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Chapter 5: Problem 77

Bacteria swim using a corkscrew-like helical flagellum that rotates. For a bacterium with a flagellum that has a pitch of \(1.0 \mu \mathrm{m}\) that rotates at 110 rev/s, how fast could it swim if there were no "slippage" in the medium in which it is swimming? The pitch of a helix is the distance between "threads."

Short Answer

Expert verified
Answer: The swimming speed of the bacterium without slippage would be 110 μm/s.

Step by step solution

01

Determine the distance per rotation of the flagellum

The pitch of flagellum is given as \(1.0 \mu \mathrm{m}\), which means the bacterium moves forward by \(1.0 \mu \mathrm{m}\) per revolution of its flagellum.
02

Calculate the speed of the bacterium

Now, since the flagellum rotates at 110 revolutions per second (rev/s), we need to multiply the pitch by the rotation speed to find the speed at which the bacterium swims without slippage. Speed = Distance per rotation × Rotation speed Speed = \((1.0 \mu \mathrm{m})(110 \, \text{rev/s})\) Speed = \(110 \, \mu \mathrm{m/s}\) The bacterium could swim at a speed of \(110 \, \mu \mathrm{m/s}\) without any slippage in the medium it's swimming in.

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