Incorrect Physics Problems

A particle moves from position n1 to position 2 and back in 2 second. Its average velocity during the trip is:

0 m/s -- velocity = displacement/time

If the position of the spheres approximates two uniformly dense rings, which of the following is true concerning a mass placed at position D? A. The net gravitational force due to the spheres of the larger ring would be 0 B. The net gravitational force due to the spheres of the smaller ring would be 0 C. The net gravitational force due to the spheres of both rings would be 0 D. If the smaller ring were removed, the mass would move toward the center of the larger ring

A. The gravitational force everywhere inside a uniformly dense sphere or ring, due to that sphere or ring, is 0. However, the uniformly dense ring or sphere does not block the gravitational field due to other sources, so the force due to the smaller ring still affects a particle at position D.

(R)-2-Methylcyclohexanone was placed in a flask containing 0.9 equivalents of LDA at −78°C. The flask was then warmed to 0°C and benzaldehyde was added to the solution. How many stereoisomeric products can be isolated from the mixture? A) 4 B) 3 C) 2 D) 1

A. (R)-2-Methylcyclohexanone contains a single stereocenter. When the ketone is added to LDA, the less substituted carbon is deprotonated generating the kinetic enolate. This does NOT affect the stereocenter on the more substituted carbon. However, there is more ketone than base, and as a result not all of the ketone is deprotonated. Thus when the solution is warmed to 0°C, the kinetic enolate tautomerizes to the more stable thermodynamic enolate. This racemizes the stereocenter of the enolate as it adds to the benzaldehyde. Since both the top and bottom of the π system of the enolate can react with either the top or bottom of the π system of the benzaldehyde, the products formed will each have two stereocenters, and all possible combinations of configurations will be produced. Therefore, four possible stereoisomeric products can be isolated.

If the ionizing gas were evacuated from the Geiger counter tube while it was in operation, which of the following would occur? A) The charge on the cylindrical shell would decrease. B) The charge on the cylindrical shell would increase. C) The voltage between the cylindrical shell and the wire would increase. D) The voltage between the cylindrical shell and the wire would decrease.

A. According to the passage, the cylindrical shell and wire act as a capacitor. The gas in between the two therefore acts as a dielectric. Removal of that gas (replacing it with vacuum) reduces the dielectric constant to 1 between the wire and the cylinder, thus reducing capacitance. If capacitance decreases while the high voltage battery remains connected (as it does while the Geiger counter is in operation, thereby maintaining the voltage between the cylindrical shell and the wire, eliminating choices C and D), the charge must also decrease according to Q = CV.

The system shown in Figure 1 is an example of which of the following types of assay? (Attaching a gene for G418 resistance to a promoter) A) Reporter gene system B) Two-hybrid screen C) Microarray D) Immunohistochemical screening

A. In attaching a gene for G418 resistance to a promoter which indicates a transformation to an ES cell, researchers created a reporter gene system in which neomycin resistance reports on iPS cell transformation (choice A is correct). Two-hybrid screening is another type of reporter gene screening usually performed in yeast and is used to study protein-protein interaction (choice B is incorrect). Microarray is a high-throughput screening method for biological molecules (usually nucleic acids; choice C is incorrect), and immunohistochemical methods involve antibody detection of a molecule (choice D is incorrect).

Which of the following are true of the transporter studied in the passage? The binding site is stereospecific. Na+ enhances the rate of amino acid transport. The transporter shows a 100-fold preference for D-alanine over L-alanine. A) I and II only B) I and III only C) I, II, and III D) II only

A. Item II appears in three answer choices so let's begin there. Item II is true: in all cases, the Vmax is greater in the presence of Na+ than in the absence of Na+ (eliminate answer choice B). Item I is true: looking at L-alanine and D-alanine, we see that L-alanine shows both a greater Vmax and smaller KT than D-alanine, both of which indicate an improved binding preference of the transporter for the L amino acid compared to the D amino acid (eliminate answer choice D). Item III is false: although there is a 100-fold difference in KT values between D-alanine and L-alanine (10/0.1 =100), recall that a lower Km (analogous to KT) indicates a higher binding preference, so the transporter shows a 100-fold preference for L-alanine over D-alanine, not vice versa (choice A is correct and choice C is wrong).

The concentration of ketone bodies in the blood is expected to be increased the most in: A) a person with untreated type 1 diabetes. B) a runner who is eating extra carbohydrates for a few days before a race. C) a person with untreated epilepsy. D) an individual who has started a low fat diet in an attempt to lose weight.

A. Ketone bodies are increased when carbohydrates are lacking and the body switches to utilizing fats for energy. A runner who is eating extra carbohydrates would not have high levels of ketones (choice B can be eliminated). Untreated epilepsy has no impact on ketosis; it's the other way around...a ketogenic diet can be used to treat epilepsy (choice C can be eliminated). A low fat diet does not increase ketone levels, as fats are the essentially the substrate for ketogenesis (choice D is wrong). An individual with untreated type 1 diabetes would have decreased glucose inside of cells, leading to cellular "starvation," even though blood glucose is high. This would cause the breakdown of fats for energy, leading to ketosis, and often dangerous increases in ketosis (choice A is correct).

Which of the following changes would decrease the sensitivity of the Geiger counter? A) Using a gas with a higher ionization energy B) Using a gas with a lower dielectric constant C) Increasing the radius of the wire D) Increasing the voltage of the voltage source

A. The Geiger counter emits clicks when radiation from decaying radioactive nuclei ionizes the gas in the tube and initiates a cascade of emitted electrons. If the gas had a high ionization energy, then decaying radioactive nuclei which emit radiation of low energy would be unable to ionize the gas, so the decay would go undetected (no click). Therefore, using ages with a higher ionization energy would decrease the sensitivity of the Geiger counter.

Although not all areas of the hypothalamus are protected by the blood-brain barrier, research performed in a laboratory suggests that the area expressing leptin receptors is protected. To bind to the leptin receptor, the scientist in the laboratory hypothesizes that leptin will most likely have to: A) bind to a specific transport protein in the blood-brain barrier. B) use simple diffusion to cross the blood-brain barrier. C) filter across the blood-brain barrier with the help of increased hydrostatic pressure. D) use facilitated diffusion in the form of leptin-specific channels to cross the blood brain barrier

A. The blood brain barrier (BBB) is designed to help control what molecules pass into the cerebral spinal fluid. Leptin, is a protein that is too large to cross the BBB by simple or facilitated diffusion and will require a specific transport protein to cross the BBB (choice A is correct). Only small lipid-soluble compounds, such as O2 and CO2, can cross the BBB by simple diffusion (choice B is wrong). The lumen of channels is too small to allow diffusion of even small proteins to enter. Channels are used for smaller solutes such as ions to cross the membrane (choice D is wrong). The hydrostatic pressure in the brain is not any greater than in most areas of the body preventing the use of filtration for protein transport at the BBB. Even in capillary beds with high filtration pressure such as the kidney glomerulus, the filtration pores are too small to allow most proteins to cross (choice C is wrong).

2-Fluoro-2-deoxy glucose (shown above as the β-anomer) exists as an equilibrium mixture of α- and β-anomers that are interconverted by mutarotation via the open-chain form. Which of the following statements is true regarding the concentrations of the open-chain forms of glucose and 2-fluoro-2-deoxy glucose in 0.1 M aqueous solutions? A) The concentration of the open-chain form is greater for glucose because the fluorine atom destabilizes the aldehyde. B) The concentration of the open-chain form is lower for glucose because the fluorine atom stabilizes the hemiacetal. C) The concentrations of open-chain forms cannot be determined because they are transition states of mutarotation. D) The concentrations will be the same because both fluorine and oxygen are electronegative atoms.

