MCAT Physics Q Pack
A ray of light in air strikes the flat surface of a liquid, resulting in a reflected ray and a refracted ray. If the angle of reflection is known, what additional information is needed in order to determine the relative refractive index of the liquid compared to air? Angle of incidence Angle of refraction Refractive index of air Wavelength of the light
Angle of incidence = Angle of reflection Solution: The correct answer is B. To find the relative refractive index to air one needs both the incident and refracted angles. Since the angle of incidence is equal to the angle of reflection we will know the angle of reflection. We still must know the angle of refraction and this is answer B. None of the other answers will allow one to know the angle of refraction.
How do the pressures Pw and Pm compare, measured at the bottom of two identical containers filled to the levels shown in the figure with water and mercury? (Note: Density of water = 1 g/cm3; density of mercury = 14 g/cm3.) Pm = 2Pw Pm = 7Pw Pm = 14Pw Pm = 28Pw
Divide 14 by 1/2 because the height of the mercury is half of water in the photo!!! Solution: The correct answer is B. Pressure is given by (density) • g • (height). Here, mercury's density is given as 14 times that of water but the column is only half as high, making the pressure of the mercury (14)(0.5) = 7 times more than that of the water. Thus, B is the best answer.
The wavelength of the detected sound when the projectile is at h is: 0.50 m. 1.00 m. 2.00 m. 170 m. Given: Speed of Sound 240 m/s, Sounds emits at a frequency of 170 Hz
Don't forget WAVELENGTH = SPEED OF SOUNDS/LIGHT/FREQUENCY Solution: The correct answer is C. Because the projectile has stopped moving, there is no Doppler shift. The wavelength is λ = vsound/f = 340/170 = 2 m. Thus, C is the best answer.
When a downward force is applied at a point 0.60 m to the left of a fulcrum, equilibrium is obtained by placing a mass of 10-7 kg at a point 0.40 m to the right of the fulcrum. What is the magnitude of the downward force? 1.5 × 10-7 N 6.5 × 10-7 N 9.8 × 10-7 N 1.5 × 10-6 N
Don't forget to use mg to solve for the weight of the mass on the right first. .40m is just the distance from the fulcrum on the right. Solution: The correct answer is B. The system will be in equilibrium only if the right and left forces exert torques that cancel each other. One force's torque will attempt to twist the "see-saw" supported by the fulcrum clockwise, while the other force provides a counterclockwise torque. The torque of a force F applied at a distance L from the fulcrum is given by FLsinθ, with θ is the angle between the beam and the force. At equilibrium the left and right forces point vertically downward with the beam horizontal, hence θ = 90° and sin90° = 1. At equilibrium the torque magnitudes are equal so: Fleft Lleft = Fright Lright. The force at Lright = 0.40 m is the weight of the given mass, mg = (10^-7 kg)(9.8 m/s2) = 9.8 × 10^-7 N. The unknown force is at Lleft = 0.60 m, so Fleft (0.60 m) = (9.8 × 10-7 N)(0.40 m) which may be solved to yield Fleft = 6.5 × 10-7 m. Thus B is the best answer.
Ball 2 is in the water 20 cm above Ball 3. What is the approximate difference in pressure between the 2 balls? 2 × 10^2 N/m2 5 × 10^2 N/m2 2 × 10^3 N/m2 5 × 10^3 N/m2
Easy just plug in to equation, do not get overwhelmed. Don't forget to convert cm to m (20 cm = .2 m) Solution: The correct answer is C. The examinee must determine the pressure difference between two locations in water at depths separated by 20 cm. The absolute pressure p at depth h below the surface of a fluid is: p = p0 + ρgh, where p0 is the atmospheric pressure above the liquid, ρ, is the density of the liquid, and g = 9.8 m/s2. Thus, the pressure difference Δp between two locations in the water would be Δp = ρgΔh, where Δh is the difference in depth. Here, Δh = 20 cm = 0.20 m. Thus, Δp = (1000 kg/m3)(9.8 m/s2)(0.20 m) = 1960 Pa = 2.0 x 103 Pa. Thus, C is the best answer.
If the magnitude of a positive charge is tripled, what is the ratio of the original value of the electric field at a point to the new value of the electric field at that same point? 1:2 1:3 1:6 1:9
KNOW THIS CHARGE EQUATION Solution: The correct answer is B. The magnitude of the electric field E of a point charge is given by E = kq/r^2 If q is tripled, E also will be tripled. Thus, B is the best answer.
An object that is totally immersed in benzene (specific gravity = 0.7) is subject to a buoyancy force of 5 N. When the same object is totally immersed in an unknown liquid, the buoyancy force is 12 N. What is the approximate specific gravity of the unknown liquid?
My logic - must be greater than 0.7 Solution: The correct answer is C. The buoyant force on an immersed object is the product of: (density of the liquid) × (volume of the object) × (acceleration of gravity). Forming the ratio of buoyant forces in the two cases gives: 12/5 = (density of the unknown liquid)/(density of benzene, 0.7). Solving for the specific gravity of the unknown liquid, which is the ratio of its density to that of water, gives (12/5) × 0.7 = 1.7. Answer C is correct.
As the light passes from the air into the glass, it makes an angle θa in air and an angle θl in the lens material, relative to the normal at the surface. What equation relates the angles θl and θa? A) θa = θl B) 1/θa = 1/θl C) nasin θa = nlsin θl D) na/sin θa = nl/sin θl
SNELL'S LAW!!! Solution: The correct answer is C. This item requires knowledge of Snell's law of refraction, which is given by C. The other options are erroneous. Thus, C is the best answer.
The mathematical expression for h is: mv2/2. v2/(2g). mg. mv.
Set PE = KE E(total) = mgh = 1/2mv^2 Solution: The correct answer is B. Conservation of energy implies KEinitial = PEfinal, or mv2/2 = mgh; therefore, h = v2/2g. Thus, B is the best answer.
