The vocal chords close when we are inhaling. True or False

Answers

Answer 1

When the cords vibrate more quickly, they produce a higher-pitched sound. Conversely, when they vibrate more slowly, they produce a lower-pitched sound. It's the voice box's shape and the way the vocal cords vibrate that creates the unique sound of each person's voice.

The statement "The vocal chords close when we are inhaling" is False.Vocal cords or vocal folds are two bands of elastic muscle tissue. They are located side by side in the voice box, also known as the larynx. When we inhale, the vocal cords open up to allow the air to enter the lungs. And when we exhale, the air passes between the vocal cords, and they vibrate to produce sound. They don't close when we inhale.The opening and closing of the vocal cords regulate the flow of air that passes through them. This airflow is essential in producing sounds, which are then shaped into words. The vocal cords open and close in a precise manner that determines the pitch and tone of the voice. When the cords vibrate more quickly, they produce a higher-pitched sound. Conversely, when they vibrate more slowly, they produce a lower-pitched sound. It's the voice box's shape and the way the vocal cords vibrate that creates the unique sound of each person's voice.

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Related Questions

Special Topic: COVID-19 as an Environmental Health Challenge
1. In mid-1800s London, what did a doctor named John Snow do
that helped gain a new understanding of the cholera pandemic of the
time? A. H

Answers

In the mid-1800s London, a doctor named John Snow's action that helped gain a new understanding of the cholera pandemic of the time was that he mapped the locations of cholera cases.

In the mid-1800s London, a doctor named John Snow did the following things that helped gain a new understanding of the cholera pandemic of the time:He mapped the locations of cholera cases. His research helped locate the source of the outbreak to contaminated water in the public water pump on Broad Street.He was able to understand the epidemiology of cholera. Snow's study confirmed that cholera was a waterborne disease. He concluded that ingestion of contaminated water was responsible for the spread of the disease.

He advised the removal of the water pump handle. Once the authorities removed the pump handle, the epidemic stopped immediately. Cholera pandemic is an epidemic of cholera that has spread across a large region like several countries or worldwide. It occurs when a new strain of the cholera bacterium emerges that is resistant to all the current treatments available. This makes it challenging to treat the disease and control its spread.

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Plot the phosphorylation of the carboxyl terminal domain of Rpb1 and describe the processes associated with the phosphorylation.

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The C-terminal domain (CTD) undergoes phosphorylation at serine residues during transcription initiation. Multiple protein kinases have the ability to phosphorylate this sequence.

What happens to the CTD extension at the C-terminus

The CTD extension at the C-terminus of RNA polymerase 2 serves as a flexible binding site for numerous factors, and this binding is determined by the phosphorylation of the CTD repeats.

Phosphorylation is a process that involves the addition of a phosphate group to a molecule derived from ATP. In glycolysis, one molecule of ATP is consumed during this process. It is a chemical modification that adds a phosphoryl group to an organic compound.

Dephosphorylation refers to the removal of the phosphoryl group. Phosphorylation plays a vital role in biochemistry and molecular biology, as it is a crucial reaction for protein and enzyme function, sugar metabolism, and energy storage.

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The vertical gaze center contains premotor neurons that project to lower motor neurons and interneurons in the abducens nucleus. True False

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The statement is false. The vertical gaze center does not contain premotor neurons that project to lower motor neurons and interneurons in the abducens nucleus.

The vertical gaze center, which is responsible for controlling eye movements in the vertical direction, does not directly contain premotor neurons that project to lower motor neurons and interneurons in the abducens nucleus. Instead, the vertical gaze center involves the integration of multiple brain regions and neural pathways.

The primary brain structure involved in vertical eye movements is the rostral interstitial nucleus of the medial longitudinal fasciculus (riMLF). The riMLF receives input from the superior colliculus, a midbrain structure involved in eye movements, and it projects to the oculomotor nucleus, which controls the extraocular muscles responsible for vertical eye movements. The abducens nucleus, on the other hand, primarily controls horizontal eye movements. Thus, there is no direct connection between the premotor neurons of the vertical gaze center and the lower motor neurons and interneurons in the abducens nucleus.

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what is the typical route of exposure to pesticides? (5 marks)
its a toxicology question and I need help asap

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The typical route of exposure to pesticides can vary depending on the circumstances and the specific pesticide involved. This can occur during mixing, spraying, or other activities involving pesticide handling without adequate protective measures.

The following are common routes of pesticide exposure:

Dermal Exposure: Skin contact with pesticides is a significant route of exposure. People who handle or apply pesticides directly may come into contact with the chemicals through their skin

Inhalation Exposure: Inhalation of pesticide vapors, dust, or aerosols is another route of exposure. It can happen when pesticides are sprayed or applied in areas with poor ventilation, or when working with volatile pesticides. Inhalation exposure can occur not only for applicators but also for bystanders in the vicinity of pesticide application.

Oral Exposure: Accidental ingestion of pesticides is a potential route of exposure, especially in cases of improper storage or handling of pesticide containers. Children may be at higher risk due to accidental ingestion, as they may mistake pesticide containers for food or beverages.

Ocular Exposure: Exposure to pesticides through the eyes can occur if pesticides come into contact with the eyes, leading to irritation or absorption through the conjunctiva. This can happen during pesticide application, particularly if protective eyewear is not used.

