MCAT Physics Problems

Consider a positively-charged particle that is experiencing a force due to an external electric field. Which of the following are conserved for the particle?

Since the system is experiencing an outside force, the only quantity conserved is the total energy; total energy is conserved except in extremely rare cases like nuclear fusion reactions. The other quantities like potential energy, kinetic energy, and momentum are not conserved.

Monochromatic light with a wavelength of 640 nm is directed onto a screen with two slits. A lab assistant measures the distance from one slit to a bright band, then measures the distance from the other slit to the same bright region. Which of the following could be the calculated difference between these two values?

A = 1920 nm The area is a bright region on the optical screen which must be resulted from constructive interference; this tells that the two waves must be completely in phase or differ by exact *multiples* of a wavelength given as 640nm; thus, the answer is 1920nm.

An object weighs 150g in air, 75g when fully submerged in water and 60g when fully submerged in an unknown fluid. What can be concluded about the specific gravity of the unknown fluid and the order of layers when water is mixed with the unknown fluid (assuming the unknown fluid is immiscible in water)?

A. The specific gravity of the unknown fluid is 1.2, and water will be the top layer The buoyant force is equal to the weight of the displaced fluid. The buoyant mass (mass of displaced water) is 150g-75g = 75g. Since the specific density of water is 1g/cm^3, the volume of displaced water is 75cm^3. Since the object is fully submerged in the unknown fluid, Vfluid is also 75cm^3. The buoyant mass and therefore the mass of the unknown fluid is 150g-60g= 80g. Using the mass and the volume of the unknown fluid, specific gravity = 90g/75cm^3 = 1.2g/cm^3. The specific gravity of the unknown fluid is the ratio of the density of the unknown fluid to that of water: 1.2/1 = 1.2. Since the unknown fluid is denser than water, it will occupy the bottom layer.

Which of the following expressions gives the amount of energy lost to friction by the toboggan and rider between points A (top of ramp) to B (bottom)?

A: (60N)(l) Frictional forces dissipate an amount of energy given by the product of the force times the distance over which the force acts. - The coefficient of friction is defined as the frictional force divided by the normal force

A horizontal force of 100N is applied to a 50Kg box that is accelerating at 1m/s^2 on a rough, horizontal surface. What is the work done by kinetic friction if the box is moved 4m?

A: -200J Fnet = ma = Fapplied - Fk, where Fk is the force of kinetic friction. Using this formula, (50 kg)(1 m/s2) = 100 N - Fk 50 N = 100 N - Fk Fk = 50 N To calculate work, we use W = Fd cos θ. W = (50 N)(4 m)(cos 180º) = -200 Nm = -200 J

The function -cos(t) is how many degrees out of phase with sin(t)?

A: -270 degree The function cos(t) is 90 degree behind sin(t). -cos(t) is perfectly out of phase with cos(t); they are separated by 180 degrees. Adding these two phases lags yields a total of 270 degrees. The answer is negative since it's a phase lag (-cos(t) is staggered 270 degree behind sin(t)).

The moon's orbit around the Earth is approximately circular. How much work is done on the moon by the force of gravity when the moon completes half an orbit around the Earth?

A: 0 J For an object moving in a circle, the force is always towards the center of the circle but the velocity vector points at a tangent out from the circle, which makes the angle between the force and velocity vectors 90. The equation for work is W = Fd cos θ. The cosine of 90º is zero, so the work done on any object moving in a circle by the centripetal force is always zero.

What is the closest distance the electrodes used in an NCV test can be placed on a nerve in order to measure the voltage change as a response to the stimulus?

A: 0.1m The shortest time duration that can be detected is not less than the action potential time response of 1msec (according to the figure). The distance travelled by the impulse during this time is 1 msec * 100m/s = 0.1m, which is the closest distance the electrodes must be placed on the nerve

If the average pitcher is releasing the ball from a height of 1.8 m above the ground, and the pitcher's mound is 0.2 m higher than the rest of the baseball field, at what height would the catcher need to hold his glove to catch the pitched ball? (Note: neglect air resistance, estimate the acceleration due to gravity as 10 m/s2, and assume the pitcher is only throwing the ball horizontally.)

