Physics: Principles and Problems
Physics: Principles and Problems
9th Edition
Elliott, Haase, Harper, Herzog, Margaret Zorn, Nelson, Schuler, Zitzewitz
ISBN: 9780078458132
Table of contents
Textbook solutions

All Solutions

Page 55: Standardized Test Practice

Exercise 1
Step 1
1 of 2
C- The dots would be close together to start with, and get farther apart as the plane accelerated. As the plane starts to move, it starts off slow, then begins to pick up speed (acceleration) in order to take lift. So in the particle model motion diagram, the gaps are close at first, then starts to spread out more.
Result
2 of 2
The answer is C.
Exercise 2
Solution 1
Solution 2
Step 1
1 of 2
A vector diagram is need to solve all Physics problems
Result
2 of 2
A
Step 1
1 of 6
In this problem, we are asked to analyze a number of statements and determine which are false.
Step 2
2 of 6
**Statement A**

**A** states that to solve any physics problem we need to draw a vector diagram first.

This is incorrect, there are many problems concerning only vector intensities and scalar quantities where vector diagrams are not needed.

So statement **A** is false.

Step 3
3 of 6
**Statement B**

**B** states that the length of a vector should be proportional to the data.

The drawn length of a vector must adequately represent the vector’s intensity. To accomplish this, the data describing the vector must be considered.

So statement **B** is true.

Step 4
4 of 6
**Statement C**

**C** states that vectors can be added by simply adding their intensities(their lengths).

This is very incorrect, the fact that vectors cannot be added by just adding their intensities is, in a sense, what makes them vectors and not scalars.

Their direction has to be considered, and figures in the calculation.

So statement **C** is false.

Step 5
5 of 6
**Statement D**

**D** states that vectors can be added in either triangles or straight lines.

The statement is quite simply true.

When adding vectors, they can be drawn on the same straight line if they are parallel(or antiparallel) but must be drawn in a triangle if they point in different directions.

So statement **D** is true.

Step 6
6 of 6
**Conclusion**

Statements **A** and **C** are false.

Statements **B** and **D** are true.

Exercise 3
Step 1
1 of 3
In a position vs. time diagram, velocity is the slope of the graph at a given time t.
Step 2
2 of 3
Since the slope is the STEEPEST in the interval III, the answer is
Result
3 of 3
B
Exercise 4
Step 1
1 of 4
In a position vs. time diagram, velocity is the slope of the graph at a given time t.
Step 2
2 of 4
Since the slope is positive all the way to point C, this means that the direction of motion was positive this whole time.

The cyclist was increasing his displacement (moving away from initial position) up to then.

At C, the velocity becomes negative, meaning that the cyclist turns around,

and in doing so, decreases his displacement as he heads back to initial position.

Answer: C

Step 3
3 of 4
A simpler conclusion could be reached by observing that the graph gives the greatest value for C in terms of displacement.

So it is the farthest from inital point.

Result
4 of 4
C
Exercise 5
Step 1
1 of 3
The distance traveled is the change in position in each interval.
Step 2
2 of 3
From least to greatest the change in position (distance traveled):
II, IV, III, I

(Section III and section I are very close, but section I seems to have a larger change in position on the graph)

Result
3 of 3
A
Exercise 6
Step 1
1 of 3
Graph A, because it is the only graph
Step 2
2 of 3
that has initial vertical displacement equal to the final displacement.

(the squirrel returns to initial height)

Result
3 of 3
A
Exercise 7
Step 1
1 of 3
Observing vertical (north-south) and horizontal (east-west) components separately,

the rat displaces first 1.0m to the north, then later, 0.8m to the south.

Vertical component: 0.2m north.

Step 2
2 of 3
Horizontally: first displaces 0.3m east, then later, again east for 0.4m.

Total horizontal displacement: 0.7 m east.

Step 3
3 of 3
The final position in regard to the inital can be written as:

0.7m east, 0.2m north.

By placing the origin of a coordinate system at the initial point, (+x) being East, (+y) north, this position would represent the point with coordinates (0.7,0.2)m.