Environmental Exposure: People can also be exposed to pesticides indirectly through environmental contamination. Pesticides may enter water bodies, soil, or crops, and individuals may come into contact with them by consuming contaminated food or water, or through direct contact with contaminated surfaces.

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Can you simplify and summarize the meaning of shortsighted
evolution hypothesis with examples. Please help me understand this
topic hope you can explain it clearly.

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The shortsighted evolution hypothesis, also known as the "Red Queen hypothesis," suggests that in a changing environment, organisms must constantly adapt and evolve in order to survive and reproduce.

This hypothesis is based on the idea that species must continuously evolve just to maintain their current fitness levels relative to other species they interact with. It implies that evolutionary changes are driven by interactions and competition between species, rather than simply adapting to the environment.

For example, in the predator-prey relationship between cheetahs and gazelles, as cheetahs evolve to become faster and more efficient hunters, gazelles must also evolve to become faster and more agile to avoid predation. This constant adaptation and counter-adaptation create a "evolutionary arms race" between the two species.

Another example is the coevolution between parasites and their hosts. Parasites evolve strategies to exploit their hosts, such as developing drug resistance, while hosts evolve defenses to combat the parasites, like immune system adaptations.

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Sensations of dynamic equilibrium are registered in the Select one: a. cochlea. b. vestibule. c. organ of Corti. d. semicircular ducts. e. tympanic membrane.

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The sensations of dynamic equilibrium are registered in the semicircular ducts. The semicircular ducts are filled with fluid that moves in response to changes in the orientation of the head. This movement is detected by hair cells that are located within the ampulla of each semicircular duct.

The hair cells are stimulated when the fluid moves and this triggers the sensation of dynamic equilibrium.The vestibule is responsible for detecting sensations of static equilibrium and linear acceleration, while the cochlea is responsible for detecting sound waves. The organ of Corti is a structure within the cochlea that contains the hair cells responsible for detecting sound waves.

The tympanic membrane, or eardrum, is a thin layer of tissue that separates the outer ear from the middle ear and vibrates in response to sound waves.The semicircular ducts are part of the vestibular system, which is responsible for maintaining balance and spatial orientation. When the head moves, the fluid within the semicircular ducts moves in a way that corresponds to the movement of the head. This movement is detected by the hair cells, which then send signals to the brain to help us maintain our balance and stay oriented in space.In summary, the correct answer to this question is d. Semicircular ducts. This is because the sensations of dynamic equilibrium, which are responsible for maintaining balance during movement, are detected by hair cells located within the semicircular ducts.

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describe how the structure of DNA is correlated with its role as
the molecular basis of inheritance. In detail please!

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DNA (Deoxyribonucleic acid) is a double-stranded helix that contains the genetic code that is used to store genetic information in all living cells. The double helix structure of DNA has been an important factor in determining its role as the molecular basis of inheritance.
The structure of DNA comprises of a sugar-phosphate backbone and nitrogenous bases that protrude from the backbone, perpendicular to it. The nitrogenous bases in DNA can be of four different types: adenine (A), guanine (G), cytosine (C), and thymine (T). They form hydrogen bonds between complementary base pairs, A with T and G with C, that hold the two strands of DNA together.
The double helix structure of DNA enables it to carry genetic information through DNA replication, which is a process that duplicates DNA before cell division. During DNA replication, the double helix separates, and each strand serves as a template for the formation of new complementary strands by the base pairing rule.
The structure of DNA also enables it to store a large amount of genetic information, as the number of possible base combinations is very high, and the sequence of bases on one strand is complementary to that on the other strand. This ensures that the genetic information is stored in a stable and reproducible manner, as the base pairs remain unchanged over multiple generations.
In conclusion, the double helix structure of DNA is an essential feature that allows it to store and transmit genetic information accurately. Its structure is closely linked to its role as the molecular basis of inheritance, which is crucial for the continuity of life and the evolution of organisms.

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3. Fill in each of the blanks below with the correct term:
a) The presence of fat and acid in chyme stimulates the
__________________ to release ______________ and _______________
into the bloodstream
Major Component of Food (macromolecule) at ingestion: End product of chemical digestion (ie, absorbed as): Transported away from digestive system by 2. On the back of the page or on a separate page, c

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The presence of fat and acid in chyme stimulates the small intestine to release secretin and cholecystokinin into the bloodstream.Secretin and cholecystokinin are hormones released by the small intestine

. These hormones are stimulated by the presence of fat and acid in chyme. Secretin stimulates the pancreas to release bicarbonate ions into the small intestine. Bicarbonate ions neutralize the acidic chyme, which helps protect the small intestine from damage. Cholecystokinin stimulates the gallbladder to release bile into the small intestine.

Bile is important for the digestion and absorption of fat.Major Component of Food (macromolecule) at ingestion:FatEnd product of chemical digestion (i.e., absorbed as):Fatty Acids and GlycerolTransported away from the digestive system by:Lymphatic System.

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Prokaryotic cells must integrate numerous metabolic signals to balance catabolic and anabolic processes in the cell.For example,when nitrogen levels are low and amino acids are scarce, compounds called alpha-ketoacids inhibit the synthesis of cyclic AMP(cAMP).What impact would this have on the lac operon?(choose all that apply) increase expression of the lac operon inhibit binding of repressdr protein to the operator inhibit binding of CAP protein to the Cap Binding Site allow only some lac operon genes to be expressed,but not all inhibit expression of the lac operon 3points A transcription factor recruits a histone deacetylase enzyme to bind to a particular gene. The likely consequence of this is(choose all that apply Chromatin structure will become less compact due to removal of acetyl groups from histones Gene expressionwill decrease DNA nucleotides in this gene will have acetyl groups removed Chromatin structure will become more compact due to removal of acetyl groups from histones Geneexpression will increase

Answers

Impact on the lac operon when alpha-ketoacids inhibit the synthesis of cyclic AMPProkaryotic cells must integrate numerous metabolic signals to balance catabolic and anabolic processes in the cell.

