Problem: Assuming one nerve impulse must end before another can begin, what is the maximum firing rate of a nerve in impulses per second?
Solution:
One nerve impulse lasts for 10-3 s.
\text{max firing rate}=\frac{1\:\text{nerve impulse}}{10^{-3}\:\text{sec}}=10^3\:\text{impulses/sec} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)College Physics Chapter 1 Problems
College Physics 2nd Edition Solutions Table of Contents
Chapter 1: Introduction: The Nature of Science and Physics
Chapter 3: Two-Dimensional Kinematics
Chapter 4: Dynamics: Force and Newton’s Law of Motion
Chapter 5: Further Applications of Newton’s Laws: Friction, Drag, and Elasticity
Chapter 6: Uniform Circular Motion and Gravitation
Chapter 7: Work, Energy, and Energy Resources
Chapter 8: Linear Momentum and Collisions
Chapter 10: Rotational Motion and Angular Momentum
Chapter 12: Fluid Dynamics and Its Biological and Medical Applications
Chapter 13: Temperature, Kinetic Theory, and the Gas Laws
Chapter 14: Heat and Heat Transfer Methods
Chapter 16: Oscillatory Motion and Waves
Chapter 17: Physics of Hearing
Chapter 18: Electric Charge and Electric Field
Chapter 19: Electric Potential and Electric Field
Chapter 20:
Electric Current, Resistance, and Ohm’s Law
Chapter 21: Circuits and DC Instruments
Chapter 23: Electromagnetic Induction, AC Circuits, and Electrical Technologies
Chapter 24: Electromagnetic Waves
Chapter 26: Vision and Optical Instrument
Chapter 28: Special Relativity
Chapter 29: Introduction to Quantum Physics
Chapter 31: Radioactivity and Nuclear Physics
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