Which circuit elements store energy? Capacitors Resistors Batteries
Solution: CAPACITORS AND BATTERIES The capacitor charges up and stores energy in the electric field between the places. The energy stored is ½C(Vc)^2, where Vc is the voltage across the capacitor. The battery is the source of energy for the circuit and thus is a store of energy. The resistor is not a storage device for energy and answer C is the correct answer.
When fewer than 3.75 × 10^6 americium nuclei remain, the ionization smoke detector will not operate due to insufficient ionization. How much time will pass before there are this many nuclei remaining? A) 1720 years B) 2150 years C) 4300 years D) 6880 years
Solution: The correct answer is A. Each half-life reduces the number of americium nuclei by half. Because it takes 4 half-lives to go from 60,000,000 nuclei to 3,750,000 nuclei (1/16 of the original amount), the elapsed time is 4(430) = 1720 years. Thus, A is the best answer.
Visible light travels more slowly through an optically dense medium than through a vacuum. A possible explanation for this could be that the light: is absorbed and re-emitted by the atomic structure of the optically dense medium. is absorbed and re-emitted by the nucleus of the material in the optically dense medium. bounces around randomly inside of the optically dense medium before emerging. loses amplitude as it passes through the optically dense medium.
Solution: The correct answer is A. A is known to occur and is the reason for the slowing down of light. B is incorrect because the nucleus is involved. C is incorrect because the motion of the photons is certainly not random. D is true but does not answer the question. Thus, A is the best answer.
What is the magnitude of the detected sound frequency shift from 170 Hz during the projectile flight described in the passage? It falls to zero, then increases. It is constant throughout the flight. It rises continuously. It falls continuously.
Solution: The correct answer is A. As the object moves up and slows down, the frequency shift is negative and falls to zero at the peak of the object's flight; as the object falls, the shift becomes increasingly positive. However, to find the magnitude of the frequency shift from 170 Hz, the negative or positive sign of the shift can be ignored and only the absolute value matters. Thus, A is the best answer.
A receiver is in a jet flying alongside another jet that is emitting 2.0 x 106 Hz radio waves. If the jets fly at 268 m/s, what is the change in frequency detected at the receiver? 0 Hz 0.90 Hz 1.79 Hz 3.58 Hz
Solution: The correct answer is A. Because there is no relative motion between the jets, there is no frequency shift.
Which of the following changes to the circuit will decrease the electric field between the electrodes by the greatest amount? Increasing L by a factor of 2 Decreasing L by a factor of 2 Increasing R by a factor of 2 Decreasing R by a factor of 2
Solution: The correct answer is A. For a fixed voltage between cathode and anode, the electric field is inversely proportional to the distance between them. Increasing the circuit resistance for a fixed current will decrease the electric field, but not by as much as does the length change. E = (V − IR)/L
A student has a thin copper beaker containing 100 g of a pure metal in the solid state. The metal is at 215°C, its exact melting temperature. If the student lights a Bunsen burner and holds it for a fraction of a second under the beaker, what will happen to the metal? A small amount of the metal will turn to liquid, with the temperature remaining the same. All the metal will turn to liquid, with the temperature remaining the same. The temperature of the metal at the top of the beaker will increase. The temperature of the whole mass of the metal will increase slightly.
Solution: The correct answer is A. Melting occurs at a constant temperature because a certain amount of energy, the latent heat of fusion, is needed to convert a substance from its solid to liquid state. The temperature of the metal will not increase above its melting point until all of the metal has melted. The small amount of heat supplied by the bunsen burner is insufficient to melt 100 g of the metal but could melt a small amount of the metal at the constant temperature of the melting point. Thus, answer choice A is the best answer.
Why are the percentages of the change in frequency and wavelength much greater when sound waves are used instead of radio waves in these experiments? Sound waves travel more slowly. Sound waves have a much higher frequency. Sound waves have a much shorter wavelength. Interference in the atmosphere affects sound waves much more.
Solution: The correct answer is A. The Doppler equation for frequency is Δf/f = -v/c for a given relative velocity v between source and detector. Thus, the frequency shift Δf depends inversely on the speed of the wave in the medium in which it propagates, c. The velocity of sound is much smaller than that of electromagnetic radiation, so for the same relative velocity the frequency and wavelength shifts are much greater for sound than for radio waves.
The glass that is used as a beam splitter is replaced with glass that is identical except that it has a 10% higher index of refraction. Which of the following changes will occur to the pinhole image? It will move. It will become larger. It will become smaller. It will become more clear.
Solution: The correct answer is A. The examinee is asked what an increase in the index of refraction of the beam splitter will do to the image of a pinhole on the viewing screen. The pinhole image is formed by light that: (1) passes through the glass of the beam splitter, (2) reflects off the retroflector, (3) returns to the surface of beam splitter nearest the retroflector, (4) reflects off that beam splitter surface, and (5) falls on the viewing screen, forming an image. The only refraction occurs in step (1), when the entering light is refracted. This refraction shifts the rays of light over a bit. If the index of refraction is increased, this increases the shift. This will cause a shift in the position of the light striking the retroflector which will persist through the steps (3), (4), and (5) above, causing the image on the screen to shift in position. Thus, A is the best answer.