For example, when nitrogen levels are low and amino acids are scarce, compounds called alpha-ketoacids inhibit the synthesis of cyclic AMP (cAMP).

The impact on the lac operon would be: inhibit binding of CAP protein to the Cap Binding Siteinhibit expression of the lac operonWhen cyclic AMP levels are low, CAP protein is not able to bind to the CAP Binding Site, which is upstream of the promoter region of the lac operon.

CAP protein is required for RNA polymerase to efficiently bind to the promoter, which results in the high-level transcription of the structural genes of the lac operon.When cAMP is low, the binding of CAP to the CAP site is inhibited, and there is less expression of the lac operon.

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Which of the following secretes citric acid, and what is the function of this molecule?
Nurse cells, is a source of nutrient for sperm
Cowper’s gland, helps sperm motility
Bulbourethral gland, has an antimicrobial effect
Prostate gland, is used by sperm for ATP production
Seminal vesicle, acts as a lubricant

Answers

The prostate gland secretes citric acid, which is used by sperm for ATP production.

Citric acid is secreted by the prostate gland. It is an important molecule for sperm function and plays a role in energy production. Citric acid is utilized by the mitochondria in the sperm cells to generate ATP (adenosine triphosphate), which is the main energy currency in cells.

ATP provides the energy required for various cellular processes, including sperm motility and fertilization.

The prostate gland, located in the male reproductive system, contributes to the seminal fluid. Along with other components of semen, such as seminal vesicle secretions, it provides the necessary nutrients, enzymes, and fluids to support the survival and function of sperm.

Citric acid, as one of the components of prostate secretions, serves as a substrate for ATP production in sperm mitochondria. This ATP production is vital for sperm motility, allowing them to swim and reach the site of fertilization.

In summary, the prostate gland secretes citric acid, which acts as a source of energy for sperm by being utilized in ATP production.

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Check all that occur during localized inflammation. Check All That Apply Chemical mediators cause vasodilation. Blood flow is decreased to the area. Vascular permeability is decreased. Fibrin walls of

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Localized inflammation involves the response of the body to tissue injury. The signs and symptoms of localized inflammation include redness, swelling, heat, pain, and loss of function. Check all that occur during localized inflammation.

Chemical mediators cause vasodilation: The chemical mediators released from cells including histamine, prostaglandins, and leukotrienes, cause vasodilation, which leads to an increase in blood flow to the injured area.Blood flow is decreased to the area: Blood flow is not decreased to the area, but rather, it is increased due to vasodilation.

Vascular permeability is increased: The increased permeability of the blood vessels at the site of injury allows leukocytes and plasma proteins to move from the blood into the tissues. The result of the increased vascular permeability is the accumulation of fluid in the interstitial spaces.

Fibrin walls off the area: Fibrinogen is converted to fibrin, which forms a clot around the area of injury. The clot helps to prevent the spread of infection and protects the tissue during the healing process.

Therefore, Fibrin does not wall off the area but rather, it helps to protect the area from further damage.

So, the correct options are:Chemical mediators cause vasodilation.Vascular permeability is increased.

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1,200 lb Average mature weight of cows at BCS of 5: Target calving season of mature cows: September 1 to November 1 annually May 1 annually Target weaning date (calves from mature cows): Average weight of heifers at weaning (from mature cows): 520 lb 7. What are the (a) target breeding weights and (b) calving weights from typical heifers produced in this herd? a. b. 8. Your mature cows need to go from BCS 5 at weaning time to a BCS 6 at the start of the upcoming calving season. Assume that these cows need to gain 75 lb of weight just from the pregnancy. Calculate: (a) the total weight gain needed per cow and (b) the average daily gain needed per cow. a. 8. Your mature cows need to go from BCS 5 at weaning time to a BCS 6 at the start of the upcoming calving season. Assume that these cows need to gain 75 lb of weight just from the pregnancy. Calculate: (a) the total weight gain needed per cow and (b) the average daily gain needed per cow. a. b. 9. Relative to this scenario: (a) what breeding season is needed for the cows, and (b) what breeding season is needed for the heifers for them to begin and end calving three weeks earlier than the cows? a. b. Name: 10. Using the above information, calculate the average daily gain (ADG) needed on heifers from (a) the weaning date to the start of the breeding season, and (b) from start of breeding season to the start of the calving season. a. b. In regard to mature cows, there were 180 cows exposed to bulls during the previous breeding season. There were 168 cows palpated pregnant, 163 cows that calved, and 158 cows that weaned calves; the average weaning weight of all the calves was 535 lb. 11. For this scenario what were: (a) the percent pregnant, and (b) the pounds of calf weaned per cow exposed? a. b.

Answers

The target breeding weight for heifers would be approximately 780-840 lb. the target calving weight for heifers would be approximately 1,020-1,080 lb. The total weight gain needed per cow is 75 lb.

To calculate the target breeding weights and calving weights for typical heifers produced in this herd, we need to consider the average mature weight of cows, the average weight of heifers at weaning, and the desired calving season.