A. While both fluorine and oxygen have greater electronegativity than carbon, the electronegativity of fluorine is significantly greater than that of oxygen. The chemical properties of the molecule are therefore modified by replacing an oxygen atom with a fluorine atom (eliminate choice D). The open-chain forms of sugars are not transition states, but intermediates in the mutarotation process (eliminate choice C). The carbon atoms of hemiacetals have partial positive charges due to the electron withdrawing effects of two bound oxygen atoms via induction. The greater electronegativity of the fluorine substituent increases the positive charge on the adjacent carbon, which destabilizes the hemiacetal. Choice B can thus be eliminated without evaluating the first part of the statement. Choice A is correct because the carbon atoms of aldehydes have a greater partial positive charge than hemiacetals due to electron withdrawal by the double-bonded oxygen atom via induction and resonance. The increase in partial positive charge on the adjacent carbon due to the presence of the fluorine destabilizes the aldehyde, which reduces the concentration of the open-chain form of 2-fluoro-2-deoxy glucose relative to glucose.

The total area between the cure and the zero time axis on any velocity versus time graph represents: A. distance B. displacement C. velocity D. acceleration

A. distance The total area would assume that area beneath the x axis is positive. This indicates distance, not displacement.

All of the following increase the force of air resistance on a projectile EXCEPT: A. greater mass of the projectile B. greater SA of the projectile C. greater velocity of the projectile D. greater density of air

A. greater mass of the projectile The greater mass won't change the force of air resistance, but it will add inertia, which increases the force required to change its motion.

Two balls with exactly the same size & shape are launched with the same initial velocity from the surface of a perfectly flat plane. When air resistance is considered, the ball with the greater mass will have a: A. longer flight time and a greater maximum height B. longer flight time and a lower maximum height C. shorter fight time and a greater maximum height D. shorter flight time and a lower maximum height

A. longer flight time and a greater maximum height Air resistance acts against motion. The air resistance is based on size and shape and will be the same for both balls, so the ball with the greater inertia will resist the change in its motion the most.

A particle moving at 10 m/s reverses its direction to move at 20 m/s in the opposite direction. If its acceleration is -10m/s(squared), what is its displacement from its original position? A. 10m B. 15m C. 20m D. 30m

B. 15m Use x = v(initial)t + 0.5at(squared), where t = [v(final)-v(initial)]/a

A lead ball is dropped from a height of 20 m. When air resistance is considered, approximately how long does it take for the ball to strike the ground? A. 1.4s B. 2s C. 3s D. 4s

B. 2s This question is designed to remind you that the effect f air resistance on heavy objects with a spherical shape is small. IN other words, no air resistance is often a good approximation.

A particle starts from rest and travels in a straight line for 4 s. If the particle is accelerating at a constant rate, which of the following could be the distances traveled by the particle during each consecutive second? A. 10m, 20m, 30m, 40m B. 5m, 15m, 25m, 35m C. 5m, 25m, 125m, 625m D. 2m, 4m, 8m, 16m

B. 5m, 15m, 25m, 35m Since x = 0.5at(squared), the question asks for distance with each consecutive second. this would be the total d at t=2 minus the d at t=2, or 4x-x

Sky Lab with its weightless astronauts orbited the earth 444 km above its surface. The radius of the earth is approximately 6370 km. If g = 9.8 m/s(squared), what was the correct value for g on SkyLab? A. 0 m/s(squared) B. 8.5 m/s(squared) C. 9.8 m/s(squared) D. 10.4 m/s(squared)

B. 8.5 m/s(squared) Since r is the only change, and r increases, g decreases, but not to 0.

How many urea groups are present in a caffeine molecule? A) 0 B) 1 C) 2 D) 3

B. A urea group is a carbonyl carbon bonded to two nitrogens, and caffeine has exactly one such group.

All of the following are true of the sialylated receptor described in the passage EXCEPT: A) after being modified in the ER, the protein is transported to the plasma membrane. B) the signal recognition particle recognizes the fully-folded sialylated protein. C) translation of the protein begins in the cytosol. D) molecular chaperones located in the lumen of the ER may assist in correct folding of the receptor.

B. All proteins destined for the plasma membrane pass through the secretory pathway, including the sialic acid receptor described in the first paragraph. Posttranslational modification occurs in the ER before transportation to the plasma membrane via vesicles (choice A is true and can be eliminated). The signal recognition particle identifies partially translated mRNA-ribosome complexes with a targeting sequence in the cytosol and targets them to the rough ER, at which point translation continues and post-translational modifications (such as sialic acid) are added (choice B is false and the correct answer choice). Translation of all proteins, regardless of their cellular fate, generally begins in the cytosol (choice C is true and can be eliminated). Chaperones are often involved in assisting with protein folding in the ER (choice D is true and can be eliminated).

Which of the following would be increased the most in an individual with uncontrolled diabetes? A) Acetyl-CoA B) Ketone bodies C) Lactic acid D) NADH

B. An individual with uncontrolled diabetes would have hyperglycemia and a relative (or complete) deficiency of insulin. This means that glucose levels would be elevated in the blood and unable to get inside of insulin-responsive tissues. Thus, any molecule that is produced mainly through glucose catabolism and cannot be made from alternative pathways (such as lactic acid), would be decreased, not increased (choice C is incorrect). In addition to being produced during cellular respiration, NADH can be made in the Krebs cycle and acetyl-CoA can be made from fatty acid oxidation would enter the Krebs cycle; however the greatest increase would be in the production of ketone bodies (choices A and D are eliminated, choice B is correct). Ketone bodies represent an alternate energy molecule in the absence of glucose.

Ball A & B have the same radius & mass. However, Ball A is hollow while Ball B has uniform density. Both balls are rolled down the same incline. Which of the following is true? A. Ball A will accelerate at a greater rate because it has greater rotational inertia B. Ball A will accelerate more slowly because it has greater rotational inertia C. Both balls will accelerate down the plane at the same rate D. Both balls will roll down the plane at a constant velocity

B. Ball A will accelerate more slowly because it has greater rotational inertia since the hollow ball has greater rotational inertia, it accelerates at a lesser rate, and the static friction not the hollow ball must be greater than the static friction on the solid ball.

When a gas atom (or molecule) is ionized, in which direction are the electrons accelerated? A) Away from the insulator B) Toward the cylinder C) Toward the wire D) The electrons remain stationary.

B. Being negatively-charged, electrons would be accelerated toward the cylinder, since the cylinder is at a higher potential than the wire. [Note that the cylinder is connected to the (+) terminal of the battery, while the wire is connected to the (-) terminal.]

What kind of amino acid binding site does the transporter studied in the experiment described in the passage most likely use to uptake the amino acids from the intestinal lumen? A) Hydrophilic B) Hydrophobic C) Acidic D) Basic

B. Examining the data provided in the passage, we see that the Vmax values for the hydrophobic residues, alanine and isoleucine, are relatively high, especially when compared to the Vmax values for the polar residue, serine. Given this preference for the hydrophobic, relatively nonpolar amino acids, we would expect the binding site to also be nonpolar or hydrophobic (choice B is correct). The low Vmax values for serine indicate that this is a poor hydrophilic transporter (choice A is incorrect), and there are no acidic or basic amino acid residues tested (choices C and D are incorrect because they cannot be determined without additional information).