Assume that the side of the water tank is punctured 5.0 m below the top of the water, and that atmospheric pressure is 1.0 × 10^5 N/m2. What is the approximate speed of the water flowing from the hole? 10 m/s 12 m/s 14 m/s 17 m/s
Solution: The correct answer is A. The examinee must calculate the speed of water exiting a small puncture in the side of the tank 5.0 m below the surface of the water. The passage provides the equation needed, Bernoulli's equation: P2 + ½ ρv2^2 + ρgy2 = P1 + ½ ρv1^2 + ρgy1. Take point 2 to be at the location of the puncture and point 1 to be at the upper surface of the fluid. At point 1 above the fluid the pressure P1 is the atmospheric pressure and at point 2 outside the puncture the pressure P2 is also atmospheric pressure, so these terms cancel in Bernoulli's equation for this situation. Also note that the speed of water exiting the small puncture will be much larger than the speed of the upper water level falling at point 1. Thus, v2 >>v1 and v1 may be ignored. Canceling the pressures and setting v1 = 0 leaves the approximate result: ½ ρv2^2 + ρgy2 = ρgy1, which may be solved to yield v2 = [2g(y1 - y2)]1/2. The examinee is given that the height difference y1 - y2 is 5.0 m, hence, v2 = [2 (9.8 m/s^2) (5.0 m)]^1/2 = [98 m^2/s^2]1/2 ≈ 10 m/s. Thus, A is the best answer.
How will W change if the angle of the ramp to the horizontal is increased? W will decrease, because the normal force to the surface of the ramp will decrease. W will not change, because the coefficient of friction between the box and the ramp will not change. W will not change, because the gravitational force is always constant and the length of the ramp is not changed. W will increase, because the height from which the box falls increases.
Solution: The correct answer is A. The examinee must determine the effect that increasing the ramp's angle will have on the work done by the friction, W, on the mass. The definition of the work of a force is W = Fdcosθ, where F is the magnitude of the force, d is the distance traveled, and θ is the angle between the force and the displacement. Kinetic friction F acts up the ramp and d is down the ramp, so θ= 180°, hence W = −Fd.The kinetic friction force F is the product of the normal force N that the ramp exerts on the mass and the coefficient of kinetic friction μkh. Thus W = − μkhNd. Increasing the tilt of the ramp has no effect on d or μkh, but it does decrease normal force N. For a very steep ramp, the mass is nearly in free fall and pushes down on the ramp very little, hence N is small. Therefore, W decreases when the ramp's tilt is increased. Thus, A is the best answer.
The densities of the balls, D1, D2, and D3, are related by which of the following? D1 < D2 < D3 D1 < D2 = D3 D1 = D2 < D3 D1 = D2 > D3
Solution: The correct answer is A. The examinee must rank the densities D1, D2, and D3 of the three balls with identical volumes at rest at different locations in the stationary water. In the diagram provided, Ball 3 rests on the bottom of the tank, Ball 2 is suspended half-way between the surface the water and the bottom, and Ball 1 floats partially submerged. Objects denser than water sink to the bottom of the tank, objects with density equal to that of the water remain suspended (or float) wherever they are placed, provided that they are completely submerged, and objects less dense than water float at the surface partially submerged. Ball 3 sinks and touches the bottom of the tank, so it must be denser than water because the bottom of the tank also supports it. Ball 2 is suspended in the water without touching the bottom, so its density equals that of the water. Ball 1 floats at the surface, so its density is smaller than that of the water. Using Archimedes' formula of the buoyant force that depends on density and volume, the densities are ranked D1 < D2 < D3.
A charged particle with a mass of m and a charge of q is injected midway between the plates of a capacitor that has a uniform electric field of E. What is the acceleration of this particle due to the electric field? Eq/m Em/q mq/E Emq
Solution: The correct answer is A. The force on the charge is qE and force is also ma. Setting qE = ma and solving for acceleration a yields
Which of the following graphs best illustrates the relationship between speed of the transmitter away from the receiver and the increase in wavelength of the received signal?
Solution: The correct answer is A. The linear relation between relative speed v and change in detected wavelength is a consequence of the Doppler equation, Δλ/λ = v/c where c is the velocity of the radio wave. Furthermore, Graph A is the only one that shows an increase in wavelength Δλ with an increase in speed as the jet flies away from the receiver.
When the number of photons incident on the cathode with energies above the value of the work function increases, which of the following quantities also increases? Number of electrons ejected Potential energy of each ejected electron Magnitude of the electric field between the electrodes Speed of electrons at the anode
Solution: The correct answer is A. The number of incident photons affects only the number of electrons, not their energies. The electron energies depend on photon energy, the cathode work function, and the potential difference between the cathode and anode.
If the frequency of the first harmonic in Figure 2a is 100 Hz, what is the period of the second harmonic? 0.005 sec 0.01 sec 50.0 sec 200.0 sec
Solution: The correct answer is A. The period T and frequency f of a tone are related by T = 1/f. If the first harmonic has a frequency of 100 Hz, then the second harmonic has a frequency of 200 Hz. The period corresponding to 200 Hz is 1/200 s-1 = 0.005 s. Thus, answer choice A is the best answer.
Suppose that a ball is thrown vertically upward from earth with velocity v, and returns to its original height in a time t. If the value of g were reduced to g/6 (as on the moon), then t would: increase by a factor of 6. increase by a factor of 6^1/2. decrease by a factor of 6. decrease by a factor of 6^1/2.
Solution: The correct answer is A. The round-trip time t for a ball thrown vertically is given by t = 2v/g. If g is replaced by g/6, then t is increased by a factor of 6. Choice A states this fact.
A plasma wave moving through a plasma has a frequency of 10^9 Hz and a speed of 3.0 × 10^7 m/s. What is the wavelength of this wave? 3.0 cm 3.0 m 3.3 cm 3.3 m
Solution: The correct answer is A. Wavelength is given by speed/frequency = 3 × 10^7/10^9 or 3 × 10^-2 m. This is the same as 3 cm or answer A.
Another capacitor, identical to the original, is added in series to the circuit described in the passage. Compared to the original circuit, the equivalent capacitance of the new circuit is: 1/2 as great. the same. 2 times as great. 4 times as great.
Solution: The correct answer is A. When in series, capacitors C1 and C2 add by the inverse rule 1/Ceq = 1/C1 + 1/C2 Therefore, if C1 = C2 = C, then 1/Ceq = 1/C + 1/C = 2/C Ceq = C/2
Sound of a known frequency, wavelength, intensity, and speed travels through air and bounces off an imperfect reflector which is moving toward the source. Which of the following properties of the sound remains the same before and after reflection? Speed Intensity Frequency Wavelength
Solution: The correct answer is A. Within still air, the speed of sound remains constant. Thus, A is the best answer.