(a) Target Breeding Weights for Heifers: The target breeding weight for heifers is typically around 65-70% of their projected mature weight. Assuming a mature cow weight of 1,200 lb, the target breeding weight for heifers would be approximately 780-840 lb.

(b) Calving Weights for Heifers: The calving weight for heifers can vary, but a common target is around 85-90% of their mature weight. Using the mature cow weight of 1,200 lb, the target calving weight for heifers would be approximately 1,020-1,080 lb.

Moving on to the calculations for mature cows, assuming they need to gain 75 lb of weight just from pregnancy to reach a BCS of 6 at the start of the calving season:

(a) Total Weight Gain Needed per Cow: The total weight gain needed per cow is 75 lb.

(b) Average Daily Gain Needed per Cow: To determine the average daily gain needed, we need to consider the duration of pregnancy. If the calving season starts 9 months after the weaning time (assuming 280 days of pregnancy), the average daily gain needed would be 75 lb divided by 280 days, resulting in approximately 0.27 lb/day.

For the breeding and calving seasons to begin and end three weeks earlier for both cows and heifers:

(a) Breeding Season for Cows: The breeding season for cows would need to be adjusted to ensure a three-week earlier start, typically around late November to early January.

(b) Breeding Season for Heifers: Similarly, the breeding season for heifers would also need to be adjusted to achieve a three-week earlier start, usually in late November to early January.

To calculate the average daily gain (ADG) needed for heifers:

(a) ADG from Weaning to the Start of Breeding Season: To determine the ADG needed, we would divide the weight gain from weaning to the start of the breeding season by the number of days between those two time points.

(b) ADG from Start of Breeding Season to the Start of Calving Season: Similarly, we would divide the weight gain from the start of the breeding season to the start of the calving season by the number of days in that period.

Lastly, for the scenario with 180 cows exposed to bulls, 168 cows palpated pregnant, and 163 cows that calved and weaned:

(a) Percent Pregnant: The percent pregnant would be calculated by dividing the number of cows palpated pregnant (168) by the number of cows exposed to bulls (180) and multiplying by 100. This would result in approximately 93.3% pregnant.

(b) Pounds of Calf Weaned per Cow Exposed: The pounds of calf weaned per cow exposed would be calculated by dividing the total pounds of calf weaned (from 163 cows) by the number of cows exposed to bulls (180). With an average weaning weight of 535 lb, the pounds of calf weaned per cow exposed would be approximately 481 lb.

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41​The site of the formation of the primary structure for protein synthesis in animal cells is the
a) mitochondrion​b) nucleus​c) SER ​d) RER​​e) vacuole
42. ​Phospholipids can form all of the following structures in water except which one?
a) cell membranes​ b) bilayers​c)nuclear membranes ​d) vesicles​e) Bones cell membranes

Answers

The site of the formation of the primary structure for protein synthesis in animal cells is the ribosome. The site of the formation of the primary structure for protein synthesis in animal cells is the ribosome.

Ribosomes, the site of protein synthesis in cells, are composed of two subunits that are unequal in size. Both ribosomal subunits contain ribosomal RNA (rRNA) molecules and a number of ribosomal proteins that help to maintain the structure and function of the ribosome.
Therefore, option D is the answer.
Phospholipids can form all of the following structures in water except bones cell membranes. Phospholipids are the main structural component of cell membranes in living organisms. When in contact with water, these amphipathic molecules spontaneously self-organize into a bilayer to form a cell membrane. The two layers of a bilayer have opposing orientations of the phospholipid molecules that create a hydrophobic interior sandwiched between two hydrophilic surfaces.
They can also form vesicles or liposomes when a bilayer spontaneously closes to create an isolated compartment. However, bones cell membranes is not a structure that can be formed by phospholipids in water.
Therefore, option E is the answer.

Ribosomes are the site of the formation of the primary structure for protein synthesis in animal cells, while phospholipids can form all of the following structures in water except bones cell membranes.

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9 38 Question 2 (1 point) Which of the following is true about post-translational modifications? They are encoded in the DNA They can alter the protein structure Acetylation is the most common They ca

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Answer: Post-translational modifications can alter the protein structure.

Post-translational modifications (PTMs) are covalent modifications that occur to proteins after they are synthesized. These changes can alter the protein's structure, localization, activity, or interaction with other molecules, among other things. PTMs are essential for protein function in a wide range of biological processe

s. Some of the most common types of PTMs include phosphorylation, acetylation, glycosylation, and ubiquitination. These modifications can occur at specific amino acids in the protein sequence and are mediated by specific enzymes. Unlike DNA, which encodes the primary structure of proteins, PTMs are dynamic and can respond to changes in the environment or other cellular signals. They are essential for many biological processes, including signaling pathways, gene expression, and cell division.

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Research one genetic disorder, either something you have
experience with or something interesting, and discuss how it is
tied to meiosis. Include your understanding of how this disorder
occurs in the

Answers

One genetic disorder that is tied to meiosis is Down syndrome, also known as trisomy 21. It is caused by the presence of an extra copy of chromosome 21, which disrupts the normal chromosomal distribution during meiosis.

During meiosis, the process of cell division that produces gametes (sperm and eggs), chromosomes undergo recombination and segregation to create genetically diverse and haploid cells. However, in individuals with Down syndrome, there is an error in meiosis called nondisjunction, where chromosome 21 fails to separate properly. This results in one of the resulting gametes having two copies of chromosome 21 instead of one.