If the front mirror in Figure 1, which is made of glass of refractive index 3/2, were repositioned so that the laser beam strikes at an angle of 60° to its surface, what would be the angle of reflection? A) sin-1(1/3) B) 30° C) sin-1(1/31/2) D) 60°

B. If the laser beam strikes the mirror at an angle of 60°, the angle that the beam makes with the normal to the mirror is 30°. This is the angle of incidence, and, by the Law of Reflection, it is also the angle of reflection. The refractive index of the mirror is irrelevant.

If the active neuron in Figure 2 is clamped at a constant voltage and the total current through the Na+ channels plus the leakage channels is 2.5 µA, what is the voltage drop across the membrane? A) 1 mV B) 10 mV C) 62.5 mV D) 62.5 V

B. In Figure 2, there are two parallel resistors: one of 25 million ohms and one of only 4 thousand ohms. Since 25 million is so much greater than 4 thousand, the equivalent resistance for this parallel combination will be so close to 4 thousand ohms that the difference is negligible. Taking Req = 4000 Ω gives V = IR = (2.5 × 10-6 A)(4 × 103 Ω) = 10 × 10-3 V = 10 mV.

The acidity of hydrogen halides increases as one moves down the periodic table. Which of the following properties of halides is most useful in explaining this trend? A) Boiling point B) Ionic radius C) Density D) Atomic weight

B. Moving down the periodic table, the ionic radii of the atoms increase. As the ionic radius increases, the stability of the corresponding anion also increases (since the negative charge is distributed over a larger volume). Therefore, since larger halides are more stable as anions, larger hydrogen halides are more likely to undergo deprotonation, and are stronger acids.

The tertiary structure of a hydrophilic sodium-hydrogen transport channel would best accommodate which of the following substances? A) Anions B) Cations C) Phosphoric acids D) Neutral compounds

B. Proteins transporting ions through the plasma membrane are typically highly selective in the ions they transport. Both H+ and Na+ are cations, so choice B is the best option.

Which of the following represents the maximum height h reached by the projectile? A. v(squared)/2 B. (1/8)gt(squared) C vt D. vtsin(theta)

B. Remember that t is the time for the entire flight of ht projectile. If we examine only the second half of flight, the projectile drops from maximum height, and starts from an initial velocity of 0 in 0.5t seconds. Using the equation: x = 0.5at(squared), but plugging in 0.5t for t, g for a, and h for x, we have: h = 0.5a(0.5t)(squared)

A chemist mixes 4.0 g AgI(s) with 25 mL of deionized water and allows the system to come to equilibrium. Which of the following changes to the mixture will NOT affect the equilibrium [Ag+]? A) Addition of NaOH(aq) B) Dilution with 10 mL deionized water C) Addition of NaCN(aq) D) Addition of AgNO3(s)

B. Since AgI is a relatively insoluble compound, the equilibrium system described should be a saturated solution of AgI, which should contain some undissolved solid AgI at the bottom of the container. Anything added to the mixture that will react with either the Ag+ or I- ions in solution will change the ion concentrations. Addition of NaOH will precipitate silver ions to form a different insoluble compound, AgOH (eliminate choice A). Addition of a NaCN solution will form a soluble silver cyanide complex ion, dissolving more of the AgI(s) in the mixture (eliminate choice C). Addition of solid AgNO3 will increase the silver ion concentration since all nitrate compounds are highly soluble (eliminate choice D). Dilution with a small amount of water will initially reduce the ion concentrations, but the same equilibrium concentrations will be reestablished after a small amount of time by dissolving additional AgI solid.

The lab which produced the iPS cell lines also worked with normal ES cells. In order to prove that their iPS cells were not simply ES cells which had contaminated the medium, which of the following experiments could they perform? A) A northern blot of the MEF cell genome to show that the genomic integration pattern of the virally-transduced transgenes in iPS cells differs from normal ES cells B) A southern blot of the MEF cell to show that the genomic integration pattern of the virally-transduced transgenes in iPS cells differs from normal ES cells C) A northern blot to show that the mRNA expression profile of iPS cells differs from normal ES cells D) A southern blot to show that the mRNA expression profile of iPS cells differs from normal ES cells

B. Since the transgenes are individually transfected into the fibroblast cells by retroviruses, which randomly insert their genome into the host genome, the four transgenes in MEF-4 cells will be found at different locations in the host genome than they would normally be found in wildtype ES cells. This allows differentiation of wildtype ES cells from lab-produced iPS cells. Southern blotting of a host genome digest can thus show the integration location of a piece of DNA in a genome by using a DNA probe (choice B is correct). Northern blotting looks at mRNA, not genomic DNA (choice A is incorrect). Since the mRNA profile of an iPS cell would be expected to be similar or identical to an ES cell, examination of the mRNA expression profile by northern blot would not differentiate an iPS from an ES cell (choice C is incorrect). Southern blots do not measure mRNA levels (choice D is incorrect).

In Experiment 1, several products are formed. One of the reactions, which involves unreacted carbonyl compounds and their conjugate bases, is an example of: A) oxidation. B) nucleophilic addition. C)nucleophilic substitution, unimolecular. D) nucleophilic substitution, bimolecular.

B. The only reaction that will occur between unreacted carbonyl compounds and their conjugate bases (enolate ions) is an aldol condensation. An aldol condensation is an example of nucleophilic addition.

Creatine phosphate is a high-energy buffer that helps maintain the level of available high-energy phosphates such as ATP. During intense muscular exertion, creatine phosphate replenishes the muscle's ATP by transferring its phosphate group to ADP. ATP is a convenient carrier of energy because: A) it releases more energy upon hydrolysis than any other molecule in the body. B) it has phosphate groups with an intermediate transfer potential. C) it has such a specific and limited role. D) it is present in selected cells.

B. What makes ATP a particularly good energy carrier is that it has an intermediate transfer potential. This allows higher energy phosphate carriers (like creatine phosphate) to drive the synthesis of ATP by transferring their phosphate groups to ADP (choice B is correct). ATP's hydrolysis does not release the largest amount of energy of any molecule in the body, since creatine phosphate can drive ATP synthesis (choice A is wrong). ATP has a varied role in the body, as an energy carrier and also as a nucleotide for RNA and DNA (choice C is wrong), and ATP is present in every cell in the body (choice D is wrong)

A particle starts from rest and travels in a straight line for 4s. IF the particle is accelerating at a constant rate, which of the following could be the total distance traveled by the particle at the end of each consecutive second? A. 10m, 20m, 30m, 40m B. 5m, 15m, 25m, 35m C. 5m, 20m, 45m, 80m D. 2m, 4m, 8m, 16m

C 5m, 20m, 45m, 80m Since x=0.5at(squared), we know that x increases by a factor of four when t doubles

A particle moving at 10 m/s reverses its direction to move at 10 m/s in the opposite direction. If its acceleration is -10 m/s(squared), what is the total distance that it travels? A. 0m B. 5m C. 10m D. 20m

C. 10m x = v(initial)t + 0.5at(squared), where t = [v(final)-v(initial)]/a. Or, since the particle is reversing its direction to reveal the same distance in the opposite direction, calculate the distance for one direction and double it for the total distance traveled. The distance in one direction is the average velocity times the time. The average velocity i exactly between the initial (10m/s) and the final (0), or 5 m/s. The 10m/s(squared) acceleration tells us that ta change of 10 m/s in velocity requires one second. 5m/s average velocity over one seconds gives a distance of 5m in one direction. We double this for a total distance of 10 m.