A: 0.2m above the ground The passage tells us that pitchers throw the ball at an average of 30m/s and the ball travels 18m horizontally. Thus, the ball's flight time is: 18m/30m = 0.6s. The ball is released from a position 2m off the group (0.2m from the pitcher's mound and 1.8m from the pitcher). To calculate the distance the ball falls during 0.6s, d=v0t + 1/2at^2 = 1.8 The ball has fallen 1.8m from an initial height of 2m. Thus, the catcher must hold his glove 0.2m above the group to catch the pitch

Given a 5.0-ampere current running for a total of four hours and ten minutes, what will be the amount of hydrogen gas produced in an electrolysis reaction? (Faraday's constant = 96,485 C per mole of electrons)

A: 0.39 moles We know that the current is 5.0A (or 5 coulombs/s). To find coulombs, we simply need to multiply by seconds. (5 C/s)(15,000 s) = 75,000 C of charge transmitted (75,000 C) / (96,485 C/mol) ~ 0.8 moles of electrons transferred. However, according to the reduction half-reaction, only one molecule of H2 is created for every two electrons. Thus, the amount of hydrogen gas produced must be ~ 0.4 moles.

If the average bond mineral density is 3.88g/cm^3, which of the following is a reasonable estimation of the specific gravity of fats in the body?

A: 0.8 Specific gravity is the ratio of a material's density to the density of water. Since fats are less dense than bone, we can conclude that the specific gravity of fats is less than 3.88.

The stress and strain values for four elastic solid materials were used to obtain the following graphs. Which material would have the greatest resistance to linear deformation (strain)?

A: 1 (material with the highest stress/strain slope) In solid-state physics, Young's modulus is a measure of the stiffness of an elastic solid material. The slope of the graph (stress/strain) is equal to the Young's modulus. The higher the Young's modulus, the stiffer the solid material is. Graph '1' has the highest slope and therefore is the stiffest.

A hypothetical neuron possesses a high number of voltage-gated potassium channels. Imagine that one of these ion channels is open for 10 ms, during which it lets 1mmol of K+ ions out of the cell. what amount of current passes through the channel during this interval?

A: 1*10^4 A 1 mmol is approximately equal to 6 × 1020 molecules; this value represents the number of charged particles that flow outward during the specified time period (1 × 10-2 s). However, our answers are in amps or milliamps, which have seconds (s) as the denominator. We thus need to calculate the amount of charge that crosses the membrane per second: (6 × 1020 molecules) / (1 × 10-2 s) = 6 × 1022 ions crossing the membrane per second To translate this to the correct answer, simply multiply by the amount of charge held by each ion. (6 × 1022 ions/s)(1.6 × 10-19 C/ion) = about 10 × 103 C/s. Since 1 C/s = 1 ampere, the current must be 10,000 A.

In the passage, the researchers attempted to model chemical bonds and intermolecular forces as springs. Suppose the chemical potential energy in a single disulfide bond was modeled as elastic potential energy stored in a stretched 10 Å spring, with a spring constant of 2 N/m. What would the length of the stretched spring be? (A disulfide bond has a bond dissociation enthalpy of 54 kJ/mol.)

A: 1.3 x 10-9 m We first figure out how much chemical potential energy is contained in a single disulfide bond. To do this, we divide 54kJ/mol by Avogadro's number. 54 kJ/mol / 6 × 1023 bonds/mol = 9 × 10-23 kJ/bond = 9 × 10-20J/bond Recall that elastic potential energy has: PE (spring) = ½ kx2. By rearranging and substitution, we get x = 3*10^-10m. The passage tells us that 1 Å = 10-10 m, so our spring has a length of 10-9 m. Summing that value and the displacement of the spring gives us a grand total of 1.3 x 10-9 m for the length of the stretched spring.

An object that is totally immersed in benzene (specific gravity = 0.7) is subjected to a buoyant force of 5N. When the same object is totally immersed in an unknown liquid, the buoyancy force is 12N. What's the approximate specific gravity of the unknown liquid?

A: 1.8 The buoyant force on an immersed object is the product of: (density of the liquid)*(volume of object)*(acceleration of gravity). Thus, 12/5=(density of unknown molecule)/(density of benzene)

If the energy of a photon is 2*10^-20J and planck's constant is 6.62*10^-34 m^2kg/s, the photon's wavelength is

A: 10*10^-6m We know that E = hf and c=hf where E = hc/λ- answer obtained by plugging the values.

Assuming that the side of the water tank is punctured 5.0m below the top of the water, and that atmospheric pressure is 1*10^5 N/m^2. What is the approximate speed of the water flowing from the hole?

A: 10m/s We solve this using Bernoulli's equation. P2+ 1/2ρv2^2 + ρgy2 = P1 + 1/2ρv1^2 + 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. The speed of water exiting the small puncture will be much larger than the speed of the upper water level falling at 1; therefore, v2>>v1 and we can ignore v1 by v1=0. Thus, equation simplifies and we solve v2=[2g(y1-y2)]^1/2.

