/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 7 Blo Bioluminescence Some species... [FREE SOLUTION] | 91Ó°ÊÓ

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Blo Bioluminescence Some species of dinoflagellate (a type of unicellular plankton) can produce light as the result of biochemical reactions within the cell. This light is an example of bioluminescence. It is found that bioluminescence in dinoflagellates can be triggered by deformation of the cell surface with a pressure as low as one dyne \(\left(10^{-5} \mathrm{N}\right)\) per square centimeter. What is this pressure in (a) pascals and (b) atmospheres?

Short Answer

Expert verified
The pressure is 1 × 10^{-6} Pa and approximately 9.87 × 10^{-12} atm.

Step by step solution

01

Convert Dynes to Pascals

First, we need to convert the given pressure in dynes per square centimeter to pascals (Pa), the SI unit for pressure. Since 1 dyne/cm² equals 0.1 pascal, we can calculate as follows:\[1 \text{ dyne/cm}^2 = 0.1 \text{ Pa}\] Thus,\[1 \times 10^{-5} \text{ N/cm}^2 = 0.1 \times 10^{-5} \text{ Pa}\]Therefore, the pressure is \[1 \times 10^{-6} \text{ Pa}\]
02

Convert Pascals to Atmospheres

Next, we need to convert the calculated pressure in pascals to atmospheres (atm). The conversion is based on the fact that 1 atmosphere equals 101,325 pascals. Use the conversion factor:\[1 \text{ atm} = 101325 \text{ Pa}\]Therefore:\[\frac{1 \times 10^{-6} \text{ Pa}}{101325 \text{ Pa/atm}} \approx 9.87 \times 10^{-12} \text{ atm}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Dinoflagellates
Dinoflagellates are fascinating creatures that thrive in aquatic environments. These single-celled organisms are a type of plankton frequently found in oceanic and freshwater ecosystems. What makes dinoflagellates unique is their ability to exhibit bioluminescence, the phenomenon where organisms emit light.
Bioluminescence in dinoflagellates occurs through biochemical reactions inside the cells. This light emission can serve various purposes, such as repelling predators or communicating with other organisms. The emission of light is often triggered by external stimuli, such as changes in pressure or physical disturbances.
The glow of the dinoflagellates is so intriguing that it can make the ocean water appear to sparkle, especially noticeable at night. Watching this natural light show can be a mesmerizing experience.
Biochemical Reactions
Biochemical reactions in organisms are crucial for processes like energy conversion, growth, and response to stimuli. In the context of dinoflagellates, these reactions are responsible for producing light through bioluminescence.
The light in bioluminescence is produced through a reaction involving a light-emitting molecule called luciferin and an enzyme called luciferase. When energy is provided, typically by the oxidation of luciferin, the reaction emits photons, resulting in visible light.
This process is highly efficient, meaning it produces more light and less heat compared to many other forms of light production. The efficiency allows organisms to use it as a means of communication or defense without wasting energy, demonstrating the brilliant design of nature's chemical reactions.
Pressure Conversion
Pressure conversion is an important skill when dealing with physics problems, especially in scientific calculations. Pressure is the force exerted per unit area, and it can be measured in various units like pascals (Pa), atmospheres (atm), and dynes.
In the original exercise, the pressure is given in dynes per square centimeter, which needs to be converted to more commonly used units like pascals. To do this, we use conversion factors. For instance, 1 dyne per square centimeter is equivalent to 0.1 pascal.
Converting between units requires careful attention to the conversion factors, as mistakes can lead to incorrect results. Once the pressure is expressed in pascals, it can then be converted to atmospheres using the relationship that 1 atm equals 101,325 pascals. Accurate conversions are essential for precise scientific measurements and calculations.
Physics Problem Solving
Solving physics problems effectively involves a series of steps and clear understanding of underlying principles. Whether it's bioluminescence or pressure conversion exercises, the approach usually starts with understanding the given data and the relationships between different physical quantities.
In the given exercise, we start with the pressure expressed in dynes, which is an uncommon unit outside of some specialized fields. The challenge is to convert it into standard units, which requires knowing conversion factors and applying them correctly. Each step builds on the previous one, reinforcing the understanding of units and conversion principles.
Practice with such problems not only improves problem-solving skills but also enhances one's ability to apply physics concepts in real-world scenarios. By mastering these skills, students can tackle a wide range of problems, from simple conversions to complex applications of physical laws.

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Most popular questions from this chapter

Doughnuts are cooked by dropping the dough into hot vegetable oil until it changes from white to a rich, golden brown. One popular brand of doughnut automates this process; the doughnuts are made on an assembly line that customers can view in operation as they wait to order. Watching the doughnuts cook gives the customers time to develop an appetite as they ponder the physics of the process. First, the uncooked doughnut is dropped into hot vegetable oil, whose density is \(\rho=919 \mathrm{kg} / \mathrm{m}^{3} .\) There it browns on one side as it floats on the oil. After cooking for the proper amount of time, a mechanical lever flips the doughnut over so it can cook on the other side. The doughnut floats fairly high in the oil, with less than half of its volume submerged. As a result, the final product has a characteristic white stripe around the middle where the dough is always out of the oil, as indicated in Figure \(15-40\). The connection between the density of the doughnut and the height of the white stripe is illustrated in Figure \(15-41\). On the \(x\) axis we plot the density of the doughnut as a fraction of the density of the vegetable cooking oil; the \(y\) axis shows the height of the white stripe as a fraction of the total height of the doughnut. Notice that the height of the white stripe is plotted for both positive and negative values. Assuming the doughnut has a cylindrical shape of height \(H\) and diameter \(D\), and that the height of the white stripe is \(0.22 H,\) what is the density of the doughnut? A. \(260 \mathrm{kg} / \mathrm{m}^{3}\) B. \(360 \mathrm{kg} / \mathrm{m}^{3}\) \(\mathrm{C} .720 \mathrm{kg} / \mathrm{m}^{3}\) D. \(820 \mathrm{kg} / \mathrm{m}^{3}\)

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The weight of your \(1420-\mathrm{kg}\) car is supported equally by its four tires, each inflated to a gauge pressure of \(35.0 \mathrm{lb} / \mathrm{in}^{2}\). (a) What is the area of contact each tire makes with the road? (b) If the gauge pressure is increased, does the area of contact increase, decrease, or stay the same? (c) What gauge pressure is required to give an area of contact of \(116 \mathrm{cm}^{2}\) for each tire?

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