Its distance from the origin is calculated with the Pythagorean theorem:

$$
d=sqrt{0.7^2+0.2^2}=0.73m=73cm
$$

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Chapter 1: A Physics Toolkit
Section 1.1: Mathematics and Physics
Section 1.2: Measurement
Section 1.3: Graphing Data
Page 24: Assessment
Page 29: Standardized Test Practice
Chapter 3: Accelerated Motion
Section 3.1: Acceleration
Section 3.2: Motion with Constant Acceleration
Section 3.3: Free Fall
Page 80: Assessment
Page 85: Standardized Test Practice
Chapter 4: Forces in One Dimension
Section 4.1: Force and Motion
Section 4.2: Using Newton’s Laws
Section 4.3: Interaction Forces
Page 112: Assessment
Page 117: Standardized Test Practice
Chapter 5: Forces in Two Dimensions
Section 5.1: Vectors
Section 5.2: Friction
Section 5.3: Force and Motion in Two Dimensions
Page 140: Assessment
Page 145: Standardized Test Practice
Chapter 6: Motion in Two Dimensions
Section 6.1: Projectile Motion
Section 6.2: Circular Motion
Section 6.3: Relative Velocity
Page 164: Assessment
Page 169: Standardized Test Practice
Chapter 7: Gravitation
Section 7.1: Planetary Motion and Gravitation
Section 7.2: Using the Law of Universal Gravitation
Page 190: Assessment
Page 195: Standardized Test Practice
Chapter 8: Rotational Motion
Section 8.1: Describing Rotational Motion
Section 8.2: Rotational Dynamics
Section 8.3: Equilibrium
Page 222: Assessment
Page 227: Standardized Test Practice
Chapter 9: Momentum and Its Conservation
Chapter 10: Energy, Work, and Simple Machines
Section 10.1: Energy and Work
Section 10.2: Machines
Page 278: Assessment
Page 283: Standardized Test Practice
Chapter 11: Energy and Its Conservation
Section 11.1: The Many Forms of Energy
Section 11.2: Conservation of Energy
Page 306: Assessment
Page 311: Standardized Test Practice
Chapter 13: State of Matter
Section 13.1: Properties of Fluids
Section 13.2: Forces Within Liquids
Section 13.3: Fluids at Rest and in Motion
Section 13.4: Solids
Page 368: Assessment
Page 373: Standardized Test Practice
Chapter 14: Vibrations and Waves
Section 14.1: Periodic Motion
Section 14.2: Wave Properties
Section 14.3: Wave Behavior
Page 396: Assessment
Page 401: Section Review
Chapter 15: Sound
Section 15.1: Properties of Detection of Sound
Section 15.2: The Physics of Music
Page 424: Assessment
Page 429: Standardized Test Practice
Chapter 17: Reflections and Mirrors
Section 17.1: Reflection from Plane Mirrors
Section 17.2: Curved Mirrors
Page 478: Assessment
Page 483: Standardized Test Practice
Chapter 18: Refraction and lenses
Section 18.1: Refraction of Light
Section 18.2: Convex and Concave Lenses
Section 18.3: Applications of Lenses
Page 508: Assessment
Page 513: Standardized Test Practice
Chapter 21: Electric Fields
Section 21.1: Creating and Measuring Electric Fields
Section 21.2: Applications of Electric Fields
Page 584: Assessment
Page 589: Standardized Test Practice
Chapter 22: Current Electricity
Section 22.1: Current and Circuits
Section 22.2: Using Electric Energy
Page 610: Assessment
Page 615: Standardized Test Practice
Chapter 23: Series and Parallel Circuits
Section 23.1: Simple Circuits
Section 23.2: Applications of Circuits
Page 636: Assessment
Page 641: Standardized Test Practice
Chapter 24: Magnetic Fields
Section 24.1: Magnets: Permanent and Temporary
Section 24.2: Forces Caused by Magnetic Fields
Page 664: Assessment
Page 669: Standardized Test Practice
Chapter 25: Electromagnetic Induction
Section 25.1: Electric Current from Changing Magnetic Fields
Section 25.2: Changing Magnetic Fields Induce EMF
Page 690: Assessment
Page 695: Standardized Test Practice
Chapter 30: Nuclear Physics
Section 30.1: The Nucleus
Section 30.2: Nuclear Decay and Reactions
Section 30.3: The Building Blocks of Matter
Page 828: Assessment
Page 831: Standardized Test Practice