Prob. 54PCh. 3 - Prob. 55PCh. 3 - Prob. 56PCh. 3 - Prob. 57PCh. 3 - Prob. 58PCh. 3 - Prob. 59PCh. 3 - Prob. 60PCh. 3 - Prob. 61PCh. 3 - Prob. 62PCh. 3 - Prob. 63PCh. 3 - Prob. 64PCh. 3 - Prob. 65PCh. 3 - Prob. 66PCh. 3 - Prob. 67PCh. 3 - Prob. 68PCh. 3 - Prob. 69PCh. 3 - Prob. 70PCh. 3 - Prob. 71PCh. 3 - Prob. 72PCh. 3 - 75. A Nile cruise ship takes 20.8 h to go upstream...Ch. 3 - Prob. 74PCh. 3 - Prob. 75PCh. 3 - Prob. 76PCh. 3 - Prob. 77PCh. 3 - Prob. 78PCh. 3 - Prob. 79PCh. 3 - Prob. 80PCh. 3 - Prob. 81PCh. 3 - Prob. 82PCh. 3 - Prob. 83PCh. 3 - Prob. 84PCh. 3 - Prob. 85PCh. 3 - Prob. 86PCh. 3 - Prob. 87PCh. 3 - Prob. 88PCh. 3 - Prob. 89PCh. 3 - Prob. 90PCh. 3 - Prob. 91PCh. 3 - Prob. 92PCh. 3 - Prob. 93PCh. 3 - Prob. 94PCh. 3 - Prob. 96PCh. 3 - Prob. 95PCh. 3 - Prob. 97PCh. 3 - Prob. 98PCh. 3 - Prob. 99PCh. 3 - Prob. 100PCh. 3 - Prob. 101PCh. 3 - Prob. 102PCh. 3 - Prob. 103PCh. 3 - Prob. 104PCh. 3 - Prob. 105PCh. 3 - Prob. 106PCh. 3 - Prob. 107PCh. 3 - Prob. 108PCh. 3 - 111. A ball is thrown horizontally off the edge of...Ch. 3 - 112. A marble is rolled so that it is projected...Ch. 3 - Prob. 111PCh. 3 - Prob. 112PCh. 3 - Prob. 113PCh. 3 - Prob. 114P
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.
Similar questions
SEE MORE QUESTIONS
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
Introduction to Vectors and Their Operations; Author: Professor Dave Explains;https://www.youtube.com/watch?v=KBSCMTYaH1s;License: Standard YouTube License, CC-BY