12.1: Prelude to Static Equilibrium and Elasticity
In earlier chapters, you learned about forces and Newton’s laws for translational motion. You then studied torques and the rotational motion of a body about a fixed axis of rotation. You also learned that static equilibrium means no motion at all and that dynamic equilibrium means motion without acceleration.
In this chapter, we combine the conditions for static translational equilibrium and static rotational equilibrium to describe situations typical for any kind of construction. What type of cable will support a suspension bridge? What type of foundation will support an office building? Will this prosthetic arm function correctly? These are examples of questions that contemporary engineers must be able to answer.
The elastic properties of materials are especially important in engineering applications, including bioengineering. For example, materials that can stretch or compress and then return to their original form or position make good shock absorbers. In this chapter, you will learn about some applications that combine equilibrium with elasticity to construct real structures that last.


29.3: Amphibians
Skills to Develop
- Describe the important difference between the life cycle of amphibians and the life cycles of other vertebrates
- Distinguish between the characteristics of Urodela, Anura, and Apoda
- Describe the evolutionary history of amphibians
Amphibians are vertebrate tetrapods. Amphibia includes frogs, salamanders, and caecilians. The term amphibian loosely translates from the Greek as “dual life,” which is a reference to the metamorphosis that many frogs and salamanders undergo and their mixture of aquatic and terrestrial environments in their life cycle. Amphibians evolved during the Devonian period and were the earliest terrestrial tetrapods.
Link to Learning
Watch this series of Animal Planet videos on tetrapod evolution:
1: The evolution from fish to earliest tetrapod
Characteristics of Amphibians
As tetrapods, most amphibians are characterized by four well-developed limbs. Some species of salamanders and all caecilians are functionally limbless; their limbs are vestigial. An important characteristic of extant amphibians is a moist, permeable skin that is achieved via mucus glands that keep the skin moist; thus, exchange of oxygen and carbon dioxide with the environment can take place through it ( cutaneous respiration ). Additional characteristics of amphibians include pedicellate teeth—teeth in which the root and crown are calcified, separated by a zone of noncalcified tissue—and a papilla amphibiorum and papilla basilaris, structures of the inner ear that are sensitive to frequencies below and above 10,00 hertz, respectively. Amphibians also have an auricular operculum, which is an extra bone in the ear that transmits sounds to the inner ear. All extant adult amphibians are carnivorous, and some terrestrial amphibians have a sticky tongue that is used to capture prey.
Evolution of Amphibians
The fossil record provides evidence of the first tetrapods: now-extinct amphibian species dating to nearly 400 million years ago. Evolution of tetrapods from fishes represented a significant change in body plan from one suited to organisms that respired and swam in water, to organisms that breathed air and moved onto land; these changes occurred over a span of 50 million years during the Devonian period. One of the earliest known tetrapods is from the genus Acanthostega . Acanthostega was aquatic; fossils show that it had gills similar to fishes. However, it also had four limbs, with the skeletal structure of limbs found in present-day tetrapods, including amphibians. Therefore, it is thought that Acanthostega lived in shallow waters and was an intermediate form between lobe-finned fishes and early, fully terrestrial tetrapods. What preceded Acanthostega ?
In 2006, researchers published news of their discovery of a fossil of a “tetrapod-like fish,” Tiktaalik roseae , which seems to be an intermediate form between fishes having fins and tetrapods having limbs (Figure \(\PageIndex{1}\)). Tiktaalik likely lived in a shallow water environment about 375 million years ago. 1
The early tetrapods that moved onto land had access to new nutrient sources and relatively few predators. This led to the widespread distribution of tetrapods during the early Carboniferous period, a period sometimes called the “age of the amphibians.”
Modern Amphibians
Amphibia comprises an estimated 6,770 extant species that inhabit tropical and temperate regions around the world. Amphibians can be divided into three clades: Urodela (“tailed-ones”), the salamanders; Anura (“tail-less ones”), the frogs; and Apoda (“legless ones”), the caecilians.
Urodela: Salamanders
Salamanders are amphibians that belong to the order Urodela. Living salamanders (Figure \(\PageIndex{1}\)) include approximately 620 species, some of which are aquatic, other terrestrial, and some that live on land only as adults. Adult salamanders usually have a generalized tetrapod body plan with four limbs and a tail. They move by bending their bodies from side to side, called lateral undulation, in a fish-like manner while “walking” their arms and legs fore and aft. It is thought that their gait is similar to that used by early tetrapods. Respiration differs among different species. The majority of salamanders are lungless, and respiration occurs through the skin or through external gills. Some terrestrial salamanders have primitive lungs; a few species have both gills and lungs.
Unlike frogs, virtually all salamanders rely on internal fertilization of the eggs. The only male amphibians that possess copulatory structures are the caecilians, so fertilization among salamanders typically involves an elaborate and often prolonged courtship. Such a courtship allows the successful transfer of sperm from male to female via a spermatophore. Development in many of the most highly evolved salamanders, which are fully terrestrial, occurs during a prolonged egg stage, with the eggs guarded by the mother. During this time, the gilled larval stage is found only within the egg capsule, with the gills being resorbed, and metamorphosis being completed, before hatching. Hatchlings thus resemble tiny adults.
Link to Learning
View River Monsters: Fish With Arms and Hands? to see a video about an unusually large salamander species.
Anura: Frogs
Frogs are amphibians that belong to the order Anura (Figure \(\PageIndex{3}\)). Anurans are among the most diverse groups of vertebrates, with approximately 5,965 species that occur on all of the continents except Antarctica. Anurans have a body plan that is more specialized for movement. Adult frogs use their hind limbs to jump on land. Frogs have a number of modifications that allow them to avoid predators, including skin that acts as camouflage. Many species of frogs and salamanders also release defensive chemicals from glands in the skin that are poisonous to predators.
Frog eggs are fertilized externally, and like other amphibians, frogs generally lay their eggs in moist environments. A moist environment is required as eggs lack a shell and thus dehydrate quickly in dry environments. Frogs demonstrate a great diversity of parental behaviors, with some species laying many eggs and exhibiting little parental care, to species that carry eggs and tadpoles on their hind legs or backs. The life cycle of frogs, as other amphibians, consists of two distinct stages: the larval stage followed by metamorphosis to an adult stage. The larval stage of a frog, the tadpole , is often a filter-feeding herbivore. Tadpoles usually have gills, a lateral line system, long-finned tails, and lack limbs. At the end of the tadpole stage, frogs undergo metamorphosis into the adult form (Figure \(\PageIndex{4}\)). During this stage, the gills, tail, and lateral line system disappear, and four limbs develop. The jaws become larger and are suited for carnivorous feeding, and the digestive system transforms into the typical short gut of a predator. An eardrum and air-breathing lungs also develop. These changes during metamorphosis allow the larvae to move onto land in the adult stage.
Apoda: Caecilians
An estimated 185 species comprise caecilians , a group of amphibians that belong to the order Apoda. Although they are vertebrates, a complete lack of limbs leads to their resemblance to earthworms in appearance. They are adapted for a soil-burrowing or aquatic lifestyle, and they are nearly blind. These animals are found in the tropics of South America, Africa, and Southern Asia. They have vestigial limbs, evidence that they evolved from a legged ancestor.
Evolution Connection: The Paleozoic Era and the Evolution of Vertebrates
The climate and geography of Earth was vastly different during the Paleozoic Era, when vertebrates arose, as compared to today. The Paleozoic spanned from approximately 542 to 251 million years ago. The landmasses on Earth were very different from those of today. Laurentia and Gondwana were continents located near the equator that subsumed much of the current day landmasses in a different configuration (Figure \(\PageIndex{5}\)). At this time, sea levels were very high, probably at a level that hasn’t been reached since. As the Paleozoic progressed, glaciations created a cool global climate, but conditions warmed near the end of the first half of the Paleozoic. During the latter half of the Paleozoic, the landmasses began moving together, with the initial formation of a large northern block called Laurasia. This contained parts of what is now North America, along with Greenland, parts of Europe, and Siberia. Eventually, a single supercontinent, called Pangaea, was formed, starting in the latter third of the Paleozoic. Glaciations then began to affect Pangaea’s climate, affecting the distribution of vertebrate life.
During the early Paleozoic, the amount of carbon dioxide in the atmosphere was much greater than it is today. This may have begun to change later, as land plants became more common. As the roots of land plants began to infiltrate rock and soil began to form, carbon dioxide was drawn out of the atmosphere and became trapped in the rock. This reduced the levels of carbon dioxide and increased the levels of oxygen in the atmosphere, so that by the end of the Paleozoic, atmospheric conditions were similar to those of today.
As plants became more common through the latter half of the Paleozoic, microclimates began to emerge and ecosystems began to change. As plants and ecosystems continued to grow and become more complex, vertebrates moved from the water to land. The presence of shoreline vegetation may have contributed to the movement of vertebrates onto land. One hypothesis suggests that the fins of aquatic vertebrates were used to maneuver through this vegetation, providing a precursor to the movement of fins on land and the development of limbs. The late Paleozoic was a time of diversification of vertebrates, as amniotes emerged and became two different lines that gave rise, on one hand, to mammals, and, on the other hand, to reptiles and birds. Many marine vertebrates became extinct near the end of the Devonian period, which ended about 360 million years ago, and both marine and terrestrial vertebrates were decimated by a mass extinction in the early Permian period about 250 million years ago.
Link to Learning
View Earth’s Paleogeography: Continental Movements Through Time to see changes in Earth as life evolved.
Summary
As tetrapods, most amphibians are characterized by four well-developed limbs, although some species of salamanders and all caecilians are limbless. The most important characteristic of extant amphibians is a moist, permeable skin used for cutaneous respiration. The fossil record provides evidence of amphibian species, now extinct, that arose over 400 million years ago as the first tetrapods. Amphibia can be divided into three clades: salamanders (Urodela), frogs (Anura), and caecilians (Apoda). The life cycle of frogs, like the majority of amphibians, consists of two distinct stages: the larval stage and metamorphosis to an adult stage. Some species in all orders bypass a free-living larval stage.
Footnotes
- 1 Daeschler, E. B., Shubin, N. H., and Jenkins, F. J. “A Devonian tetrapod-like fish and the evolution of the tetrapod body plan,” Nature 440 (2006): 757–763, doi:10.1038/nature04639, http://www.nature.com/nature/journal/v440/n7085/abs/nature04639.html.
Glossary
- Acanthostega
- one of the earliest known tetrapods
- Amphibia
- frogs, salamanders, and caecilians
- Anura
- frogs
- Apoda
- caecilians
- caecilian
- legless amphibian that belongs to the clade Apoda
- cutaneous respiration
- gas exchange through the skin
- frog
- tail-less amphibian that belongs to the clade Anura
- salamander
- tailed amphibian that belongs to the clade Urodela
- tadpole
- larval stage of a frog
- Urodela
- salamanders


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5.4.13: Critical Thinking Case
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Uber Pays the Price
Nine years ago, Uber revolutionized the taxi industry and the way people commute. With the simple mission “to bring transportation—for everyone, everywhere,” today Uber has reached a valuation of around $70 billion and claimed a market share high of almost 90% in 2015. However, in June 2017 Uber experienced a series of bad press regarding an alleged culture of sexual harassment, which is what most experts believe caused their market share to fall to 75%.
In February of 2017 a former software engineer, Susan Fowler, wrote a lengthy post on her website regarding her experience of being harassed by a manager who was not disciplined by human resources for his behavior. In her post, Fowler wrote that Uber’s HR department and members of upper management told her that because it was the man’s first offense, they would only give him a warning. During her meeting with HR about the incident, Fowler was also advised that she should transfer to another department within the organization. According to Fowler, she was ultimately left no choice but to transfer to another department, despite having specific expertise in the department in which she had originally been working.
As her time at the company went on, she began meeting other women who worked for the company who relayed their own stories of harassment. To her surprise, many of the women reported being harassed by the same person who had harassed her. As she noted in her blog, “It became obvious that both HR and management had been lying about this being his 'first offense.'” Fowler also reported a number of other instances that she identified as sexist and inappropriate within the organization and claims that she was disciplined severely for continuing to speak out. Fowler eventually left Uber after about two years of working for the company, noting that during her time at Uber the percentage of women working there had dropped to 6% of the workforce, down from 25% when she first started.
Following the fallout from Fowler’s lengthy description of the workplace on her website, Uber’s chief executive Travis Kalanick publicly condemned the behavior described by Fowler, calling it “abhorrent and against everything Uber stands for and believes in.” But later in March, Uber board member Arianna Huffington claimed that she believed “sexual harassment was not a systemic problem at the company.” Amid pressure from bad media attention and the company’s falling market share, Uber made some changes after an independent investigation resulted in 215 complaints. As a result, 20 employees were fired for reasons ranging from sexual harassment to bullying to retaliation to discrimination, and Kalanick announced that he would hire a chief operating officer to help manage the company. In an effort to provide the leadership team with more diversity, two senior female executives were hired to fill the positions of chief brand officer and senior vice president for leadership and strategy.
Critical Thinking Questions:
- Based on Cox’s business case for diversity, what are some positive outcomes that may result in changes to Uber’s leadership team?
- Under what form of federal legislation was Fowler protected?
- What strategies should have been put in place to help prevent sexual harassment incidents like this from happening in the first place?
Sources: Uber corporate Website, https://www.uber.com/newsroom/company-info/ (February, 2017); Marco della Cava, “Uber has lost market share to Lyft during crisis,” USA Today , June 13, 2017, https://www.usatoday.com/story/tech/...als/102795024/ ; Tracey Lien, “Uber fires 20 workers after harassment investigation,” Los Angeles Times , Jun 6, 2017, http://www.latimes.com/business/la-f...606-story.html ; Susan Fowler, “Reflecting On One Very, Very Strange Year At Uber,” February 19, 2017, https://www.susanjfowler.com/blog/20...e-year-at-uber .


2.10: §16. The Legacy of Latin- I. Old English
Although this course is in no sense a history of the English language, you should have some idea about the various historical circumstances that brought English into contact with Latin. Depending upon the chronological period when a Latin word came into English, it may be totally transformed in appearance, or, conversely, retain almost its exact original spelling. In some periods, Latin words were likely to have passed through another language before entering English; at other times, they tended to come directly from Latin to English. A knowledge of the historical context can provide insight into otherwise puzzling features of morphological change, enabling us to appreciate why Latin words assume such a diversity of forms in the English language. In three installments, therefore—here, and in parallel sections of Chapters 3 and 4—you will be given a broad summary of the interrelationship of Latin and English. The story is quite varied and fascinating: it involves military conquest; trade and commercial intercourse; invasions and other population movements; missionary activity; and a powerful cultural and intellectual influence that continues to this day.
You will have no trouble finding any number of books on the history of English, if you wish to pursue this subject in greater depth. One interesting and reliable study is C.M. Millward’s, A Biography of the English Language (Holt, Rinehart and Winston, 1989).
In the first of these three sections, let us consider the impact of Latin on English in the era prior to the Norman Conquest of AD 1066.
Although the portion of the British Isles now called England had been conquered by the ancient Romans and made part of their empire, that conquest assumes very little importance in our present study. This is not because Rome’s contact with Britain had been casual or superficial: after the Emperor Claudius decided in AD 43 to subjugate this island that had been first raided in 55 BC by his great-grandfather, Julius Caesar, the province of Britain was firmly established within the system of imperial government. It remained that way for some three and a half centuries, until the problems of a disintegrating empire forced the Romans to withdraw in AD 410. The archaeological heritage of the Roman occupation is most impressive, from the great baths and villas of southern England to Hadrian’s Wall in the north. The linguistic legacy, however, is very meagre: it consists mainly of the word mile , derived from Latin mille (“a thousand” [1] ), and a number of English place names that are derived from Latin castra (“army camp”)— Chester, Manchester, Doncaster, Lancaster, and the like. One reason is that the native population of Roman Britain appears not to have been as thoroughly assimilated as their counterparts in Roman Gaul or Spain. Also, we must remember that the Britons of this period were tribal Celts, a people later overwhelmed by successive waves of Germanic invaders. At the time of the Roman occupation, the language that we call “English” had not yet come into being.
Soon after the Romans withdrew, Britain was subjected to a series of invasions by Germanic tribes living along the north German coast, from Frisia in the Netherlands to Denmark. Mainly Angles, Saxons, and Jutes, these invaders brought with them closely related low German dialects that merged into the language popularly known as “Anglo-Saxon,” which scholars prefer to call OLD ENGLISH . ( England is “the land of the Angles.”) The earliest of the Anglo-Saxon invasions is traditionally dated to AD 449.
Having come from a region that was close to the Roman imperial frontier, these Germanic peoples had already been in contact with speakers of Latin, through trade and commerce, and they brought with them a number of Latin loan-words that filled gaps in their native vocabulary. These are among the oldest Latin words in English, having likely been part of Anglo-Saxon speech as early as the 5th or 6th centuries AD. Because the first English manuscript written in the Roman alphabet dates only from c.737 AD, it is impossible to document these early words, let alone date them precisely. Probably they were carried to Britain by Anglo-Saxon tribes from the mainland, but they may have entered Old English at a somewhat later date.
Here is a selection, presented in their Modern English form:
| ENGLISH | LATIN |
ORIGINAL LATIN MEANING
|
| street |
[via] strata
|
“layered” or “paved” [road] |
| wall | vallum | “rampart,” “wall” |
| wine | vinum | “wine” |
| cook | coquus | “cook” |
| kitchen | coquina | “kitchen” |
| chalk | calx | “limestone” |
| cheese | caseus | “cheese” |
| pitch | pix | “pitch” |
| post | postis | “pillar,” “post’ |
| pound |
[libra] pondo
|
[pound] “by weight” |
| mint | moneta | “coined money,” “mint” |
| inch | uncia | “twelfth part,” “inch” |
The above words all relate, in one way or another, to everyday secular life. However, there is another, rather different group of Latin derivatives that found its way into Old English at a somewhat later time; these are words that have to do with Christianity and aspects of Christian worship. In all likelihood, this terminology was adopted in Britain, during the seventh and eighth centuries, after the arrival of the first Christian missionaries around AD 600. [2] There is a considerable number of these words, many of which go back through Latin to a source in Greek—for Greek was the language in which early Christianity had spread throughout the Mediterranean world. The following is a sample only:
| ENGLISH | LATIN | ORIGINAL GREEK FORM |
| pope | papa | papas (“father”) |
| bishop | episcopus | episkopos (“overseer”) |
| hymn | hymnus | hymnos (“hymn”) |
| school | schola | scholē (“leisure”) |
| monk | monachus | monachos (“monk”) |
| minster | monasterium | monastērion |
| nun | nonna | (Latin: “child’s nurse,” “nun”) |
| creed | credo | (Latin: “I believe”) |
By the time of the Norman Invasion in 1066, there may have been as many as 300 Latin derivatives in Old English. By later standards, this was still a very low number.