Retention time in gas chromatography is the time it takes a compound to be eluted and detected after injection into the system. Of the following compounds, which will display both a short retention time and a high Rf in analysis by thin-layer chromatography?

C. A low retention time in gas chromatography is associated with a volatile (low boiling point) compound, while a high Rf in thin-layer chromatography is associated with a nonpolar compound. Thus, the best answer for this question is a nonpolar compound with a low boiling point. Choices A and B, being alcohols, will have high boiling points due to their ability to hydrogen bond, so are incorrect answers. Choices C and D are both nonpolar compounds, but they differ in size by one carbon. The compound with more carbons (and less branching) in choice D, 2,3-dimethylpentane, will have a higher boiling point because it has greater London dispersion forces, so is also incorrect.

Of the following, which is the strongest oxidizing agent, based on Table 1? A) Ag(s) B) Zn2+(aq) C) Cu2+(aq) D) Ni(s)

C. An oxidizing agent is a species that is reduced in a redox reaction, so the strongest oxidizing agent among the given choices must be the species that is the most easily reduced. Choices A and D can be eliminated because neutral metals tend to be oxidized, not reduced. Zn2+ has an Eº < 0, so its reduction is not spontaneous, making it a poor oxidizing agent. (eliminate choice B). The reduction of Cu2+, however, is spontaneous (because E > 0), so this is the best answer.

The tertiary structure of proteins is partially due to disulfide bonds between cysteine residues. These bonds can be broken by treatment with a mild reducing agent. Which one of the following could be that agent? A) CrO3 B) H2SO4 C) NaBH4 D) CH3CH2OH

C. CrO3 is a good oxidizing agent (eliminate choice A) while H2SO4 (sulfuric acid) is a strong acid (eliminate choice B). CH3CH2OH (ethanol) is a poor nucleophile and has no reducing abilities (eliminate choice D). NaBH4 is a reducing agent since it has hydrogen present in the form of hydride.

If the students had used propylamine in Experiment 2 in place of piperidine, what functional group would they synthesize in step 1?

C. Enamines form from the combination of an aldehyde or ketone with a secondary amine (like piperidine). Since propylamine is a primary amine (eliminate choice A), an imine is formed instead (choice C is correct). Amides are carboxylic acid derivatives and form from the reaction of an amine with any other derivative (eliminate choice B). A lactam is a cyclic amide, so choices B and D are effectively the same answer (eliminate choice D).

A human tibia with a cross-sectional area of 2 cm2 undergoes a 1% change in length when compressed by a force of 20,000 N. What is the approximate elastic energy density within the bone while compressed if the bone has not reached its yield point? A) 5 × 103 J/m3 B) 1 × 104 J/m3 C) 5 × 105 J/m3 D) 1 × 106 J/m3

C. From Figure 1, the elastic energy density is equal to the area under the stress vs. strain graph. If the yield point has not been reached, then the region under the graph has the shape of a right triangle, whose area is (1/2) × base × height = (1/2) × strain × stress. In this case, the strain is 1% = 10-2, and the stress is F = 2 × 104 N divided by A = 2 cm2 = 2 × 10-4m2, which equals 108 N/m2. Thus, the elastic energy density is (1/2)(10-2)(108 N/m2) = 5 × 105 J/m3.

Which of the following lenses would correct for myopia (nearsightedness), in which the image is formed in front of the retina? I. Concave lens II. Convex lens III. Diverging lens A) I only B) II only C) I and III only D) II and III only

C. Item I is correct: concave lenses diverge the light rays to project the image correctly onto the retina (choices B and D can be eliminated). Note that neither of the remaining choices include Item II so it must be false and we can look at Item III. Item III is correct: concave lenses are also known as diverging lenses (choice A can be eliminated and choice C is correct). Note that Item II is in fact incorrect: convex lenses would converge the rays, causing the image to be focused even further in front of the retina instead of on the retina.

Which of the following statements is plausible regarding changes in the epigenetic regulation of ES cell-associated genes in MEF cells following transfection? A) Histone acetyl transferase (HAT) is activated, leading to decreased transcription of genes for ES cell-associated proteins. B) Histone deacetylase (HDAC) is activated, leading to decreased transcription of genes for ES cell-associated proteins. C) DNA is demethylated, leading to increased transcription of genes for ES cell-associated proteins D) Nucleotide methyl transferases are activated, leading to increased transcription of genes for ES cell-associated proteins.

C. Since the goal of iPS cells is to mimic ES cells, they can be expected to express more ES cell-associated genes and thus display epigenetic changes associated with increased expression of these genes (choices A and B, discussing decreased transcription of ES genes, are incorrect). DNA methylation (caused by nucleotide methyl transferases) decreases transcription of gene products (choice D is incorrect). DNA demethylation would be expected to increase transcription of the genes for ES cell-associated proteins, since methylation decreases transcription (choice C is correct).

A sample of 222Rn is collected in a sealed glass container and allowed to stand for 11.5 days. After this time, approximately what percentage of the original 222Rn remains? A) 3% B) 6% C) 12% D) 25%

C. Since the passage states that the half-life of 222Rn is a little less than 4 days, a time period of 11.5 days is approximately 3 half-lives. After 3 half-lives have elapsed, the amount of radioisotope remaining is (1/2)3 = 1/8 = 12.5%, so choice C is best.

If the electric field within the chamber is 1 × 1010 V/m, and a newly released electron, originally at rest, travels one nanometer before colliding with a gas atom and ionizing it, what is the highest possible value for ionization energy of that atom? A) 1 × 10-19 eV B) 0.1 eV C) 10 eV D) 1 × 1019 eV

C. The energy of the released electron would be ΔPE = qV = qEd = (1 e)(1010 V/m)(10-9 m) = 10 eV. If this electron is to ionize a gas atom, then the ionization energy of the gas can be no greater than 10 eV.

The change in Q, the magnitude of charge on each surface of the axon membrane that is depicted in the graph above, could be produced by which of the following? (graph exponentially decreases) A) A decrease in the resistance of the Na+ channels B) A steady decrease in the voltage across the axon membrane C) Inactivity of the Na+/K+ pump D) An increase in the resistance of the leakage channels

C. The graph shows how the charge decreases as a capacitor discharges, which occurs when the voltage between the plates is no longer maintained. If the Na+/K+ pump (the voltage source) stopped working, the voltage between the sides of the membrane would no longer be maintained, so C is the best choice. B is wrong since it says "a steady decrease in the voltage"; if V decreased steadily, then, since Q ∝ V, Q would also decrease steadily and its graph would be a straight line.

The passage describes the simplest interpretation of the basic properties of a laser, but some modifications are necessary to build a workable laser. Which of the following, if true, are possible problems with the simple model described in the passage? I. It is difficult to keep a collection of atoms in their excited state long enough to be stimulated to emit the induced photon. II. Atomic energy levels are quantized. III. Atoms that happen to be in their ground state will undergo absorption, thereby removing photons from the beam as it builds up. A) I only B) III only C) I and III only D) II and III only

C. The laser acquires its intense energy by induced emission, so if Statement I were true (and it is), then too many atoms could emit energy by spontaneous emission, thereby decreasing the efficiency of the laser. If Statement III were true (and it is), this would also represent a mechanism by which the laser's power would be compromised. But Statement II, although true, does not represent a problem; laser light is designed to be virtually monochromatic.

