Biological characteristics of a species that influence how quickly a population can potentially increase in number is life-history strategies.
What are life history strategies?The life history of a species gives us an idea of how a species survives and reproduces during its lifetime.This includes factors which affect the fitness of an organism like life span, maturity rate, death rate, parental care.Two types of life history strategies are Semelparous (those that can reproduce only once in their lifetime) and Iteroparous (those that can reproduce multiple times over their lifetime e.g. humans).Life history strategies relate to natural selection, adaptation and fitness of an organism.Factors like the size, number and sex ratio of offspring, time of reproduction, age and size at which an organism matures affect the life history of an organism.Learn about reproductive strategy here:
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what does the model represent?
Answer:
D) a molecule of oxygen
Answer:
An oxygen atom
Explanation:
its not a neutron, nor the oxygen's atomic mass (I hope you'll know what that is), and a molecule of oxygen is two oxygens atoms together, 02. Hope this helps.
please help me! Quick!
Answer:
b, i think
During mitosis, the nucleus of a eukaryote cell splits into two. The parent cell goes through other stages of division before forming two genetically identical daughter cells, which are also able to duplicate. There are five stages in the process of mitosis: prophase, metaphase, anaphase, telophase and cytokinesis.
Answer:
D. Formation of two diploid daughter cells
Explanation:
Meiosis forms 4 haploid daughter cells, not 2 diploid. It's mitosis that creates 2 diploid. Crossing over, formation of 2 haploid nuclei, and pairing of homologous chromosomes all occur.
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Human impacts on biomes are best characterized by _______. a. increased fossil fuel usage b. industrial pollution of landfills c. interruption of natural biome processes d. sewage contamination of groundwater please select the best answer from the choices provided a b c d
Human impacts on biomes are best characterized by interruption of natural biome process.
What is a Biome?An area of the globe that may be categorized based on the kind of plants and animals that inhabit there is known as a biome. What life there is in a biome depends on factors like temperature, soil, light, and water availability.
An ecosystem is distinct from a biome. An ecosystem is a habitat where living and nonliving things coexist. A biome is a particular geographical region distinguished by the organisms that call it home. Many different ecosystems can make up a biome. For instance, habitats like coral reefs and kelp forests can be found in aquatic biomes.
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A carbohydrate sample weighing 0. 235 g was found to have a fuel value of 3. 84 kj. What is the fuel value of one gram of this carbohydrate, in nutritional calories?
The fuel value of one gram of this carbohydrate, in nutritional calories is 3910 cal.
A vast variety of both good and bad foods, including bread, beans, milk, popcorn, potatoes, cookies, spaghetti, soft drinks, corn, and cherry pie, include carbohydrates. Because they work quickly and provide energy as soon as they are taken, carbohydrates—which have 4 kcal per gram—are the foods that are used as an energy source the most frequently.From the given data,
0.235 g carbohydrate = 3.84 kj = 3840 joule
1g carbohydrate = 3840/0.235 = 16340.43 joule
(1 Cal = 4.1841)
1 g carbohydrate = 16340.43/ 4.184 = 3905.5 Cal
fuel value of 1 g carbohydrate = 3905.5 Cal
answer: 3910 Cal
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A carbohydrate sample weighing 0. 235 g was found to have a fuel value of 3. 84 Kilojoule. The fuel value of one gram of this carbohydrate, in nutritional calories is 3910 Cal.
What is Carbohydrates ?In our daily life we usually uptake Carbohydrates as the primary source of energy. It contains 4 KCal/gm.Examples of Carbohydrates are ; Starch, Bread, Rice, Milk, Soft drinks, Corn etc.Carbohydrates are fast acting fuel because it breaks down into Sugar and Sugar is the main energy source of our brain and cells.
Now,
0.235 g Carbohydrate = 3.84 kilojoule
= 3840 joule
1g Carbohydrate = 3840/0.235
= 16340.43 joule [1 Cal = 4.1841]
1g carbohydrate = 16340.43/ 4.184 = 3905.5 Cal
Fuel value of 1 g Carbohydrate = 3905.5 Cal
= 3910 Cal
Thus from the above conclusion we can say that The fuel value of one gram of this carbohydrate, in nutritional calories is 3910 Cal.
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Name the three types of muscle tissue. What is the difference between the three of them?
Answer:Skeletal muscle, Smooth muscle, Cardiac muscle –
Explanation:
Skeletal muscle – the specialised tissue that is attached to bones and allows movement.