A sound wave with a frequency of 350 Hz is emitted from a speaker placed at x = 0 m. Another speaker, emitting a noise with the same frequency and decibel level, is positioned at x = d m. What must the value of d be, if an observer standing at x > d hears a sound that has twice the amplitude of either individual sound? Assume that sound travels at 300 m/s.

A: 12/7 m These waves display *maximum constructive interference*; they are either 0 or 360 degrees out of phase. Also, the period of each wave is 1/350s from given information. Assuming that the speaker are speakers are spaced a full wavelength apart; we determine how far this is in space by multiplying (the period)*(velocity of wave)=(1/350s)(300m/s) = 6/7. This is the distance representing a 360 degree phase gap. In actuality, the speakers separated by ANY MULTIPLE of this distance.

If a 3kg rabbit's leg muscles act as imperfectly elastic springs, how much energy will they hold if the rabbit lands from a height of 0.5m and its legs are compressed by 0.2m?

A: 13J Although imperfectly elastic, this doesn't mean all the energy is lost. If the energy were completely conserved, potential energy would be PE = (3)(10)(0.5) =15J. The answer has to be a bit less than this because some energy is lost

If the majority of the baseball's kinetic energy comes from power generation in the legs and hips, approximately how much energy do the lower extremities produce in the pitch?

A: 140J The KE of the baseball can be calculated using the equation 1/2mv^2. In the given information, the ball has 67.5 J of KE. But the passage asks for how much energy the lower extremities generates. Using the overall efficiency of each energy transfer shown in the table, the overall efficiency of the kinetic chain can be calculated by multiplying efficiencies of each step: Efficiency ~ 0.5. Thus, the total energy generated in the lower extremities to ensure that 67.5J makes it to the ball is 67.5J/0.5 = 135J.

What is the kinetic energy of a photoelectron produced in the energy meter of the PAC device when the frequency of an incident photon is not absorbed in the solution is f=5*10^5 Hz (use h=4.1*10^-15eV*s)

A: 17.1eV (Picked 20.5eV) The kinetic energy of a photoelectric is equal to hf-3.4eV (=work function) = 20.5eV - 3.4 eV = 17.1eV

Destructive interference occurs in photodiode detectors when direct and scattered light rays take paths to the photocell that differ in phase by

A: 180 degrees The phase difference corresponding to a half of a wave is 180 degrees. Half a wave difference in phase between two waves corresponds to destructive interference

What is the kinetic energy of 60cm^3 of pure water if its flowing at a velocity of 12m/s?

A: 4.3 J Recall that the kinetic energy for fluid is ½ ρv2 * volume. Make sure to memorize the density of pure water, which is 1000 kg/m3 (or 1 g/cm3, or 1 kg/L; you should know this value in multiple different units). Finally, ensure that your units align by converting volume to cubic meters. 1 cm3 = 1 × 10-6 m3, so 60 cm3 = 60 × 10-6, or 6 × 10-5, m3. [½ (1000 kg/m3)(12 m/s)2](6 × 10-5 m3) = (7.2 × 104)(6 × 10-5), or 4.32 J. *Unit*

A woman is standing two meters away from a speaker. The speaker is turned up to 11 on its dial, and the power of the sound hitting her tympanic membrane increases 100-fold. How far would the woman need to move to reduce the sound back to its original decibel level?

A: 18m The decibel level of sound is given by dB = 10log (I/Io), where I denotes intensity and Io is the reference intensity, which represents the lowest intensity audible to the human ear. In order to have the same decibel level, the same intensity, I, must reach the tympanic membrane. The intensity of sound is given by the equation I = P/A, where P = power and A = area. If the power increases 100-fold, then the area would need to increase by a factor of 100 to maintain the same intensity. The area, A, of the sound wave is given by the surface area of a sphere as the sound travels away from the speaker. The surface area of a sphere is proportional to the radius squared, or r2 (A = 4πr2). So, in order to increase the area 100-fold, the radius, r, must increase by a factor of 10 (102 = 100). Thus, if the woman originally stands two meters away then to maintain the same decibel level, she needs to stand 2 x 10 = 20 m away from the speaker. Standing 2 m away, the woman must move 18 m to reach a total final distance of 20 m

A 0.5-kg uniform meter stick is suspended by a single string at the 30cm mark. A 0.2kg mass hangs at the 80cm mark. What mass hung at the 10cm mark will produce equilibrium?