To keep the current constant during the discharge cycle: the resistance R must be continually increased. the resistance R must be continually decreased. the resistance r must be continually increased. the resistance r must equal R.
Solution: The correct answer is B. As the capacitor discharges the voltage across it falls, thus to maintain a constant current, R must be proportionately reduced. This is so from Ohm's law, I = V/R. To keep I fixed, R must fall with V. This is answer B.
What best describes changes that occur as the electron sea moves from position A to position B in Figure 1? Kinetic energy is transformed into potential energy. Potential energy is transformed into kinetic energy. Power is dissipated as heat. Turbulence brings the electron sea to rest.
Solution: The correct answer is B. As the electrons move from A to B they convert potential energy (½kx2) into kinetic energy (½mv2), where x is the displacement from point B. It is conservation of energy.
The intensity level in decibels is defined as 10 log10(I/I0), where I0 is a reference intensity equal to the human threshold of hearing, 10^-12 W/m^2. What is the intensity of the threshold of pain, 120 decibels? 10^12 W/m2 10^° W/m2 10^-2 W/m2 10^-12 W/m2
Solution: The correct answer is B. Because the intensity level is 120 dB = 10 log (I/I0), log (I/I0) must equal 12 and I/I0 must equal 1012. Therefore I = 10° W/m2, because I0 = 10-12 W/m2. Thus, B is the best answer.
A beam of light shines into a transparent medium with parallel surfaces. Part of the beam is reflected back into the air as diagrammed above. (The figure is NOT to scale.) The index of refraction of the medium is 1.5. Which of the following is true? θ < θ' and θ < α θ = θ' and θ > α θ = θ' and θ < α θ > θ' and θ > α
Solution: The correct answer is B. Because the medium's surfaces are parallel, a perpendicular line drawn to the lower surface of the medium will be parallel to both of the perpendiculars shown in the figure. This means that the angle of incidence at the lower surface will also be α, as will the angle of reflection at the lower surface, and the beam reflecting from the lower surface of the medium will then be a mirror image of the incoming beam, so θ' = θ. Further, because air's index of refraction is about 1.0, Snell's law would show that θ > α. Thus, B is the best answer.
If the speed of the charged particle described in the passage is increased by a factor of 2, the electrical force on the particle will: decrease by a factor of 2. remain the same. increase by a factor of 2. increase by a factor of 4.
Solution: The correct answer is B. Electrical force depends on the particle's charge and the strength of the electric field experienced by the particle, not on the particle's speed. Thus, B is the best answer.
Why can the positive ions be considered to be fixed during the electrons' oscillations? The ions are bound together with strong nuclear forces. An ion is much more massive than an electron. The ions experience no force when the electron sea is displaced. Coulomb's law prohibits the motion of the ions.
Solution: The correct answer is B. Since the ions are thousands of times more massive than the electron, answer B is justified (the hydrogen ion is the lightest ion and is nearly 2000 times more massive than the electron). An ion with so much mass compared to an electron will not be able to respond quickly because of its inertia.
A small negatively charged particle is placed near a fixed positively charged particle (Q). Which of the following describes the motion of the negatively charged particle? It accelerates away from Q. It accelerates toward Q. It moves with constant speed away from Q. It moves with constant speed toward Q.
Solution: The correct answer is B. The Coulomb force between the negative and positive charges is attractive. That force accelerates the negatively-charged particle toward the positively charged particle. Choice B is the correct answer.
A sample of radiosodium has a half-life of 15 hr. If the sample's activity is 100 millicuries after 24 hr, approximately what must its original activity have been? 200 millicuries 300 millicuries 600 millicuries 1,000 millicuries
Solution: The correct answer is B. The examinee must take the given half life of 15 hours and the current activity of 100 millicuries and infer what the activity was 24 hours earlier. After one half life the activity decreases by one-half. Thus, it can be inferred that 15 hours earlier the activity was twice as much, 2 × 100 = 200 millicuries. Another 15 hours earlier, the activity would have been 2 × 200 = 400 millicuries. Carefully note the sequence: present yields 100 millicuries, 15 hours before the present yields 200 millcuries, 30 hours before the present yields 400 millicuries. This implies that 24 hours before the present the activity was in the range between 200 and 400 millicuries and only one choice, 300 millicuries, is in this range. Thus, B is the best answer.
An astronaut on Earth notes that in her soft drink an ice cube floats with 9/10 of its volume submerged. If she were instead in a lunar module parked on the Moon where the gravitation force is 1/6 that of Earth, the ice in the same soft drink would float: with more than 9/10 submerged. with 9/10 submerged. with 6/10 submerged. totally submerged.
Solution: The correct answer is B. The floating ice cube implies that its weight is balanced by the buoyant force on it W(ice) = mg = r(fluid)V(submerged)g Note that both the weight and the buoyant force are proportional to g, making the numerical value of g irrelevant to the volume of the ice cube that is submerged. Thus, B is the best answer.
Changing which of the following will change the focal length of the convex mirror in Figure 2? Index of refraction of the mirror Radius of curvature of the mirror Position of the lens at B Focal length of the lens at B
Solution: The correct answer is B. The focal length of the mirror depends only on the radius of the curvature. Because there is no refraction, the index of refraction is irrelevant and the properties of Lens B will not affect the focal length of the mirror. Thus, B is the best answer.
The initial decay activity of a given quantity of a radioactive element is 240 counts/min. After 24 min, the activity is 60 counts/min. What is the half-life of the element? 4 min 12 min 24 min 48 min
Solution: The correct answer is B. The half-life is the time for the activity to drop by a factor of two. In two half-lives the activity drops by a factor 22 = 4. The activity of the sample dropped by a factor of (240/60) = 4 in 24 minutes. Therefore, the sample has a half-life of 12 minutes. Answer B is the correct choice.