An object with a height of 6 cm is placed at a distance of 20 cm from a concave mirror with a focal length of 12 cm. If the mirror produces a real, inverted image at a distance of 30 cm, the height of the image is: A) 4 cm. B) 6 cm. C) 9 cm. D) 12 cm.

C. The linear magnification m is given by the expression -i / o. In this case, m = -(30 cm) / (20 cm) = -1.5, so the image is 1.5 times taller than the object. Therefore, if the object is 6 cm tall, then the image is 1.5(6 cm) = 9 cm tall. (Note: The minus sign on m simply signifies that the image is inverted, as stated in the question.)

Which one of the following graphs best illustrates the relationship between a, an oil drop's acceleration perpendicular to the plates, and E, the strength of the electric field between the plates?

C. The net force on an oil drop (of charge -q) between Plates A and B is the difference between the electric force and the gravitational force: Fnet = qE - mg. Since Fnet = ma, the acceleration of the oil drop is given by the equation a = (q/m)E - g. This is an equation of the form "y = mx + b," whose graph is a straight line. The answer must therefore be graph C.

Which of the following is the anticodon sequence on the tRNA for the start codon? A) 5'-AUG-3' B) 5'-UAC-3' C) 5'-CAU-3' D) 5'-GUA-3'

C. The start codon on the mRNA would be 5'-AUG-3', therefore, the anticodon on the tRNA would be the complementary sequence: 3'-UAC-5' (or alternatively written, 5'-CAU-3').

If the speed of the air flowing through the cylinder is found to be 15 m/s, what is the mass of the air moving past any point in the cylinder each second? A) 75 grams B) 75π grams C) 90 grams D) 90π grams

C. The volume flow rate (volume of air flowing per second) is given by the equation f = Av, where A is the cross-sectional area of the cylinder and v is the average speed of the flow. Since density times volume equals mass, multiplying f by ρair will give the mass moving past any point (per unit time) through the cylinder. This gives ρairAv = (1.2 × 103 g/m3)(5 × 10-3 m2)(15 m/s) = 90 g/s.

The negligible percent yield of ammonia at low pressure and high temperature suggests that the decomposition of ammonia gas is an: A) exothermic reaction with a negative ΔG°. B) exothermic reaction with a positive ΔG°. C) endothermic reaction with a negative ΔG°. D) endothermic reaction with a positive ΔG°.

C. The ΔH of the formation of ammonia is given as -91.8 kJ/mol. Therefore, the decomposition of ammonia must have a ΔH of +91.8 kJ/mol, and must be endothermic. Since the formation of ammonia is said to be negligible under the quoted conditions, ΔG° for dissociation must be negative. Note that the decomposition of ammonia produces 4 molecules (1 nitrogen gas molecule and 3 hydrogen gas molecules) for every two molecules of ammonia, so the change in entropy (ΔS) of the decomposition is positive. Therefore, at high temperatures, the reaction will be entropy driven, and the Gibbs free energy (ΔG = ΔH - TΔS) will be negative due to the TΔS term.

What single change to the experimental system would reduce the electric field strength necessary to maintain the oil drop in static equilibrium? A) Increasing the density of the oil B) Decreasing the distance between the plates C) Decreasing the mass of the drop D) Decreasing the charge magnitude on the drop

C. When the oil drop (with charge -q) is in static equilibrium, the electric force it feels, qE upward, is balanced by the gravitational force, mg downward. Thus, qE = mg, so E = mg / q. Choices A and D would necessitate an increase in E, and choice B would have no effect on E. The answer is C.

Which of the following is NOT one of the four fundamental forces in nature? A. gravitational B. electromagnetic C. chemical D. strong nuclear

C. chemical There are only 4 forces in nature: strong nuclear, weak nuclear, gravity, & electromagnetic

An object with mass ms its on a plane inclined from the horizontal at an angle theta. Which of the following represents the force on the object due to gravity? A. mgsin(theta) B. mgcos(theta) C. mg D. gsin(theta)

C. mg The force due to gravity near the surface of the earth is always mg.

The investigators transfected neuronal cells plated in small culture dishes and 48 hours later measured the levels of Aβ peptide. Which of the following procedures should be implemented to measure the levels of Aβ? A) Place the culture dishes under a microscope and directly count the amount of Aβ B) Harvest the cells, break down the cell membrane, and measure the levels of cellular Aβ using a specific antibody. C) Harvest the cells, break down the cell membrane, and measure the levels of cellular Aβ using PCR. D) Harvest the condition media from the dishes and use an ELISA (enzyme-linked immunosorbent assay) coated with anti-Aβ peptide to measure the levels of Aβ released.

D. Aβ peptide is released into the extracellular space (Figure 1 and Experiment 1) so it should be found in the condition media of the culture dishes. ELISA is the preferred method for secreted proteins found in the media, blood or extracellular space, therefore the harvested media can be used with an ELISA coated with anti-Aβ to measure the levels of Aβ (choice D is correct). Since the protein is secreted, any answer that suggests the measurement of cellular Aβ should be eliminated (choices B and C are wrong). Further, it is impossible to visualize and measure any protein or peptide with a microscope (choice A is wrong).

Which technique would be the most appropriate for separating acetone from hexane? Please choose from one of the following options. thin layer chromatography fractional distillation column chromatography simple distillation

Column chromatography and thin layer chromatography separate compounds based on polarity. Hint #2 Since acetone and hexane do not have a large difference in polarity, chromatography also would not be the best separation technique. Hint #3 Distillation separates compounds based on their boiling points. Hint #4 Simple distillation works best when the difference in boiling point between compounds is greater than 30 °C. Hint #5 Since acetone and hexane have boiling points that are less than 30 °C, it would be appropriate to do fractional distillation instead of simple distillation.

Two planets have the same mass but different radii. Each has a moon with the same orbital radius. Compared to the smaller planet, the larger planet attracts its moon with a gravitational force that is: A. greater B. smaller C. the same D. the gravitational force depends on upon the masses of the moons

D.

An object with mass m is held near the surface of the earth. IF r is the distance between the centers of gravity of the earth and the object, and M is the mass of the earth, which of the following represents the gravitational force on the earth due to the object? A. Gm/r(squared) B. GM/r(squared) C. Mg D. mg

D. The gravitational force on the earth due to the object is equal and opposite to the force not eh object due to the earth, which is mg.

Within animal cells, the transport of Na+ out of the cell by the Na+/K+ ATPase involves: A) a symport. B) an antiport. C) facilitated diffusion. D) active transport.

D. An ATPase uses the energy of ATP hydrolysis to drive the transport of Na+ out of the cell against its electrochemical gradient, thus it is an example of active transport (choice D is correct). Symports, antiports, and facilitated diffusion (choices A, B, and C) all involve the movement of ions down a gradient, which is not active transport.