Smooth muscle – located in various internal structures including the digestive tract, uterus and blood vessels such as arteries.
Cardiac muscle – the muscle specific to the heart.
Which structure is highlighted?
a) mesometrium suspensory
b) ligament mesosalpinx
c) round ligament
d) broad ligament
Using the image attached that pertains to the question. The highlighted structure is known to be option d) broad ligament.
What is a broad ligament?The term broad ligament is known to be an aspect or a layer of tissue that is said to function as it links the sides of the uterus to the walls and floor of the pelvis.
Note that The broad ligament is known to be one that is said to covers the uterus, ovaries, and also the fallopian tubes and it is one that is made up of the nerves and also some blood vessels.
The broad ligament is made up of also the visceral and parietal peritoneum that is said to contain both smooth muscle and connective tissue.
Hence, Using the image attached that pertains to the question. The highlighted structure is known to be option d) broad ligament.
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Polysaccharide digestion and glycogen breakdown involve sequential cleavage from ________ ends of glucose polymers.
Sequential cleavage from the non-reducing terminals of glucose molecules is required for both glycogen degradation and polysaccharides hydrolysis.
Why non-reducing end is selected for digestion?
A polysaccharide's non-reducing end is the one where an anomeric carbon participates in the glycosidic connection. The elimination of carbohydrate remnants one at a time out from the non-reducing terminal occurs during glycogenolysis and polysaccharides hydrolysis.
For example, several enzymes are involved in glycogenolysis in the liver and muscle. An example of such an enzyme is glycogen phosphorylase, which catalyzes the successive dissociation of the alpha 1->4 glycosidic bond that connects two glucose molecules at a non-reducing terminal of glycogen. The last glucose residue is eliminated as alpha-D-glucose 1-phosphate.That is why non-reducing end of glucose is chosen for digestion or breakdown of the carbohydrate polymer.
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The contribution of b-cells is mainly in:_______
a) humoral immunity.
b) phagocytosis.
c) inflammation.
d) complement activity.
e) cell-mediated immunity
The contribution of b-cells is mainly in Humoral immunity. Thus Option A is correct.
What is immune system?
Two types of immunity such as innate immunity is an inbuilt nonspecific immune system protect us from all antigens.
Secondly, we have one more sophisticated adoptive immune system which is up two types humoral immunity and cell mediated immunity.
The adoptive response majorly done by two types of cells such as B and T cells.
This type of immune response use past experience identify a specific foreign threats and counteract them when these threats reappear.
The humoral immune system is activated by external macromolecules antibodies, antimicrobial peptides etc. and protect the extracellular spaces of body.
Hence, Option A is correct.
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Meiosis is the process by which cells are produced from a cell that was originally diploid. True or false?.
The statement "Meiosis is the process by which cells are produced from a cell that was originally diploid" is false.
This is because meiosis is the process by which cells are produced from a cell that was originally haploid.
What is meant by meiosis?Meiosis is a type of cell division that results in the production of four gamete cells and a 50% reduction in the number of chromosomes in the parent cell.To develop egg and sperm cells for sexual reproduction, this process is necessary.We observe the following stages in meiosis:Meiosis Iinterphase ITwo identical complete sets of chromosomes are produced as a result of the cell's DNA being duplicated.Two centrosomes, each comprising a pair of centrioles, are located outside the nucleus and are essential for the process of cell division.These centrosomes produce microtubules during interphase.prophase IUnder a microscope, the copied chromosomes condense into distinct X-shaped structures.Two sister chromatids with identical genetic material make up each chromosome.So that both copies of chromosome 1 are together, both copies of chromosome 2 are together, and so on, the chromosomes couple up.The chromosomal pairs may then exchange DNA strands through a process known as recombination or crossing over.The membrane around the cell's nucleus melts at the conclusion of Prophase I, freeing the chromosomes.Microtubules and other proteins make up the meiotic spindle, which spans the cell between the centrioles.metaphase IAlong the cell's center (equator), the chromosomal pairs are positioned adjacent to one another.The meiotic spindles are now extending from the centrioles, which are now located at the cell's opposite poles.One chromosome from each pair is attached by the meiotic spindle fibers.anaphase IThe meiotic spindle then separates the pair of chromosomes by pulling one chromosome to one pole and the other chromosome to the opposite