An electron is ejected from the cathode by a photon with an energy slightly greater than the work function of the cathode. How will the final kinetic energy of the electron upon reaching the anode compare to its initial potential energy immediately after it has been ejected? It will be 2 times as large. It will be approximately equal. It will be 1/4 as large. It will be 0.
Solution: The correct answer is B. The near equality of the photon energy and the work function means that little initial kinetic energy will be left for the electron. This initial kinetic energy is small compared to the 50 eV it will gain from the potential difference between electrodes.
What is the approximate number of wavelengths of light that can travel in 1 direction within a retroreflecting bead that has a diameter of 5 × 10-5 m? (Note: The speed of light = 3 × 108 m/s, and its frequency is approximately 1015Hz.) 0.6 1.7 × 10^2 1.5 × 10^4 3.3 × 10^6
Solution: The correct answer is B. This question requires the examinee to envision light waves traveling along the diameter of a retroflecting bead and to determine the total number of waves that could be present along that bead diameter at a given time. Thus, the examinee must determine how many wavelengths will fit within the width of a bead, because there is one wavelength, λ of spatial length per wave. The item gives the speed of light to be v = 3.0 × 10^8 m/s and the frequency to be f = 10^15 Hz. The wavelength λ of the light may be found from the relationship for wave speed v = λf, thus, λ= v/f = (3.0 × 10^8 m/s)/( 10^15 Hz) = 3.0 × 10^-7 m. The number of waves that will fit along a diameter is found by dividing this wavelength into the diameter D = 5.0 × 10^-5 m, so the number of waves is D/λ=(5.0 × 10^-5 m)/( 3.0 × 10^-7 m) = 1.7 × 10^2. Thus, B is the best answer.
What is the volume flow rate of blood that moves at 0.20 m/s through an artery with a diameter of 1.0 × 10-2 m? 5.0 × 10^-6 m^3/s 5π × 10^-6 m^3/s π × 10^-5 m^3/s 2π × 10^-5 m^3/s
Solution: The correct answer is B. Volume flow rate is the product of blood speed and artery cross-sectional area: (0.20 m/s)·(π/4)·(1.0 × 10^-2 m)^2 = 5π × 10^-6 m^3/s. Therefore, answer choice B is the best answer.
Which of the following best describes the movement of an electron after it is ejected from the cathode? It is stationary until collisions propel it toward the anode. It moves with constant speed toward the anode. It accelerates toward the anode. It exits through a side of the vacuum photodiode.
Solution: The correct answer is C. A charged particle accelerates in an electric field. The electron starts with a velocity that increases as it approaches the anode through the vacuum.
Which of the following will occur when the magnet used in the flowmeter discussed in the passage is replaced with a stronger magnet? The electric field will reverse polarity. The electric field will decrease. The voltage will increase. Blood will flow faster.
Solution: The correct answer is C. A stronger B field increases the magnetic force, Fm = q v B^. The electric force must also increase to achieve equilibrium. This implies a larger electric field in the artery and a larger voltage across the artery. Thus, answer choice C is the best answer.
How much work is done in launching the object? 9.80 kg m/s2 19.6 J 9.8 J 19.6 N
Solution: The correct answer is C. In a one-dimensional problem such as this, work is force times distance. Hence, (19.6 N)(0.5 m) = 9.8 J. Thus, C is the best answer.
An artery is constricted at one location to 1/2 its normal cross-sectional area. How does the speed of blood past the constriction compare to the speed of blood flow in the rest of the artery? (Note: Assume ideal fluid flow.) It is 1/4 as fast. It is 1/2 as fast. It is 2 times as fast. It is 4 times as fast.
Solution: The correct answer is C. In the artery, the product of blood speed and artery cross-sectional area is everywhere constant because the volume flow rates are equal. At the point with half the normal area, the speed must double so that the same volume of blood passes through the constriction as does through the normal part of the artery. Thus, answer choice C is the best answer.
Which of the following best describes the motion of a negatively charged particle after it has been injected between the plates of a charged, parallel-plate capacitor? (Note: Assume that the area between the plates is a vacuum.) It moves with constant speed toward the positive plate. It moves with constant speed toward the negative plate. It accelerates toward the positive plate. It accelerates toward the negative plate.
Solution: The correct answer is C. Opposites attract, so the negatively charged particle will move toward the positive plate. Because there is a constant force qE on the particle, it will accelerate toward the positive plate. Thus, C is the best answer.
Which of the following statements best explains why the intensity of sound heard is less when a wall is placed between a source of sound and the listener? Sound travels more slowly in a solid than in air. The frequency of sound is lower in a solid than in air. Part of the sound energy is reflected by the solid. The wavelength of sound is shorter in a solid than in air.
Solution: The correct answer is C. Sound wave are reflected and absorbed when incident on materials just as light waves are. In doing so the waves are reduced in their forward intensity (units of J/m2). When a wall is placed between a source of sound and the listener, some of the sound energy is transmitted to the listener, but some is reflected and sent backward toward the source or absorbed in the material. The listener hears less sound, but at the same frequency and wavelength as the unimpeded sound.
Which action requires a larger absolute value of work: lifting the weight from A to B with constant speed, or lowering the weight from B to A with the same constant speed? a) Lifting from A to B b) Lowering from B to A c) Equal absolute value of work in both actions d) No work is required using a pulley.
Solution: The correct answer is C. The absolute value of the work done is mgΔh where Δh is positively defined, and because none of these values changes in magnitude when the mass goes up or down, these actions involve the same amount of work. Thus, C, is the best answer.
Which of the following occurs when electrons are ejected from the cathode? The voltage across the electrodes reverses polarity. The voltage difference between the electrodes increases. Current flows through the circuit. The total resistance of the circuit increases.