When a fertilized egg with an extra copy of chromosome 21 (trisomy) is formed, it leads to the development of Down syndrome. Individuals with Down syndrome typically exhibit physical characteristics such as distinct facial features, intellectual disabilities, and various health issues.

The occurrence of Down syndrome is directly linked to the abnormal distribution of chromosomes during meiosis, specifically the failure of proper separation of chromosome 21, resulting in an additional copy of this chromosome in the resulting offspring.

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What are the seven chordate Classes that have survived to the
present day, what are their evolutionary relationships between
these chordate classes and events lead to their orgin and
success?

Answers

The seven chordate classes that have survived to the present day are:

Class Myxini (Hagfishes)Class Petromyzontida (Lampreys)Class Chondrichthyes (Cartilaginous Fishes)Class Actinopterygii (Ray-finned Fishes)Class Amphibia (Amphibians)Class Reptilia (Reptiles)Class Mammalia (Mammals)

What are chordates?

Chordates are thought to have originated from a common ancestor that had certain key features, such as a notochord and a dorsal nerve cord.

Over time, evolutionary events such as genetic mutations, natural selection, and environmental changes led to the divergence and diversification of these chordate classes.

Major events in chordate evolution include the transition from water to land, the development of jaws and paired fins, the evolution of amniotic eggs in reptiles, and the development of mammary glands and other mammalian adaptations.

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pitenesin 6. In this lab, we reviewed numerous fossil species and their defining characteristics. To help you make compari- sons across these species and understand larger trends in our evolutionary history, complete the Australopith and Early Homo Chart on pp. 446-447. AUSTRALOPITH AND EARLY HOMO CHART Fossil Species Dates and Geographic Region Cranial and Dental Traits Postcranial Traits Suggested Behavior Australopithecus anamensis Australopithecus afarensis LAB 15 | The Australopiths and Early Members of the Australopithecus africanus Australopithecus garhi Australopithecus sediba Australopithecus (Paranthropus) aethiopicus AUSTRALOPITH AND EARLY HOMO CHART (continued) Fossil Species Dates and Geographic Region Cranial and Dental Traits Postcranial Traits Suggested Behavior Australopithecus (Paranthropus) boisei Australopithecus (Paranthropus) robustus Australopithecus deyiremeda Homo habilis (including H. rudolfensis)
Previous question

Answers

In this lab, we have examined many fossil species and their defining characteristics. To help you make comparisons across these species and understand larger trends in our evolutionary history.

let us complete the Australopith and Early Homo Chart. The Australo pith and Early Homo Chart is a tabular presentation of some Australopith and Early Homo fossils. This chart allows you to make comparisons across these fossils, to identify some of their similarities and differences.

Understand some of the significant trends in the evolution of these hominins.The following is a sample of the Australopith and Early Homo Chart that we have completed in this lab: Fossil Species Dates and Geographic Region Cranial and Dental Traits Postcranial Traits Suggested Behavior Australopithecus anamensis .

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Hey,
I need help with the following question from physiology, Thank you!
The question:
The heart's pumping ability is of course central to the circulatory system and the heart's chambers pump out a certain amount of blood during each heartbeat.
a) What is this volume called?
b) It is important that the two halves of the heart pump an equal amount of blood per unit of time. What is the law that describes this phenomenon called?
c) Describe the mechanism that causes the two halves of the heart to always pump the same amount of blood per unit of time.

Answers

a) The volume of blood pumped out by the heart during each heartbeat is called stroke volume.

b) The law that describes the phenomenon of the two halves of the heart pumping an equal amount of blood per unit of time is known as Starling's law of the heart.

c) The mechanism that ensures the two halves of the heart pump the same amount of blood per unit of time is based on the principle of cardiac output. The cardiac output is the product of stroke volume (the volume of blood pumped out by each ventricle per beat) and heart rate (the number of beats per minute). To maintain equal cardiac output, the heart adjusts the stroke volume and heart rate based on the body's needs. The two halves of the heart work in coordination through electrical signals and the timing of contractions to ensure that the blood is pumped in a synchronized manner and in equal amounts. The heart's electrical system, including the sinoatrial (SA) node and the atrioventricular (AV) node, plays a crucial role in coordinating the contraction of the atria and ventricles to achieve this balance.

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Opisthotonus, a condition where all the muscles in the body
contract simultaneously, is associated with which of the following
toxins?
a. diphtheria
b. cholera
c. tetanus
d. botulinum

Answers

Opisthotonus, a condition where all the muscles in the body contract simultaneously, is associated with the toxin produced by the bacterium Clostridium tetani, which causes tetanus. Therefore, the correct answer is c. tetanus.

Botulinum toxin, commonly known as Botox, is a potent neurotoxin produced by the bacterium Clostridium botulinum. It is considered one of the most powerful toxins known to humankind. Botulinum toxin blocks the release of acetylcholine, a neurotransmitter responsible for muscle contraction, leading to muscle paralysis. This property has made it valuable in medical and cosmetic applications. It is used to treat various medical conditions such as muscle spasms, chronic migraines, excessive sweating, and certain eye disorders. In cosmetic treatments, it is employed to reduce the appearance of wrinkles and fine lines. However, due to its extreme potency, botulinum toxin requires careful handling and administration to prevent potential adverse effects.