What is the free energy change associated with Reaction 2 if 1 mole of zinc reacts to completion with silver in aqueous solution? A) -7,720 J B) -146,680 J C) -154,400 J D) -301,080 J

D. By adding the half-reaction Ag+(aq) + e- Ag(s), whose E° = +0.80 V, to the half-reaction Zn(s) Zn2+(aq) + 2e-, whose E° = +0.76 V (note the reversal in sign from Table 1), we conclude that the value of E° for Reaction 2 is 0.80 + 0.76 = +1.56 V. Furthermore, the stoichiometry of Reaction 2 implies that for every mole of Zn(s) that reacts, there will be 2 moles of electrons transferred. Thus, ΔG° = -nFE° = -(2)(96,500)(1.56) -300,000 J.

Which of the following techniques would be useful in driving Step 3 to completion? A) Column chromatography B) Size exclusion chromatography C) 1H NMR spectroscopy D) Distillation

D. Chromatography methods cannot be utilized on reaction mixtures while the reaction is still progressing (eliminate choices A and B), so cannot affect the outcome of the reaction. NMR spectroscopy likewise has no effect on the reaction (eliminate choice C). The passage states that the dehydration can be shifted toward completion by the addition of heat, and Figure 1 indicates that the reaction is reversible. Therefore, by Le Châtelier's principle, distillation will have a two-fold effect. The heightened temperature will shift the equilibrium to the right as the passage states, and as either the aldol product or the water is distilled away, removal of a product from the equilibrium system will cause the reaction to shift right as well.

Two blocks are suspended from the ends of a massless meter stick as shown below (one 4kg, the other 1kg): How far from the center of the stick must the rope be attached in order to maintain rotational equilibrium? A) 10 cm B) 20 cm C) 25 cm D) 30 cm

D. Intuitively, the rope must be closer to the heavier block to balance the stick. Since the block on the left is 4 times heavier, the rope must be attached at a point which is 4 times closer to the left-hand block than to the right-hand block to balance the torques due to the weights of the blocks. So, letting x denote the distance from the 4 kg block to the rope's suspension point, the distance from the 1 kg block to the suspension point is 4x. Thus, x + 4x = 100 cm, which implies x = 20 cm. The rope is therefore suspended at a point which is 50 - 20 = 30 cm from the center of the stick.

#38

D. The hydroxide ion can deprotonate both the 3-pentanone and the 2-methyl-3-pentanone (eliminate choice A). Aldol condensation can occur between the enolates formed from the deprotonations described above and any ketones in solution (eliminate choice B). Nucleophilic substitutions can occur between any enolate and methyl iodide (eliminate choice C). Hydride shifts cannot occur from an enolate, but only from a carbocation intermediate, which cannot form under the basic conditions of the reaction (chioce D is correct)

Which one of the following diagrams best illustrates the magnetic field that would be produced by a negatively-charged oil droplet moving downward with velocity v?

D. The magnetic field created by a moving charge circles around the direction of motion of the charge (in accordance with the right-hand rule). This eliminates choices A and B. Choice C illustrates the magnetic field generated by a positively-charged particle moving downward, so the answer must be D.

Which of the following would be expected in an individual with ketoacidosis? A) Decreased respiratory rate B) Increased reabsorption of H+ ions in the kidney C) Decreased reabsorption of HCO3- ions in the kidney D) Increased respiratory rate

D. The passage states that ketoacidosis is a decrease in blood pH due to the buildup of ketone bodies. An individual with ketoacidosis would be expected to exhibit compensatory mechanisms that can return blood pH to normal. If blood pH is low, reabsorption of H+ ions would only make the problem worse and decrease blood pH even more (choice B can be eliminated). Likewise, decreaseing the reabsorption of bicarbonate ions would not help, if anything bicarbonate reabsorption should be increased (choice C can be eliminated). Since blood pH is low, the body would drive towards expelling as much CO2 as possible by hyperventilating (choice D is correct); decreasing the respiratory rate would retain CO2 and further decrease the pH (choice A is wrong).

What impact do the results shown in Figure 1 have on the leptin resistance hypothesis? A) Supports the hypothesis due to the increase in neural activity with injection of 30 µg of leptin in obese mice B) Weakens the hypothesis due to the increase in neural activity with injection of 30 µg of leptin in obese mice C) Weakens the hypothesis due to the lack of impact of leptin on renal sympathetic nerve activity D) Neither supports nor weakens the hypothesis due to the differential effect of hormones on differing tissues

D. The passage states that no significant difference is observed between the obese and lean mice. While this may seem to contradict the leptin resistance hypothesis, leptin is a hormone which will be expected to affect target tissues differently. In the passage, the leptin resistance hypothesis specifically relates to satiety and it is unclear here how renal sympathetic nerve activity (RSNA) would relate to this. Therefore the lack of a significant difference between obese and lean mice does not provide significant evidence to strengthen or weaken the leptin resistance hypothesis (choice D is correct). Given the dramatic increase in nerve activity with increasing leptin dosage, leptin does have an impact on RSNA (choice C is wrong).

The screening process of SELEX for high affinity oligonucleotides incorporates which of the following? A) Size-exclusion chromotography B) Gel electrophoresis C) Ion-exchange chromatography D) Affinity chromatography

D. The second and third step of SELEX involves selection of oligonucleotides that bind specifically to the target protein, which is affinity chromatography (choice D is correct). As stated by the passage, all oligonucleotides in the library have the same length. As a result, they cannot be separated by size or charge (choices A, B, and C are wrong).

A CO2 laser produces a beam of laser light with a wavelength of 10.6 µm. Compared to the excited electrons in the laser shown in Figure 1, the excited electrons in the CO2 laser most likely have: A) more total energy. B) less total energy. C) a greater energy difference between their normal state and their excited state. D) a smaller energy difference between their normal state and their excited state.

D. The wavelength of the CO2 laser is 10 times greater than the wavelength of the laser shown in Figure 1. Therefore, the frequency and the energy of the CO2 laser are 10 times lower. Since the laser light is the energy released by transitions of electrons dropping to a lower energy level, less emitted energy implies a smaller difference between the energy levels.

Digoxin is a well-studied drug known to be a Na+/K+ ATPase pump inhibitor. Assuming digoxin primarily affects the Na+/K+ activity in the intestinal lumen cells, what effect would the consumption of digoxin have on the rate of absorption of L-alanine in the small intestine? A) Increased because of increased intestinal lumen Na+ levels. B) Increased because of decreased intestinal lumen Na+ levels. C) Decreased because of increased intestinal lumen Na+ levels. D) Decreased because of decreased intestinal lumen Na+ levels.

D. This is a 2x2 question, which means we can eliminate two answer choices initially. Recall that the Na+/K+ pump is responsible for pumping 3 Na+ ions out of the cell and 2 K+ ions into the cell. Therefore we expect an inhibitor of the pump to result in an increase in intracellular Na+ levels and a decrease in intestinal lumen Na+ levels, since Na+ will no longer be pumped out of the cell (choices A and C can be eliminated). Because there will be an decreased concentration of Na+ in the intestinal lumen, there will be a decreased driving force for any kind of Na+-coupled transport from the small intestine into the intestinal lumen cells. Since the absorption of L-alanine relies on the movement of Na+ from the lumen into the intestinal cells, there will be a decrease in the rate of absorption of L-alanine (choice D is correct and choice B is wrong).

Based on information in the passage, which of the following is the most likely product of the reaction between propanal and 2-propanone, in the presence of KOH?