Solution: The correct answer is C. The electrons ejected from the cathode are replaced by electrons from the battery and anode. The only effect on the apparatus of electron ejection from the cathode is the current in the circuit.
Consecutive resonances occur at wavelengths of 8 m and 4.8 m in an organ pipe closed at one end. What is the length of the organ pipe? (Note: Resonances occur at L = nλ/ 4, where L is the pipe length, λ is the wavelength, and n = 1, 3, 5,...) 3.2 m 4.8 m 6.0 m 8.0 m
Solution: The correct answer is C. The examinee is given two wavelengths (8 m and 4.8 m) that yield resonance in an organ pipe closed at one end. From this information the examinee must determine the length of the pipe. The question reminds the examinee that the resonant wavelengths λ are related to the length of this pipe via: L = nλ /4, where n = 1, 3, 5... Since n increases from n to n + 2 for two consecutive wavelengths one may write for the two given wavelengths, 4L = n(8 m) and 4L = (n + 2)(4.8 m). Equating the right sides of these two equations: n(8 m) = (n + 2)(4.8 m) which is easily solved to show that n = 3. Putting n = 3 into 4L = n(8 m) yields 4L = 24 m, so L = 6.0 m. Thus C is the best answer.
What physical quantity is represented by the area of quadrilateral OABC in Figure 2? Average speed of the box Average acceleration of the box Distance traveled by the box Work done on the box
Solution: The correct answer is C. The examinee must determine what the area of Figure 2 represents physically. Fig 2 is a plot of speed versus time. A clue as to what the area represents may be found in the units of the area. The area units would be the vertical axis units of speed (m/s) times the horizontal axis units of time (s) to yield distance units (m). This leads one to conclude that the total area yields the total distance traveled by the mass. Thus, the best answer is C.
The density of a typical laboratory plasma is 10^18 m-3. This value leads to plasma oscillations at: 9 × 10^18 Hz. 9 × 10^12 Hz. 9 × 10^9 Hz. 9 × 10^6 Hz.
Solution: The correct answer is C. The frequency is given by 9n^0.5. With n = 10^18, n^0.5 = 10^9. Thus, frequency is 9 × 10^9, which is answer C.
As the speed of the jet flying away from the receiver increases, what happens to the distance between adjacent peaks of the transmitted waves, as measured at the receiver? It decreases. It remains constant. It increases. It changes, but is not dependent on the speed.
Solution: The correct answer is C. The increase in distance between radio wave peaks due to the jet speed causes the received wavelength to be lengthened. This effect is the essence of the Doppler effect.
Destructive interference occurs in photodiode detectors when direct and scattered light rays take paths to the photocell that differ in phase by: A) 0 degrees. B) 90 degrees. C) 180 degrees. D) 360 degrees.
Solution: The correct answer is C. The phase difference corresponding to a half of a wave is 180°. Half a wave difference in phase between two waves corresponds to destructive interference. Thus, C is the best answer.
The table above gives pressure measured at various depths below the surface of a liquid in a container. A second liquid, whose density is twice that of the first liquid, is poured into a second container. Similar pressure measurements are taken for the second liquid at various depths below the surface of the second liquid. What is the pressure at a depth of 10 cm for the second liquid? 250 N/m2 450 N/m2 850 N/m2 1650 N/m2
Solution: The correct answer is C. The pressure in a liquid due to the gravitational force of the liquid above a given depth is proportional to the density and the depth. Although the pressures shown in the table change linearly with depth (an increase of 200 N/m2 per 5 cm of depth), the data also imply an extra pressure of 50 N/m2 at zero depth. For the same depth, the liquid with twice the density will create twice as much pressure as the first liquid, but the zero-depth pressure must be added to get the total pressure (2 × 400 +50) N/m2.
When a sound source moves away from an observer, the observer has the impression that the sound source is: rotating. louder than it actually is. lower in frequency than it actually is. higher in frequency than it actually is.
Solution: The correct answer is C. The question requires knowledge of the Doppler-shift phenomenon. A sound source moving away from an observer appears to have its source frequency shifted to a lower frequency when detected. Choice C is the correct answer. Perceived frequency increases moving towards observer and decreases moving away from observer
At the second position where the three curves intersect in Figure 1a, the curves are all: in phase. out of phase. at zero displacement. at maximum displacement.
Solution: The correct answer is C. The three curves in Figure 1a intersect at three points in time. The second intersection occurs in the middle of the time axis. At that point all three curves have zero displacement. Therefore, answer choice C is the best answer.
Which of the waveforms shown in Figure 1 has the shortest period? First harmonic Second harmonic Third harmonic The waveform in Figure 1c
Solution: The correct answer is C. The tone with the shortest period has the shortest wavelength. In Figure 1a, the period of the third harmonic (the curve with the smaller dashes) is seen to be shorter than the other two harmonics. Thus, answer choice C is the best answer.
When switch S is closed to the left, charge begins to accumulate on the capacitor. Charge cannot accumulate indefinitely because: a) the variable resistor inhibits the current flow. b) the battery continually loses charge. c) successive charges brought to the plates are repelled by charges accumulated earlier. d) the fixed resistor loses energy to heat.
Solution: The correct answer is C. The two plates of the capacitor collect charges of opposite sign. As more charge arrives it is harder and harder to fill the plates until finally an equilibrium occurs, thus C is correct.
What is the magnitude of the electric field between the two electrodes in ionization type detectors? A) 1.5 N/C B) 1.66 N/C C) 15 N/C D) 166 N/C
Solution: The correct answer is D. An electric field can be given in volts per meter. This means the electric field value is 5/(0.03) = 166.7 V/m. Because 1 V ≡ 1 J/C, the electric field value 166.7 V/m = 166.7 N/C. Thus, D is the best answer.
Advanced photodiode detectors have a second light-emitting diode, operating at a wavelength of 2.0 × 10-7 m, to detect even smaller smoke particles from smoldering flames. What is the frequency difference between the two light beams? A) 12.0 × 10^15 Hz B) 3.0 × 10^15 Hz C) 2.0 × 10^15 Hz D) 1.0 × 10^15 Hz
Solution: The correct answer is D. Light with a wavelength of 2.0 × 10-7 m has a frequency of f = c/λ = (3.0 ×10^8 m/s)/(2.0 × 10^-7 m) = 1.5 × 10^15 Hz and light with a wavelength of 6.0 × 10^-7 m has a frequency of f = c/λ = (3.0 × 10^8 m/s)/(6.0 × 10^-7 m) = 5.0 × 10^14 Hz. The frequency difference is 10.0 × 10^14 Hz, or 1.0 × 10^15 Hz. Thus, D is the best answer.
The researchers devised a second procedure to measure static friction. They removed the string from a block, placed the block at rest on a board, and raised one end of the board until the block began to slide. To determine the static friction force on the block when sliding began, which of the following measurements did they make? Time it took for the block to slide down the board Distance the block slid down the board before coming to rest Mass of the board Angle of the board with respect to the horizontal
Solution: The correct answer is D. A board tilted at an angle with respect to the horizontal causes a component of the gravitational force on the block to point down the length of the board. The coefficient of static friction is the ratio of the force along the board to the normal force (which also changes because of the angle). The mass of the block cancels in the ratio. Only the angle of tilt determines the static friction result.
If a fourth harmonic exists for the tone graphed in Figure 1, then, compared to the third harmonic, the fourth harmonic will have: lower amplitude. higher amplitude. lower frequency. higher frequency.
Solution: The correct answer is D. A fourth harmonic would have a shorter period than the other three. Since T = 1/f, the fourth harmonic would have a higher frequency than the third harmonic. Therefore, answer choice D is the best answer.
Which of the following describes the direction of the magnetic force on an ion moving in an artery past a flowmeter? Parallel to both the direction of v and the direction of B Parallel to the direction of v and perpendicular to the direction of B Perpendicular to the direction of v and parallel to the direction of B Perpendicular to both the direction of v and the direction of B
Solution: The correct answer is D. A magnetic force acts on a moving charge in a direction that is perpendicular to both the velocity of the charge and the direction of the magnetic field. This is a basic law of the interaction of electric currents and magnetic fields. Thus, answer choice D is the best answer.
A block of weight W is pulled across a rough floor by a rope that exerts a force T on the block. The frictional force between the floor and the block is F. Which of the following expressions equals the frictional force F when the block moves with a constant speed? T W - T T sin θ T cos θ
Solution: The correct answer is D. Because the object moves at a constant speed, the net force on the object is zero. In the horizontal direction the sum of the forces must be zero. This will occur if the horizontal component of T, T cos θ, has the same magnitude as the frictional force F. Thus, D is the best answer.
Making which of the following changes to a circuit element will increase the capacitance of the capacitor described in the passage? Replacing the 500-Ω resistor with a 250-Ω resistor Replacing the 10-V battery with a 20-V battery Increasing the separation of the capacitor plates Increasing the area of the capacitor plates
Solution: The correct answer is D. Capacitance C depends on geometric factors only, and in the case of parallel plates, C is proportional to the plate area and inversely proportional to the separation distance of the plates. Thus, D is the best answer.
The frequency of the light used in photodiode detectors is: A) 1.8 × 10^12 Hz. B) 5 × 10^12 Hz. C) 1.8 × 10^14 Hz. D) 5 × 10^14 Hz.
Solution: The correct answer is D. Frequency is the speed of light divided by wavelength. Hence, f = c/λ = (3.0 × 10^8 m/s)/(6.0 × 10^-7 m) = 5 × 10^14 Hz
In order to determine the relative speed of approach of a sound source by Doppler measurements, three of the following items of data are necessary. Which one is NOT required? The speed of sound in the medium The frequency of the emitted sound The frequency of the observed sound The distance between source and observer
Solution: The correct answer is D. Given that v is the speed of sound in the medium, the Doppler equation for a source that is approaching (receding from) an observer can be written as f' = (v / v +/- v(source) ) f
After a block began to slide, how did its speed vary with time? (Note: Assume that the tension and kinetic friction forces on the block were constant in magnitude.) It was constant in time. It increased exponentially with time. It was first constant, then increased linearly with time. It increased linearly with time.
Solution: The correct answer is D. The coefficient of kinetic friction is always lower than that of static friction. Therefore there is a net accelerating force on the block once it starts to slide. A constant force on a mass produces a constant acceleration (Newton's second law). Thus, the velocity of the block increases linearly with time.
The intensity level of Sound B is 20 dB greater than the intensity level of Sound A. How many times greater is the intensity level of Sound B than the intensity level of Sound A? 2 10 20 100
Solution: The correct answer is D. The examinee is asked to compare the intensities of two sounds, given a comparison of their intensity levels in decibels, dB. The intensity level β in dB is defined by: β = 10Log(I/Io), where I is the intensity of the sound and Io is the minimum intensity audible to the human ear. "Log" indicates the base 10 logarithm. Two sounds A and B would have intensity levels βA = 10Log(IA/I0) and βB = 10Log(IB/I0) which may be subtracted to yield: βB- βA = 10Log(IB/IA). The question states that βB is 20 dB larger than βA, hence βB- βA = 20 dB. Thus, 20 = 10Log(IB/IA) or Log(IB/IA) = 2. This yields IB/IA = 10^2 = 100. Thus D is the best answer.
Which of the following describes the direction of motion of alpha, beta, and gamma rays in the presence of an external magnetic field? a) They all travel straight. b) They are all bent in the same direction. c) Gamma rays travel straight; alpha and beta rays are bent in the same direction. d) Gamma rays travel straight; alpha and beta rays are bent in opposite directions.
