×
Log in to StudySoup
Get Full Access to University Physics - 13 Edition - Chapter 23 - Problem 73p
Join StudySoup for FREE
Get Full Access to University Physics - 13 Edition - Chapter 23 - Problem 73p

Already have an account? Login here
×
Reset your password

Charge Q = +4.00 µC is distributed uniformly over the

University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman ISBN: 9780321675460 31

Solution for problem 73P Chapter 23

University Physics | 13th Edition

  • Textbook Solutions
  • 2901 Step-by-step solutions solved by professors and subject experts
  • Get 24/7 help from StudySoup virtual teaching assistants
University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman

University Physics | 13th Edition

4 5 1 373 Reviews
24
3
Problem 73P

Charge Q = +4.00 µC is distributed uniformly over the volume of an insulating sphere that has radius R = 5.00 cm. What is the potential difference between the center of the sphere and the surface of the sphere?

Step-by-Step Solution:
Step 1 of 3

Exam 3 Study Guide Momentum, Rotational Motion, Torque Momentum moves in the same direction as velocity and is given by the equation: p=mv The initial momentum is always the same as the final momentum. In the problems that ask for recoil motion use the following formula often times in momentum problems they are paired with translational kinematic equations: m1v 1m v2 2 An open systems is a system that gains or loses mass. This is the formula to us for the rocket or a similar system: ∆ M F thrust ∆t ) The angular position is given in polar coordinates and is found with the following: s θ= r Angular displacement is the change in angular position and is given by the following: ∆ θ=θ fθ i The average angular speed is shown by: θ −θ ∆θ ω av f = tf−ti ∆t Angular Acceleration is given in the following equation: a= ω fω i= ∆ω tf−ti ∆t Similar to translational kinematic equations there are rotational kinematic equations. For these velocity is replaced by ω and x is replaced by θ. These act the same way as translational kinematic equations. To find tangential velocity use: v=ωr Where r is the radius. To find tangential velocity a similar formula is used: a=ar Don’t forget the formula for centripetal acceleration is: 2 a = (ωr) =+ω r2 c r Torque is a part of Newtons second law. It is given by the following: τ=rFsinφ φ is the angle between F and r. The unit for torque is an Nm. Torque can also be found by taking the cross product of two vectors. The magnitude of Torque is given by: R=ABsinφ Inertia is the torque divided by the angular acceleration. In translational motion the more massive particle has the most rotational inertia Mass distribution with respect to the rotational axis also impacts rotational inertia; the farther away the mass is from the rotational axis the more rotational inertia will exist. Rotational inertia can be found by: n I= ∑ m r2 i=1 ii M is the mass of the particle r is the distance from the rotational axis On a continuous object the rotational inertia is: I= ∫ dm The parallel axis theorem where M is mass, h is the perpendicular distance between the new axis and the axis through the center of mass and I CMs the rotational inertia around the center of mass I=ICM +M h 2 The center of mass is found by: m xCM=( 2 )x2 m 1m 2 For center of mass multiply the associated mass with the associated x coordinate. Kinetic Energy for rotating object: 1 2 K r I ω 2 Conservation of energy holds true for rotational motion similar to translational motion K iK +U ri=K +i +U +∆E f rf f th K­ is the translational kinetic K ­ris rotational kinetic Newtons second law is shown through torque and is the sum of all torque. If the system is conserved torque initial is equal to torque final

Step 2 of 3

Chapter 23, Problem 73P is Solved
Step 3 of 3

Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

The full step-by-step solution to problem: 73P from chapter: 23 was answered by , our top Physics solution expert on 05/06/17, 06:07PM. This textbook survival guide was created for the textbook: University Physics, edition: 13. Since the solution to 73P from 23 chapter was answered, more than 312 students have viewed the full step-by-step answer. The answer to “Charge Q = +4.00 µC is distributed uniformly over the volume of an insulating sphere that has radius R = 5.00 cm. What is the potential difference between the center of the sphere and the surface of the sphere?” is broken down into a number of easy to follow steps, and 39 words. This full solution covers the following key subjects: sphere, potential, difference, distributed, Insulating. This expansive textbook survival guide covers 26 chapters, and 2929 solutions. University Physics was written by and is associated to the ISBN: 9780321675460.

Other solutions

People also purchased

Related chapters

Unlock Textbook Solution

Enter your email below to unlock your verified solution to:

Charge Q = +4.00 µC is distributed uniformly over the