
Concept explainers
(a)
The free-fall acceleration on the surface of the satellite.
(a)

Answer to Problem 9P
The free-fall acceleration on the surface of the satellite is
Explanation of Solution
Write the expression for the force on the test object.
Here,
Write the expression for the gravitational force.
Here,
Equate equation (I) and (II) to solve for
Conclusion:
Substitute
Therefore, the free-fall acceleration on the surface of the satellite is
(b)
The time taken by the athlete to climb the cliff vertically.
(b)

Answer to Problem 9P
The time taken by the athlete to climb the cliff vertically is
Explanation of Solution
Write the expression for the equation of motion in vertical direction.
Here,
Set the acceleration in the vertical direction,
Use equation (V) in (IV) to solve for
If the object starts from the rest, the equation (VI) becomes,
Use equation (VII) to solve for
Conclusion:
Substitute
Therefore, the time taken by the athlete to climb the cliff vertically is
(c)
The distance covered by the athlete from the base of the vertical cliff to the icy surface of the satellite.
(c)

Answer to Problem 9P
The distance covered by the athlete from the base of the vertical cliff to the icy surface of the satellite is
Explanation of Solution
Write the expression for the equation of motion in horizontal direction
Here,
Conclusion:
Substitute
Therefore, the distance covered by the athlete from the base of the vertical cliff to the icy surface of the satellite is
(d)
The vector impact velocity of the athlete in climbing the cliff.
(d)

Answer to Problem 9P
The vector impact velocity of the athlete in climbing the cliff is
Explanation of Solution
Write the expression for the velocity vector.
Here, is the final velocity vector,
Write the expression for
Here,
Write the expression for the magnitude of the
Write the expression for the angle of the cliff making with the
Here,
Conclusion:
Substitute
Substitute
Substitute
Therefore, the vector impact velocity of the athlete in climbing the cliff is
Want to see more full solutions like this?
Chapter 11 Solutions
Bundle: Principles of Physics: A Calculus-Based Text, 5th + WebAssign Printed Access Card for Serway/Jewett's Principles of Physics: A Calculus-Based Text, 5th Edition, Multi-Term
- A capacitor with a capacitance of C = 5.95×10−5 F is charged by connecting it to a 12.5 −V battery. The capacitor is then disconnected from the battery and connected across an inductor with an inductance of L = 1.55 H . At the time 2.35×10−2 s after the connection to the inductor is made, what is the current in the inductor? At that time, how much electrical energy is stored in the inductor?arrow_forwardCan someone help me with this question. Thanks.arrow_forwardCan someone help me with this question. Thanks.arrow_forward
- Principles of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningPhysics for Scientists and Engineers: Foundations...PhysicsISBN:9781133939146Author:Katz, Debora M.Publisher:Cengage LearningGlencoe Physics: Principles and Problems, Student...PhysicsISBN:9780078807213Author:Paul W. ZitzewitzPublisher:Glencoe/McGraw-Hill
- Classical Dynamics of Particles and SystemsPhysicsISBN:9780534408961Author:Stephen T. Thornton, Jerry B. MarionPublisher:Cengage LearningUniversity Physics Volume 1PhysicsISBN:9781938168277Author:William Moebs, Samuel J. Ling, Jeff SannyPublisher:OpenStax - Rice UniversityPhysics for Scientists and Engineers, Technology ...PhysicsISBN:9781305116399Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning