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The Roman mile comprised a thousand double paces of an army on the march; this worked out to 1,618 English yards, a little less than the later English mile of 1,760 yards. Even the word mile is not a certain legacy of the Roman conquest; many scholars think it entered Old English as a Germanic loan-word. ↵
-
The missionary St. Augustine arrived at Canterbury in AD 597, bringing also the Roman alphabet. Before the end of the seventh century, England could be described as a Christian country. ↵


7.4: How do we Eradicate Infectious Diseases?
To completely eradicate an infectious disease from the planet, the pathogen that causes that disease must be completely gone or unable to cause the disease in humans anymore. Although this is a possibility for many diseases, eradication has only happened once - with smallpox. Other diseases have been eliminated from some geographical areas (where there is no longer transmission of that disease), but haven’t been completely eradicated. Often, poorer countries are burdened with highly preventable, curable, or treatable diseases that are no longer problematic in wealthier countries. And just because a disease has been eliminated or even eradicated, doesn’t mean we don’t still need to be prepared for the possibility of it re-emerging, particularly if it is a zoonotic or vector-borne disease (Roser et al., 2024).


15.1: Embryogenesis
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Learning Objectives
- Explain the steps of embryogensis in eudicots.
- Compare embryogenesis in eudicots versus monocots.
The Seed Plants chapter discussed the fertilization of the egg within the ovule. The zygote ultimately divides to produce the mature embryo, the ovule develops into a seed, and the ovary that contained one or more ovules develops into a fruit. A typical seed contains a seed coat, cotyledons, endosperm, and a single embryo. The development of the embryo occurs through a process called embryogenesis .
Embryogenesis in Eudicots
After fertilization in eudicots, the zygote (Figure \(\PageIndex{1-I}\)) divides to form two cells: the upper apical cell and the lower basal cell . The division of the basal cell gives rise to the suspensor , which attaches the embryo to the micropyle (the pore through which the pollen tube original entered). The suspensor provides a route for nutrition to be transported from the mother plant to the growing embryo. The apical cell also divides, initially producing a proembryo (figure \(\PageIndex{1-II}\)).
As the proembryo continues to divide, it takes a spherical form, called the globular stage (Figures \(\PageIndex{1-III}\) and \(\PageIndex{2a}\)). The globular is the first stage that is considered the embryo proper . Next, cotyledons arise from the embryo proper, forming the heart stage (Figures \(\PageIndex{1-IV}\) and \(\PageIndex{2-4}\)). Cotyledons are embryonic leaf-life structures that function in food storage, food absorption and/or photosynthesis. As the cotyledons elongate, and the base of the embryo thickens, it results a torpedo. (This stage is called the torpedo stage; Figures \(\PageIndex{1-V}\) and \(\PageIndex{2c}\)). Cell division is concentrated at the shoot apical meristem , located at the shoot tip in between the cotyledons, and the root apical meristem at the most basal (bottom) part of the embryo. Most of the suspensor deteriorates during the torpedo stage.
The final stage of embryogenesis results in the mature embryo (Figures \(\PageIndex{1-VI}\) and \(\PageIndex{2d}\)). The mature embryo includes an embryonic root called the radicle . The embryo becomes dormant at this point, halting metabolic activity and cell division. At this point, the seed is ready for dispersal. Growth resumes when the seed germinates and the embryo develops into a seedling.
In some eudicots, the endosperm (triploid tissue that was formed when a sperm fertilized the two polar nuclei) cells divide, and endosperm fills a substantial portion of the mature seed. The endosperm stores nutrients. In other (non-endospermic) eudicots, such as Capsella bursa-pastoris , the endosperm develops initially, but is then digested, and the nutrients are moved into the two cotyledons (Figure \(\PageIndex{2}\)). After germination, the developing seedling relies on these food reserves stored in the endosperm or cotyledons until the first set of leaves begin photosynthesis.
Embryogenesis in Monocots
The process of embryogenesis in monocots is similar to that of eudicots, but as there is only a single cotyledon, no heart stage occurs. Instead, the embryo proper of the monocot becomes cylindrical at this point in development. The shoot apical meristem, while still present at the shoot tip, is not in between cotyledons in the monocot (because there is only a single cotyledon).
Attribution
Curated and authored by Melissa Ha using 32.2 Pollination and Fertilization from Biology 2e by OpenStax (licensed CC-BY ). Access for free at openstax.org .


3.E: Interference (Exercises)
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Conceptual Questions
3.1 Young's Double-Slit Interference
1. Young’s double-slit experiment breaks a single light beam into two sources. Would the same pattern be obtained for two independent sources of light, such as the headlights of a distant car? Explain.
2. Is it possible to create a experimental setup in which there is only destructive interference? Explain.
3. Why won’t two small sodium lamps, held close together, produce an interference pattern on a distant screen? What if the sodium lamps were replaced by two laser pointers held close together?
3.2 Mathematics of Interference
4. Suppose you use the same double slit to perform Young’s double-slit experiment in air and then repeat the experiment in water. Do the angles to the same parts of the interference pattern get larger or smaller? Does the color of the light change? Explain.
5. Why is monochromatic light used in the double slit experiment? What would happen if white light were used?
3.4 Interference in Thin Films
6. What effect does increasing the wedge angle have on the spacing of interference fringes? If the wedge angle is too large, fringes are not observed. Why?
7. How is the difference in paths taken by two originally in-phase light waves related to whether they interfere constructively or destructively? How can this be affected by reflection? By refraction?
8. Is there a phase change in the light reflected from either surface of a contact lens floating on a person’s tear layer? The index of refraction of the lens is about 1.5, and its top surface is dry.
9. In placing a sample on a microscope slide, a glass cover is placed over a water drop on the glass slide. Light incident from above can reflect from the top and bottom of the glass cover and from the glass slide below the water drop. At which surfaces will there be a phase change in the reflected light?
10. Answer the above question if the fluid between the two pieces of crown glass is carbon disulfide.
11. While contemplating the food value of a slice of ham, you notice a rainbow of color reflected from its moist surface. Explain its origin.
12. An inventor notices that a soap bubble is dark at its thinnest and realizes that destructive interference is taking place for all wavelengths. How could she use this knowledge to make a nonreflective coating for lenses that is effective at all wavelengths? That is, what limits would there be on the index of refraction and thickness of the coating? How might this be impractical?
13. A nonreflective coating like the one described in Example 3.3 works ideally for a single wavelength and for perpendicular incidence. What happens for other wavelengths and other incident directions? Be specific.
14. Why is it much more difficult to see interference fringes for light reflected from a thick piece of glass than from a thin film? Would it be easier if monochromatic light were used?
3.5 The Michelson Interferometer
15. Describe how a Michelson interferometer can be used to measure the index of refraction of a gas (including air).
Problems
3.2 Mathematics of Interference
16. At what angle is the first-order maximum for 450-nm wavelength blue light falling on double slits separated by 0.0500 mm?
17. Calculate the angle for the third-order maximum of 580-nm wavelength yellow light falling on double slits separated by 0.100 mm.
18. What is the separation between two slits for which 610-nm orange light has its first maximum at an angle of \(\displaystyle 30.0°\)?
19. Find the distance between two slits that produces the first minimum for 410-nm violet light at an angle of \(\displaystyle 45.0°\).
20. Calculate the wavelength of light that has its third minimum at an angle of \(\displaystyle 30.0°\) when falling on double slits separated by \(\displaystyle 3.00μm\). Explicitly show how you follow the steps from the Problem-Solving Strategy: Wave Optics, located at the end of the chapter.
21. What is the wavelength of light falling on double slits separated by \(\displaystyle 2.00μm\) if the third-order maximum is at an angle of \(\displaystyle 60.0°\)?
22. At what angle is the fourth-order maximum for the situation in the preceding problem?
23. What is the highest-order maximum for 400-nm light falling on double slits separated by \(\displaystyle 25.0μm\)?
24. Find the largest wavelength of light falling on double slits separated by \(\displaystyle 1.20μm\) for which there is a first-order maximum. Is this in the visible part of the spectrum?
25. What is the smallest separation between two slits that will produce a second-order maximum for 720-nm red light?
26. (a) What is the smallest separation between two slits that will produce a second-order maximum for any visible light?
(b) For all visible light?
27. (a) If the first-order maximum for monochromatic light falling on a double slit is at an angle of \(\displaystyle 10.0°\), at what angle is the second-order maximum?
(b) What is the angle of the first minimum?
(c) What is the highest-order maximum possible here?
28. Shown below is a double slit located a distance x from a screen, with the distance from the center of the screen given by y. When the distance d between the slits is relatively large, numerous bright spots appear, called fringes. Show that, for small angles (where \(\displaystyle sinθ≈θ\), with \(\displaystyle θ\) in radians), the distance between fringes is given by \(\displaystyle Δy=xλ/d\)
Picture shows a double slit located a distance x from a screen, with the distance from the center of the screen given by y. Distance between the slits is d.
