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In Exercises 920, start by drawing a number line that

Introductory & Intermediate Algebra for College Students | 4th Edition | ISBN: 9780321758941 | Authors: Robert F. Blitzer ISBN: 9780321758941 177

Solution for problem 1.1.289 Chapter 1.3

Introductory & Intermediate Algebra for College Students | 4th Edition

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Introductory & Intermediate Algebra for College Students | 4th Edition | ISBN: 9780321758941 | Authors: Robert F. Blitzer

Introductory & Intermediate Algebra for College Students | 4th Edition

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Problem 1.1.289

In Exercises 920, start by drawing a number line that shows integers from 5 to 5. Then graph each real number on your number line. 9. 2

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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

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Chapter 1.3, Problem 1.1.289 is Solved
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Textbook: Introductory & Intermediate Algebra for College Students
Edition: 4
Author: Robert F. Blitzer
ISBN: 9780321758941

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In Exercises 920, start by drawing a number line that