pole of the cell.The sister chromatids remain together throughout meiosis I. In contrast to mitosis and meiosis II, this occurs.telophase I and cytokinesisThe chromosomes reach the opposite poles of the cell in whole.A complete pair of chromosomes congregates at each pole of the cell.Each set of chromosomes is surrounded by a membrane that divides into two new nuclei.The single cell then divides into two independent daughter cells, each of which has a nucleus and a complete complement of chromosomes. Cytokinesis is the name for this process.Meiosis IIprophase IITwo daughter cells with 23 chromosomes each are present at this time (23 pairs of chromatids).The chromosomes once more condense into clear, X-shaped structures that are simple to observe under a microscope in each of the two daughter cells.Each daughter cell's nucleus membrane disintegrates, releasing the chromosomes.Centrioles are duplicative.Once more, the meiotic spindle forms.metaphase IIChromosomes (a pair of sister chromatids) are arranged end to end along the cell's equator in each of the two daughter cells.In each of the daughter cells, the centrioles are now located at opposing poles.The sister chromatids are connected by meiotic spindle fibers that are attached to each pole of the cell.anaphase IIChromosomes (a pair of sister chromatids) are arranged end to end along the cell's equator in each of the two daughter cells.In each of the daughter cells, the centrioles are now located at opposing poles.The sister chromatids are connected by meiotic spindle fibers that are attached to each pole of the cell.telophase II and cytokinesisThe chromosomes reach the opposite poles of the cell in whole.A complete pair of chromosomes congregates at each pole of the cell.Each set of chromosomes is surrounded by a membrane to produce two new cell nuclei.Even though this is the final stage of meiosis, cell division is not finished without one more cycle of cytokinesis.Four granddaughter cells, each with half a set of chromosomes (haploid), are produced once cytokinesis is finished.In men, these four cells are all sperm cells, whereas in females, one of the cells is an egg cell and the other three are polar bodies (small cells that do not develop into eggs).To learn more about meiosis visit:
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write the type of interrelation that exists between sea sponges and corals?
Answer: sponges are the cleaners of the water around the corals , coral feeders , and healers in a sens
Explanation:
Which tissue layer deep to the integument acts as a shock and thermal insulator?
Hypodermis deep to the integument acts as a shock and thermal insulator
The connective tissue that joins the dermis to underlying structures is the hypodermis, which is deep to the dermis of the skin. The hypodermis also contains adipose tissue for fat storage and defence.
The hypodermis performs similar protective roles to the skin: it binds the skin to the underlying structures so that the skin can move virtually freely over them; it stores fat to help minimise heat loss; and it serves as a shock absorber. This characteristic makes sure that blows only hurt the surface tissues of the body.
Therefore, Hypodermis deep to the integument acts as a shock and thermal insulator
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Saturated fatty acids are different to unsaturated fatty acids because they?
Saturated fatty acids are different to unsaturated fatty acids because they have no C=C double bonds.
What are the differences between saturated and unsaturated fatty acids?Saturated fatty acids do not have C=C double bonds, while in unsaturated fatty acids there is at least one C=C double bond in the fatty acid chain.Saturated fats are usually solid at room temperature and are derived from animal sources, whereas unsaturated fats are usually liquid at room temperature and are from plant sources.One must reduce the amount of saturated fat in their diet and consume more unsaturated fats like olives, seeds and nuts.Unsaturated fats can be monounsaturated (contains only C=C double bond) or polyunsaturated (contains two or more C=C double bonds).Learn more about fats here:
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In an isolated population of 1000 individuals, 444 are genotype mn, 478 are mm, and 78 are nn. What are the frequencies of the m and n alleles in this population?.
f(M) = 0.7, f(N) = 0.3 are the frequencies of the m and n alleles in this population.
What is the relative fitness of genotype BB What is the relative fitness of genotype BB?
A relative fitness of 1.0 can occur for more than one genotype. B.The relative fitness of genotype BB in a population of chickens used as a sample is 1.0.The selection coefficients for genotype Bb and genotype Bb are respectively 0.3 and 0.4.Which of the following processes tends to increase genetic variation in a population?
Genetic diversity is typically increased by mutations while being decreased by the other three mechanisms.Migration tends to homogenize genetic variation, reducing the genetic diversity between populations, whereas natural selection and genetic drift tend to increase it.Where are alleles?
