
Concept explainers
(a)
To identify an action-reaction pair of forces.
(a)

Answer to Problem 1Q
The force the person exerts on the wall and the force the wall exerts on the person is an action-reaction pair.
Explanation of Solution
Action – reaction pair of forces always acts on separate object.
When a person pushes on a wall, the wall pushes back on the person. Here the person exerts force on the wall and wall in turn exerts force on the person. The force exerted by the person on the wall is equal in magnitude and opposite in direction to the force exerted by the wall on the person and they acts on different objects.
Conclusion:
Therefore, the force the person exerts on the wall and the force the wall exerts on the person is an action-reaction pair.
(b)
To identify an action-reaction pair of forces, book resting on the table.
(b)

Answer to Problem 1Q
The gravitational force exerted the book on the table and the normal force exerted by the table on the book is action-reaction pair force.
Explanation of Solution
The gravitational force exerted by the book on the table and the normal force exerted by the table on the book is action-reaction pair force and they acts on different objects. The gravitational force exerted by the book on the table is equal in magnitude and opposite in direction to the force exerted by the table on the book.
Conclusion:
Therefore, the gravitational force exerted the book on the table and the normal force exerted by the table on the book is action-reaction pair force.
(c)
To identify an action-reaction pair of forces, a hockey puck sliding across an icy surface.
(c)

Answer to Problem 1Q
The
Explanation of Solution
The frictional force exerted by the moving hockey puck on the ice and the frictional force exerted by the ice on the hockey puck is action-reaction pair force and they acts on different objects. The frictional force exerted by the moving hockey puck on the ice is equal in magnitude and opposite in direction to the frictional force exerted by the ice on the hockey puck.
Conclusion:
Therefore, the frictional force exerted by the moving hockey puck on the ice and the frictional force exerted by the ice on the hockey puck is action-reaction pair force.
(d)
To identify an action-reaction pair of forces, a car accelerating from rest.
(d)

Answer to Problem 1Q
The frictional force exerted by the tire on the road and the force exerted by the road on the tire is action-reaction pair force.
Explanation of Solution
The frictional force exerted by the tire on the road and the force exerted by the road on the tire is action-reaction pair force and they acts on different objects. The frictional force exerted by the rotating tire on the road is equal in magnitude and opposite in direction to the frictional force exerted by the road on the tire.
Conclusion:
Therefore, the frictional force exerted by the tire on the road and the force exerted by the road on the tire is action-reaction pair force.
(e)
To identify an action-reaction pair of forces, an object undergoing free fall in a vacuum.
(e)

Answer to Problem 1Q
The gravitational force exerted on the object by the object on which the object is falling and gravitational force on the falling object on which the object is falling is an action-reaction pair.
Explanation of Solution
The gravitational force exerted on the object by the object on which the object is falling is equal in magnitude and opposite in direction to gravitational force on the falling object on which the object is falling is an action-reaction pair and they acts on different objects.
Conclusion:
Therefore, the gravitational force exerted on the object by the object on which the object is falling and gravitational force on the falling object on which the object is falling is an action-reaction pair.
(f)
To identify an action-reaction pair of forces, a basketball player jumping to dunk a basketball.
(f)

Answer to Problem 1Q
The force exerted by the basketball player against the floor and the force the floor exerts on the basket ball are action-reaction pairs.
Explanation of Solution
The force exerted by the basketball player against the floor while he dunks the ball and the force the floor exerts on the basket ball are equal in magnitude and opposite in direction and they acts on different objects. The force exerted by the basketball player against the floor while and the force the floor exerts on the basket ball are action-reaction pairs.
Conclusion:
Therefore, the force exerted by the basketball player against the floor while and the force the floor exerts on the basket ball are action-reaction pairs.
(g)
To identify an action-reaction pair of forces, a person throwing a baseball.
(g)

