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A gene responsible for production of hair pigment in dogs is called Fursilla (frsl). A map of the Fursilla locus is shown in Figure 1, below. When expressed, it results in darkly pigmented dog fur. When unexpressed, the hair is devoid of pigmentation and appears pure white. Expression of Fursilla (frsl) depends on the binding/lack of binding of several proteins: Nefur (NEFR), Lesfur (LSFR), and Dirkfur (DRKFR). Figure 2 shows the relative levels of frsl transcript as measured via RT-qPCR when each protein is overexpressed or unexpressed in the cell.Scientists identified a mutant allele of the Fursilla gene caused by a deletion of bases 1231-1295 on chromosome 8. What is the likely effect of this mutation?
A. No transcription and no translation of Fursilla will occur.
B. Transcription will not occur, but translation will be unaffected.
C. Transcription of Fursilla will occur, but no translation will occur.
D. Both transcription and translation of Fursilla will be unaffected.
|
A The location appears to be upstream of the transcription start site. This is the promoter region and is likely important for polymerase binding and initiation of transcription. Without it, there would be no transcription and subsequently there could not be any translation.
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Biology/188
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Biology
| null |
Figure 1 shows a cell at the onset of meiosis (just prior to initiation of prophase I) and the resulting 4 gametes achieved at the end of meiosis. Select the statement that is likely true.
A. Meiosis occurred as it should.
B. There was a nondisjunction event in Meiosis I.
C. There was a nondisjunction event in Meiosis II.
D. Two Barr bodies were produced instead of four gametes.
|
B If meiosis occurred as it should, there should be 4 haploid gametes, each with one long and one short chromosome. None of the gametes look correct. There was a nondisjunction at Meiosis I, and the two long homologous chromosomes failed to separate. This caused two gametes to get no copies of the long chromosome and two gametes to get two copies. If the nondisjunction was in Meiosis II, then two of the gametes would look normal and two of the gametes would have incorrect amounts of chromosomes. Barr bodies are small and unviable, but they still have the correct number of chromosomes in them.
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Biology/189
|
Biology
| null |
The active site of the enzyme Ritzolinine (RZN45) contains three positively charged lysine residues. When ascorbic acid is present, binding of JB-76, the substrate of RZN45, decreases. The reaction rate is affected by the presence of ascorbic acid as shown in the figure below. It is thought that a daily supplement of Vitamin C might aid those suffering from Ritzolierre's Disease, which is caused by elevated levels of RZN45.If Figure 1 were to include additional data for higher concentrations of JB-76, how would the reaction rate change?
A. The reaction rate would be lower than any reaction rate shown in Figure 1.
B. The reaction rate would be higher than any reaction rate shown in Figure 1.
C. The reaction rate would be equal to the highest reaction rate shown in Figure 1.
D. The reaction rate would increase gradually for each concentration of JB-76 added.
|
C As shown in the graph, the amount of substrate has reached a saturation point and adding additional substrate will not increase the rate further. The reaction rate will remain at the same high level.
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Biology/190
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Biology
| null |
A patient with Ritzolierre's Disease would likely benefit from which of the following:
I. Injections of RZN45
II. Injections of JB-76
The active site of the enzyme Ritzolinine (RZN45) contains three positively charged lysine residues. When ascorbic acid is present, binding of JB-76, the substrate of RZN45, decreases. The reaction rate is affected by the presence of ascorbic acid as shown in the figure below. It is thought that a daily supplement of Vitamin C might aid those suffering from Ritzolierre's Disease, which is caused by elevated levels of RZN45.III. Injections of ascorbic acid
A. I only
B. II only
C. III only
D. II and III
|
C Ritzolierreâs Disease is caused by elevated enzyme levels. Only option III would decrease the enzyme activity levels. Options I and II would raise the levels and make the disease worse.
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Biology/191
|
Biology
| null |
The active site of the enzyme Ritzolinine (RZN45) contains three positively charged lysine residues. When ascorbic acid is present, binding of JB-76, the substrate of RZN45, decreases. The reaction rate is affected by the presence of ascorbic acid as shown in the figure below. It is thought that a daily supplement of Vitamin C might aid those suffering from Ritzolierre's Disease, which is caused by elevated levels of RZN45.Which is the best explanation for the differing effects on reaction rate caused by ascorbic acid at low and high concentrations of JB-76?
A. At high concentrations of JB-76, there are fewer free active sites for the ascorbic acid to bind to.
B. At low concentrations of JB-76, there are fewer free allosteric sites for the RZN45 to bind to.
C. At high concentrations of JB-76, there are more free allosteric sites for the ascorbic acid to bind to.
D. At low concentrations of JB-76, there are more free active sites for the RZN45 to bind to.
|
A The graph shows that at low/medium levels of substrate the reaction rate is low in the presence of ascorbic acid and then at high levels of substrate the reaction rate is high, even with ascorbic acid present. This is because with lots of substrate, the ascorbic acid does not stand a chance at finding a free active site to bind to because the superfluous amount of substrate is likely to be bound at most of the sites.
|
Biology/192
|
Biology
| null |
The active site of the enzyme Ritzolinine (RZN45) contains three positively charged lysine residues. When ascorbic acid is present, binding of JB-76, the substrate of RZN45, decreases. The reaction rate is affected by the presence of ascorbic acid as shown in the figure below. It is thought that a daily supplement of Vitamin C might aid those suffering from Ritzolierre's Disease, which is caused by elevated levels of RZN45.Which of the following is likely true?
A. RZN45 and JB-76 have similarly charged amino acids at their active sites.
B. Ritzolinine is stabilized in the presence of Vitamin C.
C. Ascorbic acid and JB-76 each have a pocket of negatively charged amino acids.
D. JB-76 and ascorbic acid have an identical number of amino acids.
|
C RZN45 is the enzyme and it binds to JB-76. Ascorbic acid (vitamin C) is an inhibitor of this reaction and binding to JB-76 is reduced, which suggests the inhibitor binds to the same place as the substrate. The graph confirms that the inhibition is competitive since it shows a decreased reaction rate when ascorbic acid is present. Choice (A) is not true because similarly charged amino acids would repel each other and RZN45 and JB-76 bind to each other. There is nothing to suggest that (B) is correct. Choice (C) would make sense since both ascorbic acid and JB-76 bind to the same place, so they probably have similar structures. Choice (D) could be true, but it is unlikely. Even things that are the same shape do not necessarily have the same number of amino acids. Proteins are often hundreds of amino acids long, so having the exact same number is like a needle in a haystack.
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Biology/193
|
Biology
| null |
A proponent of the vegan diet writes a blog post that claims that the vegan diet is the only way to "reduce your carbon footprint, with respect to food, because it cuts your carbon footprint in half" and shows the graph above. Which of the following directly supports this claim?
A. According to the 10% rule, only 10% of energy is passed from producers to consumers, so when humans consume animals, the energy harvested is greatly reduced.
B. Livestock produce methane emissions and consume large amounts of plants every year.
C. Deforestation for grazing livestock contributes to the loss of plants for carbon sequestration.
D. Beef and lamb livestockânot chicken, fish, or porkâhas the most dramatic impact on carbon footprint.
|
D If (D) is a correct statement, then simply cutting out beef and lamb from your diet can dramatically decrease your carbon footprint-even if you do not decide to eat vegan. Choices (A), (B), and (C) are all true statements with respect to reducing the carbon footprint of food.
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Biology/194
|
Biology
| null |
Viruses are obligate intracellular parasites, requiring their host cells for replication. Consequently, viruses generally attempt to reproduce as efficiently and quickly as possible in a host. Below is a graph depicting the initial growth pattern of a bacteriophage within a population of E. coli. This reproductive strategy is most similar to which of the following?
A. An r-strategist, because it aims to produce a large abundance of offspring to ensure survival
B. A k-strategist, because it aims to produce a large abundance of offspring to ensure survival
C. An r-strategist, because it is best suited to thrive in stable environments and over a long life span
D. A k-strategist, because it is best suited to thrive in stable environments and over a long life span
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A Viruses would display reproductive strategies most similar to r-strategists because they aim to reproduce as fast as possible and create as many progeny as possible in order to increase their odds of transmission to other hosts.
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Biology/195
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Biology
| null |
During the Industrial Revolution, a major change was observed in many insect species due to the mass production and deposition of ash and soot around cities and factories. One of the most famous instances was within the spotted moth population. An ecological survey was performed in which the number of spotted moths and longtail moths were counted in 8 different urban settings over a square kilometer in 1802. A repeat experiment was performed 100 years later in 1902.How would the results of this study have been different if factories produced white or light gray ash and soot rather than black?
A. There would be no change to the results of the experiment.
B. There would have been added selection pressure for more white-bodied spotted moths and against black-bodied spotted moths.
C. There would have been added selection pressure for more black-bodied spotted moths and against white-bodied spotted moths.
D. There would have been an increase in the frequency of both black-bodied and white-bodied spotted moths.
|
B If the color of ash or soot produced by the Industrial Revolution were white or light gray, this would likely reverse the trend observed, applying additional selection against the black moths.
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Biology/196
|
Biology
| null |
During the Industrial Revolution, a major change was observed in many insect species due to the mass production and deposition of ash and soot around cities and factories. One of the most famous instances was within the spotted moth population. An ecological survey was performed in which the number of spotted moths and longtail moths were counted in 8 different urban settings over a square kilometer in 1802. A repeat experiment was performed 100 years later in 1902.Why was the population of longtail moths also surveyed in this study?
A. Variations in the environment were expected to alter the population of longtail moths.
B. Longtail moths were included as a control because they were not expected to change appreciably due to changes associated with the Industrial Revolution.
C. The peppered moth did not have a long enough tail to visualize.
D. Longtail moths were poisoned by the coal dust and suffered drastic population loss.
|
B Longtail moths were included in the experiment as a control to compare the effects that are not associated with color.
|
Biology/197
|
Biology
| null |
During the Industrial Revolution, a major change was observed in many insect species due to the mass production and deposition of ash and soot around cities and factories. One of the most famous instances was within the spotted moth population. An ecological survey was performed in which the number of spotted moths and longtail moths were counted in 8 different urban settings over a square kilometer in 1802. A repeat experiment was performed 100 years later in 1902.Which of the following statements best explains the data?
A. As time passed from 1802 to 1902, the frequency of white-bodied pepper moths increased and black-bodied pepper moths decreased.
B. As time passed from 1802 to 1902, the frequency of white-bodied pepper moths decreased and black-bodied pepper moths increased.
C. As time passed from 1802 to 1902, the frequency of white-bodied pepper moths and black-bodied pepper moths both increased.
