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Solutions for Physics, Volume 1, 12th Edition Cutnell (Chapters 1 to 17 included) £23.86   Add to cart

Exam (elaborations)

Solutions for Physics, Volume 1, 12th Edition Cutnell (Chapters 1 to 17 included)

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  • Module
  • Physics - General Relativity
  • Institution
  • Physics - General Relativity

Complete Solutions Manual for Physics, Volume 1, 12th Edition by John D. Cutnell, Kenneth W. Johnson, David Young, Shane Stadler ; ISBN13: 9781119803690. Full Chapters included Chapter 1 to 17. Practice Problems included. 1.Introduction and Mathematical Concepts. 2.Kinematics in One Dimension. 3...

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  • December 4, 2023
  • 1257
  • 2021/2022
  • Exam (elaborations)
  • Questions & answers
  • Physics - General Relativity
  • Physics - General Relativity
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Physics
12th Edition
by John D. Cutnell


Volumn - 1

Vol 1 Complete Chapter Solutions
Manual are included (Ch 1 to 17)


** Immediate Download
** Swift Response
** All Chapters included
** Practice Problems

,CHAPTER 1 INTRODUCTION AND
MATHEMATICAL CONCEPTS

ANSWERS TO FOCUS ON CONCEPTS QUESTIONS


1. (d) The resultant vector R is drawn from the tail of the first vector to the head of the
last vector.

2. (c) Note from the drawing that the magnitude R of the resultant vector R is equal to
the shortest distance between the tail of A and the head of B. Thus, R is less than the
magnitude (length) of A plus the magnitude of B.

3. (b) In this drawing the vector –C is reversed relative to C, while vectors A and B are
not reversed.

4. (c) In this drawing the vectors –B and –C are reversed relative to B and C, while vector
A is not reversed.

5. (c) When the two vector components Ax and Ay are added by the tail-to-head method,
the sum equals the vector A. Therefore, these vector components are the correct ones.

6. (b) The displacement vector A points in the –y direction. Therefore, it has no scalar
component along the x axis (Ax = 0 m) and its scalar component along the y axis is
negative.

7. (e) The scalar components are given by Ax′ = −(450 m) sin 35.0° = −258 m and
Ay′ = −(450 m) cos 35.0° = −369 m.

8. (d)

9. Rx = 0 m, Ry = 6.8 m

10. R = 7.9 m, θ = 21 degrees

11. (a) The triangle in the drawing is a right triangle. The lengths A and B of the two sides
are known, so the Pythagorean theorem can be used to determine the length R of the
hypotenuse.

⎛ 4.0 km ⎞
12. (b) The angle is found by using the inverse tangent function, θ = tan −1 ⎜ ⎟ = 53° .
⎝ 3.0 km ⎠

, 2 INTRODUCTION AND MATHEMATICAL CONCEPTS




13. (e) These vectors form a closed four-sided polygon, with the head of the fourth vector
exactly meeting the tail of the first vector. Thus, the resultant vector is zero.

14. (b) The three vectors form a right triangle, so the magnitude of A is given by the
Pythagorean theorem as A = Ax2 + Ay2 . If Ax and Ay double in size, then the magnitude

( 2A ) + ( 2A )
2 2 2 2 2 2
of A doubles: x y
= 4 Ax + 4 Ay = 2 Ax + Ay = 2 A.


⎛A ⎞
15. (a) The angle θ is determined by the inverse tangent function, θ = tan −1 ⎜ y ⎟ . If Ax and
⎝ Ax ⎠
Ay both become twice as large, the ratio does not change, and θ remains the same.

16. (d) The distance (magnitude) traveled by each runner is the same, but the directions
are different. Therefore, the two displacement vectors are not equal.

17. (c) Ax and Bx point in opposite directions, and Ay and By point in the same direction.

18. Ay = 3.4 m, By = 3.4 m

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