29. Using the result of the preceding problem,
(a) calculate the distance between fringes for 633-nm light falling on double slits separated by 0.0800 mm, located 3.00 m from a screen.
(b) What would be the distance between fringes if the entire apparatus were submersed in water, whose index of refraction is 1.33?
30. Using the result of the problem two problems prior, find the wavelength of light that produces fringes 7.50 mm apart on a screen 2.00 m from double slits separated by 0.120 mm.
31. In a double-slit experiment, the fifth maximum is 2.8 cm from the central maximum on a screen that is 1.5 m away from the slits. If the slits are 0.15 mm apart, what is the wavelength of the light being used?
32. The source in Young’s experiment emits at two wavelengths. On the viewing screen, the fourth maximum for one wavelength is located at the same spot as the fifth maximum for the other wavelength. What is the ratio of the two wavelengths?
33. If 500-nm and 650-nm light illuminates two slits that are separated by 0.50 mm, how far apart are the second-order maxima for these two wavelengths on a screen 2.0 m away?
34. Red light of wavelength of 700 nm falls on a double slit separated by 400 nm.
(a) At what angle is the first-order maximum in the diffraction pattern?
(b) What is unreasonable about this result?
(c) Which assumptions are unreasonable or inconsistent?
3.3 Multiple-Slit Interference
35. Ten narrow slits are equally spaced 0.25 mm apart and illuminated with yellow light of wavelength 580 nm. (a) What are the angular positions of the third and fourth principal maxima? (b) What is the separation of these maxima on a screen 2.0 m from the slits?
36. The width of bright fringes can be calculated as the separation between the two adjacent dark fringes on either side. Find the angular widths of the third- and fourth-order bright fringes from the preceding problem.
37. For a three-slit interference pattern, find the ratio of the peak intensities of a secondary maximum to a principal maximum.
38. What is the angular width of the central fringe of the interference pattern of
(a) 20 slits separated by \(\displaystyle d=2.0×10^{−3}mm\)?
(b) 50 slits with the same separation? Assume that \(\displaystyle λ=600nm\).
3.4 Interference in Thin Films
39. A soap bubble is 100 nm thick and illuminated by white light incident perpendicular to its surface. What wavelength and color of visible light is most constructively reflected, assuming the same index of refraction as water?
40. An oil slick on water is 120 nm thick and illuminated by white light incident perpendicular to its surface. What color does the oil appear (what is the most constructively reflected wavelength), given its index of refraction is 1.40?
41. Calculate the minimum thickness of an oil slick on water that appears blue when illuminated by white light perpendicular to its surface. Take the blue wavelength to be 470 nm and the index of refraction of oil to be 1.40.
42. Find the minimum thickness of a soap bubble that appears red when illuminated by white light perpendicular to its surface. Take the wavelength to be 680 nm, and assume the same index of refraction as water.
43. A film of soapy water (\(\displaystyle n=1.33\)) on top of a plastic cutting board has a thickness of 233 nm. What color is most strongly reflected if it is illuminated perpendicular to its surface?
44. What are the three smallest non-zero thicknesses of soapy water (\(\displaystyle n=1.33\)) on Plexiglas if it appears green (constructively reflecting 520-nm light) when illuminated perpendicularly by white light?
45. Suppose you have a lens system that is to be used primarily for 700-nm red light. What is the second thinnest coating of fluorite (magnesium fluoride) that would be nonreflective for this wavelength?
46. (a) As a soap bubble thins it becomes dark, because the path length difference becomes small compared with the wavelength of light and there is a phase shift at the top surface. If it becomes dark when the path length difference is less than one-fourth the wavelength, what is the thickest the bubble can be and appear dark at all visible wavelengths? Assume the same index of refraction as water.
(b) Discuss the fragility of the film considering the thickness found.
47. To save money on making military aircraft invisible to radar, an inventor decides to coat them with a nonreflective material having an index of refraction of 1.20, which is between that of air and the surface of the plane. This, he reasons, should be much cheaper than designing Stealth bombers.
(a) What thickness should the coating be to inhibit the reflection of 4.00-cm wavelength radar?
(b) What is unreasonable about this result?
(c) Which assumptions are unreasonable or inconsistent?
3.5 The Michelson Interferometer
48. A Michelson interferometer has two equal arms. A mercury light of wavelength 546 nm is used for the interferometer and stable fringes are found. One of the arms is moved by \(\displaystyle 1.5μm\). How many fringes will cross the observing field?
49. What is the distance moved by the traveling mirror of a Michelson interferometer that corresponds to 1500 fringes passing by a point of the observation screen? Assume that the interferometer is illuminated with a 606 nm spectral line of krypton-86.
50. When the traveling mirror of a Michelson interferometer is moved \(\displaystyle 2.40×10^{−5}m\), 90 fringes pass by a point on the observation screen. What is the wavelength of the light used?
51. In a Michelson interferometer, light of wavelength 632.8 nm from a He-Ne laser is used. When one of the mirrors is moved by a distance D, 8 fringes move past the field of view. What is the value of the distance D?
52. A chamber 5.0 cm long with flat, parallel windows at the ends is placed in one arm of a Michelson interferometer (see below). The light used has a wavelength of 500 nm in a vacuum. While all the air is being pumped out of the chamber, 29 fringes pass by a point on the observation screen. What is the refractive index of the air?
Picture shows a schematics of a set-up utilized to measure the refractive index of a gas. The glass chamber with a gas is placed in the Michelson interferometer between the half-silvered mirror M and mirror M1. The space inside the container is 5 cm wide.
Additional Problems
53. For 600-nm wavelength light and a slit separation of 0.12 mm, what are the angular positions of the first and third maxima in the double slit interference pattern?
54. If the light source in the preceding problem is changed, the angular position of the third maximum is found to be \(\displaystyle 0.57°\). What is the wavelength of light being used now?
55. Red light (\(\displaystyle λ=710.nm\)) illuminates double slits separated by a distance \(\displaystyle d=0.150mm\). The screen and the slits are 3.00 m apart.
(a) Find the distance on the screen between the central maximum and the third maximum.
(b) What is the distance between the second and the fourth maxima?
56. Two sources as in phase and emit waves with \(\displaystyle λ=0.42m\). Determine whether constructive or destructive interference occurs at points whose distances from the two sources are
(a) 0.84 and 0.42 m,
(b) 0.21 and 0.42 m,
(c) 1.26 and 0.42 m,
(d) 1.87 and 1.45 m,
(e) 0.63 and 0.84 m and
(f) 1.47 and 1.26 m.
57. Two slits \(\displaystyle 4.0×10^{−6}m\) apart are illuminated by light of wavelength 600 nm. What is the highest order fringe in the interference pattern?
58. Suppose that the highest order fringe that can be observed is the eighth in a double-slit experiment where 550-nm wavelength light is used. What is the minimum separation of the slits?
59. The interference pattern of a He-Ne laser light (\(\displaystyle λ=632.9nm\)) passing through two slits 0.031 mm apart is projected on a screen 10.0 m away. Determine the distance between the adjacent bright fringes.
60. Young’s double-slit experiment is performed immersed in water (\(\displaystyle n=1.333\)). The light source is a He-Ne laser, \(\displaystyle λ=632.9nm\) in vacuum.
(a) What is the wavelength of this light in water?
(b) What is the angle for the third order maximum for two slits separated by 0.100 mm.
61. A double-slit experiment is to be set up so that the bright fringes appear 1.27 cm apart on a screen 2.13 m away from the two slits. The light source was wavelength 500 nm. What should be the separation between the two slits?
62. An effect analogous to two-slit interference can occur with sound waves, instead of light. In an open field, two speakers placed 1.30 m apart are powered by a single-function generator producing sine waves at 1200-Hz frequency. A student walks along a line 12.5 m away and parallel to the line between the speakers. She hears an alternating pattern of loud and quiet, due to constructive and destructive interference. What is (a) the wavelength of this sound and (b) the distance between the central maximum and the first maximum (loud) position along this line?
63. A hydrogen gas discharge lamp emits visible light at four wavelengths, \(\displaystyle λ=\)410, 434, 486, and 656 nm. (a) If light from this lamp falls on a N slits separated by 0.025 mm, how far from the central maximum are the third maxima when viewed on a screen 2.0 m from the slits? (b) By what distance are the second and third maxima separated for \(\displaystyle l=486nm\)?
64. Monochromatic light of frequency \(\displaystyle 5.5×10^{14}Hz\) falls on 10 slits separated by 0.020 mm. What is the separation between the first and third maxima on a screen that is 2.0 m from the slits?
65. Eight slits equally separated by 0.149 mm is uniformly illuminated by a monochromatic light at \(\displaystyle λ=523nm\). What is the width of the central principal maximum on a screen 2.35 m away?
66. Eight slits equally separated by 0.149 mm is uniformly illuminated by a monochromatic light at \(\displaystyle λ=523nm\). What is the intensity of a secondary maxima compared to that of the principal maxima?
67. A transparent film of thickness 250 nm and index of refraction of 1.40 is surrounded by air. What wavelength in a beam of white light at near-normal incidence to the film undergoes destructive interference when reflected?
68. An intensity minimum is found for 450 nm light transmitted through a transparent film (\(\displaystyle n=1.20\)) in air.
(a) What is minimum thickness of the film?