D. As time passed from 1802 to 1902, the frequency of white-bodied pepper moths and black-bodied pepper moths both decreased.
|
B Based on the data, the number of white-bodied pepper moths decreased between 1802 and 1902, and the number of black-bodied pepper moths increased during the same period.
|
Biology/198
|
Biology
| null |
During the Industrial Revolution, a major change was observed in many insect species due to the mass production and deposition of ash and soot around cities and factories. One of the most famous instances was within the spotted moth population. An ecological survey was performed in which the number of spotted moths and longtail moths were counted in 8 different urban settings over a square kilometer in 1802. A repeat experiment was performed 100 years later in 1902.What type of selection is represented by the results of this study?
A. Stabilizing selection
B. Directional selection
C. Disruptive selection
D. Divergent selection
|
B The data provided show a transition toward one extreme (black) and away from another (white). This is an example of directional selection.
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Biology/199
|
Biology
| null |
An experiment is performed to evaluate the amount of DNA present during a complete cell cycle. All of the cells were synced prior to the start of the experiment. During the experiment, a fluorescent chemical was applied to cells, which would fluoresce only when bound to DNA. The results of the experiment are shown above. Differences in cell appearance by microscopy or changes in detected DNA were determined to be phases of the cell cycle and are labeled with the letters A-D.During which of the labeled phases of the experiment would the cell undergo anaphase?
A. Phase A
B. Phase B
C. Phase C
D. Phase D
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D Anaphase represents the cell division stage of the cell cycle and would be the phase that occurs right before the amount of genetic material should decrease. Phase D is the phase right before the genetic material would drop, so (D) is the correct answer.
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Biology/200
|
Biology
| null |
An experiment is performed to evaluate the amount of DNA present during a complete cell cycle. All of the cells were synced prior to the start of the experiment. During the experiment, a fluorescent chemical was applied to cells, which would fluoresce only when bound to DNA. The results of the experiment are shown above. Differences in cell appearance by microscopy or changes in detected DNA were determined to be phases of the cell cycle and are labeled with the letters A-D.Approximately how long does S phase take to occur in these cells?
A. 15 min
B. 20 min
C. 30 min
D. 40 min
|
C The synthesis, or S phase, of the cell cycle represents the step in which the genetic material is duplicated. The only phase labeled in the experiment that represents an increase is phase B. Based on the time scale on the x-axis, this phase lasts approximately 30 minutes.
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Biology/201
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Biology
| null |
A scientist is testing new chemicals designed to stop the cell cycle at various stages of mitosis. Upon applying one of the chemicals, she notices that all of the cells appear as shown below. Which of the following best explains how the chemical is likely acting on the cells?
A. The chemical has arrested the cells in prophase and has prevented attachment of the spindle fibers to the kinetochore.
B. The chemical has arrested the cells in metaphase and has prevented dissociation of the spindle fibers from the centromere.
C. The chemical has arrested the cells in metaphase and is preventing the shortening of the spindle fibers.
D. The chemical has arrested the cells in anaphase and is preventing the formation of a cleavage furrow.
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C During anaphase, the chromatids are separated by shortening of the spindle fibers. Chemically blocking the shortening of these fibers would arrest the cell in metaphase. The cells are arrested in metaphase as indicated by the alignment of the chromosomes in the center of the cell and their attachment to spindle fibers, eliminating (A) and (D). The chromosomes still seem to be attached to the fibers, so there doesn't appear to be dissociation of the fibers, eliminating (B).
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Biology/202
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Biology
| null |
Glycolysis (shown below) is a critical metabolic pathway that is utilized by nearly all forms of life. The process of glycolysis occurs in the cytoplasm of the cell and converts 1 molecule of glucose into 2 molecules of pyruvic acid.Which of the following most accurately describes the net reaction of glycolysis?
A. It is an endergonic process because it results in a net increase in energy.
B. It is an exergonic process because it results in a net increase in energy.
C. It is an endergonic process because it results in a net decrease in energy.
D. It is an exergonic process because it results in a net decrease in energy.
|
B Glycolysis results in the production of ATP (energy), so it is considered an exergonic process.
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Biology/203
|
Biology
| null |
Glycolysis (shown below) is a critical metabolic pathway that is utilized by nearly all forms of life. The process of glycolysis occurs in the cytoplasm of the cell and converts 1 molecule of glucose into 2 molecules of pyruvic acid.Glycolysis does not require oxygen to occur in cells. However, under anaerobic conditions, glycolysis normally requires fermentation pathways to occur to continue to produce ATP. Which best describes why glycolysis is dependent on fermentation under anaerobic conditions?
A. Glycolysis requires fermentation to produce more glucose as a substrate.
B. Glycolysis requires fermentation to synthesize lactic acid and restore NADH to NAD+.
C. Glycolysis requires fermentation to generate ATP molecules to complete the early steps of the pathway.
D. Glycolysis requires fermentation to generate pyruvate for a later step in the pathway.
|
B In fermentation, pyruvic acid is converted into either ethanol or lactic acid. During this process, NADH is recycled into NAD+.
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Biology/204
|
Biology
| null |
Glycolysis (shown below) is a critical metabolic pathway that is utilized by nearly all forms of life. The process of glycolysis occurs in the cytoplasm of the cell and converts 1 molecule of glucose into 2 molecules of pyruvic acid.How many net ATP would be generated directly from glycolysis from the breakdown of 2 glucose molecules?
A. 2
B. 4
C. 8
D. 12
|
B Based on the pathway provided, consumption of one glucose and two ATP results in production of four ATP. In other words, each glucose results in a net gain of two ATP. Therefore, two glucose molecules would result in a net gain of four ATP.
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Biology/205
|
Biology
| null |
Binding of Inhibitor Y as shown below inhibits a key catalytic enzyme by inducing a structural conformation change. Which of the following accurately describes the role of Inhibitor Y?
A. Inhibitor Y competes with substrates for binding in the active site and functions as a competitive inhibitor.
B. Inhibitor Y binds allosterically and functions as a competitive inhibitor.
C. Inhibitor Y competes with substrates for binding in the active site and functions as a noncompetitive inhibitor.
D. Inhibitor Y binds allosterically and functions as a noncompetitive inhibitor.
|
D Inhibitor Y is binding at a site outside the active site and is inducing a conformational change in the enzyme structure. By binding outside the active site, it must be an allosteric inhibitor, eliminating (A) and (C). Because the inhibitor is binding outside the active site, it is not competing with the substrate for binding, so it is considered a noncompetitive inhibitor.
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Biology/206
|
Biology
| null |
The mitochondrion is a critical organelle structure involved in cellular respiration. Below is a simple schematic of the structure of a mitochondrion. Which of the structural components labeled below in the mitochondrion is the primary location of the Krebs cycle?
A. Inner membrane
B. Matrix
C. Intermembrane space
D. Outer membrane
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B The Krebs cycle occurs primarily in the matrix of the mitochondria. The inner membrane and the intermembrane space-(A) and (C), respectively-are used in oxidative phosphorylation.
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Biology/207
|
Biology
| null |
Vibrio cholerae (shown below) are highly pathogenic bacteria that are associated with severe gastrointestinal illness and are the causative agent of cholera. In extreme cases, antibiotics are prescribed that target bacterial structures that are absent in animal cells. Which of the following structures is most likely targeted by antibiotic treatment?
A. Cytoplasm
B. Plasma membrane
C. Ribosomes
D. Cell wall
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D The cell wall is a structure that is present in bacteria but absent in animal cells. Consequently, this structure is targeted by several leading classes of antibiotics and would be an effective target of therapeutics against V. cholerae. Cytoplasm, plasma membrane, and ribosomes-(A), (B), and (C), respectively-are all structures that are present in animal cells.
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Biology/208
|
Biology
| null |
After one day of exposure, the mixture in the flask had turned pink in color, and later analysis showed that at least 10% of the carbon had been transformed into simple and complex organic compounds including at least 11 different amino acids and some basic sugars. No nucleic acids were detected in the mixture.
In 1953, Stanley Miller and Harold Urey performed an experiment at the University of Chicago to test the hypothesis that the conditions of the early Earth would have favored the formation of larger, more complex organic molecules from basic precursors. The experiment, shown in Figure 1, consisted of sealing basic organic chemicals (representing the atmosphere of the primitive Earth) in a flask, which was exposed to electric sparks (to simulate lightning) and water vapor.A scientist believes that the Miller-Urey experiment failed to yield the remaining amino acids and the nucleic acids because of the absence of critical chemical substrates that would have existed on the primordial Earth due to volcanism. Which of the following basic compounds, which are associated with volcanism, would NOT need to be added in a follow-up Miller-Urey experiment?
A. H2S (gas)
B. SiO2 (silica)
C. SO2
D. H3PO4 (phosphoric acid)
|
B Silica is a mineral form of glass, is not a common component of life-forms, and is largely chemically inert. Since oxygen is already present in several compounds included in the experiment, the addition of this compound does not provide any additional elements or chemical substrates, which would permit generation of additional amino acids or synthesis of nucleic acids. The addition of sulfur compounds and phosphorus is necessary to generate some amino acids and all nucleic acids, which eliminates (A), (C), and (D).
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Biology/209
|
Biology
| null |
After one day of exposure, the mixture in the flask had turned pink in color, and later analysis showed that at least 10% of the carbon had been transformed into simple and complex organic compounds including at least 11 different amino acids and some basic sugars. No nucleic acids were detected in the mixture.
In 1953, Stanley Miller and Harold Urey performed an experiment at the University of Chicago to test the hypothesis that the conditions of the early Earth would have favored the formation of larger, more complex organic molecules from basic precursors. The experiment, shown in Figure 1, consisted of sealing basic organic chemicals (representing the atmosphere of the primitive Earth) in a flask, which was exposed to electric sparks (to simulate lightning) and water vapor.Some amino acids, such as cysteine and methionine, could not be formed in this experiment. Which of the following best explains why these molecules could not be detected?
A. The chemical reactions necessary to create amino acids such as cysteine and methionine require more energy than the simulated lightning provided in the experiment.
B. The chemical reactions necessary to create amino acids such as cysteine and methionine require enzymes for catalysis to occur, which were not included in the experiment.
C. Sulfur-based compounds were not included in the experiment.
D. Nitrogen-based compounds were not included in the experiment.
|
C The amino acids cysteine and methionine contain the element sulfur (as indicated by the S in the amino acid structure shown). However, no sulfur-based compounds were included in the Miller-Urey experiment, so it was impossible to form these two amino acids under the conditions of the experiment.
|
Biology/210
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Biology
| null |
After one day of exposure, the mixture in the flask had turned pink in color, and later analysis showed that at least 10% of the carbon had been transformed into simple and complex organic compounds including at least 11 different amino acids and some basic sugars. No nucleic acids were detected in the mixture.
