EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
6th Edition
ISBN: 9781319321710
Author: Mosca
Publisher: VST
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Chapter 12, Problem 44P
To determine

Tocalculate: The resulting change in the length of this composite wire if an object with a mass of 5.00 kg is hung vertically from one of its ends.

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A steel wire of length 1.40 m and diameter 1.00 mm is joined to an aluminum wire of identical dimensions to make a composite wire of length 2.80 m. What is the resulting change in length AL of this composite wire if an object with a mass of 5.00 kg is hung vertically from one of its ends? (Neglect any effect the masses of two wires have on the changes in their lengths.) AL = 1.751 Incorrect mm
A rigid bar of mass M is supported symmetrically by three wires each of length l. Those at each end are ofcopper and the middle one is of iron. The ratio of their diameter, if each is to have the same tension, is equal to (a)YcopperYiron(b) √YironY copper(c)Yiron2Ycopper2(d)YironYcopper
(A-(/2) (A-/2)(ptq)L)202 be 2. In engineering, the modulus of elasticity Is a way to measure how much an object deforms along an axis when opposing forces are applied among that axis. The formula can be expressed as E = ddsof elasticity, F is the axial force, & is a cdeformation measure, L s the original length of the object, and A is the croSs-sectional area. (F2-F,)Lo where E is (82-81)A' Part A: Rearrange the elasticity formula to find the cross-sectional area, A Part B: Rearrange the elasticity formula to find the original length, Lo, of the object.
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