Answer to Problem 1Q
The force the person exerts against the basket ball to accelerate it and the force the basketball exerts on the person is action-reaction pair.
Explanation of Solution
The force the person exerts against the basket ball to accelerate it and the force the basketball exerts on the person is equal in magnitude and opposite in direction and they acts on different objects. The force the person exerts against the basket ball to accelerate it and the force the basketball exerts on the person is action-reaction pair.
Conclusion:
Therefore, the force the person exerts against the basket ball to accelerate it and the force the basketball exerts on the person is action-reaction pair.
Want to see more full solutions like this?
Chapter 3 Solutions
College Physics, Volume 1
- Two complex values are z1=8 + 8i, z2=15 + 7 i. z1∗ and z2∗ are the complex conjugate values. Any complex value can be expessed in the form of a+bi=reiθ. Find r and θ for (z1-z∗2)/z1+z2∗. Find r and θ for (z1−z2∗)z1z2∗ Please show all stepsarrow_forwardAn electromagnetic wave is traveling through vacuum in the positive x direction. Its electric field vector is given by E=E0sin(kx−ωt)j^,where j^ is the unit vector in the y direction. If B0 is the amplitude of the magnetic field vector, find the complete expression for the magnetic field vector B→ of the wave. What is the Poynting vector S(x,t), that is, the power per unit area associated with the electromagnetic wave described in the problem introduction? Give your answer in terms of some or all of the variables E0, B0, k, x, ω, t, and μ0. Specify the direction of the Poynting vector using the unit vectors i^, j^, and k^ as appropriate. Please explain all stepsarrow_forwardAnother worker is performing a task with an RWL of only 9 kg and is lifting 18 kg, giving him an LI of 2.0 (high risk). Questions:What is the primary issue according to NIOSH?Name two factors of the RWL that could be improved to reduce risk.If the horizontal distance is reduced from 50 cm to 30 cm, how does the HM change and what effect would it have?arrow_forward
- Two complex values are z1=8 + 8i, z2=15 + 7 i. z1∗ and z2∗ are the complex conjugate values. Any complex value can be expessed in the form of a+bi=reiθ. Find r and θ for z1z2∗. Find r and θ for z1/z2∗? Find r and θ for (z1−z2)∗/z1+z2∗. Find r and θ for (z1−z2)∗/z1z2∗ Please explain all steps, Thank youarrow_forwardAn ac series circuit consists of a voltage source of frequency 60 Hz and voltage amplitude V, a 505-Ω resistor, and a capacitor of capacitance 7.2 μF. What must be the source voltage amplitude V for the average electrical power consumed in the resistor to be 236 W? There is no inductance in the circuit.arrow_forwardAn L−R−C series circuit has R= 280 Ω . At the frequency of the source, the inductor has reactance XLL= 905 Ω and the capacitor has reactance XC= 485 Ω . The amplitude of the voltage across the inductor is 445 V . What is the amplitude of the voltage across the resistor and the capacitor? What is the voltage amplitude of the source? What is the rate at which the source is delivering electrical energy to the circuit?arrow_forward
- A 0.185 H inductor is connected in series with a 98.5 Ω resistor and an ac source. The voltage across the inductor is vL=−(12.5V)sin[(476rad/s)t]vL. Derive an expression for the voltage vR across the resistor. Express your answer in terms of the variables L, R, VL (amplitude of the voltage across the inductor), ω, and t. What is vR at 2.13 ms ? Please explain all stepsarrow_forwardA worker lifts a box under the following conditions:Horizontal distance (H): 30 cmInitial height (V): 60 cmVertical travel (D): 50 cmTorso rotation (A): 30°Frequency: 3 times/minute for 1 hourGrip: Good Question:What is the RWL for this task?What does this value mean in terms of occupational safety?arrow_forwardCan someone helparrow_forward
- Can someone help mearrow_forward3. Four identical small masses are connected in a flat perfect square. Rank the relative rotational inertias (IA, IB, IC) about the three axes of rotation shown. Axes A and B are in the plane of the square, and axis C is perpendicular to the plane, through mass m1. ΙΑ IB m2 m1 m3 Ic m4 (a) IAarrow_forwardConsider the circuit shown in the figure below. (Assume L = 5.20 m and R2 = 440 Ω.) (a) When the switch is in position a, for what value of R1 will the circuit have a time constant of 15.4 µs? (b) What is the current in the inductor at the instant the switch is thrown to position b?arrow_forwardarrow_back_iosSEE MORE QUESTIONSarrow_forward_ios
- Principles of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningCollege PhysicsPhysicsISBN:9781305952300Author:Raymond A. Serway, Chris VuillePublisher:Cengage LearningCollege PhysicsPhysicsISBN:9781285737027Author:Raymond A. Serway, Chris VuillePublisher:Cengage Learning
- Physics for Scientists and Engineers with Modern ...PhysicsISBN:9781337553292Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningPhysics for Scientists and Engineers: Foundations...PhysicsISBN:9781133939146Author:Katz, Debora M.Publisher:Cengage LearningClassical Dynamics of Particles and SystemsPhysicsISBN:9780534408961Author:Stephen T. Thornton, Jerry B. MarionPublisher:Cengage Learning