In 1953, Stanley Miller and Harold Urey performed an experiment at the University of Chicago to test the hypothesis that the conditions of the early Earth would have favored the formation of larger, more complex organic molecules from basic precursors. The experiment, shown in Figure 1, consisted of sealing basic organic chemicals (representing the atmosphere of the primitive Earth) in a flask, which was exposed to electric sparks (to simulate lightning) and water vapor.Which of the following contradicts the hypothesis of the experiment that life may have arisen from the formation of complex molecules in the conditions of the primitive Earth?
A. Complex carbon-based compounds were generated after only one day of exposure to simulated primitive Earth conditions.
B. Nucleic acid compounds such as DNA and RNA were not detected in the mixture during the experiment.
C. Over half of known amino acids involved in life were detected in the mixture during the experiment.
D. Basic sugar molecules were generated and detected in the mixture during the experiment.
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B The hypothesis that life may have arisen from formation of complex molecules from the primordial "soup" of Earth is not supported by the absence of nucleic acids. All life is DNA-based, yet no nucleic acid molecules were detected. The presence of carbon molecules, amino acids, and sugars, which are common compounds and compose life, supports the hypothesis, so eliminate (A), (C), and (D). Choice (B) is correct.
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Biology/211
|
Biology
| null |
Amino acids are the basic molecular units which compose proteins. All life on the planet forms proteins by forming chains of amino acids. Which component of Figure 1 will vary from amino acid to amino acid?
A. Label A
B. Label B
C. Label C
D. Label D
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C All amino acids share a carboxylic acid group, COOH, labeled (B), and an amino-group, NH2, labeled (D). They also share a hydrogen atom bound to the central carbon, (A). Differences in amino acids are defined by variations in the fourth position called the R-group, labeled (C).
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Biology/212
|
Biology
| null |
At which position in the above circuit will the charge passing that position in one second be largest?
A. A
B. B
C. C
D. D
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A: Kirchoff's junction rule, a statement of conservation of electric charge in a circuit, says that current entering a junction equals current leaving a junction. Current is charge flowing per second. The charge passing point A in one second must be equal to the sum of the charge passing B, C, and D in one second; A must have the greatest charge flow per second.
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Physics/1
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Physics
| null |
In the laboratory, a 0.5-kg cart collides with a fixed wall, as shown in the preceding diagram. The collision is recorded with a video camera that takes 20 frames per second. A student analyzes the video, placing a dot at the center of mass of the cart in each frame. The analysis is shown above.
About how fast was the cart moving before the collision?
A. 0.25 m/s
B. 4.0 m/s
C. 0.20 m/s
D. 5.0 m/s
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B: The dots divide the 1-meter distance into five parts. In the time between dots, the cart travels 1/5 of a meter, or 0.2 m. The time between dots is 1/20 of a second, or 0.05 s. At constant speed, the speed is given by distance/time: 0.20 m/0.05 s = 4 m/s.
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Physics/2
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Physics
| null |
In the laboratory, a 0.5-kg cart collides with a fixed wall, as shown in the preceding diagram. The collision is recorded with a video camera that takes 20 frames per second. A student analyzes the video, placing a dot at the center of mass of the cart in each frame. The analysis is shown above.
Which of the following best estimates the change in the cart’s momentum during the collision?
A. 27 N·s
B. 13 N·s
C. 1.3 N·s
D. 2.7 N·s
|
D: Initially, the cart's mass is 0.5 kg and speed is 4 m/s, so the cart's momentum is mv = 2 N·s. In the collision, the cart loses that 2 N·s in order to stop briefly and then gains more momentum in order to speed up again. So the momentum change must be more than 2 N·s. How much more? After collision, the cart is moving slower than 4 m/s because the dots are closer together, so the cart's momentum is less than 2 N·s. The cart's momentum change is (2 N·s) + (something less than 2 N·s); the only possible answer is 2.7 N·s.
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Physics/3
|
Physics
| null |
In the laboratory, a 60-Hz generator is connected to a string that is fixed at both ends. A standing wave is produced, as shown in the preceding figure. In order to measure the wavelength of this wave, a student should use a meterstick to measure from positions
A. B to C
B. B to D
C. D to E
D. A to F
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D: On a standing wave, the wavelength is measured from node-to-node-to-node (i.e., across two "humps").
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Physics/4
|
Physics
| null |
Four identical lead balls with large mass are connected by rigid but very light rods in the square configuration shown in the preceding figure. The balls are rotated about the three labeled axes. Which of the following correctly ranks the rotational inertia I of the balls about each axis?
A. $I_B > I_A = I_C$
B. $I_A > I_C = I_B$
C. $I_C > I_A > I_B$
D. $I_C > I_A = I_B$
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D: The rotational inertia of a point mass is $MR^2$, where $R$ is the distance from the mass to the axis of rotation. Pretend the side of the square is of length 2 m, and that each mass is 1 kg. For axis A, each mass has rotational inertia $(1 kg) (1 m)^2 = 1 kg·m^2$. With four masses total, that's $4 kg·m^2$. For axis B, each mass is $\sqrt 2$ m from the axis (the diagonal of the square is $2\sqrt 2$ m, each mass is half a diagonal from the axis). Each mass has $(1 kg)(\sqrt 2 m)^2 = 2 kg·m^2$. Two masses make a total of $4 kg·m^2$. And for axis C, the masses are each 2 m from the axis, so they each have $(1 kg) × (2 m)^2 = 4 kg·m^2$. With two masses, that's a total of $8 kg·m^2$. So this would be ranked axis C, followed by equal axes A and B.
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Physics/5
|
Physics
| null |
In the laboratory, a 3-kg cart experiences a varying net force. This net force is measured as a function of time, and the data collected are displayed in the graph above. What is the change in the cart’s momentum during the interval t = 0 to t = 2 s?
A. 5 N·s
B. 10 N·s
C. 15 N·s
D. 30 N·s
|
A: Change in momentum is also known as impulse and is equal to force times time interval. On this graph, the multiplication of the axes means to take the area under the graph. Each segment of the data looks like it represents a straight line, making a big triangle. The area of a triangle is ½(base)(height). That's ½(5 N)(2 s) = 5 N·s.
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Physics/6
|
Physics
| null |
A block is attached to a vertical spring. The block is pulled down a distance A from equilibrium, as shown above, and released from rest. The block moves upward; the highest position above equilibrium reached by the mass is less than A, as shown. When the mass returns downward, how far below the equilibrium position will it reach?
A. Greater than the distance A below equilibrium
B. Less than the distance A below equilibrium
C. Equal to the distance A below equilibrium
D. No distance—the block will fall only to the equilibrium position.
|
B: If friction and air resistance are negligible, a mass on a spring oscillates about the equilibrium position, reaching the same maximum distance above and below. In this case, since the mass doesn't get all the way to position A at the top, mechanical energy was lost (to friction or air resistance or some nonconservative force). Thus, without some external energy input, the mass won't reach its maximum position at the bottom, either-at the bottom it will have no kinetic energy, so all the energy will be potential, and we've already established that some total mechanical energy was lost.
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Physics/7
|
Physics
| null |
Two charged Styrofoam balls are brought a distance d from each other, as shown. The force on Ball B is 2 μN to the right. When the distance between the balls is changed, the force on Ball B is 8 μN to the right.
Which of the following can indicate the sign of the charges of balls A and B?
Ball A Ball B
A. positive nagative
B. neutral positive
C. negative negative
D. positive neutral
|
C: If Ball B is forced to the right by Ball A, then Ball A must be forced to the left by Ball B-that's Newton's third law. These balls repel-only like-signed charges repel.
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Physics/8
|
Physics
| null |
Two charged Styrofoam balls are brought a distance d from each other, as shown. The force on Ball B is 2 μN to the right. When the distance between the balls is changed, the force on Ball B is 8 μN to the right.
When the force on Ball B is 8 μN, what is the distance between the centers of the two balls?
A. $d/4$
B. $d/2$
C. $d/16$
D. $d/\sqrt 2$
|
B: Coulomb's law gives the force of one charge on another, $k\frac{Q_1Q_2}{r^2}$. We haven't changed the charges $Q$; it's the force of A on B that's quadrupled. Since the force $F$ has increased, the distance between charges $r$ has decreased. Since the force has quadrupled, and since the r in the denominator is square, the distance has been cut in half $(1/2)^2 = 1/4$; and making the denominator four times smaller makes the whole fraction four times bigger.
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Physics/9
|
Physics
| null |
A disk of radius 1 m and rotational inertia I = 0.5 kg.m2 is free to rotate, but initially at rest. A blob of putty with mass 0.1 kg is traveling toward the disk with a speed of 10 m/s, as shown in the preceding figure. The putty collides with the outermost portion of the disk and sticks to the disk. What is the angular momentum of the combined disk-putty system after the collision?
A. $5 kg·m^2/s$
B. $1 kg·m^2/s$
C. $0.5 kg·m^2/s$
D. $0 kg·m^2/s$
|
B: In a collision, momentum-including angular momentum-is conserved. The question might as well be asking, "What is the angular momentum of the two objects before the collision?" And since the disk is at rest initially, the question is asking the even easier question, "What is the angular momentum of the putty before collision?" The axis of rotation is the center of the disk. The putty is a point mass; the angular momentum of a point mass is mvr with r the distance of closest approach to the axis. That's $(0.1 kg)(10 m/s)(1 m) = 1 kg·m^2/s$.
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Physics/10
|
Physics
| null |
What is the current in the 4 Ω resistor in the circuit in the preceding illustration?
A. 1.5 A
B. 2.0 A
C. 3.0 A
D. 6.0 A
|
A: First simplify the 4 Ω and 12 Ω parallel combination to a 3 Ω equivalent resistance. In series with the other 3 Ω resistance, that gives a total resistance for the circuit of 6 Ω. By Ohm's law used on the whole circuit, 12 V = I(6 Ω). The total current in the circuit is thus 2 A. This current must split between the two parallel resistors. The only possible answer, then, is 1.5 Ω-the current in the 4 Ω resistor must be less than the total current.
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Physics/11
|
Physics
| null |
Two sounds are played in the laboratory. A microphone is connected to an oscilloscope, which displays the traces shown above for each sound. On these traces, the horizontal axis is time; the vertical axis is related to the distance the microphone’s diaphragm is displaced from its resting position. The scales are identical for each diagram. Which of the following is correct about the sounds that produce the traces above?
