FlipIt for College Physics (Algebra Version - Six Months Access)
FlipIt for College Physics (Algebra Version - Six Months Access)
17th Edition
ISBN: 9781319032432
Author: Todd Ruskell
Publisher: W.H. Freeman & Co
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 11, Problem 50QAP
To determine

(a)

Convert 1500kPa to N/m2

Expert Solution
Check Mark

Answer to Problem 50QAP

  1500kPa is equal to 15×105N/m2

Explanation of Solution

Given info:

Pressure, P=1500kPa

Calculation:

The value of 1500kPa in N/m2 is calculated as

  1500kPa=1500×103Pa(1kPa=1×103Pa)=15×105N/m2

Conclusion:

  1500kPa is equal to 15×105N/m2

To determine

(b)

Convert 35psi to N/m2

Expert Solution
Check Mark

Answer to Problem 50QAP

  35psi is equal to 241317N/m2

Explanation of Solution

Given info:

Pressure, 35psi

Calculation:

The value of 35psi in N/m2 is calculated as

  35psi=35×6894.76N/m2(1psi=6894.76N/m2)=241317N/m2

Conclusion:

  35psi is equal to 241317N/m2

To determine

(c)

Convert 2.85atm to N/m2

Expert Solution
Check Mark

Answer to Problem 50QAP

  2.85atm is equal to 2.87×105N/m2

Explanation of Solution

Given info:

Pressure, 2.85atm

Calculation:

The value of 2.85atm in N/m2 is calculated as

  2.85atm=2.85×1.01×105Pa(1atm=1.01×105Pa)=2.87×105N/m2

Conclusion:

  2.85atm is equal to 2.87×105N/m2

To determine

(d)

Convert 883torr to N/m2

Expert Solution
Check Mark

Answer to Problem 50QAP

  883torr is equal to 117721.52N/m2.

Explanation of Solution

Given info:

Pressure, 883torr

Calculation:

The value of 883torr in N/m2 is calculated as

  883torr=883×133.32Pa(1torr=133.32Pa)=117721.52N/m2

Conclusion:

  883torr is equal to 117721.52N/m2.

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
The rectangular loop of wire shown in the figure (Figure 1) has a mass of 0.18 g per centimeter of length and is pivoted about side ab on a frictionless axis. The current in the wire is 8.5 A in the direction shown. Find the magnitude of the magnetic field parallel to the y-axis that will cause the loop to swing up until its plane makes an angle of 30.0 ∘ with the yz-plane. Find the direction of the magnetic field parallel to the y-axis that will cause the loop to swing up until its plane makes an angle of 30.0 ∘ with the yz-plane.
Give a more general expression for the magnitude of the torque τ. Rewrite the answer found in Part A in terms of the magnitude of the magnetic dipole moment of the current loop m. Define the angle between the vector perpendicular to the plane of the coil and the magnetic field to be ϕ, noting that this angle is the complement of angle θ in Part A. Give your answer in terms of the magnetic moment mm, magnetic field B, and ϕ.
Calculate the electric and magnetic energy densities at thesurface of a 3-mm diameter copper wire carrying a 15-A current. The resistivity ofcopper is 1.68×10-8 Ω.m.Prob. 18, page 806, Ans: uE= 5.6 10-15 J/m3 uB= 1.6 J/m3

Chapter 11 Solutions

FlipIt for College Physics (Algebra Version - Six Months Access)