A: 1kg The examinee is given: (1) it's suspended by a single string at the 30cm mark, (2) a 0.2kg mass hangs from it at the 80cm mark (d=50cm from the suspension point), and (3) to achieve equilibrium an unknown mass m is hung at the 10cm mark (d=20cm from the suspension). The examinee seeks mass m. At equilibrium the net torque left of the suspension must equal the net torque to the right of it. The torque of force mg applied at a distance d from the suspension is mgdsinθ. At equilibrium, θ = 90. Thus, the torque left of the suspension is mg(20cm). There are TWO torques right of the suspension, the hanging mass torque (0.2kg)g(50cm) and the weight of the meterstick's torque (0.5kg)g(20cm) where 20cm represents the distance from the suspension to the center of mass of the meter stick. Thus, the net right torque is (0.2kg)g(50cm)+(0.5kg)g(20cm). Thus, m=1

What is the mechanical advantage of using the ramp as opposed to lifting the gurney straight up (no friction involved)?

A: 2 For an inclined plane, this advantage is represented using the formula: MA = hypotenuse / height From trigonometry, we know this as sin-1. We're told the ramp makes a 30º angle with the ground. This means it is a 30º-60º-90º right triangle, with side ratio of x : x√3 : 2x. Thus: MA = 2x / x = 2 That is, by putting in some measure of force (e.g. 1000 N) you're able to lift an object that would otherwise take twice as much force (e.g. 2000 N) to lift straight up.

Ball 2 is in the water 20cm above Ball 3. What is the approximate difference in pressure between the 2 balls?

A: 2*10^3 N/m^2 The absolute pressure p at depth h below the surface of a fluid is p=p0+ρgh. Since h is 0.2m, we get 1960Pa by plugging in.

The masses of a proton and a neutron are 1.0072 amu and 1.0087 amu, respectively. If the measured mass of a deuterium nucleus (2H) is 2.0140 amu, how much energy must be released in the synthesis of one such nucleus? (1 amu = 1.66 × 10-27 kg)

A: 2.8*10^-13 J First, calculate the mass defect. A deuterium nucleus has one proton and one neutron, meaning that the total predicted mass of the nucleus is 2.0159 amu. Subtracting the actual mass of deuterium yields a mass defect of roughly 0.002 amu. The next step is to convert the mass to kilograms, the units required for the equation E = mc2. 0.002 amu × (1.66 × 10-27 kg/amu) is equal to about 3.3 × 10-30 kg. Finally, plug in to E = mc2. E = (3.3 × 10-30) (3 × 108)2 = (3.3 × 10-30) (9 × 1016) = 30 × 10-14, or 3 × 10-13, J, which is closest to choice B.

An engine is supplied with 1 kg of a hydrocarbon fuel with a heat of combustion of 50 MJ/kg. The engine can consume exactly 1 kg of fuel per minute. If this engine can produce 150 kW of power for one minute, the efficiency of the engine is:

A: 20% The basic formula is: efficiency = (energy out)/(energy in). First, then, we must find the energy output, or the amount of work that was actually done by the engine. To do so, multiply the power (1.5 × 10^5 J/s) by the time interval given (60 s), yielding 9 × 10^6 J or 9 MJ. Since the engine used 1 kg of combustion material, the energy input is exactly 50 MJ. Dividing by this value yields 9 / 50, or 18%.

A force F is used to raise a 4kg mass M from the ground to a height of 5m. What is the work done by the force F? (in pulley)

A: 200J Work is the product of force and distance. The easiest way to calculate the work in this pulley is to multiply the net force on the weight mg by the distance raised: 4kg*10m/s^2*5m = 200J

**A prosthetics researcher is conducting tests on a 40-cm-long prosthetic forearm, which is positioned according to the simplified diagram below. He exerts an upward force of 5 N on the forearm 30 cm to the right of the point of rotation (the elbow joint). If the forearm has a weight of 15 N that is uniformly distributed, which of the forces listed below will allow the system to exist in rotational equilibrium?

A: 20N upward force exerted 7.5cm to the right of the point of rotation When dealing with multiple torques, separate them based on whether they rotate the system clockwise or counterclockwise. To exist in rotational equilibrium, the system must have no net torque (the counterclockwise = clockwise torques). The only counterclockwise torque shown is due to the upward 5-N force. Here, then, torque_counterclockwise = (5 N)(0.3 m)(sin 90°) = (1.5 Nm)(1) = 1.5 Nm. There's a clockwise torque from the prosthetic forearm- its weight will pull it downward! Since the forearm has a *uniformly-distributed mass*, we position this torque at *its center of mass*. Thus, Torque_clockwise = (15 N)(0.2 m)(sin 90°) = (3 Nm)(1) = 3 Nm. So we see that we have a net torque of 1.5N clockwise; therefore, we need an upward torque of 1.5N, which can be accomplished by the answer.