A. Sound 1 is louder, and sound 2 is higher pitched.
B. Sound 2 is louder, and sound 2 is higher pitched.
C. Sound 1 is louder, and sound 1 is higher pitched.
D. Sound 2 is louder, and sound 1 is higher pitched.
|
C: The amplitude of a sound is related to the loudness. Sound 1 has higher amplitude, so it is louder. The pitch of a sound is related to the frequency. Since the horizontal axis is time, the peak-to-peak distance corresponds to a period, which is the inverse of the frequency. Sound 1 has a smaller period, so it has a larger frequency and a larger pitch.
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Physics/12
|
Physics
| null |
In an experiment, a marble rolls to the right at speed v, as shown in the top diagram. The marble rolls under a canopy, where it is heard to collide with marbles that were not initially moving. Such a collision is known to be elastic. After the collision, two equal-mass marbles are observed leaving the canopy with velocity vectors directed as shown. Which of the following statements justifies why the experimenter believes that a third marble was involved in the collision under the canopy?
A. Before collision, the only marble momentum was directed to the right. After the collision, the combined momentum of the two visible marbles is still to the right. Another marble must have a leftward momentum component to conserve momentum.
B. Before collision, the only marble momentum was directed to the right. After the collision, the combined momentum of the two visible marbles has a downward component; another marble must have an upward momentum component to conserve momentum.
C. Before collision, the only marble kinetic energy was directed to the right. After the collision, the combined kinetic energy of the two visible marbles is still to the right. Another marble must have a leftward kinetic energy component to conserve kinetic energy.
D. Before collision, the only marble kinetic energy was directed to the right. After the collision, the combined kinetic energy of the two visible marbles has a downward component; another marble must have an upward kinetic energy component to conserve kinetic energy.
|
B: Choices C and D are wrong because kinetic energy doesn't have a direction. Choice A is wrong because momentum conservation does not require a leftward momentum component-since the initial momentum was all to the right, the final momentum should be to the right. It's the vertical momentum that's the problem. Since the vertical momentum was zero to start with, any vertical momentum after collision must cancel out.
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Physics/13
|
Physics
| null |
In the laboratory, two carts on a track collide in the arrangement shown in the preceding figure. Before the collision, the 2-kg cart travels through photogate 1, which measures its speed; the 0.25-kg cart is initially at rest. After the collision, the carts bounce off one another. Photogate 2 measures the speed of each cart as it passes.
A student is concerned about his experimental results. When he adds the momentum of both carts after collision, he gets a value greater than the momentum of the 2-kg cart before collision. Which of the following is a reasonable explanation for the discrepancy?
A. The track might have been slanted such that the carts were moving downhill.
B. Human error might have been involved in reading the photogates.
C. Friction might not have been negligible.
D. The collision might not have been elastic.
|
A: Choice B is ridiculous-scientists should never refer generically to "human error." Significant friction should reduce, not increase, the speed (and thus the momentum) measured by Photogate 2. The elasticity of a collision refers to kinetic energy conservation, not momentum conservation-even inelastic collisions must conserve momentum. If the track is slanted downhill to the right, then the carts speed up; conservation of momentum won't be valid between Photogates 1 and 2 because the downhill component of the gravitational force is a force external to the two-cart system.
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Physics/14
|
Physics
| null |
Three wagons each have the same total mass (including that of the wheels) and four wheels, but the wheels are differently styled. The structure, mass, and radius of each wagon’s wheels are shown in the preceding chart. In order to accelerate each wagon from rest to a speed of 10 m/s, which wagon requires the greatest energy input?
A. Wagon A
B. Wagon B
C. Wagon C
D. All require the same energy input
|
B: The energy input must be enough to change the translational kinetic energy of the cart and to change the rotational kinetic energy of the wheels. Since all carts have the same mass and change speed by the same amount, they all require the same energy input to change the translational KE. Whichever wheels have the largest rotational inertia will require the largest energy input to get to the same speed. Calculating, wagon B has the largest rotational inertia of 0.004 $kg·m^2$.
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Physics/15
|
Physics
| null |
A 10-kg wagon moves horizontally at an initial speed of 5 m/s. A 30-N force is applied to the wagon by pulling the rigid handle, which is angled 60° above the horizontal. The wagon continues to move horizontally for another 20 m. A negligible amount of work is converted into thermal energy. By how much has the wagon’s kinetic energy increased over the 20 m?
A. 300 J
B. 600 J
C. 125 J
D. 63 J
|
A: The work-energy theorem says that the work done by a nonconservative force is equal to the change in potential energy plus the change in kinetic energy. Since the wagon is on a horizontal surface, the potential energy change is zero; the work done by the 30-N pulling force is the change in the wagon's KE. Work is force times parallel displacement, so we don't use 30 N in this formula, we use the component of the 30-N force parallel to the 20-m displacement. That's (30 N)(cos 60)(20 m) = 300 N.
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Physics/16
|
Physics
| null |
Four identical resistors are connected to a battery in the configuration shown in the preceding figure.
Which of the following ranks the current I through each resistor?
A. $I_1 = I_4 > I_2 > I_3$
B. $I_1 = I_4 > I_2 = I_3$
C. $I_1 = I_2 = I_3 = I_4$
D. $I_1 > I_2 = I_3 > I_4$
|
B: Current can only travel through a wire. At the junction after Resistor 1, the current splits; the current comes back together after resistors 2 and 3. Thus, resistors 1 and 4 have the same current that's equal to the total coming from the battery. Since resistors 2 and 3 are identical, they split the current evenly.
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Physics/17
|
Physics
| null |
A force probe is used to stretch a spring by 20 cm. The graph of the force as a function of distance shown in the preceding figure is produced and used to determine the amount of work done in stretching the spring 20 cm. The experimenter reports the result as 3,000 N·cm. Which of the following is a reasonable estimate of the experimental uncertainty on this measurement?
A. 3,000 ± 3 N.cm
B. 3,000 ± 30 N.cm
C. 3,000 ± 300 N.cm
D. 3,000 ± 3,000 N.cm
|
C: The work done by the spring is the area under a force-distance graph because work = force times distance. Using the best-fit line drawn as the top of a triangle, the area is (1/2) × (300 N)(20 cm) = 3,000 N.cm.* Now, put your ruler along the data points. Try to draw another line that's still a reasonable best fit, but is a bit shallower. Where does that line intersect the 20-cm position? It intersects at a point probably not much below 280 N, maybe even 290 N. The smallest possible work done, given this data, would be area = (1/2)(280 N)(20 cm) = 2,800 N.cm, which is 200 N.cm short of the 3,000 N.cm original estimate. That's closest to Choice C. If you've done a lot of in-class lab work, you might have noticed that your data often look about as scattered as shown in the graph; and that anything you calculate is never much closer to a known value or to your classmates' calculations than 5 or 10 percent. Here, Choice B works out to an uncertainty of 1 percent; Choice D is 100 percent. So C is the reasonable choice.
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Physics/18
|
Physics
| null |
In the laboratory, a long platform of negligible mass is free to rotate on a fulcrum. A force probe is placed a fixed distance x from the fulcrum, supporting the platform. An object of fixed mass is placed a variable distance d from the fulcrum. For each position d, the force probe is read. It is desired to determine the mass of the object from a graph of data. Which of the following can determine the object’s mass?
A. Plot the reading in the force probe times x on the vertical axis; plot the gravitational field times d on the horizontal axis. The mass is the slope of the line.
B. Plot the reading in the force probe on the vertical axis; plot the distance d on the horizontal axis. The mass is the area under the graph.
C. Plot the reading in the force probe on the vertical axis; plot the distance d multiplied by the distance x on the horizontal axis. The mass is the y-intercept of the graph.
D. Plot the reading in the force probe times d on the vertical axis; plot the distance x on the horizontal axis. The mass is the slope of the line divided by the gravitational field.
|
A: Since the platform itself is of negligible mass, only two torques act on the platform: counterclockwise by the force of the force probe ($F_P$), and clockwise by the downward force of the object (mg). Torque is force times distance from a fulcrum, so set these torques equal: $F_P(x) = mg(d)$. Solving for the mass, we get $\frac{F_P x}{gd}$. Plot the numerator on the vertical, the denominator on the horizontal, and the slope is the mass m.
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Physics/19
|
Physics
| null |
A rigid rod is pivoted at its right end. Three forces of identical magnitude but different directions are applied at the positions 1, 2, and 3 as shown. Which of the following correctly ranks the torques $τ_1$, $τ_2$, and $τ_3$ provided by the forces $F_1$, $F_2$, and $F_3$?
A. $τ_1 > τ_2 > τ_3$
B. $τ_3 > τ_2 > τ_1$
C. $τ_2 > τ_1 > τ_3$
D. $τ_2 > τ_1 = τ_3$
|
A: Torque is force times distance from a fulcrum; but that force must be perpendicular to the rod, so in this case the force used in the equation will be the vertical component, which includes a sin 45° term for $F_1$ and $F_3$. The sine of 45° is 0.7; call the length of the rod L, so $F_2$ is a distance L/2 from the pivot, and $F3$ is about L/4 from the pivot. So $τ_1 = 0.7FL$. $τ_2 = 0.5FL$. $τ_3 = (0.7·0.25)FL$.
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Physics/20
|
Physics
| null |
The preceding diagram represents a photograph of three transverse waves, each of which is moving to the right in the same material as the others. Which of the following ranks the waves by their amplitudes?
A. A = B > C
B. B > C > A
C. A > C > B
D. A = B = C
|
A: The amplitude is measured from the midpoint to the peak or trough of a wave. Waves A and B are each two "bars" above the midpoint, while C is only one bar above the midpoint.
|
Physics/21
|
Physics
| null |
Question below refers to the following information:
Block B is at rest on a smooth tabletop. It is attached to a long spring, which in turn is anchored to the wall. Identical block A slides toward and collides with block B. Consider two collisions, each of which occupies a duration of about 0.10 s:
Collision I: Block A bounces back off of block B.
Collision II: Block A sticks to block B.
In which collision, if either, is the period and frequency of the ensuing oscillations after the collision larger?
A. Period and frequency are the same in both.
B. Period is greater in collision II, and frequency is greater in collision I.
C. Period and frequency are both greater in collision I.
D. Period and frequency are both greater in collision II.
|
B: The period of a mass-on-a-spring oscillator is $T=2\pi\sqrt{\frac{m}{k}}$. The important part here is that the mass term is in the numerator-a larger mass means a larger period. More mass oscillates on the spring in collision II, so collision II has a greater period. Frequency is the inverse of the period, so the period is smaller in collision II.
