Chapter 8
CHAPTER 8 - Conservation of Energy 1. The potential energy of the spring is zero when the spring is not compressed (x= 0). For the stored potential energy, we have U = ! kxf2 0; 35.0 J = ! (82.0 N/m)xf2 0, which gives xf = 2. 3. 4. For t
4
Motion in Two Dimensions
CHAPTER OUTLINE
4.1 4.2 4.3 4.4 4.5 4.6 The Position, Velocity, and Acceleration Vectors Two-Dimensional Motion with Constant Acceleration Projectile Motion Uniform Circular Motion Tangential and Radial Acceleration Relativ
6
Circular Motion and Other Applications of Newton's Laws
CHAPTER OUTLINE
6.1 Newton's Second Law Applied to Uniform Circular Motion Nonuniform Circular Motion Motion in Accelerated Frames Motion in the Presence of Resistive Forces Numerical Modeling
1.63: Assume each person sees the dentist twice a year for checkups, for 2 hours. Assume 2 more hours for restorative work. Assuming most dentists work less than 2000 hours per year, this gives 2000 hours 4 hours per patient 500 patients per dentist.
Uniform and nonuniform motion
Motion Diagram
3. The person is motionless. b. Equally spaced imaes show her moving
at a constant speed.
if? a}; m;
eeﬁL M
6. She is speeding up. d. She is slowing down.
Figure 1 The distance the jogger moves in each time i
Terminal velocity
Multiple Choice
Identify the choice that best completes the statement or answers the question.
_
1. Which of the following is the tendency of an object to maintain its state of motion?
a. acceleration
c. force
b. inertia
d. velocity
_
2.
Practice: Static and Kinetic Friction
Name:_ Date:_
1. Gwen exerts a 36-N horizontal force as she pulls a 52-N sled across a cement sidewalk at
constant speed. What is the coefficient of kinetic friction between the sidewalk and the metal
sled runners? Ig
Term 2- Quiz 1
Grade/Cluster
Section
Name
ID
10/ES
Subject
ANSWER KEY
Physics
Date
Duration 45 minutes
Score/30
I-Choose the best answer
[6 marks, 1 each]
1.
2.
T2-Q1
Page 1 of 6
G10-Adv-Physics
3. Ball X is released to fall down the same time as ball Y i
terminal velocity
Describe the motion at
each stage of the skydivers
jump:
Explain why her motion is the
way it is (think about the forces):
Stage 1: at the moment of
leaving the plane
Stage 1:
Stage 2:
Stage 2:
Stage 3: terminal velocity
Stage 3:
Stage 4
3
Vectors
CHAPTER OUTLINE
3.1 3.2 3.3 3.4 Coordinate Systems Vector and Scalar Quantities Some Properties of Vectors Components of a Vector and Unit Vectors
ANSWERS TO QUESTIONS
Q3.1 No. The sum of two vectors can only be zero if they are in opposit
1
Physics and Measurement
CHAPTER OUTLINE
1.1 1.2 1.3 1.4 1.5 1.6 1.7 Standards of Length, Mass, and Time Matter and Model-Building Density and Atomic Mass Dimensional Analysis Conversion of Units Estimates and Order-ofMagnitude Calculations Signific
Chapter 6
CHAPTER 6 - Gravitation and Newton's Synthesis 1. Because the spacecraft is 2 Earth radii above the surface, it is 3 Earth radii from the center. The gravitational force on the spacecraft is F = GMEM/r2 = (6.67 1011 N m2/kg2)(5.98 1024 kg
Chapter 5
CHAPTER 5 - Further Applications of Newton's Laws 1. The friction is kinetic, so Ffr = kFN. With constant velocity, the acceleration is zero. Using the force diagram for the crate, we can write ?F = ma: x-component: F kFN = 0; y-component
Chapter 9
CHAPTER 9 - Linear Momentum and Collisions 1. We find the force on the expelled gases from F = ?p/?t = (?m/?t)v = (1200 kg/s)(50,000 m/s) = 6.0 107 N. An equal, but opposite, force will be exerted on the rocket: 6.0 107 N, up. For the mome
Chapter 2
CHAPTER 2 - Describing Motion: Kinematics in One Dimension 1. We find the time from average speed = d/t; 15 km/h = (75 km)/t , which gives
t = 5.0 h. 88 km/h.
2. 3.
We find the average speed from average speed = d/t = (280 km)/(3.2 h) =
Chapter 3
CHAPTER 3 - Kinematics in Two Dimensions; Vectors 1. We choose the west and south coordinate system shown. D 2x For the components of the resultant we have W RW = D1 + D2 cos 45 D 2y D 2 = (200 km) + (80 km) cos 45 = 257 km; RS = 0 + D2 =
Chapter 4
1
CHAPTER 4 - Dynamics: Newton's Laws of Motion 1. We convert the units: # lb = (0.25 lb)(4.45 N/lb) ~ 1 N. If we select the bike and rider as the object, we apply Newton's second law to find the mass: ?F = ma; 255 N = m(2.20 m/s2), which
Chapter 7
CHAPTER 7 - Work and Energy 1. 2. 3. The displacement is in the direction of the gravitational force, thus W = Fh cos 0 = mgh = (250 kg)(9.80 m/s2)(2.80 m) = 6.86 103 J. The displacement is opposite to the direction of the retarding force,
Chapter 11
CHAPTER 11 General Rotation 1.
z (a) For the magnitudes of the vector products we have i i = i i sin 0 = 0; k j j = j j sin 0 = 0; j k k = k k sin 0 = 0. y (b) For the magnitudes of the vector products we have i j = i j sin 90 = (1)(1)(1
Chapter 12
CHAPTER 12 Static Equilibrium; Elasticity and Fracture 1. From the force diagram for the sapling we can write ?Fx = F1 F2 sin 20 F3 cos = 0; 380 N (255 N) sin 20 F3 cos = 0, or F3 cos = 293 N. ?Fy = F2 cos 20 F3 sin = 0; F3 sin
Falling Objects
or, how cats survive falls from
tall buildings
Lesson objectives
Know: what terminal velocity means
Understand: why falling things reach a terminal
velocity
Be able to: describe and explain the stages in a
parachute jump (including the for