Problem:
(a) Calculate the number of cells in a hummingbird assuming the mass of an average cell is ten times the mass of a bacterium.
(b) Making the same assumption, how many cells are there in a human?
Solution:
Part A
The mass of a hummingbird is 10-2 kg, while the mass of a cell is 10-15 kg. The number of cells in the hummingbird is
\frac{10^{-2}}{10\left(10^{-15}\right)}=10^{12}\:\text{cells} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)Part B
The mass of a person is 102 kg.
\frac{10^2}{10\left(10^{-15}\right)}=10^{16}\:\text{cells}\ \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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