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Physics/22
|
Physics
| null |
Three equal-mass objects (A, B, and C) are each initially at rest horizontally on a pivot, as shown in the figure. Object A is a 40 cm long, uniform rod, pivoted 10 cm from its left edge. Object B consists of two heavy blocks connected by a very light rod. It is also 40 cm long and pivoted 10 cm from its left edge. Object C consists of two heavy blocks connected by a very light rod that is 50 cm long and pivoted 20 cm from its left edge. Which of the following correctly ranks the objects’ angular acceleration about the pivot point when they are released?
A. A = B > C
B. A > B = C
C. A < B < C
D. A > B > C
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D: Angular acceleration is net torque divided by rotational inertia, $\alpha=\frac{\tau_{net}}{I}$. Imagine each object has total mass of 2 kg. Begin by comparing objects A and B: To find the net torque on object A, assume the entire 20 N weight is concentrated at the dot representing the rod's center of mass. That's located 10 cm from the pivot, giving a net torque of 200 N·cm. For object B, consider the torques provided by each block separately. The right block provides a torque of (10 N)(30 cm) = 300 N·cm clockwise; the left block provides a torque of (10 N)(10 cm) = 100 N·cm counterclockwise. That makes the net torque 200 N·cm, the same as for object A. But object B has more rotational inertia, since its 2 kg of mass are concentrated farther away from the pivot than object A's mass. So the denominator of the angular acceleration equation is bigger for object B with the same numerator, which means A > B.
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Physics/23
|
Physics
| null |
An object of mass m is attached to an object of mass 3m by a rigid bar of negligible mass and length L. Initially, the smaller object is at rest directly above the larger object, as shown in the figure. How much work is necessary to flip the object 180°, such that the larger mass is at rest directly above the smaller mass?
A. 2πmgL
B. 4mgL
C. 4πmgL
D. 2mgL
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D: Consider the system consisting of the objects and Earth, with the location of the 3m mass being the zero of gravitational energy. The initial gravitational energy of the system is mgL. After the rotation, the final gravitational energy of the system is 3mgL. That extra gravitational energy of 2mgL came from the work done on the system, meaning choice D. If you want instead to think of work on the objects as force times distance, remember that the force of Earth on the objects acts straight down, not along a circle. So the distance term to use here is just L, not πL.
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Physics/24
|
Physics
| null |
A person pulls on a string, causing a block to move to the left at a constant speed. The free body diagram shows the four forces acting on the block: the tension ($T$) in the string, the normal force ($F_n$), the weight ($W$), and the friction force ($F_f$). The coefficient of friction between the block and the table is 0.30.
Which is the Newton’s third law force pair to $T$?
A. The force of the block on the string
B. The force of the block on the table
C. The force of the table on the block
D. The force of friction on the block
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A: The tension is the force of the string acting on the block. Newton's third law says that since the string pulls on the block, the block pulls equally on the string. (The answer is NOT for force of friction: Newton's third law force pairs can never act on the same object.)
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Physics/25
|
Physics
| null |
A person pulls on a string, causing a block to move to the left at a constant speed. The free body diagram shows the four forces acting on the block: the tension ($T$) in the string, the normal force ($F_n$), the weight ($W$), and the friction force ($F_f$). The coefficient of friction between the block and the table is 0.30.
Which of the following correctly ranks the four forces shown?
A. $T > F_f > W = F_n$
B. $W = F_n > T = F_f$
C. $W = F_n > T > F_f$
D. $W = F_n = T = F_f$
|
B: Since the block has no acceleration, left forces equal right forces, and up forces equal down forces. The equation for a force of friction is $F_f = μF_n$. Since the coefficient of friction μ is less than 1, the friction force must be less than the normal force.
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Physics/26
|
Physics
| null |
Question below refers to the circuit shown in the figure, which includes a 9 V battery and three resistors.
Which of the following ranks the resistors by the charge that flows through each in a given time interval?
A. 300 Ω > 100 Ω = 50 Ω
B. 50 Ω > 100 Ω > 300 Ω
C. 300 Ω > 100 Ω > 50 Ω
D. 50 Ω = 100 Ω > 300 Ω
|
D: "Charge that flows through each in a given time interval" is a complicated way of saying "current." Current through series resistors must always be the same through each, so the 50 Ω resistor and the 100 Ω resistor should rank equally. Then simplify the circuit to two parallel branches. The left branch has an equivalent resistance of 150 Ω and the right branch of 300 Ω. With the same voltage across each branch, the larger current goes through the path with smaller resistance by Ohm's law.
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Physics/27
|
Physics
| null |
Question below refers to the circuit shown in the figure, which includes a 9 V battery and three resistors.
What is the voltage across the 50 Ω resistor?
A. 9.0 V
B. 6.0 V
C. 3.0 V
D. 1.0 V
|
C: Consider just the two series resistors, which have an equivalent resistance of 150 Ω and are connected to the 9 V battery. The current through that branch of the circuit is $\frac{9V}{150\Omega}=0.06$. Now consider just the 50 Ω resistor. The voltage across it is (0.06 A)(50 Ω) = 3 V. As a sanity check, we know that for two resistors in series (which take the same current), V = IR says that the smaller resistor takes less voltage; 3 V across the 50 Ω resistor leaves 6 V across the 100 Ω resistor.
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Physics/28
|
Physics
| null |
A cart attached to a spring initially moves in the x direction at a speed of 0.40 m/s. The spring is neither stretched nor compressed at the cart’s initial position (x = 0.5 m). The figure shows a graph of the magnitude of the net force experienced by the cart as a function of x, with two areas under the graph labeled. Is it possible to analyze the graph to determine the change in the cart’s kinetic energy as it moves from its initial position to x = 1.5 m?
A. No, the cart’s mass must be known.
B. Yes, subtract area 2 from area 1.
C. Yes, add area 2 to area 1.
D. Yes, determine area 1.
|
D: The work done by the net force is the area under the graph. Since the cart only moved from position x = 0.5 m to x = 1.5 m, area 1 is the work done by the net force. By the work-energy theorem, work done by the net force is the change in an object's kinetic energy. (Yes, the mass must be known to determine the values of the initial and final kinetic energy; however, the question asks only for the change in kinetic energy.)
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Physics/29
|
Physics
| null |
Question below refers to the following information: A bicycle wheel of known rotational inertia is mounted so that it rotates clockwise around a vertical axis, as shown in the first figure. Attached to the wheel’s edge is a rocket engine, which applies a clockwise torque τ on the wheel for a duration of 0.10 s as it burns. A plot of the angular position θ of the wheel as a function of time t is shown in the graph.
In addition to the wheel’s rotational inertia and the duration of time the engine burns, which of the following information from the graph would allow determination of the net torque the rocket exerts on the wheel?
A. The area under the graph between t = 0 s and t = 3 s
B. The change in the graph’s slope before and after t = 2 s
C. The vertical axis reading of the graph at t = 3 s
D. The vertical axis reading of the graph at t = 2 s
|
B: Newton's second law for rotation says $\tau_{net} = I\alpha$, where α is the angular acceleration, or change in the wheel's angular velocity per time. To find the wheel's angular velocity, look at the slope of the angular position versus time graph. The slope changes after the torque is applied; so the change in the slope is the change in the angular velocity, which (when divided by the 0.10 s duration of the rocket firing) gives the angular acceleration.
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Physics/30
|
Physics
| null |
Question below refer to the following information:
Two metal balls of equal mass (100 g) are separated by a distance (d). In state 1, shown in the figure, the left ball has a charge of +20 µC, while the right ball has a charge of –20 µC. In state 2, both balls have an identical charge of –40 µC.
Which of the following statements about the magnitudes of the electrostatic and gravitational forces between the two balls is correct?
A. The electrostatic force is greater in state 2 than in state 1. The electrostatic force is greater than the gravitational force in state 1, but the gravitational force is greater than the electrostatic force in state 2.
B. The electrostatic force is greater in state 1 than in state 2. The electrostatic force is greater than the gravitational force in state 1, but the gravitational force is greater than the electrostatic force in state 2.
C. The electrostatic force is greater in state 2 than in state 1. The electrostatic force is greater than the gravitational force in both states.
D. The electrostatic force is greater in state 1 than in state 2. The electrostatic force is greater than the gravitational force in both states.
|
C: The magnitude of the electrostatic force depends on the product of the charges, regardless of sign. In state 2, the charges are both bigger than in state 1, giving a bigger electrostatic force in state 2. Without even doing the calculation, you can recognize that since Newton's gravitational constant G is orders of magnitude less than the coulomb's law constant k, the electrostatic force will be much greater than the gravitational force between two objects in virtually any laboratory situation. The only time these forces become comparable is when objects the size of planets exert gravitational forces.
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Physics/31
|
Physics
| null |
Two metal balls of equal mass (100 g) are separated by a distance (d). In state 1, shown in the figure, the left ball has a charge of +20 µC, while the right ball has a charge of –20 µC. In state 2, both balls have an identical charge of –40 µC.
An experimenter claims that he took the balls from state 1 to state 2 without causing them to contact anything other than each other. Which of the following statements provides correct evidence for the reasonability of this claim?
A. The claim is not reasonable, because there was more net charge in the two-ball system in state 2 than in state 1.
B. The claim is not reasonable, because there was identical charge on each ball in state 2.
C. The claim is reasonable, because in both states, each ball carried the same magnitude of charge as the other.
D. The claim is reasonable, because the positive charge in state 1 could have been canceled out by the available negative charge.
|
A: Charge is conserved, meaning that while negative charge can neutralize positive charge, the net charge of a system cannot change. The net charge of state 1 is zero; thus, the net charge of state 2 must also be zero, regardless of whether the balls touched or not. State 2 has a net charge of –80 µC, not zero; thus, the claim is not reasonable due to this violation of charge conservation.
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Physics/32
|
Physics
| null |
When a 0.20 kg block hangs at rest vertically from a spring of force constant 4 N/m, the spring stretches 0.50 m from its unstretched position, as shown in the figure. Subsequently, the block is stretched an additional 0.10 m and released such that it undergoes simple harmonic motion. What is the maximum kinetic energy of the block in its harmonic motion?
A. 0.50 J
B. 0.02 J
C. 0.72 J
D. 0.20 J
|
B: You can look at this two ways. The hard way is to consider the spring energy gained and the gravitational energy lost in stretching the spring the additional 0.10 m separately. The block-earth system loses $mgh = (0.20 kg)(10 N/kg)(0.10 m) = 0.20 J$ of gravitational energy; but the block-spring system gains $0.5kx_2^2 – 0.5kx_1^2 = {0.5(4 N/m)(0.60 m)^2 – 0.5(4 N/m)(0.50 m)^2} = 0.22 J$ of spring energy. Thus, the net work done on the block in pulling it the additional 0.10 m is 0.02 J. That's what is converted into the block's maximum kinetic energy. You can also look at it the easy way. With a vertical spring, consider the block-earth-spring system as a whole. Define the hanging equilibrium as the zero of the whole system's potential energy; then the potential energy of the whole system can be written as $0.5kx^2$, where x is the distance from this hanging equilibrium position. That's $0.5(4 N/m)(0.10 m)^2 = 0.02 J$.