(b) If this wavelength is the longest for which the intensity minimum occurs, what are the next three lower values of \(\displaystyle λ\) for which this happens?
69. A thin film with \(\displaystyle n=1.32\) is surrounded by air. What is the minimum thickness of this film such that the reflection of normally incident light with \(\displaystyle λ=500nm\) is minimized?
70. Repeat your calculation of the previous problem with the thin film placed on a flat glass (\(\displaystyle n=1.50\)) surface.
71. After a minor oil spill, a think film of oil (\(\displaystyle n=1.40\)) of thickness 450 nm floats on the water surface in a bay. (a) What predominant color is seen by a bird flying overhead? (b) What predominant color is seen by a seal swimming underwater?
72. A microscope slide 10 cm long is separated from a glass plate at one end by a sheet of paper. As shown below, the other end of the slide is in contact with the plate. he slide is illuminated from above by light from a sodium lamp (\(\displaystyle λ=589nm\)), and 14 fringes per centimeter are seen along the slide. What is the thickness of the piece of paper? Picture shows a microscope slide that touches the glass plate at one end and is separated from it at another end by a sheet of paper.
73. Suppose that the setup of the preceding problem is immersed in an unknown liquid. If 18 fringes per centimeter are now seen along the slide, what is the index of refraction of the liquid?
74. A thin wedge filled with air is produced when two flat glass plates are placed on top of one another and a slip of paper is inserted between them at one edge. Interference fringes are observed when monochromatic light falling vertically on the plates are seen in reflection. Is the first fringe near the edge where the plates are in contact a bright fringe or a dark fringe? Explain.
75. Two identical pieces of rectangular plate glass are used to measure the thickness of a hair. The glass plates are in direct contact at one edge and a single hair is placed between them hear the opposite edge. When illuminated with a sodium lamp (\(\displaystyle λ=589nm\)), the hair is seen between the 180th and 181st dark fringes. What are the lower and upper limits on the hair’s diameter?
76. Two microscope slides made of glass are illuminated by monochromatic (\(\displaystyle λ=589nm\)) light incident perpendicularly. The top slide touches the bottom slide at one end and rests on a thin copper wire at the other end, forming a wedge of air. The diameter of the copper wire is 29.45μm . How many bright fringes are seen across these slides?
77. A good quality camera “lens” is actually a system of lenses, rather than a single lens, but a side effect is that a reflection from the surface of one lens can bounce around many times within the system, creating artifacts in the photograph. To counteract this problem, one of the lenses in such a system is coated with a thin layer of material (\(\displaystyle n=1.28\)) on one side. The index of refraction of the lens glass is 1.68. What is the smallest thickness of the coating that reduces the reflection at 640 nm by destructive interference? (In other words, the coating’s effect is to be optimized for \(\displaystyle λ=640nm\).)
78. Constructive interference is observed from directly above an oil slick for wavelengths (in air) 440 nm and 616 nm. The index of refraction of this oil is \(\displaystyle n=1.54\). What is the film’s minimum possible thickness?
79. A soap bubble is blown outdoors. What colors (indicate by wavelengths) of the reflected sunlight are seen enhanced? The soap bubble has index of refraction 1.36 and thickness 380 nm.
80. A Michelson interferometer with a He-Ne laser light source (\(\displaystyle λ=632.8nm\)) projects its interference pattern on a screen. If the movable mirror is caused to move by \(\displaystyle 8.54μm\), how many fringes will be observed shifting through a reference point on a screen?
81. An experimenter detects 251 fringes when the movable mirror in a Michelson interferometer is displaced. The light source used is a sodium lamp, wavelength 589 nm. By what distance did the movable mirror move?
82. A Michelson interferometer is used to measure the wavelength of light put through it. When the movable mirror is moved by exactly 0.100 mm, the number of fringes observed moving through is 316. What is the wavelength of the light?
83. A 5.08-cm-long rectangular glass chamber is inserted into one arm of a Michelson interferometer using a 633-nm light source. This chamber is initially filled with air (\(\displaystyle n=1.000293\)) at standard atmospheric pressure but the air is gradually pumped out using a vacuum pump until a near perfect vacuum is achieved. How many fringes are observed moving by during the transition?
84. Into one arm of a Michelson interferometer, a plastic sheet of thickness \(\displaystyle 75μm\) is inserted, which causes a shift in the interference pattern by 86 fringes. The light source has wavelength of 610 nm in air. What is the index of refraction of this plastic?
85. The thickness of an aluminum foil is measured using a Michelson interferometer that has its movable mirror mounted on a micrometer. There is a difference of 27 fringes in the observed interference pattern when the micrometer clamps down on the foil compared to when the micrometer is empty. Calculate the thickness of the foil?
86. The movable mirror of a Michelson interferometer is attached to one end of a thin metal rod of length 23.3 mm. The other end of the rod is anchored so it does not move. As the temperature of the rod changes from \(\displaystyle 15°C\) to \(\displaystyle 25C\), a change of 14 fringes is observed. The light source is a He Ne laser, \(\displaystyle λ=632.8nm\). What is the change in length of the metal bar, and what is its thermal expansion coefficient?
87. In a thermally stabilized lab, a Michelson interferometer is used to monitor the temperature to ensure it stays constant. The movable mirror is mounted on the end of a 1.00-m-long aluminum rod, held fixed at the other end. The light source is a He Ne laser, \(\displaystyle λ=632.8nm\). The resolution of this apparatus corresponds to the temperature difference when a change of just one fringe is observed. What is this temperature difference?
88. A 65-fringe shift results in a Michelson interferometer when a \(\displaystyle 42.0-μm\) film made of an unknown material is placed in one arm. The light source has wavelength 632.9 nm. Identify the material using the indices of refraction found in Table 1.1.
Challenge Problems
89. Determine what happens to the double-slit interference pattern if one of the slits is covered with a thin, transparent film whose thickness is λ/[2(n−1)]λ/[2(n−1)], where λλ is the wavelength of the incident light and n is the index of refraction of the film.
90. Fifty-one narrow slits are equally spaced and separated by 0.10 mm. The slits are illuminated by blue light of wavelength 400 nm. What is angular position of the twenty-fifth secondary maximum? What is its peak intensity in comparison with that of the primary maximum?
91. A film of oil on water will appear dark when it is very thin, because the path length difference becomes small compared with the wavelength of light and there is a phase shift at the top surface. If it becomes dark when the path length difference is less than one-fourth the wavelength, what is the thickest the oil can be and appear dark at all visible wavelengths? Oil has an index of refraction of 1.40.
92. Figure 3.14 shows two glass slides illuminated by monochromatic light incident perpendicularly. The top slide touches the bottom slide at one end and rests on a 0.100-mm-diameter hair at the other end, forming a wedge of air. (a) How far apart are the dark bands, if the slides are 7.50 cm long and 589-nm light is used? (b) Is there any difference if the slides are made from crown or flint glass? Explain.
93. Figure 3.14 shows two 7.50-cm-long glass slides illuminated by pure 589-nm wavelength light incident perpendicularly. The top slide touches the bottom slide at one end and rests on some debris at the other end, forming a wedge of air. How thick is the debris, if the dark bands are 1.00 mm apart?
94. A soap bubble is 100 nm thick and illuminated by white light incident at a \(\displaystyle 45°\) angle to its surface. What wavelength and color of visible light is most constructively reflected, assuming the same index of refraction as water?
95. An oil slick on water is 120 nm thick and illuminated by white light incident at a \(\displaystyle 45°\) angle to its surface. What color does the oil appear (what is the most constructively reflected wavelength), given its index of refraction is 1.40?


6.7: Review Questions
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1.
After performing a community assessment, a nurse determines that a neighborhood with a higher concentration of BIPOC clients lacks access to medical care. Which term describes a major factor contributing to this finding?
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Structural racism
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Personally mediated racism
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Internalized racism
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Scientific racism
2.
Which of these nurse actions would be considered an example of microaggression?
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Asking a transgender client their pronouns
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Telling an Asian client that they speak perfect English
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Determining the dietary preferences of a Jewish client
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Obtaining a medical interpreter for a client who does not speak English
3.
Which of the following is an example of racial profiling?
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Suspecting a person of criminal behavior based on race
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Segregating communities based on race
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Assuming all men of a certain racial group are dangerous
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Community-wide disinvestment based on race
4.
Which statement best highlights the continuing prevalence of racism in the U.S. health care system?
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A university researcher has built a relationship with the local community.
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Some states have attempted to impose work-reporting requirements as a means of restricting Medicaid expansion.
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The city council is considering a summer jobs program for local teenagers.
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The police department has established a community advisory board in each precinct.
5.
Which client being seen in the community health clinic does the nurse anticipate is most affected by allostatic load?
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A 52-year-old Native American construction worker with osteoarthritis of the knees
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A 24-year-old Black man with persistent hypertension
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A 62-year-old White woman with newly diagnosed hypothyroidism
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A 38-year-old Hispanic woman concerned about a change in the color of a mole
6.
Which environmental condition is a consequence of the redlining of neighborhoods that has negatively affected health outcomes of BIPOC individuals?
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Decreased urban heat
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Less green space
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Improved air quality
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Increased access to clean water
7.
A public health nurse running a cost-free vaccination clinic for COVID-19 in an underserved community notes that the turnout is lower than expected. Which factor is the most likely explanation for the low turnout?