Alleles are matching genes; one from our biological mother, one from our biological father.We have two copies of every gene (strings of code that drive some biological function on our chromosomes).They can be identical, but they can often have slight differences.Learn more about alleles
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As soon as alcohol and other toxins are absorbed into the body, the _____ begins eliminating them as alcohol cannot be stored in tissue.
As soon as alcohol and other toxins are absorbed into the body, the liver begins eliminating them as alcohol cannot be stored in tissue.
What is detoxification by the liver?Detoxification by the liver is one of the major functions of the lover that involves the removal of toxins from the body sinusoid channels of the liver.
The liver helps to detoxify used drugs and alcohol which are not stored in the body.
The liver begins eliminating alcohol from the body through the action of the enzyme alcohol dehydrogenase which breaks alcohol into ketones.
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Testing positive for the epstein-barr virus would most likely indicate a diagnosis of _____.
Testing positive for the Epstein-Barr virus would most likely indicate a diagnosis of infectious mononucleosis.
Why is Epstein-Barr virus (EBV) test done?The Epstein-Barr virus is a disease-causing agent that also causes infectious mononucleosis. It is mainly transmitted through saliva when people kiss, share food, or use the same utensils. Many people get EBV infection as children and show no symptoms. Most often affecting adults and teenagers, mononucleosis.
For the diagnosis of infectious mononucleosis, an EBV antibody test is typically not required. If you don't have a normal mononucleosis case or if your doctor suspects you may have another sickness brought on by EBV infection, you might need a test.
If you have mononucleosis symptoms but a negative test, or if you're pregnant and experiencing flu-like symptoms, you might also need to be checked. You could require a test if you have been exposed to someone who has mononucleosis but don't now exhibit any symptoms. Your doctor might need to run a test to see how well your immune system is functioning.
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Between 1994 and 1997, 70 Addax were released into fenced sections of protected areas in Tunisia and Morocco. By
2003, their numbers had increased to about 200. This is an example of
OA. Ex situ conservation
OB. Coppicing
OC. Habitat management
D. In situ conservation
It is an example of Ex situ conservation. The correct option is A.
What is ex-situ conservation?It is a method of conserving biodiversities outside their natural habitats.
Ex situ conservation is in direct contrast to in situ conservation. This means that in in-situ conservation, biodiversities are conserved in their natural habitats.
In the illustration, the Addax were released into fenced sections of protected areas in two countries.
The fenced section means that the Addax were not allowed to roam freely as they would in their natural habitats. The fenced section means that their movement is limited as compared to natural habitats.
Thus, the Addax were being conserved ex-situ.
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The genus _______ is an extremely important member of the apicomplexan group because several of its species are the causative agents of malaria.
The genus plasmodium is an extremely important member of the apicomplexan group because several of its species are the causative agents of malaria.
Plasmodium is a genus of unicellular eukaryotes that are obligate parasites of vertebrates and insects. The life cycles of Plasmodium species involve development in a blood-feeding insect host which then injects parasites into a vertebrate host during a blood meal. Parasites grow within a vertebrate body tissue (often the liver) before entering the bloodstream to infect red blood cells. The ensuing destruction of host red blood cells can result in malaria. During this infection, some parasites are picked up by a blood-feeding insect (mosquitoes in majority cases), continuing the life cycle.
The life cycle of Plasmodium involves several distinct stages in the insect and vertebrate hosts. Parasites are generally introduced into a vertebrate host by the bite of an insect host (generally a mosquito, with the exception of some Plasmodium species of reptiles). Parasites first infect the liver or other tissue, where they undergo a single large round of replication before exiting the host cell to infect erythrocytes.
At this point, some species of Plasmodium of primates can form a long-lived dormant stage called a hypnozoite. It can remain in the liver for more than a year. However, for most Plasmodium species, the parasites in infected liver cells are only what are called merozoites. After emerging from the liver, they enter red blood cells. They then go through continuous cycles of erythrocyte infection, while a small percentage of parasites differentiate into a sexual stage called a gametocyte which is picked up by an insect host taking a blood meal.
In some hosts, invasion of erythrocytes by Plasmodium species can result in disease, called malaria. This can sometimes be severe, rapidly followed by death of the host (e.g. P. falciparum in humans). In other hosts, Plasmodium infection can apparently be asymptomatic.
Even when humans have such subclinical plasmodial infections, there can nevertheless be very large numbers of multiplying parasites concealed in, particularly, the spleen and bone marrow. These hidden parasites (in addition to hypnozoites) are thought to be the origin of instances of recurrent P. vivax malaria.