D. This is the product of an aldol reaction resulting from 2-propanone and propanal. In an aldol condensation the α-carbon of the nucleophile will make a new bond to the carbonyl carbon of the electrophile, so the product is always a β-hydroxy carbonyl compound. Neither answer choices B nor C are viable since they are both γ-hydroxy carbonyl compounds. Since the question doesn't specify which reactant acts as the enolate or electrophile, it is possible that either the ketone or aldehyde can act as the enolate. However, while answer choice A is a β-hydroxy aldehyde, the molecule contains seven carbons. Propanal and 2-propanone each contain three carbons, so the aldol product will have six carbons, making choice A incorrect.

If a fully saturated solution of AgI, with precipitate present, were treated with NaCl instead of NaI, which of the following observations is likely? A) As NaCl is added, all precipitates are dissolved into the aqueous solution. B) The decrease in [AgI] is even more drastic than with the addition of NaI in Figure 1. C) There is no change in the amount of undissolved AgI. D) The concentration of [I-] increases.

D. Unlike NaI, NaCl does not have a common ion with AgI and will therefore NOT cause a decrease in the solubility for AgI with increasing concentration (elminating choice B). The following will act as a competing reaction when [Cl-] concentrations become sufficiently large. Ag+ (aq) + Cl- (aq) → AgCl (s) With this in mind, there will be no situations wherein the solution is free of precipitate (eliminatinv choice A). As NaCl as the dissolved [NaCl] concentration increasess AgCl will be precipitated from solution, which will enable additional AgI to dissolve (eliminating choice C). The increased dissolution of AgI will cause the increase in [I-], even as [Ag+] levels remain low.

If the graphs shown are displacement versus time graphs, the area under the curve represents: A. displacement B. velocity C. acceleration D. none of these

D. none of these The area under the curve is the product of the x & y axis, or meters times seconds.

Simple Distillation

Distillation separates compounds based on their boiling points. Hint #2 Simple distillation most effectively separates compounds that have a boiling point difference greater than 30 °C.

Two trials are conducted with identical particles. In the first trial particles are injected at a lower speed, and in the second trial particles are sent through the region with a larger speed. What would be the most likely comparison between the two trials reported by the experimenters? Please choose from one of the following options. A. The particles with larger speed would be deflected more since the electric force on them would be larger B. The particles with larger speed would be deflected more since the voltage in the region would be larger C. The particles with larger speed would be deflected less since the electric force on them would be smaller D. The particles with larger speed would be deflected less since they spend less time in the region between the plates

Electric force does not depend on speed. Hint #2 The voltage of the battery will not change significantly due to a fast moving particle near it. Hint #3 Since the faster particles are moving faster, they spend less time in the region between the plates, and therefore the same electric force has less time to act on them, deflecting them less.

What is the volume of person C? Please choose from one of the following options. A. 0.630 B. 6.3 C. 0.00630 D. 0.0630 ​​

F(buoyant) = ρVg Hint #2 The buoyant force on person C is 684N - 54.0N = 630N F(buoyant) = ρVg, so 630N = (1000kg/m^3)(V)(9.8m/s^2), which means that V = 0.0643m^3.

Solubility Rules

Halides (Cl-, Br-, I-) are soluble except for those of mercury (Hg22+), silver (Ag+), & lead (Pb2+). Hydroxides are insoluble except for those of the Group 1A cations, calcium (Ca2+), barium (Ba2+), & strontium (Sr2+). Sulfides are insoluble except for those of the Group 1A & 2A cations & ammonium (NH4+).

If the materials were submerged in a fluid that had an index of refraction larger than the plastics, what would be seen to happen? A. The light rays would no longer bend when entering the plastic B. The light rays would bend in the other direction compared to how they bent when entering from the air C. The light ray would not enter the material at all and would completely bounce off of the plastic D. The angles would all get smaller

Now the plastic will be a faster material than the incoming material Hint #2 When light encounters a fast medium it bends away from the normal line, when light encounters a slow medium it bends towards the normal line Hint #3 Since the light would now bend away from normal line, the light rays would be bending the other direction

# 19

P1 + (1/2)p(v1)^2 = P2 + (1/2)p(v2)^2 Since the passage says that at point b the air had stopped flowing -- then that means that v2 = 0. So you can cancel that term from the right side, and to find the difference between P2 and P1, you just have to incorporate the formula they gave in the passage into v1.

If while the water was flowing someone were to place an airtight cover over the top of the container, what would they likely see and why? A. The water would immediately not flow at all out of the tank since there would be no air above the water in the tank B. The water would flow more quickly and continue to flow more quickly since the air pressure would be trapped inside the top of the tank C. The water would at first flow the same but soon stop flowing out of the tank since the dropping water level would create a partial vacuum above the water in the tank D. The water would flow the same as always and at a consistent constant rate since the air pressure will remain constant above the water

Placing a seal over the container traps the air inside No air can get in or out as the water level drops The volume of the space above the water level in the tank increases , but the amount of air remains the same The dropping water level will create a partial vacuum and the water will soon stop flowing out of the tank

If negatively charged particles are shot downwards out of the particle injector, their motion would best be described as which of the following? Please choose from one of the following options. A. The negative charges would veer to the right B. The negative charges would veer to the left C. The negative charges would continue straight through the region without deflection D. The negative charges would stop and turn around heading back towards the charge injector

Positive charges create electric fields that point away from them, negative charges create electric fields that point towards them. Hint #2 Since the right side of the battery is the positive side, since it has the larger line, the right plate is a positive plate. Hint #3 The right plate will create a field the points to the left, and the left plate will create a field that points to the left. Hint #4 Negative charges feel a force in the opposite direction of the electric field, so the electrons would feel a force to the right.

If the student placed an object at point D, what kind of image would be created and where would it be created? Please choose from one of the following options. A. A virtual image to the left of point D B. A real image to the left of point D C. A virtual image to the right of point D D. A real image to the right of point D

Since the object is placed in front of the diverging lens, it has no affect on the image. Hint #2 The object is placed within the focal length of the converging lens. Hint #3 Since the object is placed inside the focal length of the converging lens, the image will be virtual and to the left of point D.

An employee is walking up the hospital ramp. If the employee is walking with normal walking motion, what force most directly prevents the walking individual from slipping? Please choose from one of the following options. Static friction Inertia Gravitational force Kinetic friction

Static Friction In order to have kinetic friction the two surfaces must be sliding across each other. Hint #2 To prevent a slip, the person must not be slipping yet. Hint #3 Since static friction prevents slipping, it is static friction that prevents someone from starting to slip on a ramp.

Based on an absorption spectrum, how do you determine the color of the original sample?

Take note of the labels on the two axes. On the y axis, there is absorption, and a higher peak means a greater absorption. The complementary color is observed whenever a color is absorbed. On the color wheel, the complementary color is the one diametrically opposite, so if a sample absorbs blue, then orange is observed. The x-axis is in units of wavenumber, which is the reciprocal of wavelength. Use 400nm for the violet end of the spectrum. 400 nm is the same as 400 x 10^(-9)m and can be converted to 4 x 10^(-5)cm. 1/(4 x 10^(-5)) = 25000cm^(-1). Use 700nm for the red end of the spectrum. Using those two calculated values, it can be determined that absorption occurs in the red end of the spectrum. Hint #5 Since the sample is absorbing in the red end of the spectrum, the sample should appear green. Green does not appear in any of the answer choices, but blue-green is the closest answer choice. In fact, this is the spectrum of a complex ion containing copper.

Which one of the following best represents the phase diagram of water?