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Think about this! Using mice to investigate the autosomal dominant human disease, osteogenesis imperfecta (aka brittle bone disease), scientists determined that 58% of mice display incomplete penetran

Answers

Autosomal dominant inheritance involves the transmission of a dominant gene on non-sex chromosomes. Osteogenesis imperfecta (brittle bone disease) follows an autosomal dominant pattern, and using animal models like mice can provide valuable insights into the disease mechanism and potential treatments.

Autosomal dominant inheritance

Autosomal dominant inheritance is a type of inheritance that is characterized by a dominant gene that is located on one of the non-sex chromosomes (autosomes). When an individual inherits only one copy of the mutated gene from one parent, he or she develops the condition that the gene causes.

Osteogenesis imperfecta (OI), also known as brittle bone disease, is a genetic condition characterized by bones that break easily. It is inherited in an autosomal dominant pattern.

According to the question, scientists determined that 58% of mice display incomplete penetrance in investigating osteogenesis imperfecta (OI) using mice as a model. Incomplete penetrance refers to a phenomenon in which individuals with the same genotype exhibit different phenotypes. In other words, some mice may have the disease-causing gene but not show any symptoms.

Scientists use animal models to investigate human diseases because it can be challenging or unethical to conduct certain experiments in humans. Animal models can also provide insights into the disease mechanism and potential treatments.

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list of bacteria for bacterial identification assignment Here is the the "list of suspects" for the bacterial identification assignment. Again, for the bacterial identification assignment, you will design a key that allows you to identify every bacteria on this list (i.e., they key should put EACH bacteria on the list into a group all by itself). Use the same approach you used in the "building your key" exercise that you worked on over the last 2-3 weeks and turned in last friday. Bacillus cereus Citrobacter freundii Clostridium Enterobacter aerogenes Enterococcus (Streptococcus) faecalis Escherichia (E.) coli Lactococcus (Streptococcus) lactis Mycobacterium Proteus vulgaris Proteus mirabilis Serratia marcescens Staphylococcus epidermidis

Answers

In the list of bacteria for bacterial identification assignment, Bacillus cereus is an aerobic spore-forming bacterium that is gram-positive. They may be found in soil, air, water, and some foods. Citrobacter freundii is an opportunistic pathogen that is gram-negative and has peritrichous flagella.

Clostridium is a gram-positive bacterium that produces an endospore. Enterobacter aerogenes is a gram-negative bacterium that is opportunistic and may cause healthcare-associated infections. Enterococcus (Streptococcus) faecalis is a gram-positive bacterium that is a commensal of the gastrointestinal tract, but may also cause healthcare-associated infections.

Escherichia coli is a gram-negative bacterium that is a normal constituent of the gut flora but can also cause urinary tract infections. Lactococcus (Streptococcus) lactis is a gram-positive bacterium used in the dairy industry.

Mycobacterium is an acid-fast bacterium that is difficult to stain with the Gram method. Proteus vulgaris is a gram-negative bacterium that is rod-shaped and mobile. Proteus mirabilis is a gram-negative bacterium that is rod-shaped and mobile.

Serratia marcescens is an opportunistic bacterium that is gram-negative and has a prodigious pigment that gives it a reddish-orange hue. Staphylococcus epidermidis is a gram-positive bacterium that is a commensal of the skin, but can also cause healthcare-associated infections.

Thus, the list of bacteria for the bacterial identification assignment is as follows:

Bacillus cereus, Citrobacter freundii, Clostridium, Enterobacter aerogenes, Enterococcus (Streptococcus) faecalis, Escherichia (E.) coli, Lactococcus (Streptococcus) lactis, Mycobacterium, Proteus vulgaris, Proteus mirabilis, Serratia marcescens, and Staphylococcus epidermidis.

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1. Which (TWO) of the following bones would you NOT use to kick a soccer ball?
fibula humerus metacarpals metatarsals patella phalanges tarsals tibia
2. Someone has a "cervical" injury. Is this an injury to the spine in their neck, upper back, or lower back?
3. Which of the three joints affords the most range of motion?

Answers

1) The bones you would not use to kick a soccer ball are the humerus and metacarpals.

2) A "cervical" injury refers to an injury to the spine in the neck region.

3) The joint that affords the most range of motion is the ball-and-socket joint.

1) The bones you would not use to kick a soccer ball are the humerus and metacarpals. The humerus is the bone of the upper arm, and the metacarpals are the bones in the hand. These bones are not directly involved in the kicking motion.

2) A "cervical" injury refers to an injury to the spine in the neck region. The cervical spine consists of the vertebrae in the neck area, and an injury to this region can affect the neck and potentially extend to the upper back.

3) The joint that affords the most range of motion is the ball-and-socket joint. This type of joint allows for movement in multiple directions, including flexion, extension, abduction, adduction, and rotation. Examples of ball-and-socket joints in the human body are the shoulder joint and the hip joint. These joints provide a wide range of motion compared to pivot joints.

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week : BIO 1110-0L00; texam 2, Chapters 5, 6, . Help Save & Exit S 13 Katy is an aspiring tattoo artist and is doing her first tattoo, Being nervous, she doesn't adjust the needle depth setting on the tattoo machine properly. As a result, the ink is only deposited into the dermis in some spots, while much of it is deposited into the lower layers of the epidermis. What will happen to the tattoo? 3.58 points Multiple Choice 8 004729 The tattoo won't show through the layers of the skin The tattoo will be perfectly fine, The ink will bleed and the tattoo will be blurry, but the ink will be retained. In about a month or so, the ink deposited into the epidermis will be gone.

