Homework Help is Here – Start Your Trial Now!
Science
Physics
33.75 33.0+11.69 11.7 )N Part6 of 6-Analyze Finally, solving for the acceleration, - Your response differs from the correct anser by more than 10%. Double check your calculations. I+7 from the correct anser by more than 100. ) ms. Your response difers As befere. +. subatituting the x and y components of acceleration, gives m/s tan

33.75 33.0+11.69 11.7 )N Part6 of 6-Analyze Finally, solving for the acceleration, - Your response differs from the correct anser by more than 10%. Double check your calculations. I+7 from the correct anser by more than 100. ) ms. Your response difers As befere. +. subatituting the x and y components of acceleration, gives m/s tan

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
See similar textbooks
icon
Related questions
icon
Concept explainers
Question

PART 6 of 6 ONLY PLEASE

Part 4 of 6 - Analyze Now, solving for the acceleration, gives a 5.1 5.09| i+ 2 5 2.55 i) m/s?. Remembering that a = a = a 2 +. and substituting for the x and y components of acceleration, we have for the magnitude and direction a = V 5.1 5.09 m/s? 2.55 2.55 m/s? = 5.70 V 5.7 m/s 6 = tan 26.6 26.6 °. Part 5 of 6 - Analyze (b) Again, using the particle under a net force model, we calculate the net force on the object in the second figure. Finding the x and y components for F,, we have F2x = 13.5 13.5 N cos 60.0° - 6.75 F, 13.5 13.5 N sin 60.0° = 11.7 N. So in unit-vector notation F, = 6.75 i+ 11.69 11.7 j) N. For the sum of the forces on the object in (b), we have the following unit-vector notation. EF = F, +F, 33.8 i+ 11.69 11.7 j N Part 6 of 6 - Analyze Finally, solving for the acceleration, a = EF 5.7 lx = Your response differs from the correct answer by more than 10%. Double check your calculations. î + 5.7 Your response differs from the correct answer by more than 100%. j) m/s?. As before, a = a = Va 2 + a 2. Substituting the x and y components of acceleration, gives ]m/s) +( |m/s) m/s? 6 = tan- Submit Skip (you cannot come back)
Transcribed Image Text:Part 4 of 6 - Analyze Now, solving for the acceleration, gives a 5.1 5.09| i+ 2 5 2.55 i) m/s?. Remembering that a = a = a 2 +. and substituting for the x and y components of acceleration, we have for the magnitude and direction a = V 5.1 5.09 m/s? 2.55 2.55 m/s? = 5.70 V 5.7 m/s 6 = tan 26.6 26.6 °. Part 5 of 6 - Analyze (b) Again, using the particle under a net force model, we calculate the net force on the object in the second figure. Finding the x and y components for F,, we have F2x = 13.5 13.5 N cos 60.0° - 6.75 F, 13.5 13.5 N sin 60.0° = 11.7 N. So in unit-vector notation F, = 6.75 i+ 11.69 11.7 j) N. For the sum of the forces on the object in (b), we have the following unit-vector notation. EF = F, +F, 33.8 i+ 11.69 11.7 j N Part 6 of 6 - Analyze Finally, solving for the acceleration, a = EF 5.7 lx = Your response differs from the correct answer by more than 10%. Double check your calculations. î + 5.7 Your response differs from the correct answer by more than 100%. j) m/s?. As before, a = a = Va 2 + a 2. Substituting the x and y components of acceleration, gives ]m/s) +( |m/s) m/s? 6 = tan- Submit Skip (you cannot come back)
This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part. Tutorial Exercise Two forces F, andF, act on a 5.30-kg object. F, = 27.0N and F, = 13.5 N. 90.0° 60.0° (a) Find the acceleration of the object for the configuration of forces shown in Figure (a). (b) Find the acceleration of the object for the configuration of forces shown in Figure (b). Part 1 of 6 - Conceptualize Forces acting on an object add as vectors. The acceleration will be greater in magnitude in situation (b) than in situation (a). In both cases, the acceleration will be on the order of a few m/s?, directed into the first quadrant. Part 2 of 6 - Categorize First, we will determine the total force on the object. Since the object experiences acceleration, we can use the particle under net force model to determine the acceleration with Newton's second law. Part 3 of 6 - Analyze (a) Using the particle under a net force model, we calculate the net force on the object in the first figure. EF = F, +F, 27 + 13.5 13.5 )N The following is the mathematical statement of Newton's second law. = ma Expressing the sum of the forces in vector notation, and substituting the value of the mass, we have EF = (27 - 13.5 i)N = 5.3 kg)a. 13.5 5.30 Part 4 of 6 - Analyze Now, solving for the acceleration, gives a = 5.09 i+ 2.55 2.55 i) m/s?. Remembering that: and substituting for the x and y components of acceleration, we have for the magnitude and direction 5.09| 2,55 2,55 m/s? 5.70| 5.7 m/s? 5.1
Transcribed Image Text:This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part. Tutorial Exercise Two forces F, andF, act on a 5.30-kg object. F, = 27.0N and F, = 13.5 N. 90.0° 60.0° (a) Find the acceleration of the object for the configuration of forces shown in Figure (a). (b) Find the acceleration of the object for the configuration of forces shown in Figure (b). Part 1 of 6 - Conceptualize Forces acting on an object add as vectors. The acceleration will be greater in magnitude in situation (b) than in situation (a). In both cases, the acceleration will be on the order of a few m/s?, directed into the first quadrant. Part 2 of 6 - Categorize First, we will determine the total force on the object. Since the object experiences acceleration, we can use the particle under net force model to determine the acceleration with Newton's second law. Part 3 of 6 - Analyze (a) Using the particle under a net force model, we calculate the net force on the object in the first figure. EF = F, +F, 27 + 13.5 13.5 )N The following is the mathematical statement of Newton's second law. = ma Expressing the sum of the forces in vector notation, and substituting the value of the mass, we have EF = (27 - 13.5 i)N = 5.3 kg)a. 13.5 5.30 Part 4 of 6 - Analyze Now, solving for the acceleration, gives a = 5.09 i+ 2.55 2.55 i) m/s?. Remembering that: and substituting for the x and y components of acceleration, we have for the magnitude and direction 5.09| 2,55 2,55 m/s? 5.70| 5.7 m/s? 5.1
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 2 steps

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Knowledge Booster
Estimate of calculation
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
College Physics
College Physics
Physics
ISBN:
9781305952300
Author:
Raymond A. Serway, Chris Vuille
Publisher:
Cengage Learning
University Physics (14th Edition)
University Physics (14th Edition)
Physics
ISBN:
9780133969290
Author:
Hugh D. Young, Roger A. Freedman
Publisher:
PEARSON
Introduction To Quantum Mechanics
Introduction To Quantum Mechanics
Physics
ISBN:
9781107189638
Author:
Griffiths, David J., Schroeter, Darrell F.
Publisher:
Cambridge University Press
Physics for Scientists and Engineers
Physics for Scientists and Engineers
Physics
ISBN:
9781337553278
Author:
Raymond A. Serway, John W. Jewett
Publisher:
Cengage Learning
Lecture- Tutorials for Introductory Astronomy
Lecture- Tutorials for Introductory Astronomy
Physics
ISBN:
9780321820464
Author:
Edward E. Prather, Tim P. Slater, Jeff P. Adams, Gina Brissenden
Publisher:
Addison-Wesley
College Physics: A Strategic Approach (4th Editio…
College Physics: A Strategic Approach (4th Editio…
Physics
ISBN:
9780134609034
Author:
Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field
Publisher:
PEARSON
SEE MORE TEXTBOOKS