Ch. 11 - Prob. 11QAPCh. 11 - Prob. 12QAPCh. 11 - Prob. 13QAPCh. 11 - Prob. 14QAPCh. 11 - Prob. 15QAPCh. 11 - Prob. 16QAPCh. 11 - Prob. 17QAPCh. 11 - Prob. 18QAPCh. 11 - Prob. 19QAPCh. 11 - Prob. 20QAPCh. 11 - Prob. 21QAPCh. 11 - Prob. 22QAPCh. 11 - Prob. 23QAPCh. 11 - Prob. 24QAPCh. 11 - Prob. 25QAPCh. 11 - Prob. 26QAPCh. 11 - Prob. 27QAPCh. 11 - Prob. 28QAPCh. 11 - Prob. 29QAPCh. 11 - Prob. 30QAPCh. 11 - Prob. 31QAPCh. 11 - Prob. 32QAPCh. 11 - Prob. 33QAPCh. 11 - Prob. 34QAPCh. 11 - Prob. 35QAPCh. 11 - Prob. 36QAPCh. 11 - Prob. 37QAPCh. 11 - Prob. 38QAPCh. 11 - Prob. 39QAPCh. 11 - Prob. 40QAPCh. 11 - Prob. 41QAPCh. 11 - Prob. 42QAPCh. 11 - Prob. 43QAPCh. 11 - Prob. 44QAPCh. 11 - Prob. 45QAPCh. 11 - Prob. 46QAPCh. 11 - Prob. 47QAPCh. 11 - Prob. 48QAPCh. 11 - Prob. 49QAPCh. 11 - Prob. 50QAPCh. 11 - Prob. 51QAPCh. 11 - Prob. 52QAPCh. 11 - Prob. 53QAPCh. 11 - Prob. 54QAPCh. 11 - Prob. 55QAPCh. 11 - Prob. 56QAPCh. 11 - Prob. 57QAPCh. 11 - Prob. 58QAPCh. 11 - Prob. 59QAPCh. 11 - Prob. 60QAPCh. 11 - Prob. 61QAPCh. 11 - Prob. 62QAPCh. 11 - Prob. 63QAPCh. 11 - Prob. 64QAPCh. 11 - Prob. 65QAPCh. 11 - Prob. 66QAPCh. 11 - Prob. 67QAPCh. 11 - Prob. 68QAPCh. 11 - Prob. 69QAPCh. 11 - Prob. 70QAPCh. 11 - Prob. 71QAPCh. 11 - Prob. 72QAPCh. 11 - Prob. 73QAPCh. 11 - Prob. 74QAPCh. 11 - Prob. 75QAPCh. 11 - Prob. 76QAPCh. 11 - Prob. 77QAPCh. 11 - Prob. 78QAPCh. 11 - Prob. 79QAPCh. 11 - Prob. 80QAPCh. 11 - Prob. 81QAPCh. 11 - Prob. 82QAPCh. 11 - Prob. 83QAPCh. 11 - Prob. 84QAPCh. 11 - Prob. 85QAPCh. 11 - Prob. 86QAPCh. 11 - Prob. 87QAPCh. 11 - Prob. 88QAPCh. 11 - Prob. 89QAPCh. 11 - Prob. 90QAPCh. 11 - Prob. 91QAPCh. 11 - Prob. 92QAPCh. 11 - Prob. 93QAPCh. 11 - Prob. 94QAP
Knowledge Booster
Background pattern image
Algebra
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, algebra and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Principles of Physics: A Calculus-Based Text
Physics
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
Physics for Scientists and Engineers, Technology ...
Physics
ISBN:9781305116399
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
University Physics Volume 1
Physics
ISBN:9781938168277
Author:William Moebs, Samuel J. Ling, Jeff Sanny
Publisher:OpenStax - Rice University
Text book image
College Physics
Physics
ISBN:9781285737027
Author:Raymond A. Serway, Chris Vuille
Publisher:Cengage Learning
Text book image
Physics for Scientists and Engineers: Foundations...
Physics
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Cengage Learning
Text book image
College Physics
Physics
ISBN:9781938168000
Author:Paul Peter Urone, Roger Hinrichs
Publisher:OpenStax College
Fluids in Motion: Crash Course Physics #15; Author: Crash Course;https://www.youtube.com/watch?v=fJefjG3xhW0;License: Standard YouTube License, CC-BY