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The left cerebellum shows _____ activity, and the right cerebellum shows _____ activity when an individual is learning a mirror-reading task.
The left cerebellum shows decreased activity, and the right cerebellum shows increased activity when an individual is learning a mirror-reading task.
Body parts are moved by several regions of the cerebrum. While the right side of the brain directs movement on the left side of the body, the left side of the body is directed by the right side of the brain. The right hemisphere manages the left side of the body, while the left hemisphere manages the right.
The left hemisphere, for instance, has been found to be better at creating associations in memory, selective attention, and positive emotions. Contrarily, research has found that the right hemisphere performs better in terms of pitch perception, arousal, and negative emotions.
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When will there be 2,560 bacteria in the dish?
a) 40 minutes
b) 1 hour 20 minutes
c) 2 hours 40 minutes
d) 3 hours
Calculation :
It is given that in the start there are 5 bacteria and it keeps doubling after every 20 minutes we have to find the number of bacteria after 4 hours
4 hours = 240 minutes
⇒ Number of times the bacteria have doubled = 240÷20 = 12 times
It means the bacteria have doubled by 12 times
⇒Total bacteria = 5×2¹²= 20480
The time when there will be 2560 bacteria
⇒ the bacteria had doubled 9 times to reach the mark of 2560
so total time = 9 ×20= 180 minutes = 3 hours
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Disclaimer: The question was given incomplete on the portal. Here is the complete question.
Question: There are 5 bacteria in a Petri dish at the start of the experiment. This type of bacteria doubles every 20 minutes
Find the number of bacteria after 4 hours When will there be 2560 bacteria in the dish?
Determine if there is a statistically significant difference between the absorption data of
black walnut trees and maple trees
Yes, there is a statistically significant difference between the absorption data of black walnut trees and maple trees due to difference in values of absorption.
Determine if there is a statistically significant difference?Yes, there is a statistically significant difference between the absorption data of black walnut trees and maple trees because the average carbondioxide absorption of black walnut is 7.26 while on other hand, the average carbondioxide absorption of maple tree is 5.82. We can see there is big difference of absorption data between black walnut trees and maple trees.
So we can conclude that there is a statistically significant difference between the absorption data of black walnut trees and maple trees due to difference in values of absorption.
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A species of bird on the mainland can have yellow (bb), blue (bb), or green (bb) plumage. a few yellow birds land on a ship and are carried past an island, which they land on and settle. the resulting population of birds would have an allele frequency of: a. b: 1 b: 0 b. b: 0.3 b: 0.7 c. b: 0 b: 1 d. b: 0.5 b: 0.5
Birds living in this population would have an allele frequency of "b."
If the blue species of birds have homozygous dominant alleles of BB, the green species of birds have heterozygous alleles of Bb, and the yellow species of birds have homozygous recessive alleles of bb, then b is the "recessive allele" and b is the dominant allele for blue.
NOTE:The term "recessive allele" refers to the character/allele that is eclipsed by the dominant character, whereas a "dominant allele" is a character/allele whose phenotype is displayed to the complete exclusion of the other character/allele.
When a pair of alleles in a gene are homozygous, they are the same; when they are heterozygous, they are different.
Given that the yellow birds are homozygous for "b," the population of birds that results when they (only) arrive and settle on the Island will also be homozygous for "b." As a result, every bird on the island will have the gene "bb".
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Which factor can confound neuroprognostication during the post-cardiac arrest phase?
Targeted temperature management factor can confound neuroprognostication during the post-cardiac arrest phase.
What is post cardiac arrest?After a patient has been revived from a cardiac arrest, they may experience post-cardiac arrest syndrome, an inflammatory condition of pathophysiology. The body goes through a special kind of global ischemia during cardiac arrest.Global brain damage, myocardial failure, macrocirculatory dysfunction, increased susceptibility to infection, and persisting precipitating pathology are all symptoms of the clinical condition known as post-cardiac arrest syndrome.These effects, which include movement issues, memory loss or impairment, speech difficulties, weakness or immobility, and cognitive impairments including problems with attention, focus, and visual-motor skills, may be permanent or they may get better with time.Care after a cardiac arrest requires a variety of management techniques. To diagnose and treat coronary artery obstructive disease, early invasive coronary angiography should be taken into consideration. The primary shock treatment involves vasopressors like norepinephrine and dobutamine.Learn more about cardiac arrest here:
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What is the magnitude (amplitude) of an action potential?
a. 30 mv
b. 70 mv
c. 100 mv
What would happen if the operator sequence of the trp operon contained a mutation that prevented the repressor protein from binding to the operator?