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Physics/33
|
Physics
| null |
Question below refers to the following information:
A student pushes cart A toward a stationary cart B, causing a collision. The velocity of cart A as a function of time is measured by a sonic motion detector, with the resulting graph shown in the figure.
At which labeled time did the collision begin to occur?
A. A
B. B
C. C
D. D
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C: The graph represents the speed of cart A, the one that's initially moving. So right before the collision, the vertical axis of the graph must be nonzero. Right after the collision, the vertical axis must quickly either decrease or perhaps become negative if the cart changed directions. That's what happens in the tenth of a second or so after the time labeled C.
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Physics/34
|
Physics
| null |
Question below refers to the following information:
A student pushes cart A toward a stationary cart B, causing a collision. The velocity of cart A as a function of time is measured by a sonic motion detector, with the resulting graph shown in the figure.
What additional measurements, in combination with the information provided in the graph, could be used to verify that momentum was conserved in this collision?
A. The mass of each cart and cart B’s speed after the collision
B. The force of cart A on cart B, and cart B’s speed after the collision
C. The mass of each cart only
D. The force of cart A on cart B only
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A: Conservation of momentum requires that the total momentum of the two-cart system be the same before and after the collision. You already know cart A's speed and direction of motion before and after the collision by looking at the vertical axis of the graph; so the mass of cart A will give us cart A's momentum before and after the collision. You know cart B has no momentum before the collision. But you need both cart B's mass AND its velocity after the collision to finish the momentum conservation calculation.
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Physics/35
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Physics
| null |
The circuits shown in the figure contain the same three resistors connected in different configurations. Which of the following correctly explains which configuration (if either) produces the larger current coming from the battery?
A. Circuit 1; the larger resistor is closer to the battery.
B. Circuit 2; the equivalent resistance of the three resistors is smaller than in circuit 1.
C. Neither; the batteries are both the same.
D. Neither; the individual resistors are the same in each circuit, even if they are in a different order.
|
B: The equivalent resistance of the parallel resistors in circuit 1 is 12 kΩ; adding that to the 100 kΩ resistor gives a total resistance in circuit 1 of 112 kΩ. Circuit 2's parallel combination has an equivalent resistance of 23 kΩ, giving a total resistance in that circuit of 43 kΩ. By V = IR applied to both circuits in their entirety with the same total voltage, the circuit with smaller total resistance will produce the larger current. That's circuit 2.
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Physics/36
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Physics
| null |
In an experiment, a cart is placed on a flat, negligible-friction track. A light string passes over a nearly ideal pulley. An object with a weight of 2.0 N hangs from the string. The system is released, and the sonic motion detector reads the cart’s acceleration. Can this setup be used to determine the cart’s inertial mass?
A. Yes, by dividing 2.0 N by the acceleration, and then subtracting 0.2 kg.
B. No, because only the cart’s gravitational mass could be determined.
C. Yes, by dividing 2.0 N by the acceleration.
D. No, because the string will have different tensions on either side of the pulley.
|
A: The net force on the cart-object system is 2.0 N. Dividing 2.0 N by the acceleration gives the mass of the entire cart-object system, not just the mass of the cart; so subtract the 0.2 kg mass of the object from the whole system mass to get the cart mass. This is actually inertial mass, because it uses the equation $F_{net} = ma$; Newton's second law defines inertial mass as resistance to acceleration.
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Physics/37
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Physics
| null |
The diagram shown here represents the particles in a longitudinal standing wave. Which of the following is an approximate measure of the standing wave’s maximum amplitude?
A. Half the distance from 1 to 4
B. The distance from 2 to 3
C. The distance from 1 to 4
D. Half the distance from 2 to 3
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D: Positions 1 and 4 show minimum particle displacement, so these are the nodes. The maximum amplitude occurs at the antinode, which is somewhere near the positions labeled 2 and 3. The particles are moving left and right in this longitudinal wave. The amplitude is twice the peak-to-peak displacement; since the particle displacement near the antinode is about the distance from 2 to 3, the amplitude is half that.
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Physics/38
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Physics
| null |
Question below refers to the following information:
A small ball of mass m moving to the right at speed $v$ collides with a stationary rod, as shown in the figure. After the collision, the ball rebounds to the left with speed $v_1$, while the rod’s center of mass moves to the left at speed $v_2$. The rod also rotates counterclockwise.
Which of the following equations determines the rod’s change in angular momentum about its center of mass during the collision?
A. Iω where I is the rod’s rotational inertia about its center of mass, and ω is its angular speed after collision.
B. $Iv_2/r$, where I is the rod’s rotational inertia about its center of mass, and r is half the length of the rod.
C. $mv_1d$, where d is the distance between the line of the ball’s motion and the rod’s center of mass.
D. $mv_1r$, where r is half the length of the rod.
|
A: The rod starts from rest, so its final angular momentum is the same as its change in angular momentum. Although the equation ω = v/r is valid for a point object moving in a circle, it does not apply to a rotating rod; thus, choice B is wrong. Choice C gives the final angular momentum of the ball, which is not the same as the angular momentum change of the rod because the ball does NOT start from rest.
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Physics/39
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Physics
| null |
Question below refers to the following information:
A small ball of mass m moving to the right at speed $v$ collides with a stationary rod, as shown in the figure. After the collision, the ball rebounds to the left with speed $v_1$, while the rod’s center of mass moves to the left at speed $v_2$. The rod also rotates counterclockwise.
Is angular momentum about the rod’s center of mass conserved in this collision?
A. No, the ball always moves in a straight line and thus does not have angular momentum.
B. No, nothing is spinning clockwise after the collision to cancel the rod’s spin.
C. Yes, the only torques acting are the ball on the rod and the rod on the ball.
D. Yes, the rebounding ball means the collision was elastic.
|
C: Choice C states the fundamental condition for angular momentum conservation, which is correct here. The ball does have angular momentum about the rod's center of mass before and after the collision, because its line of motion does not go through the rod's center of mass. Whether or not the collision is elastic has to do with conservation of mechanical energy, not angular or linear momentum.
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Physics/40
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Physics
| null |
Question below refers to the following information:
A model rocket with a mass of 100 g is launched straight up. Eight seconds after launch, when it is moving upward at 110 m/s, the force of the engine drops as shown in the force-time graph.
Which of the following is the best estimate of the impulse applied by the engine to the rocket after the t = 8 s mark?
A. 100 N·s
B. 20 N·s
C. 5 N·s
D. 50 N·s
|
C: Impulse is the area under a force-time graph. From t = 8 s to t = 10 s, the area is an approximate rectangle of 2 N times 2 s, giving 4 N·s. Add an approximate triangle from t = 10 s to t = 11 s, which has the area ½(2 N)(1 s) = 1 N·s. That gives a total of 5 N·s.
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Physics/41
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Physics
| null |
Question below refers to the following information:
A model rocket with a mass of 100 g is launched straight up. Eight seconds after launch, when it is moving upward at 110 m/s, the force of the engine drops as shown in the force-time graph.
Which of the following describes the motion of the rocket between t = 8 s and t = 10 s?
A. The rocket moves upward and slows down.
B. The rocket moves downward and speeds up.
C. The rocket moves upward at a constant speed.
D. The rocket moves upward and speeds up.
|
D: The force of the engine on the rocket during this time is 2 N upward. The weight of the rocket is 1 N (that is, 0.1 kg times the gravitational field of 10 N/kg). So the net force is still upward during this time. Since the rocket was already moving upward, it will continue to move upward and speed up.
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Physics/42
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Physics
| null |
The velocity-time graph shown here represents the motion of a 500 g cart that initially moved to the right along a track. It collided with a wall at approximately time (t) = 1.0 s. Which of the following is the best estimate of the impulse experienced by the cart in this collision?
A. 3.6 N·s
B. 0.5 N·s
C. 0.2 N·s
D. 1.8 N·s
|
D: Impulse is change in momentum. The initial momentum was something like (0.5 kg)(1.6 m/s) = 0.8 N·s to the right. The cart came to rest, changing its momentum by 0.8 N·s, then sped back up, again changing momentum by (0.5 kg)(2 m/s) = 1.0 N·s. Thus, the total momentum change is about 1.8 N·s.
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Physics/43
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Physics
| null |
If a ball is rolling down an inclined plane without slipping, which force is responsible for causing its rotation?
A. Normal force
B. Gravity
C. Kinetic friction
D. Static friction
|
D: As a ball rolls down an inclined plane, the three forces act on the ball, as shown in the diagram below. https://img.crackap.com/ap/physics-1/cr20/Prin_9780525568292_xp_all_r1_Page_036_Image_0001.jpg In order for a force to cause an object to rotate, the force must be located off-center, and must not point directly at or away from the object's center of mass. An object's weight always acts at its center of mass, so (B) is wrong. The normal force does act at the edge of the ball, but it points directly out of the ramp, which is straight at the center of the ball. Choice (A) is wrong. Friction acts where the ball meets the inclined plane, and it points parallel to the surface, which is tangent to the ball's surface. So friction is what causes rotation. What kind of friction do we have? The phrase "without slipping" tells us that even though the ball as a whole is moving, the spot on the ball that touches the ramp is at rest with respect to the ramp. That means we have static friction, (D).
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Physics/44
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Physics
| null |
The graph above shows the velocity of an object as a function of time. What is the net displacement of the object over the time shown?
A. -23 m
B. -9 m
C. 9 m
D. 23 m
|
B: In a velocity-versus-time (also referred to as a v-versus-t) graph, the displacement of an object is the area between the curve and the horizontal axis; area below the t-axis counts as negative. In this case, the area can be broken into multiple rectangles and triangles.
https://img.crackap.com/ap/physics-1/cr20/Prin_9780525568292_xp_all_r1_Page_036_Image_0002.jpg
In order of the number labels, each shape has an area of 6, 1, 4, and 12, respectively. The last two, however, are below the t-axis, so their area represents negative displacement. Therefore, the net displacement is 6 m + 1 m - 4 m - 12 m = -9 m. As a check, it looks like most of the area is below the t-axis, so the answer should be negative.