Solution: The correct answer is D. The examinee must decide how alpha particles, beta particles, and gamma rays will be influenced when they move in a magnetic field. Magnetic fields exert a force on moving electric charge. The magnitude of the force is directly proportional to the electric charge and particles of opposite sign experience forces of opposite direction. Gamma rays possess zero charge, and thus experience no force from a magnetic field and will travel in a straight line. Alpha and beta particles possess charges of opposite sign, and thus will experience forces in opposite directions- so their trajectories will bend oppositely. Thus, D is the best answer.
Assume that the density of Ball 1 is 8.0 × 102 kg/m3. Ignoring the atmospheric pressure, what fraction of Ball 1 is above the surface of the water? a) 4/5 b) 3/4 c) 1/4 d) 1/5
Solution: The correct answer is D. The examinee must use the given density of water (ρ = 1000 kg/m3) and of ball 1 (D1 = 800 kg/m3) to determine the fraction of ball 1 above the surface of the water. As ball 1 floats, part of its volume is submerged. Call this volume VS. According to Archimedes' Principle (AP) the upward buoyant force, B, on the ball equals the weight of water displaced by the ball. The mass of displaced water would be ρVS, so the weight of displaced water would be ρVSg. Thus, according to AP we can write B = ρVSg. The ball floats, so its weight mg is balanced by the upward buoyant force, thus ρVSg = mg. The ball's mass is D1V, hence ρVSg = D1Vg which may be solved for the submerged volume: VS = (D1/ρ)V = (800/1000)V= 0.8V. This tells us that 0.8V is below the surface, thus V - 0.8V = 0.2V must be above the surface. The fraction of ball 1 above the surface is thus 2/10 = 1/5. Thus, D is the best answer.
The kinetic energy of a sliding block came from the: kinetic energy of the string. kinetic energy of the board. gravitational potential energy of the block. gravitational potential energy of the mass on the hook.
Solution: The correct answer is D. The law of conservation of mechanical energy describes the relation between the kinetic and potential energies of a system; an increase in one requires the decrease in the other. In the static friction apparatus, the increasing kinetic energy of the block sliding on the wooden board comes from the decreasing gravitational potential energy of the mass hanging on the hook as it descends toward the floor. The heat from friction also comes from the decreasing potential energy.
Increasing the frequency of each photon that is directed at the cathode will: decrease the number of photons ejected. increase the number of photons ejected. decrease the speed of the ejected electrons. increase the speed of the ejected electrons.
Solution: The correct answer is D. The only effect the photon frequency has on the ejected electron is on its kinetic energy. Photon energy equals cathode work function plus electron kinetic energy. The number of electrons ejected (the current) depends on the number of incident photons.
In Figure 1, the maximum electrical potential energy occurs at: A only. B only. C only. A and C only.
Solution: The correct answer is D. The potential energy of an oscillatory motion is ½kx2 where x is the displacement. Since the maximum displacement occurs at A and C, answer D is justified.
An astronomer observes a hydrogen line in the spectrum of a star. The wavelength of hydrogen in the laboratory is 6.563 x 10-7m, but the wavelength in the star's light is measured at 6.56186 x 10-7m. Which of the following explains this discrepancy? The star is moving away from Earth. The wavelength of light that the star is emitting changes constantly. The frequency of light that the star is emitting changes constantly. The star is approaching Earth.
Solution: The correct answer is D. The wavelength of the light detected from the star is smaller than the laboratory value on Earth. This implies a Doppler shift (a blue shift) associated with approaching relative velocity between the star and the Earth.
The resistance of the variable resistor, R, at the beginning of the discharge process is: 2000 Ω. 3000 Ω. 4000 Ω. 6000 Ω.
Solution: The correct answer is D. With 12 Volts initially across the capacitor during its discharge (the capacitor will charge to the battery voltage of 12V) and a current of 0.002 A as found in Figure 2, then the initial resistance R must have been R = V/I = 12/0.002 = 6000 W. This is answer D.
A force F is used to raise a 4-kg mass M from the ground to a height of 5 m. What is the work done by the force F? (Note: sin 60° = 0.87; cos 60° = 0.50. Ignore friction and the weights of the pulleys.) (in passage: force is at a 60 degree angle) 50 J 100 J 174 J 200 J
Solution: The correct answer is D. Work is the product of force and distance. The easiest way to calculate the work in this pulley problem is to multiply the net force on the weight mg by the distance it is raised: 4 kg × 10 m/s2 × 5 m = 200 J. Therefore, answer choice D is the best answer.
A ray of light in air is incident upon a glass plate at an angle of 45°. The angle of refraction of the ray in the glass is 30°. What is the index of refraction of the glass? Given: Sin45 = .707 Sin30 = .500 1.22 1.41 1.57 1.65
USE SNELL'S LAW The n's are the indices of refraction. n1sin(theta1) = n2sin(theta2) n2 = n1sin(theta1) / sin(theta2) Assume index of refraction is air is 1 (n1). n2 = 1sin(45) / sin(30) = .707 / .500 = 1.41 The correct answer is B.
How will W change if the initial speed of the box at Point A is increased by a factor of 2? W will decrease by 50%. W will not change. W will increase by 50%. W will increase by 100%.
Work is not related to speed? Solution: The correct answer is B. The examinee must determine the effect that doubling the initial speed of the mass at the top of the ramp has on the work done by friction, W, as the mass slides down the ramp. The definition of the work of a force is W = Fdcos θ, where F is the magnitude of the force, d is the distance traveled, and θ is the angle between the force and the displacement. Kinetic friction F acts up the ramp and d is down the ramp, so θ = 180°, hence W = -Fd. The kinetic friction force F is the product of the normal force between the ramp exerts on the mass and the coefficient of kinetic friction, and the normal force is independent of speed. Note that speed appears in none of these relationships. Doubling the speed changes none of these quantities. Therefore, doubling the speed has no effect on the total work done by friction on the ramp. Thus, B is the best answer.