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Medicaid expansion
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Low perceived personal risk for COVID
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Lack of health insurance
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Distrust of the health system
8.
A nurse is caring for four clients in the medical-surgical inpatient unit. One is a Black woman who has been admitted with pancreatitis, another is a White man who has been admitted with acute low back pain, the third is a Black man with pneumonia, and the fourth is a White woman who has been admitted with cholecystitis. The nurse medicates the Black woman with one oral Percocet tab for 8/10 pain, medicates the White man with two oxycodone tabs for 7/10 pain, does not medicate the Black man as he does not have current pain, and medicates the White woman with 2 mg of IV morphine for 8/10 pain. What is this an example of?
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Explicit racism
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Implicit bias
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Professional standard of care
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Structural racism
9.
The community health nurse is working on creating nutrition initiatives for a community with a high incidence of heart disease and diabetes. Which of the following is a driving factor in the persistence of food deserts in areas of the United States?
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Structural racism
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Implicit bias
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Explicit racism
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Mass incarceration
10.
Which of the following issues are implicated in the current maternal mortality public health crisis?
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Differences in health insurance coverage and access
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Income level
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Provider discrimination
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Distrust in the medical system


3.6: Heat Capacities of an Ideal Gas
By the end of this section, you will be able to:
- Define heat capacity of an ideal gas for a specific process
- Calculate the specific heat of an ideal gas for either an isobaric or isochoric process
- Explain the difference between the heat capacities of an ideal gas and a real gas
- Estimate the change in specific heat of a gas over temperature ranges
We learned about specific heat and molar heat capacity previously; however, we have not considered a process in which heat is added. We do that in this section. First, we examine a process where the system has a constant volume, then contrast it with a system at constant pressure and show how their specific heats are related.
Let’s start with looking at Figure \(\PageIndex{1}\), which shows two vessels A and B , each containing 1 mol of the same type of ideal gas at a temperature T and a volume V . The only difference between the two vessels is that the piston at the top of A is fixed, whereas the one at the top of B is free to move against a constant external pressure p . We now consider what happens when the temperature of the gas in each vessel is slowly increased to \(T + dT\) with the addition of heat.
Since the piston of vessel A is fixed, the volume of the enclosed gas does not change. Consequently, the gas does no work, and we have from the first law
\[dE_{int} = dQ - dW = dQ. \nonumber \]
We represent the fact that the heat is exchanged at constant volume by writing
\[dQ = C_VndT, \nonumber \] where \(C_V\) is the molar heat capacity at constant volume of the gas. In addition, since \(dE_{int} = dQ\) for this particular process,
\[dE_{int} = C_VndT. \label{3.9} \]
We obtained this equation assuming the volume of the gas was fixed. However, internal energy is a state function that depends on only the temperature of an ideal gas. Therefore, \(dE_{int} = C_VndT\) gives the change in internal energy of an ideal gas for any process involving a temperature change dT .
When the gas in vessel B is heated, it expands against the movable piston and does work \(dW = pdV\). In this case, the heat is added at constant pressure, and we write \[dQ = C_{p}ndT, \nonumber \] where \(C_p\) is the molar heat capacity at constant pressure of the gas. Furthermore, since the ideal gas expands against a constant pressure,
\[d(pV) = d(RnT) \nonumber \] becomes \[pdV = RndT. \nonumber \]
Finally, inserting the expressions for dQ and pdV into the first law, we obtain
\[dE_{int} = dQ - pdV = (C_{p}n - Rn)dT. \nonumber \]
We have found \(dE_{int}\) for both an isochoric and an isobaric process. Because the internal energy of an ideal gas depends only on the temperature, \(dE_{int}\) must be the same for both processes. Thus,
\[C_{V}ndT = (C_{p}n - Rn)dT, \nonumber \]
and
\[C_p = C_V + R. \label{eq50} \]
The derivation of Equation \ref{eq50} was based only on the ideal gas law. Consequently, this relationship is approximately valid for all dilute gases, whether monatomic like He, diatomic like \(O_2\), or polyatomic like \(CO_2\) or \(NH_3\).
In the preceding chapter, we found the molar heat capacity of an ideal gas under constant volume to be
\[C_V = \dfrac{d}{2}R, \nonumber \]
where d is the number of degrees of freedom of a molecule in the system. Table \(\PageIndex{1}\) shows the molar heat capacities of some dilute ideal gases at room temperature. The heat capacities of real gases are somewhat higher than those predicted by the expressions of \(C_V\) and \(C_p\) given in Equation \ref{eq50}. This indicates that vibrational motion in polyatomic molecules is significant, even at room temperature. Nevertheless, the difference in the molar heat capacities, \(C_p - C_V\), is very close to R , even for the polyatomic gases.
| \(C_p\) | \(C_V\) | \(C_p - C_V\) | ||
|---|---|---|---|---|
| Type of Molecule | Gas | (J/mol K) | (J/mol K) | (J/mol K) |
| Monatomic | Ideal | \(\frac{5}{2}R = 20.79\) | \(\frac{3}{2}R = 12.47\) | \(R = 8.31\) |
| Diatomic | Ideal | \(\frac{7}{2}R = 29.10\) | \(\frac{5}{2}R = 20.79\) | \(R = 8.31\) |
| Polyatomic | Ideal | \(4R = 33.26\) | \(3R = 24.04\) | \(R = 8.31\) |
Glossary
- molar heat capacity at constant pressure
- quantifies the ratio of the amount of heat added removed to the temperature while measuring at constant pressure
- molar heat capacity at constant volume
- quantifies the ratio of the amount of heat added removed to the temperature while measuring at constant volume


<IP_ADDRESS>: Active Transport
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Active transport mechanisms require the use of the cell’s energy, usually in the form of adenosine triphosphate (ATP). If a substance must move into the cell against its concentration gradient, that is, if the concentration of the substance inside the cell must be greater than its concentration in the extracellular fluid, the cell must use energy to move the substance. Some active transport mechanisms move small-molecular weight material, such as ions, through the membrane.
In addition to moving small ions and molecules through the membrane, cells also need to remove and take in larger molecules and particles. Some cells are even capable of engulfing entire unicellular microorganisms. You might have correctly hypothesized that the uptake and release of large particles by the cell requires energy. A large particle, however, cannot pass through the membrane, even with energy supplied by the cell.
Electrochemical Gradient
We have discussed simple concentration gradients—differential concentrations of a substance across a space or a membrane. However, in living systems gradients are more complex. Cells contain many proteins, most of which are negatively charged. Due to these negatively charged proteins, coupled with the movement of ions into and out of cells, there is an electrical gradient (a difference of charge) across the plasma membrane. The interior of living cells is electrically negative as compared to the extracellular fluid in which cells are bathed; at the same time, cells contain higher concentrations of potassium (K + ) and lower concentrations of sodium (Na + ) than does the extracellular fluid. Thus, in a living cell, the concentration gradient and electrical gradient of Na + promotes diffusion of the ion into the cell, and the electrical gradient of Na + (a positive ion) tends to drive it inward to the negatively charged interior. The situation is more complex, however, for other elements such as potassium. The electrical gradient of K + promotes diffusion of the ion into the cell, but the concentration gradient of K + promotes diffusion out of the cell ( Figure \(\PageIndex{1}\) ). The combined gradient that affects an ion is called its electrochemical gradient , and it is especially important to muscle and nerve cells.
Moving Against a Gradient
To move substances against a concentration or an electrochemical gradient, the cell must use energy. This energy is harvested from ATP that is generated through cellular metabolism. Active transport mechanisms, collectively called pumps or carrier proteins, work against electrochemical gradients. With the exception of ions, small substances constantly pass through plasma membranes. Active transport maintains concentrations of ions and other substances needed by living cells in the face of these passive changes. Much of a cell’s supply of metabolic energy may be spent maintaining these processes. As active transport mechanisms depend on cellular metabolism for energy, they are sensitive to many metabolic poisons that interfere with the supply of ATP.
Two mechanisms exist for the transport of small-molecular weight material and macromolecules. Primary active transport moves ions across a membrane and creates a difference in charge across that membrane. The primary active transport system uses ATP to move a substance, such as an ion, into the cell, and often at the same time, a second substance is moved out of the cell. The sodium-potassium pump, an important pump in animal cells, expends energy to move potassium ions into the cell and a different number of sodium ions out of the cell. The action of this pump results in a concentration and charge difference across the membrane.
Secondary active transport describes the movement of material using the energy of the electrochemical gradient established by primary active transport. Using the energy of the electrochemical gradient created by the primary active transport system, other substances such as amino acids and glucose can be brought into the cell through membrane channels. ATP itself is formed through secondary active transport using a hydrogen ion gradient in the mitochondrion.
Endocytosis
Endocytosis is a type of active transport that moves particles, such as large molecules, parts of cells, and even whole cells, into a cell. There are different variations of endocytosis, but all share a common characteristic: The plasma membrane of the cell invaginates, forming a pocket around the target particle. The pocket pinches off, resulting in the particle being contained in a newly created vacuole that is formed from the plasma membrane.
Phagocytosis is the process by which large particles, such as cells, are taken in by a cell. For example, when microorganisms invade the human body, a type of white blood cell called a neutrophil removes the invader through this process, surrounding and engulfing the microorganism, which is then destroyed by the neutrophil ( Figure \(\PageIndex{2}\) a).