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Physics/45
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Physics
| null |
An object is resting on a platform that rotates at a constant speed. At first, it is a distance of half the platform's radius from the center. If the object is moved to the edge of the platform, what happens to the centripetal force that it experiences? Assume the platform continues rotating at the same speed.
A. Increases by a factor of 4
B. Increases by a factor of 2
C. Decreases by a factor of 2
D. Decreases by a factor of 4
|
B: The formula for centripetal force is $F_c = mv^2/r$, which initially seems to indicate the force is inversely proportional to the radius. However, in the case of circular motion, an object's linear speed is v = Ωr. Substituting this value into the equation gives $F_c = m(Ωr)^2/r = mΩ^2r$. So it turns out that the force is directly proportional to r, which means doubling the radius will double the force on the object.
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Physics/46
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Physics
| null |
A car of mass 1000 kg is traveling at a speed of 5 m/s. The driver applies the breaks, generating a constant friction force, and skids for a distance of 20 m before coming to a complete stop. Given this information, what is the coefficient of friction between the car's tires and the ground?
A. 0.25
B. 0.2
C. 0.125
D. 0.0625
|
D: First, you know $F_f = \mu F_N = \mu (mg)$. Second, you know that $W = ΔKΕ = KE_f - KE_0 = -KE_0$ (since $KE_f = 0$ in this case) = $-0.5 mv_0^2$. Plugging in numbers, you get $W = -\frac{1}{2}(1000 kg)(5 m/s)^2 = -12,500 J$. You also know that W = Fdcosθ = [µ(mg)]dcosθ. Solving for μ, you get μ = W/[(mg)dcosθ]. Plugging in the numbers then gives $μ = -12,500 J/[(1000 kg)(10 m/s^2)(20 m)(cos180°)] = 0.0625$.
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Physics/47
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Physics
| null |
A spring-block system is oscillating without friction on a horizontal surface. If a second block of equal mass were placed on top of the original block at a time when the spring is at maximum compression, which of the following quantities would NOT be affected? Assume that the top block stays on the bottom block.
A. Frequency
B. Maximum speed
C. Amplitude
D. All of the above quantities would be affected.
|
C: The frequency of a spring-block system is $f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$ , so it would be affected by the change in mass. Furthermore, when a spring is at maximum compression or extension, all of its energy is potential energy, which is given by $U = \frac{1}{2}kx^2$. In adding this block, none of the relevant values are changed, so the spring will still extend to the same length, which means the amplitude is unchanged. Finally, maximum speed will be limited by the maximum K of the system (which will be unchanged since the maximum U was unchanged). $K = \frac{1}{2}mv^2$, so the increased mass would have to be balanced by a decrease in speed to leave the K unaltered.
If you want a physical, rather than mathematical, explanation, think about the acceleration of the pair of blocks. The spring exerts the same force as before, but we effectively raised the block's mass. Thus, the acceleration is less than before. Since the distance from the starting point (maximum compression) to equilibrium is unchanged, the blocks will pick up less speed than the single block did before. Thus, maximum speed decreases.
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Physics/48
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Physics
| null |
A certain theme park ride involves people standing against the walls of a cylindrical room that rotates at a rapid pace, making them stick to the walls without needing support from the ground. Once the ride achieves its maximum speed, the floor drops out from under the riders, but the circular motion holds them in place. Which of the following factors could make this ride dangerous for some riders but not others?
A. The mass of the individuals
B. The coefficient of friction of their clothing in contact with the walls
C. Both of the above
D. None of the above
|
B: The diagram below shows the forces involved.
https://img.crackap.com/ap/physics-1/cr20/Prin_9780525568292_xp_all_r1_Page_037_Image_0002.jpg
In this case, the person will remain suspended in the air as long as $F_f = F_g$. Furthermore, because this is an example of uniform circular motion, you know $F_N = F_C = mv^2/r$. So you can rewrite the first equation as $\mu (mv^2/r) = mg. Thus, the coefficient of friction is an important factor, but the mass of the person is not since it exists on both sides of the equation and will cancel out.
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Physics/49
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Physics
| null |
As a pendulum swings back and forth, it is affected by two forces: gravity and tension in the string. Splitting gravity into component vectors, as shown above, produces mg sinθ (the restoring force) and mg cosθ. Which of the following correctly describes the relationship between the magnitudes of tension in the string and mg cosθ?
A. Tension > mg cosθ
B. Tension = mg cosθ
C. Tension < mg cosθ
D. The relationship depends on the position of the ball.
|
D: Whenever the ball is in motion, it will be experiencing circular motion, which means there must be a centripetal force. Centripetal force is a net force toward the center, so that means tension must be the greater force whenever the ball is moving. However, the ball is not always moving. At the two extreme edges of its motion, the ball is motionless for an instant as it changes directions. At those times, the net centripetal force is zero, meaning the two given forces are equal.
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Physics/50
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Physics
| null |
In order to ensure that no current will pass through the cross path (bold in the image above), what must the resistance of $R$ be in terms of $R_1$, $R_2$, and $R_3$ ?
A. $R=\frac{R_1+R_3}{R_2}$
B. $R=\frac{R_2}{R_1+R_3}$
C. $R=\frac{R_1R_3}{R_2}$
D. $R=\frac{R_2}{R_1R_3}$
|
C: In a circuit with multiple paths, the current through each path is inversely proportional to the resistance of that path. If no current passes through the indicated segment of wire, then the ratio (current in top path)/(current in bottom path) is the same before the cross bar as it is after the cross bar. Thus, $\frac{R_1}{R_2}=\frac{R}{R_3}$, which means $R =\frac{R_1R_3}{R_2}$.
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Physics/51
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Physics
| null |
If the surface of the incline is frictionless, how long will the block take to reach the bottom if it was released from rest at the top?
A. 0.5 s
B. 1.0 s
C. 1.4 s
D. 2.0 s
|
D: The acceleration of the block is $a = g \sin \theta$, where $\theta$ is the incline angle, and the distance it must travel—the length of the incline—is $h/\sin \theta = h/\sin 30^\circ = 2h$.
Using the equation $s = \frac{1}{2} a t^2$, we find that
\[ t = \sqrt{\frac{2s}{a}} = \sqrt{\frac{2(2h)}{g \sin \theta}} = \sqrt{\frac{4(5 m)}{(10 m/s^2) \sin 30^\circ}} = 2 s \]
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Physics/52
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Physics
| null |
If the surface of the incline is frictionless, with what speed will the block reach the bottom if it was released from rest at the top?
A. 8 m/s
B. 10 m/s
C. 14 m/s
D. 18 m/s
|
B: The easiest way to answer this question is to use Conservation of Energy:
$mgh = \frac{1}{2}mv^2$ implies $v = \sqrt{2gh} = \sqrt{2(10 m/s^2)(5 m)} = 10 m/s$
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Physics/53
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Physics
| null |
If the coefficient of friction between the block and the incline is 0.4, how much work is done by the normal force on the block as it slides down the full length of the incline?
A. 0 J
B. 2.0 J
C. 4.0 J
D. 4.9 J
|
A: Since the normal force is perpendicular to the object's displacement, it does no work.
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Physics/54
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Physics
| null |
The graph above shows a wave. What quantity of the wave does the indicated segment represent?
A. Frequency
B. Wavelength
C. Period
D. Cannot be determined
|
C: Based only on the curve itself, the segment could be either wavelength or period, but the x-axis is labeled as time, which means the segment must be the period.
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Physics/55
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Physics
| null |
In the figure above, the coefficient of sliding friction between the small block and the tabletop is 0.2. If the pulley is frictionless and massless, what will be the acceleration of the blocks once they are released from rest?
A. 0.5g
B. 0.6g
C. 0.7g
D. 0.8g
|
C: The net force on the two-block system is $3mg - F_f$, where $F_f$ is the magnitude of the frictional force acting on the small block. Since $F_f = \mu mg$, we find that
\[ a = \frac{F_{\text{net}}}{\text{total mass}} = \frac{3mg - 0.2mg}{m + 3m} = \frac{2.8mg}{4m} = 0.7g \]
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Physics/56
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Physics
| null |
The graphs above show the changes in speed for two different blocks. The left graph shows the speed over time of Block 1, which has a mass of m1, and the right graph shows the speed of Block 2, which has a mass of 2m1. While speeding up, they both had to overcome some frictional force, f. Which of the two blocks had more power provided to it?
A. Block 1
B. Block 2
C. They had equal power provided.
D. It depends on the ratio of m/f.
|
A: Both objects started at rest, so finding their final kinetic energies will give the changes in kinetic energy for each block.
\[ K_1 = \frac{1}{2} m_1 v_1^2 \]
\[ K_1 = \frac{1}{2} m_1 (10 \text{ m/s})^2 \]
\[ K_1 = 50 m_1 \text{ J} \]
\[ K_2 = \frac{1}{2} m_2 v_2^2 \]
\[ K_2 = \frac{1}{2} (2 m_1)(5 \text{ m/s})^2 \]
\[ K_2 = 25 m_1 \text{ J} \]
The Work-Energy Theorem tells us that change in kinetic energy is equal to work, so Block 1 had more work done to it. Because both blocks were affected for the same time period, that means Block 1 also had more power provided.
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Physics/57
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Physics
| null |
The figure above shows a plane diagram. The figure shows four point charges arranged at the corners of a square with point B as its center. A positive charge would experience no net force if it were placed at which of the three points shown in the figure?
A. A only
B. B only
C. C only
D. A, B, or C
|
B: A point charge placed at point A, B, or C would experience no net force only if the electric field at the point were zero. Let's label the four charges 1, 2, 3, and 4, starting at the upper right and counting counterclockwise. The diagram below shows the identical electric field vectors at points A, B, and C due to the four charges. Only at point B do the four electric field vectors cancel each other to give a sum of zero, so only at point B would a charge feel zero net force.
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Physics/58
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Physics
| null |
If the rod in the figure above is uniform and has mass m, what is the tension in the supporting string? The rod is free to rotate about point P.
