By Raymond A. Serway
For Chapters 1-22, this guide includes distinctive options to nearly 20% of the issues in step with bankruptcy (indicated within the textbook with boxed challenge numbers). The guide additionally includes a abilities part, vital notes from key sections of the textual content, and an inventory of significant equations and ideas.
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Extra resources for Student Solutions Manual, Volume 1 for Serway Jewett's Physics for Scientists and Engineers, 8th Edition
2010 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Copyright 2010 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. 00 ˆj) units ( ) ■ ■ Finalize: Our calculations are as brief as possible and agree with our preliminary estimates. Brace yourself for continued successes. 36. 0 units. Find (a) the resultant in unit-vector notation and (b) the magnitude and direction of the resultant displacement.
Solution Conceptualize: The different nonzero original speeds of the two stones do not affect their accelerations, which have the same value g downward. This is a (pair of) free-fall problem(s). Categorize: Equations chosen from the standard constant-acceleration set describe each stone separately, but look out for having to solve a quadratic equation. © 2010 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
At the end of the walk, what is the person’s resultant displacement measured from the starting point? Solution Conceptualize: On our version of the diagram we have drawn in the resultant from the tail of the ﬁrst arrow to the head of the last arrow. It looks like about 200 m at 180° + 60°. 47 Analyze: We translate from the pictorial representation to a mathematical representation by writing the individual displacements in unit-vector notation: I d1 = 100 ˆi m I d 2 = –300 ˆj m I d 3 = (–150 cos 30F ) ˆi m + (–150 sin 30F ) ˆj m = –1330ˆi m – 75jˆ m I d 4 = (–200 cos 60F ) ˆi m + (200 sin 60F ) ˆj m = –1000ˆi m +173jˆ m © 2010 Cengage Learning.