A variation of endocytosis is called pinocytosis . This literally means “cell drinking” and was named at a time when the assumption was that the cell was purposefully taking in extracellular fluid. In reality, this process takes in solutes that the cell needs from the extracellular fluid ( Figure \(\PageIndex{2}\) b).
A targeted variation of endocytosis employs binding proteins in the plasma membrane that are specific for certain substances ( Figure \(\PageIndex{2}\) c). The particles bind to the proteins and the plasma membrane invaginates, bringing the substance and the proteins into the cell. If passage across the membrane of the target of receptor-mediated endocytosis is ineffective, it will not be removed from the tissue fluids or blood. Instead, it will stay in those fluids and increase in concentration.
Some human diseases are caused by a failure of receptor-mediated endocytosis. For example, the form of cholesterol termed low-density lipoprotein or LDL (also referred to as “bad” cholesterol) is removed from the blood by receptor mediated endocytosis. In the human genetic disease familial hypercholesterolemia, the LDL receptors are defective or missing entirely. People with this condition have life-threatening levels of cholesterol in their blood, because their cells cannot clear the chemical from their blood.
Exocytosis
In contrast to these methods of moving material into a cell is the process of exocytosis. Exocytosis is the opposite of the processes discussed above in that its purpose is to expel material from the cell into the extracellular fluid. A particle enveloped in membrane fuses with the interior of the plasma membrane. This fusion opens the membranous envelope to the exterior of the cell, and the particle is expelled into the extracellular space ( Figure \(\PageIndex{3}\) ).
Query \(\PageIndex{1}\)
Query \(\PageIndex{2}\)
References
Unless otherwise noted, images on this page are licensed under CC-BY 4.0 by OpenStax.
Text adapted from: OpenStax, Concepts of Biology. OpenStax CNX. May 18, 2016 http://cnx.org/contents/b3c1e1d2-839...9a8aafbdd@9.10


2.6: Non-right Triangles - Law of Cosines
In this section, you will:
- Use the Law of Cosines to solve oblique triangles.
- Solve applied problems using the Law of Cosines.
- Use Heron’s formula to find the area of a triangle.
Suppose a boat leaves port, travels \(10\) miles, turns \(20\) degrees, and travels another 8 miles as shown in Figure \(\PageIndex{1}\) How far from port is the boat?
Figure \(\PageIndex{1}\)
Unfortunately, while the Law of Sines enables us to address many non-right triangle cases, it does not help us with triangles where the known angle is between two known sides, a SAS (side-angle-side) triangle, or when all three sides are known, but no angles are known, a SSS (side-side-side) triangle. In this section, we will investigate another tool for solving oblique triangles described by these last two cases.
Using the Law of Cosines to Solve Oblique Triangles
The tool we need to solve the problem of the boat’s distance from the port is the Law of Cosines , which defines the relationship among angle measurements and side lengths in oblique triangles. Three formulas make up the Law of Cosines. At first glance, the formulas may appear complicated because they include many variables. However, once the pattern is understood, the Law of Cosines is easier to work with than most formulas at this mathematical level.
Understanding how the Law of Cosines is derived will be helpful in using the formulas. The derivation begins with the Generalized Pythagorean Theorem, which is an extension of the Pythagorean Theorem to non-right triangles. Here is how it works: An arbitrary non-right triangle \(ABC\) is placed in the coordinate plane with vertex \(A\) at the origin, side \(c\) drawn along the x -axis, and vertex \(C\) located at some point \((x,y)\) in the plane, as illustrated in Figure \(\PageIndex{2}\). Generally, triangles exist anywhere in the plane, but for this explanation we will place the triangle as noted.
Figure \(\PageIndex{2}\)
We can drop a perpendicular from \(C\) to the x- axis (this is the altitude or height). Recalling the basic trigonometric identities, we know that
\(\cos \theta=\dfrac{x(adjacent)}{b(hypotenuse)}\) and \(\sin \theta=\dfrac{y(opposite)}{b(hypotenuse)}\)
In terms of \(\theta\), \(x=b \cos \theta\) and \(y=b \sin \theta\). The \((x,y)\) point located at \(C\) has coordinates \((b \cos \theta, b \sin \theta)\). Using the side \((x−c)\) as one leg of a right triangle and \(y\) as the second leg, we can find the length of hypotenuse \(a\) using the Pythagorean Theorem. Thus,
\(\begin{array}{ll} a^2={(x−c)}^2+y^2 \\[4pt] \;\;\;\;\; ={(b \cos \theta−c)}^2+{(b \sin \theta)}^2 & \text{Substitute }(b \cos \theta) \text{ for }x \text{ and }(b \sin \theta)\text{ for }y \\[4pt] \;\;\;\;\;\; =(b^2{\cos}^2 \theta−2bc \cos \theta+c^2)+b^2 {\sin}^2 \theta & \text{Expand the perfect square.} \\[4pt] \;\;\;\;\; =b^2{\cos}^2 \theta+b^2{\sin}^2 \theta+c^2−2bc \cos \theta & \text{Group terms noting that }{\cos}^2 \theta+{\sin}^2 \theta=1 \\[4pt] \;\;\;\;\; =b^2({\cos}^2 \theta+{\sin}^2 \theta)+c^2−2bc \cos \theta & \text{Factor out }b^2 \\[4pt]
a^2=b^2+c^2−2bc \cos \theta \end{array}\)
The formula derived is one of the three equations of the Law of Cosines. The other equations are found in a similar fashion.
Keep in mind that it is always helpful to sketch the triangle when solving for angles or sides. In a real-world scenario, try to draw a diagram of the situation. As more information emerges, the diagram may have to be altered. Make those alterations to the diagram and, in the end, the problem will be easier to solve.
The Law of Cosines states that the square of any side of a triangle is equal to the sum of the squares of the other two sides minus twice the product of the other two sides and the cosine of the included angle.
Figure \(\PageIndex{3}\)
For triangles labeled as in Figure \(\PageIndex{3}\), with angles \(\alpha\), \(\beta\) and \(\gamma\), and opposite corresponding sides \(a\), \(b\), and \(c\), respectively, the Law of Cosines is given as three equations.
\[a^2=b^2+c^2−2bc \cos \alpha\]
\[b^2=a^2+c^2−2ac \cos \beta\]
\[c^2=a^2+b^2−2ab \cos \gamma\]
To solve for a missing side measurement, the corresponding opposite angle measure is needed.
When solving for an angle, the corresponding opposite side measure is needed. We can use another version of the Law of Cosines to solve for an angle.
\[\cos \alpha=\dfrac{b^2+c^2−a^2}{2bc}\]
\[\cos \beta=\dfrac{a^2+c^2−b^2}{2ac}\]
\[\cos \gamma=\dfrac{a^2+b^2−c^2}{2ab}\]
- Sketch the triangle. Identify the measures of the known sides and angles. Use variables to represent the measures of the unknown sides and angles.
- Apply the Law of Cosines to find the length of the unknown side or angle.
- Apply the Law of Sines or Cosines to find the measure of a second angle.
- Compute the measure of the remaining angle.
Find the unknown side and angles of the triangle in Figure \(\PageIndex{4}\).
Figure \(\PageIndex{4}\)
Solution
First, make note of what is given: two sides and the angle between them. This arrangement is classified as SAS and supplies the data needed to apply the Law of Cosines.
Each one of the three laws of cosines begins with the square of an unknown side opposite a known angle. For this example, the first side to solve for is side \(b\), as we know the measurement of the opposite angle \(\beta\).
\(\begin{array}{ll} b^2=a^2+c^2−2ac \cos \beta \\[4pt] b^2={10}^2+{12}^2−2(10)(12)\cos(30°) & \text{Substitute the measurements for the known quantities.} \\[4pt] b^2=100+144−240 \left(\dfrac{\sqrt{3}}{2}\right) & \text{Evaluate the cosine and begin to simplify.} \\[4pt] b^2=244−120\sqrt{3} \\[4pt] b=\sqrt{244−120\sqrt{3}} & \text{Use the square root property.} \\[4pt] b≈6.013 \end{array}\)
Because we are solving for a length, we use only the positive square root. Now that we know the length \(b\), we can use the Law of Sines to fill in the remaining angles of the triangle. Solving for angle \(\alpha\), we have
\(\begin{array}{cc} \dfrac{\sin \alpha}{a}=\dfrac{\sin \beta}{b} \\[4pt] \dfrac{\sin \alpha}{10}=\dfrac{\sin(30°)}{6.013} \\[4pt] \sin \alpha=\dfrac{10\sin(30°)}{6.013} & \text{Multiply both sides of the equation by }10. \\[4pt] \alpha={\sin}^{−1}\left(\dfrac{10\sin(30°)}{6.013}\right) & \text{Find the inverse sine of } \dfrac{10\sin(30°)}{6.013}. \\[4pt] \alpha≈56.3° \end{array}\)
The other possibility for \(\alpha\) would be \(\alpha=180°-56.3°≈123.7°\). In the original diagram,\(\alpha\) is adjacent to the longest side, so \(\alpha\) is an acute angle and, therefore, \(123.7°\) does not make sense. Notice that if we choose to apply the Law of Cosines, we arrive at a unique answer. We do not have to consider the other possibilities, as cosine is unique for angles between \(0°\) and \(180°\). Proceeding with \(\alpha≈56.3°\), we can then find the third angle of the triangle.
\[\begin{align*} \gamma&= 180^{\circ}-30^{\circ}-56.3^{\circ}\\ &\approx 93.7^{\circ} \end{align*}\]
The complete set of angles and sides is
\(\alpha≈56.3°\) \(a=10\)
\(\beta=30°\) \(b≈6.013\)
\(\gamma≈93.7°\) \(c=12\)
Find the missing side and angles of the given triangle: \(\alpha=30°\), \(b=12\), \(c=24\).