The characteristic that distinguishes the phase diagram for water from the phase diagram for virtually all other substances is that the solid-liquid boundary line in the phase diagram for water has a negative, rather than a positive slope.

If someone were to take the tank off of the table while the water is flowing out, and drop the entire tank allowing it to free fall downwards, what would be seen to happen to the water flowing out of the hole? A. The water would flow out of the hole only a little less slowly since the density of the water would decrease slightly while free falling B. The water would flow out faster since the water above the hole is moving faster while free falling C. The water would not flow out, since the water above the hole would be effectively weightless by free falling D. The water would continue to flow out the same as before dropping it since the water at the top and at the hole are effected by free fall equally

The density of water is constant Think about what would happen to the pressure of the water once it is free falling) Since the freely falling water does not exert any force on the water below it, it is effectively weightless and the water would not flow out of the hole

How would you rank the heat capacities of monoatomic gases at constant volume?

The heat capacity at constant volume of a monatomic ideal gas is C=(3/2)nR Since the heat capacities of monatomic ideal gases only depend on the number of moles, the ranking is based on the number of moles only The smaller the heat capacity, the greater the change in temperature will be when absorbing the same amount of heat.

Which of the four plastics has the largest index of refraction? Please choose from one of the following options. A. Plastic 1 B. Plastic 2 C. Plastic 3 D. Plastic 4

The material with the largest index of refraction will bend the light the most in this case since it is surrounded by air Hint #2 Remember to measure all angles from the normal line! That means you must first take 90 - θ Hint #3 Take the difference between the incoming angles and the outgoing angles when measured from normal to find out which one material bent the light most Hint #4 For material 3 the incoming angle from normal was 56 and the outgoing angle from normal was 37. The difference of 56-37=19 is the largest angle of deviation, and so material 3 has the largest index of refraction

The addition of nitric acid to the water sample produces a colorless and odorless gas. What is the likely reaction when nitric acid is added to the sample? A. MX + HNO3 --> M+ + Cl2 + NO3- B. MX + HNO3 --> M+ + NO3- + CO2 + H2O C. MX + HNO3 --> M+ + NO3- + SO2 + H2O D. MX + HNO3 --> M+ + NO2 + NO

The question stem states that a colorless, odorless gas is produced. Chlorine gas has the odor of bleach, and sulfur dioxide has the rotten odor of sulfur compounds. Both are distinctive. Carbon dioxide and hydrogen gas are both odorless, as well as most of the nitrogen oxides including nitric oxide. Nitrogen dioxide has only a faint odor as the major component of air pollution. Color is imparted to an atom whenever there are electronic transitions, absorption and emission, in the visible wavelengths. Chlorine gas has a yellow-green color, while nitrogen dioxide has a reddish-brown hue, while carbon dioxide, hydrogen gas, and nitric oxide are colorless. When nitric acid is added to the water sample, carbonate becomes carbonic acid, which quickly decomposes to produce carbon dioxide and water. Hence, carbon dioxide is the colorless and odorless gas in question.

If one were to increase the temperature of the air in the tube, how would that affect the fundamental resonant frequency observed in the tube? Please choose from one of the following options. A. The fundamental resonant frequency in the tube would not change, since the length of the tube is not changing B. The fundamental resonant frequency in the tube would decrease, since the fundamental wavelength would decrease C. The fundamental resonant frequency in the tube would increase, since the speed of sound would increase in hotter air D. The fundamental resonant frequency in the tube would decrease, since the sound takes longer to travel down the tube in hotter air

The speed of sound depends on the temperature of the medium. Hint #2 The speed of sound increases in hotter air. Hint #3 The fundamental wavelength would stay the same, since the length of the tube is not changing. Hint #4 Since f=v/λ and v increases and λ stays the same, the fundamental frequency would increase.

In one trial, the machine moves towards the patient with a constant speed v. While this is occurring, which of the following descriptions is most accurate? Please choose from one of the following options. A. The frequency continuously gets higher, and the volume continuously gets louder B. The frequency remains constant, and the volume continuously gets louder C. The frequency stays constant, and the volume stays constant D. The wavelength gets larger and larger, and the volume stays constant

The speed of the machine is constant Hint #2 Intensity gets larger as you approach the source of sound Hint #3 Since the speed is constant, the Doppler shift is constant and the frequency remains constant, but as the machine approaches the patient the sound gets louder and louder since the intensity increases

Consider the electric potential at the midway point along a line joining the two centers of the spheres. Each sphere contributes to the total electric potential by virtue of the equation V=kq/r. What can be said for sure about the contribution to the electric potential at the midway point due to the two spheres? Please choose from one of the following options. A. The contribution of both spheres to the electric potential must cancel and be equal to zero since they are equally distance from the midway point B. The contribution of both spheres to the electric potential must be equal but not necessarily different sign C. The contribution of both spheres to the electric potential must be of opposite sign, but not necessarily equal D. The contribution of both spheres to the electric potential must be of opposite signs and equal magnitude

The two charges do not have to have the same charge. Hint #2 If the two charges attract they must have opposite signs. Hint #3 The two spheres must have opposite signs but are not necessarily the same magnitude of charge.

If a car is traveling fast enough and slams on the brakes it will skid regardless of whether it is moving down a hill or up a hill. Will a car experience a larger magnitude of acceleration if it skids down the Bridger St hill or up the Bridger St hill? Please choose from one of the following options. A car will experience more acceleration when it is skidding up Bridger St hill It depends on the mass of the car A car will experience more acceleration when it is skidding down Bridger St hill A car will experience the same acceleration regardless if it is skidding up Bridger St hill or down Bridger St hill

Think about Newton's Second Law ΣF=ma. Hint #2 When the car is skidding down the hill the forces of gravity along the incline and the force of friction point in opposite direction. Hint #3 When the car is skidding up the hill the forces of gravity along the incline and the force of friction point in the same direction. Hint #4 Since both forces point in the same direction while the car is skidding up the hill, the net force will have a larger magnitude, which means the acceleration will have a larger magnitude.

The student now wants to change the lab set up to observe a total internal reflection event. Which of the following would increase the likelihood of seeing a total internal reflection event? Please choose from one of the following options. A. Increase θ3 B. Decrease θ5 C. Increase θ1 D. Decrease θ4

Total internal reflection only happens when light would refract into a faster material Hint #2 Total internal reflection first happens when the refracted angle is 90 Hint #3 To get total internal reflection you need θ5 to increase and approach 90 Hint #4 The only choice given that will increase the refracted angle θ5 is to decrease θ4

What will be detected by UV-Vis absorption spectroscopy?

UV-Vis absorption spectroscopy is only useful for detecting electronic transitions corresponding to 220-400nm in the UV range and 400-800nm in the visible range. Transition metal ions are one of the species detected because of their d electrons. Transitional metals in their elemental states will not be detected. A complex ion often contains a transition metal ion as its center. Those that do not include a transition metal ion will not be detected. Compounds with double bonds are another species detected because of their π electrons. While all double bonds absorb UV light, only species with conjugated π systems will fall into the range of detection for UV-Vis spectroscopy. The bicyclic compound cis-decalin only has sigma bonds and would not be detected. Anthracene has three conjugated benzene rings for its structure and would be detected by UV-Vis spectroscopy. In fact, anthracene exhibits blue fluorescence under UV light.

Doppler Effect

f = f of source f' = f detected vo = velocity observed