A. $ \frac{1}{2} mg \sin \theta $
B. $ mg \sin \theta $
C. $ \frac{1}{2} mg \cos \theta $
D. $ \frac{1}{2} mg $
|
D: With respect to the point at which the rod is attached to the vertical wall, the tension in the string exerts a counterclockwise (CCW) torque, and the gravitational force—which acts at the rod's center of mass—exerts a clockwise (CW) torque. If the rod is in equilibrium, these torques must balance. Letting \( L \) denote the length of the rod, this gives
\[ T_{\text{CCW}} = T_{\text{CW}} \]
\[ F_T L \sin \theta = \left( \frac{1}{2} L \right) (mg) \sin \theta \]
\[ F_T = \frac{1}{2} mg \]
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Physics/59
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Physics
| null |
In the diagram above, a positive charge and negative charge are placed at y = -2 and y = 3, respectively. If the negative charge has a greater magnitude, then the only place of 0 net electric field would be
A. Along the positive x-axis
B. Along the negative x-axis
C. Along the positive y-axis
D. Along the negative y-axis
|
D: Remember that the electric field for any given particle will point toward a negative charge and away from a positive charge. Anywhere between the two charges here (regardless of x-coordinate) would have an electric field that pointed up. The negative charge would pull it up, and the positive one would push it up. This eliminates (A) and (B). Finally, the negative charge is greater magnitude. Therefore, the fields could only be in balance somewhere closer to the positive charge. This eliminates (C)
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Physics/60
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Physics
| null |
A simple pendulum executes simple harmonic motion as it swings through small angles of oscillation. If θmax denotes the amplitude of the oscillations, which of the following statements is true?
A. When $\theta$ = 0, the tangential acceleration is 0.
B. When $\theta$ = $\theta_{max}$ , the tangential acceleration is 0.
C. When $\theta$ = 0, the speed is 0.
D. When $\theta$ = 0, the restoring force is maximized.
|
A: At the instant a simple harmonic oscillator passes through equilibrium, the restoring force is zero. Therefore, the tangential acceleration of the pendulum is also zero.
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Physics/61
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Physics
| null |
In the graph above, the instantaneous velocity at 9 seconds is closest to
A. 4/9 m/s
B. -4/9 m/s
C. 11/4 m/s
D. -11/4 m/s
|
D: The slope of the tangent line gives the instantaneous velocity. You could also use Process of Elimination. Recognize the answer must be negative, so you can eliminate (A) and (C). The point in question is (9, 4). Dividing the y-coordinate by the x-coordinate is not the same as the obtaining the slope and is there to distract you.
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Physics/62
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Physics
| null |
Two charges are placed as shown at the vertices of an equilateral triangle. What is the direction of the electric field at point P?
A. a
B. b
C. c
D. d
|
D: The two electric fields vectors are shown. The resultant can be determined as shown below. https://img.crackap.com/ap/physics-1/cr2017/00143.jpg
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Physics/63
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Physics
| null |
If a charge of +q is placed at point P, the electric field at point P would
A. increase by 1/3
B. increase by 1/2
C. decrease by 1/3
D. remain the same
|
D: Charges do not create an electric field at the location of the charge itself. Therefore, the electric field at point p is not changed by the addition of a charge at point p.
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Physics/64
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Physics
| null |
A tube with one end closed and one end open resonates for a wave with wavelength λa as shown. The next shorter wavelength at which resonance will occur is λb. The ratio of these two wavelengths λa/λb is
A. 1/4
B. 1/3
C. 3/5
D. 5/3
|
D: Resonance occurs at \(\lambda = \frac{4L}{n}\) for odd integers. This means at \(4L\), \(\frac{4L}{3}\), \(\frac{4L}{5}\), ...
We are shown a picture where a wave is in a tube of length \(L\). This means \(\lambda_a = \frac{4L}{3}\). The next shorter wavelength is \(\lambda_b = \frac{4L}{5}\). The ratio of these is \(\frac{\lambda_a}{\lambda_b}\) or \(\frac{5}{3}\).
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Physics/65
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Physics
| null |
If R1 were to burn out, the current coming out from the battery would
A. increase
B. decrease
C. stay the same
D. there is no current, because the circuit is now incomplete
|
B: If R1 were to burn out, the total resistance in the circuit would increase. Because I = V/R this increase in resistance would decrease the current.
|
Physics/66
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Physics
| null |
A Wheatstone bridge (diagram above) is a configuration of resistors and a sensitive current meter, called a galvanometer, that is used to determine the resistance of an unknown resistor. In the Wheatstone bridge shown here, find the value of Rx such that the current through galvanometer G is zero.
A. $25 \Omega$
B. $15 \Omega$
C. $10 \Omega$
D. $2.5 \Omega$
|
D: The question specifies that the current through G must be 0. Recall that in parallel circuits current will distribute itself in a manner inversely proportional to the resistance in each path. Looking at the left-hand side of the configuration, we see that the resistance on the top is half the resistance of the bottom. That means the current will distribute itself with 2/3 along the top path and 1/3 along the bottom path. For no current to pass through G, this same distribution must also be true on the right-hand side, otherwise some current would pass through G to correct any changes to the relative resistances. The 5 Ω resistor must have the same proportional resistance to the unknown resistor that the 20 Ω resistor has to the 10 Ω resistor. Therefore, the unknown must be 2.5 Ω.
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Physics/67
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Physics
| null |
In the figure above, four charges are arranged. If the magnitudes of all the charges q are all the same and the distance R between them is as shown above, what is the magnitude of the net force on the bottom right charge in terms of q, R, and k (where )?
A. $ k \left( \frac{q^2}{2r^2} \right) (1 + \sqrt{2}) $
B. $ k \left( \frac{q^2}{r^2} \right) (1 + \sqrt{2}) $
C. $ k \left( \frac{q^2}{2r'^2} \right) $
D. $ k \left( \frac{q^2}{r'^2} \right) $
|
A: Although we do not care about the direction of our net vector in the end, we do have to take orientation into account to find our magnitude. Numbering our charges from the top left and going clockwise as 1, 2, 3, and 4. Drawing out our different forces on the bottom right charge (charge 3), we get: https://img.crackap.com/ap/physics-1/cr2017/00138.jpg Now solve each of these force vectors
\[ F_{4 \text{ on } 3} = \frac{kq^2}{r^2} \]
\[ F_{2 \text{ on } 3} = \frac{kq^2}{r^2} \]
These two are the simpler force vectors to solve. In order to solve the charge 1's force on charge 3 our distance first needs to be solved. Because they are ordered in a square using a diagonal creates a 45°-45°-90° triangle. The distance in this case is \( r\sqrt{2} \). So,
\[ F_{1 \text{ on } 3} = \frac{kq^2}{(r\sqrt{2})^2} = \frac{kq^2}{2r^2} \]
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Physics/68
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Physics
| null |
Which of the following correctly ranks the change in kinetic energy for each segment from least to greatest?
A. BC, EF, DE, CD, AB
B. AB, CD, DE, EF, BC
C. BC, EF, DE, CD, AB
D. CD, DE, EF, BC, AB
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D: Mass is not changing, so we only need to consider the changes in velocity. For each segment in order, the change in velocity is +7 m/s, 0 m/s, –4 m/s, –3 m/s, and –2 m/s. The question does not specify magnitude only, so all negatives will come before the others.
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Physics/69
|
Physics
| null |
During which segment is the magnitude of average acceleration greatest?
A. AB
B. BC
C. CD
D. DE
|
C: Acceleration is the change in velocity divided by the change in time. We found the changes in velocity for each segment in the previous question, and the times for each segment are, in order, 3 s, 2 s, 1 s, 3 s, and 2 s. Divide the results in the previous question by these values, and the highest result (ignoring sign because the problem specifies magnitude only) is for segment CD.
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Physics/70
|
Physics
| null |
What is the total distance traveled by the object?
A. 32 m
B. 34 m
C. 36 m
D. 38 m
|
C: In a velocity vs. time graph, displacement is the area beneath the curve. If the question had asked for displacement, then anything beneath the x-axis would be added as a negative displacement. However, distance cannot be negative, so there is no distinction between positive and negative velocity for this question.
|
Physics/71
|
Physics
| null |
Consider the above configuration of masses attached via a massless rope and pulley over a frictionless inclined plane. What is the acceleration of the masses?
A. (m1- m2) g/(m1+ m2)
B. (m1- m2sinθ) g/(m1+ m2)
C. (m1- m2cosθ) g/(m1+ m2)
D. g
|
B: By Process of Elimination, we know we need a sine in the equation for the downward pull of the second mass down the incline of the plane, so the answer is (B).
|
Physics/72
|
Physics
| null |
In the figure above, two blocks of mass 3m and 2m are attached together. The plane is frictionless and the pulley is frictionless and massless. The inclined portion of the plane creates an angle θ with the horizontal floor. What is the acceleration of the block 2m if both blocks are released from rest (gravity = g)?
A. $ 2mg $
B. $ \frac{2}{5} g \sin \theta $
C. $ \frac{2}{3} g \sin \theta $
D. $ \frac{3}{5} g \sin \theta $
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B: Because all the masses are attached and moving as a single unit, the acceleration of any block in the system is the same as that of any other. The net force on the blocks is \( F_{\text{net}} = (2M)g \sin \theta \). Therefore,
\[ F_{\text{net}} = ma \rightarrow (5M)a = (2M)g \sin \theta = \left(\frac{2}{5}\right) g \sin \theta \]
|
Physics/73
|
Physics
| null |
The diagram above shows a top view of an object of mass m on a circular platform of mass 2m that is rotating counterclockwise. Assume the platform rotates without friction. Which of the following best describes an action by the object that will increase the angular speed of the entire system?
A. The object moves toward the center of the platform, increasing the total angular momentum of the system.
B. The object moves toward the center of the platform, decreasing the rotational inertia of the system.
C. The object moves away from the center of the platform, increasing the total angular momentum of the system.
D. The object moves away from the center of the platform, decreasing the rotational inertia of the system.
|
B: The only true statement in regards to rotational motion is (B). A mass that has greater inertia is harder to rotate. The further away the mass is from the axis of rotation, the greater the rotational inertia will be. So, when the object moves toward the center of rotation, rotational inertia is
decreased.
|
Physics/74
|
Physics
| null |
A moon has an elliptical orbit about the planet as shown above. At point A, the moon has speed vA and is at a distance rA from the planet. At point B, the moon has a speed of vB. Which of the following correctly explains the method for determining the distance of the moon from the planet at point B in the given quantities?
A. Conservation of angular momentum, because the gravitational force exerted by the moon on the planet is the same as that exerted by the planet on the moon
B. Conservation of angular momentum, because the gravitational force exerted on the moon is always directed toward the planet
C. Conservation of energy, because the gravitational force exerted on the moon is always directed toward the planet
D. Conservation of energy, because the gravitational force exerted by the moon on the planet is the same as that exerted by the planet on the moon
|
A: By Newton's Third Law, the gravitational force exerted by the planet on the moon will equal the gravitational force exerted by the moon on the planet (eliminate (B) and (C)). Between Conservation of Angular Momentum and Conservation of Energy, the correct one to use to help us find distance is Conservation of Angular Momentum. Conservation of Energy will help us find the speed of the moon. Conservation of Angular Momentum will help us find the distance of the moon from the planet.
|
Physics/75
|
Physics
| null |
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