Answers

Improper needle depth setting during the first tattoo may result in the ink being deposited into the lower layers of the epidermis, leading to a tattoo that may not show through the skin, appear blurry due to ink bleeding, and gradually fade or disappear within a month.

When the ink is only deposited into the lower layers of the epidermis due to improper needle depth setting, several outcomes can be expected for the tattoo.

Firstly, the tattoo may not show through the layers of the skin as intended. The ink needs to reach the dermis, the deeper layer of the skin, to remain visible over time.

If the ink is mainly deposited in the lower layers of the epidermis, it may not be visible or may appear faded.

Additionally, the ink may bleed and cause the tattoo to be blurry.

The lower layers of the epidermis are closer to the surface and are more prone to smudging and spreading of the ink, resulting in a less defined and clear tattoo.

Furthermore, since the epidermis is constantly renewing itself by shedding old skin cells, the ink deposited in the lower layers of the epidermis may eventually be sloughed off.

The turnover of skin cells in the epidermis occurs over a period of approximately one month, so the ink deposited in this layer may gradually fade or disappear over time.

In summary, if the ink is mainly deposited into the lower layers of the epidermis due to improper needle depth setting, the tattoo may not show through the layers of the skin, the ink may bleed and cause blurriness, and the ink deposited in the epidermis may fade or disappear within about a month.

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1) In cycle 1 and all subsequent cycles of the PCR reaction, one copy of each of the two original strands will be synthesized at the 3' end of the primer and up to the 5' end of the original strand. Write the sequence of the copies (C1 and C2) that are made of the chains O1 and O2.
(01) 3' T C G G C T A C A G C A G C A G A T G G T A C G T A 5'
(C1) 5' C C G A T _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 3'
(O2) 5' A G C C G A T G T C G T C G T C T A C C A T G C A T 3'
(C2) 3' _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ A T G G T 5'

Answers

PCR uses O1 and O2 as templates for DNA synthesis. PCR cycles involve denaturation, annealing, and extension.

The first cycle denatures O1 and O2, splitting the double-stranded DNA into single strands. Primer 3' binds to template strand complementary sequences. We don't know the primer's 3' end sequence from the sequences. The primer starts DNA synthesis by binding to a specific area. DNA polymerase uses the original strands as templates to synthesize new strands during extension. 5'-to-3' synthesis occurs. Thus, each template strand's new copy will be synthesized from the primer's 3' end to the original strand's 5' end.

Let's complete the sequences:

(C1) 5' C C G A T G G T A C G T A _ _ _ 3'

(C2) 3' _ A T G G T 5'

C1 synthesizes O1 from its 3' primer end to its 5' end. C2's synthesis begins at O2's 3' primer end and continues to its 5' end.

We can't establish C1 and C2's exact sequences without the primer sequence. The primer sequence determines DNA synthesis nucleotide order.

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Question 24 The macro-complexes formed by the intrinsic and extrinsic pathways that activate the common pathway are called: Brady-Kinen complexes Fibrinolysis complexes Factor activator complexes Tena

Answers

The macro-complexes formed by the intrinsic and extrinsic pathways that activate the common pathway are called Factor activator complexes. The correct answer is option c.

Factor activator complexes are macro-complexes formed by the intrinsic and extrinsic pathways that activate the common pathway of the blood coagulation cascade.

These complexes play a crucial role in initiating the formation of fibrin, the key component of blood clots. They involve various factors, such as factor VIII, factor IX, factor X, and factor XI, along with cofactors and other regulatory proteins.

The factor activator complexes act as catalysts to promote the conversion of factor X to its active form (factor Xa), leading to the subsequent activation of the common pathway and ultimately the formation of a stable blood clot.

The correct answer is option c.

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Complete question

Question 24 The macro-complexes formed by the intrinsic and extrinsic pathways that activate the common pathway are called:

a. Brady-Kinen complexes

b. Fibrinolysis complexes

c. Factor activator complexes

The following sequence of DNA was digested with the restriction endonuclease EcoRl.
5'-CGCCGAATTCCGGGATGTCGAATCCGCCCGGGGAATTCATATTTTAGCA-3'
3' -GCGGCTTAAGGCCCTACAGCTTAGGCGGGCCCCTTAAGTATAAAATCGT - 5'
a) ECOR recognizes the sequence GAATTC and cut(s) between the G and the A. Mark the location of All the cuts on the above sequence.
b) What type of ends does EcoRl produce?

Answers

a) Based on the recognition sequence GAATTC for EcoRI, the cuts will occur between the G and the A nucleotides within the sequence. The cuts are marked with "^" below:

5'-CGCC^GAATTC^CGGGATGTCGAATCCGCCCGGGGAATTCATATTTTAGCA-3'

3' -GCGGCTTAA^GGCCCTACAGCTTAGGCGGGCCCCTTAAGTATAAAATCGT - 5'

b) EcoRI produces sticky ends. After digestion, the DNA fragments will have overhanging ends with single-stranded regions. In this case, the sticky ends will have the sequence 5'-AATT-3' on one strand and 3'-TTAA-5' on the complementary strand.

EcoRI is a commonly used restriction endonuclease derived from the bacterium Escherichia coli. It recognizes and cuts DNA at the specific sequence GAATTC. Here are some additional details about EcoRI:

Recognition sequence: EcoRI recognizes the palindromic sequence GAATTC. The sequence reads the same on both DNA strands when read in the 5' to 3' direction.