- Answer
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\(a≈14.9\), \(\beta≈23.8°\), \(\gamma≈126.2°\).
Find the angle \(\alpha\) for the given triangle if side \(a=20\), side \(b=25\), and side \(c=18\).
Solution
For this example, we have no angles. We can solve for any angle using the Law of Cosines. To solve for angle \(\alpha\), we have
\(\begin{array}{ll} a^2=b^2+c^2−2bc \cos \alpha \\[4pt] {20}^2={25}^2+{18}^2−2(25)(18)\cos \alpha & \text{Substitute the appropriate measurements.} \\[4pt] 400=625+324−900 \cos \alpha & \text{ Simplify in each step.} \\[4pt] 400=949−900 \cos \alpha \\[4pt] −549=−900 \cos \alpha & \text{Isolate }\cos \alpha. \\[4pt] 0.61≈\cos \alpha & \text{Find the inverse cosine.} \\[4pt] \alpha≈52.4° \end{array}\)
See Figure \(\PageIndex{5}\).
Figure \(\PageIndex{5}\)
Analysis
Because the inverse cosine can return any angle between \(0\) and \(180\) degrees, there will not be any ambiguous cases using this method.
Given \(a=5\), \(b=7\), and \(c=10\), find the missing angles.
- Answer
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\(\alpha≈27.7°\), \(\beta≈40.5°\), \(\gamma≈111.8°\)
Solving Applied Problems Using the Law of Cosines
Just as the Law of Sines provided the appropriate equations to solve a number of applications, the Law of Cosines is applicable to situations in which the given data fits the cosine models. We may see these in the fields of navigation, surveying, astronomy, and geometry, just to name a few.
On many cell phones with GPS, an approximate location can be given before the GPS signal is received. This is accomplished through a process called triangulation, which works by using the distances from two known points. Suppose there are two cell phone towers within range of a cell phone. The two towers are located \(6000\) feet apart along a straight highway, running east to west, and the cell phone is north of the highway. Based on the signal delay, it can be determined that the signal is \(5050\) feet from the first tower and \(2420\) feet from the second tower. Determine the position of the cell phone north and east of the first tower, and determine how far it is from the highway.
Solution
For simplicity, we start by drawing a diagram similar to Figure \(\PageIndex{6}\) and labeling our given information.
Figure \(\PageIndex{6}\)
Using the Law of Cosines, we can solve for the angle \(\theta\). Remember that the Law of Cosines uses the square of one side to find the cosine of the opposite angle. For this example, let \(a=2420\), \(b=5050\), and \(c=6000\). Thus, \(\theta\) corresponds to the opposite side \(a=2420\).
\[\begin{align*} a^2 & =b^2+c^2−2bc \cos \theta \\[4pt] {(2420)}^2 &={(5050)}^2+{(6000)}^2−2(5050)(6000) \cos \theta \\[4pt] \cos \theta &≈ 0.9183 \\[4pt] \theta &≈ {\cos}^{−1}(0.9183) \\[4pt] \theta &≈ 23.3° \end{align*}\]
To answer the questions about the phone’s position north and east of the tower, and the distance to the highway, drop a perpendicular from the position of the cell phone, as in Figure \(\PageIndex{7}\). This forms two right triangles, although we only need the right triangle that includes the first tower for this problem.
Figure \(\PageIndex{7}\)
Using the angle \(\theta=23.3\)° and the basic trigonometric identities, we can find the solutions. Thus
\[\begin{align*} \cos(23.3°) &= \dfrac{x}{5050} \\[4pt] x &= 5050\cos(23.3°) \\[4pt] x &≈ 4638.15\, feet\\[4pt] \sin(23.3°) &= \dfrac{y}{5050} \\[4pt] y &= 5050\sin(23.3°) \\[4pt] y &≈1997.5 \, feet \end{align*}\]
The cell phone is approximately \(4638\) feet east and \(1998\) feet north of the first tower, and \(1998\) feet from the highway.
Returning to our problem at the beginning of this section, suppose a boat leaves port, travels \(10\) miles, turns \(20\) degrees, and travels another \(8\) miles. How far from port is the boat? The diagram is repeated here in Figure \(\PageIndex{8}\).
Figure \(\PageIndex{8}\)
Solution
The boat turned 20 degrees, so the obtuse angle of the non-right triangle is the supplemental angle, \(180°−20°=160°\). With this, we can utilize the Law of Cosines to find the missing side of the obtuse triangle—the distance of the boat to the port.
\[\begin{align*} x^2 &= 8^2+{10}^2−2(8)(10)\cos(160°) \\[4pt] x^2 &= 314.35 \\[4pt] x &= \sqrt{314.35} \\[4pt] x&≈17.7\, miles \end{align*}\]
The boat is about \(17.7\) miles from port.
Using Heron’s Formula to Find the Area of a Triangle
We already learned how to find the area of an oblique triangle when we know two sides and an angle. We also know the formula to find the area of a triangle using the base and the height. When we know the three sides, however, we can use Heron’s formula instead of finding the height. Heron of Alexandria was a geometer who lived during the first century A.D. He discovered a formula for finding the area of oblique triangles when three sides are known.
Heron’s formula finds the area of oblique triangles in which sides \(a\), \(b\),and \(c\) are known.
\[Area=\sqrt{s(s−a)(s−b)(s−c)}\]
where \(s=\dfrac{(a+b+c)}{2}\) is one half of the perimeter of the triangle, sometimes called the semi-perimeter.
Find the area of the triangle in Figure \(\PageIndex{9}\) using Heron’s formula.
Figure \(\PageIndex{9}\)
Solution
First, we calculate \(s\).
\[\begin{align*} s&= \dfrac{(a+b+c)}{2}\\ s&= \dfrac{(10+15+7)}{2}\\ &= 16 \end{align*}\]
Then we apply the formula.
\[\begin{align*} Area&= \sqrt{s(s-a)(s-b)(s-c)}\\ Area&= \sqrt{16(16-10)(16-15)(16-7)}\\ Area&\approx 29.4 \end{align*}\]
The area is approximately \(29.4\) square units.
Use Heron’s formula to find the area of a triangle with sides of lengths \(a=29.7\) ft, \(b=42.3\) ft, and \(c=38.4\) ft.
- Answer
-
Area = \(552\) square feet
A Chicago city developer wants to construct a building consisting of artist’s lofts on a triangular lot bordered by Rush Street, Wabash Avenue, and Pearson Street. The frontage along Rush Street is approximately \(62.4\) meters, along Wabash Avenue it is approximately \(43.5\) meters, and along Pearson Street it is approximately \(34.1\) meters. How many square meters are available to the developer? See Figure \(\PageIndex{10}\) for a view of the city property.
Figure \(\PageIndex{10}\)
Solution
Find the measurement for \(s\), which is one-half of the perimeter.
\[\begin{align*} s&= \dfrac{(62.4+43.5+34.1)}{2}\\ s&= 70\; m\\ \text {Apply Heron's formula.}\\ Area&= \sqrt{70(70-62.4)(70-43.5)(70-34.1)}\\ Area&= \sqrt{506,118.2}\\ Area&\approx 711.4 \end{align*}\]
The developer has about \(711.4\) square meters.
Find the area of a triangle given \(a=4.38\) ft , \(b=3.79\) ft, and \(c=5.22\) ft.
- Answer
-
about \(8.15\) square feet
Access these online resources for additional instruction and practice with the Law of Cosines.
Key Equations
| Law of Cosines |
\(a^2=b^2+c^2−2bc \cos \alpha\) \(b^2=a^2+c^2−2ac \cos \beta\) \(c^2=a^2+b^2−2ab \cos \gamma\) |
| Heron’s formula |
\(Area=\sqrt{s(s−a)(s−b)(s−c)}\) where \(s=\dfrac{(a+b+c)}{2}\) |
Key Concepts
- The Law of Cosines defines the relationship among angle measurements and lengths of sides in oblique triangles.
- The Generalized Pythagorean Theorem is the Law of Cosines for two cases of oblique triangles: SAS and SSS. Dropping an imaginary perpendicular splits the oblique triangle into two right triangles or forms one right triangle, which allows sides to be related and measurements to be calculated. See Example \(\PageIndex{1}\) and Example \(\PageIndex{2}\).
- The Law of Cosines is useful for many types of applied problems. The first step in solving such problems is generally to draw a sketch of the problem presented. If the information given fits one of the three models (the three equations), then apply the Law of Cosines to find a solution. See Example \(\PageIndex{3}\) and Example \(\PageIndex{4}\).
- Heron’s formula allows the calculation of area in oblique triangles. All three sides must be known to apply Heron’s formula. See Example \(\PageIndex{5}\) and See Example \(\PageIndex{6}\).
Contributors and Attributions
-
Jay Abramson (Arizona State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at https://openstax.org/details/books/precalculus .