In the absence of tryptophan, the genes trpA-E will not be transcribed.
If there were a mutation in the operator region of the trp operon, there would be a nucleotide change in the sequence of the operator. Mutations in the repressor that prevent its binding to the operator will lead to constitutive expression Mutations that prevent binding of the inducer without affecting the ability to bind to the operator lead to a non‑inducible phenotype.
The tryptophan binds to the trp repressor and causes it to change shape, converting into its active form. The trp repressor with the bound tryptophan attaches to the operator, blocking RNA polymerase from binding to the promoter and preventing transcription of the operon.
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The flexor muscles in the anterior arm (biceps brachii and brachialis) are innervated by what nerve?
Musculocutaneous nerve is responsible for innervating the flexors of the arm.
The axillary nerve innervates the deltoid and teres minor. The musculocutaneous nerve is responsible for innervating the flexors of the arm, including the biceps brachii, coracobrachialis, and the medial aspect of the brachialis. The Musculocutaneous nerve is a large branch of the Brachial Plexus. It is called musculocutaneous nerve as it supplies the muscles of the front of the arm and skin of lateral side of forearm. The Musculocutaneous nerve is a terminal branch of the lateral cord of the Brachial Plexus.
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Unrepaired single-stranded dna breaks can lead to double-stranded dna breaks during the next round of replication. Which repair pathway would be most appropriate for repairing double-stranded dna breaks?.
Nonhomologous end joining would be most appropriate for repairing double-stranded DNA breaks.
How are double stranded breaks in DNA repair?Homologous recombination and nonhomologous DNA end joining are the two main mechanisms for mending double-strand DNA breaks, which often occur in eukaryotic cells.DNA ends that need to be repaired have a different chemistry as a result of the many causes of DSBs.The enzymes of the NHEJ pathway display a remarkable level of structural tolerance in the variety of DNA end substrate configurations upon which they may function over the course of NHEJ development.The nuclease, polymerases, and ligases of NHEJ are the most versatile and mechanistically adaptable enzymes in each of their classes in vertebrate cells.To learn more about double-stranded DNA breaks from the given link
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The type of growth that a population may show when food and habitat are limitless is called?
A type of growth that a population may show when food and habitat are limitless is exponential population growth.
What is exponential population growth?The case in which resources are limitless a population can experience exponential population growth where the size increases at a greater rate.
Exponential growth occurs when there is large quantity of resources are present and few individuals are there to utilize it
When number of individuals increases the resources will depleted and growth rate will slow down.
Therefore, exponential population growth is the type of growth in which resources are plenty in comparison of population.
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The parts of the subunits of ribosomes are assembled in the ______ while the subunits are assembled into ribosomes in the ______
The parts of the subunits of ribosomes are assembled in the nucleolus while the subunits are assembled into ribosomes in the cytoplasm.
What is ribosome ?In eukaryotes, ribosomes are located in cytoplasmic organelles, other type of ribosomes present in cell are chloroplast and mitochondrial ribosomes.
Ribosome have a sedimentation coefficient such as 80 S is composed of 40 S and 60 S subunits.
In prokaryotic cells, it is 70 S type consist up 30 S and 50 S subunits.
In both prokaryote and eukaryotes majority of ribosome is composed up proteins that and the remaining part is ribonucleic acid.
Hence nucleolus and cytoplasm are correct answers.
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The portions of the mature mrna where rna-binding proteins can attach in order to change rna stability are known as?
A geologist’s compass differs from the more widely known orienteering compass in that the geologist’s compass ________.
A geologist's compass differs from the more widely known orienteering compass in that the geologist's compass can measure dip and plunge angles.
What is a compass?A compass is an instrument used to examine one’s position on the map area and to determine the relative position of various objects.It gives an idea of where you are heading to and where you need to be going.The geologist’s use a special type of compass which can not only determine the position but also make structural measurements of geological structures like dip of areas like sedimentary beds and plunge of fold axis.A clinometer is necessary in a compass to measure all the structural data. It measures features from the horizontal like dip and plunge, and gradient.Learn about magnetic compass here:
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