Cutting site: EcoRI cuts the DNA between the G and the A nucleotides within the recognition sequence. This results in the creation of two fragments with complementary sticky ends.

Sticky ends: EcoRI produces sticky ends after digestion. The sticky ends have single-stranded overhangs with the sequence 5'-AATT-3' on one strand and 3'-TTAA-5' on the complementary strand. These sticky ends can base pair with complementary sequences, facilitating the cloning and manipulation of DNA fragments.

Applications: EcoRI is commonly used in molecular biology techniques, such as DNA cloning, restriction mapping, and DNA fragment analysis. It is often used in combination with other restriction enzymes to generate compatible ends for DNA ligation.

DNA digestion: When DNA is digested with EcoRI, the enzyme cleaves the phosphodiester bonds in the DNA backbone, resulting in the fragmentation of the DNA molecule into smaller pieces.

It's important to note that EcoRI is just one of many restriction endonucleases available, each with its own recognition sequence and cutting characteristics.

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Do we have to add a chemical to see the results for the urea
tubes? protein test
Yes
No

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The urea tubes protein test is used to measure the concentration of protein in a patient's urine. There are two tubes: the protein test tube and the urea test tube.

The urea tube contains a chemical that reacts with urea, resulting in a color change. The protein test tube, on the other hand, contains a reagent that reacts with protein, resulting in a color change.The presence of protein in urine may be an indication of a variety of medical problems. These tests are used to detect and monitor these issues. As a result, it is essential to follow all of the test's instructions to achieve the desired outcome.

The chemical in the urea tube is used to make sure that the urea in the patient's urine is broken down so that the protein level can be determined accurately. In conclusion, we need to add a chemical to see the results for the urea tubes protein test. It is a critical part of the test, and if omitted, the results may not be accurate. a chemical is necessary to obtain the desired outcome.

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Which of the following statements about influenza replication
and exit is TRUE? (1.5 points)
High pH is a signal to release the viral genome into the
cytoplasm
Viral transcription and translation occ

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The statement that is TRUE about influenza replication and exit are that viral transcription and translation occur in the nucleus.

During the replication and exit of the influenza virus, several important processes take place. Influenza viruses have a segmented genome consisting of multiple RNA segments. After the virus enters the host cell, it needs to replicate its genome and produce viral proteins for the assembly of new viral particles.

In the case of influenza, viral transcription and translation occur in the nucleus of the host cell. The viral RNA segments are transcribed into messenger RNA (mRNA) by the viral RNA polymerase. These viral mRNAs are then transported out of the nucleus into the cytoplasm, where they undergo translation to produce viral proteins.

Once the viral proteins are synthesized, they are transported back into the nucleus, where viral genome replication takes place. The replicated viral RNA segments are then exported from the nucleus to the cytoplasm, where they associate with the newly synthesized viral proteins to form new viral particles.

Therefore, the statement that viral transcription and translation occur in the nucleus is true, highlighting an essential step in the replication and exit of the influenza virus.

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Question 45 Not yet graded / 7 pts Part A about the topic of nitrogen in biology. How does nitrogen come into the biosphere (including what pathway and the two important enzymes involved)? How does nitrogen come into the human body? And, what bridges the gap between how nitrogen enters the biosphere and how it enters the human body?

Answers

Nitrogen enters the biosphere primarily through the process of nitrogen fixation. In this pathway, atmospheric nitrogen (N₂) is converted into a biologically useful form, such as ammonia (NH₃), by nitrogen-fixing bacteria.

These bacteria possess the enzyme nitrogenase, which catalyzes the conversion of N₂ to NH₃. Another important enzyme involved in nitrogen fixation is nitrogen reductase, which reduces nitrate (NO₃⁻) to nitrite (NO₂⁻) during the process.

In the human body, nitrogen enters through dietary intake. We obtain nitrogen primarily through the consumption of protein-rich foods, such as meat, fish, eggs, and legumes. Proteins are composed of amino acids, and nitrogen is an essential component of amino acids. Through the digestion and breakdown of dietary proteins, the nitrogen-containing amino acids are released and utilized by the body for various biological processes.

The gap between how nitrogen enters the biosphere and how it enters the human body is bridged by the nitrogen cycle. Nitrogen compounds present in the environment, such as ammonia and nitrate, can be taken up by plants and incorporated into their tissues.

Animals then consume these plants, obtaining nitrogen in the form of dietary protein. The nitrogen cycle encompasses processes like nitrogen fixation, nitrification, assimilation, and denitrification, which ensure the cycling and availability of nitrogen in the biosphere for various organisms, including humans.

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A bacteria lives in a hydrothermal pool with an average temperature of 70 degC and a pH of 3. It's enzymes are going to function ideally at which of the following ranges? a) 65-72 degC and 2-4. b) 65-72 degC and 5-7. c) 35-40 degC and 6-8. d) 50-60 degC and 2-4.

Answers

The enzymes of the bacteria living in a hydrothermal pool with an average temperature of 70°C and a pH of 3 would function ideally within the range of a) 65-72°C and 2-4 pH.

Extreme temperature and pH conditions in the hydrothermal pool would suggest that the bacteria has adapted to survive and function optimally within those specific ranges. Therefore, the enzymes of the bacteria would be most efficient and effective within the temperature range of 65-72°C and the pH range of 2-4.

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