While playing, a 40-kg child is accidentally pushed directly forward off a ledge, causing him to fall and hit the ground at a 60° angle. If the child has 720 J of kinetic energy at the instant before impact, at what velocity was he pushed? Assume negligible air resistance.

A: 3m/s Recall KE = 1/2mv^2; we use this to calculate the child's total velocity immediately before impact. 720J = 1/2(40kg)(v^2) where v= 6 m/s. This is the child's final velocity that includes both a horizontal and a vertical component. Since air resistance is negligible and gravity acts only in the vertical direction, the child's final horizontal velocity is the same as his horizontal velocity at the moment of the push. V_horizontal = v_total * cos(60) = v_total * 0.5 = 0.5 (6m/s) = 3 m/s

Participant 1 rides a fourth roller coaster as shown below. What is the minimum ramp height H if the ride at the top of the loop maintains the minimum speed needed to stay on the track throughout the loop?

A: 20m At the top of the loop, the gravitational and the normal forces (if any) point downward toward the center of the loop. Therefore, the net force causing centripetal acceleration is the sum of the gravitational force and the normal force. When the centripetal force is the minimum amount needed for the ride to stay on track, the normal force is zero. Note: At the top, the normal force can be zero but the gravitational force cannot. Let's designate the initial height where the ride starts as H. At the very beginning, the energy of the ride is the gravitational potential energy. Therefore, Ei = mgH. At the top of the loop, the ride includes both gravitational and potential energy. Therefore, Ef = mgH + ½ mv2 = mg (2r) + ½ mv 2. The net acceleration at top (centripetal acceleration) = v2/r. Since we concluded that the normal force at the top is zero, the net acceleration at the top is the gravitational acceleration, g. Therefore, v2 = gr. Substituting this equation into the 'Ef' equation, we obtain the following: Ef = 2mgr + ½ mgr = (5/2) mgr To obey conservation of energy, Ei = Ef. Therefore: mgh = (5/2) mgr. We conclude that H is 5r/2. Since r is 8 m, H is 5(8 m)/2 = 20 m.

A large steel water storage tank with a diameter of 20 m is filled with water and is open to the atmosphere (1 atm = 101 kPa) at the top of the tank. If a small hole rusts through the side of the tank, 5.0 m below the surface of the water and 20.0 m above the ground, assuming wind resistance and friction between the water and steel are not significant factors, how far from the base of the tank will the water hit the ground?

A: 20m We first need to determine the velocity of the water that comes out of the hole, using Bernoulli's equation. P1 + ρgy1 + 1/2 ρv12 = P2 + ρgy2 + 1/2 ρv22 The atmospheric pressure exerted on the surface of the water at the top of the tank and at the hole are essentially the same. Additionally, since the opening at the top of the tank is so large compared to the hole on the side, the velocity of water at the top of the tank will be essentially zero. We can also set y1 as zero, simplifying the equation to: 0 = ρgy2 + 1/2 ρv22 -gy2 = 1/2 v22 Inserting the value for gravitational acceleration (g = -10 m/s2), and the level of the hole below the surface (y2 = 5.0 m) into the equation, we get: (10 m/s2)(5.0 m) = 1/2 v2 50 = 0.5 v2 102 = v2 10 m/s = v This velocity will be in the horizontal direction, and we will assume that the water acts like a projectile. Now, we need to determine the time that it will take the water to fall to the ground. We can assume that the initial vertical velocity of the water is zero and the displacement of the water is -20 m, because the hole is 20.0 m above the ground. The kinematic equation is: d = vit + 1/2 gt2 -20 = 0 + 1/2 (-10) t2 -20 = -5 t2 4 = t2 2 s = t Finally, we can determine the range, or displacement in the horizontal direction. Since the acceleration in the horizontal direction is zero, the equation becomes d = vit d = (10 m/s)(2 s) d = 20 m

A compound microscope contains an objective lens with a focal length of 4 mm. If the objective produces an image located 5 mm behind the lens, then what is the sample's distance from the objective?

A: 20mm The passage asks us to find the object distance produced by the objective lens in a microscope where the focal length, f, is 4mm and the image distance, i, is 5mm. Solving the thin lens equation for 1/o, we have: 1/o = 1/f - 1/i = (1/4mm) - (1/5mm) = (1/20mm); therefore, o = 20mm

A weight is suspended from the ceiling by two massless ropes. If the force exerted by tension on Rope 1 is 15 N (angle 30 degree), what force is felt by Rope 2 (angle 60)?

A: 25.5 N Since the object is not moving (static equilibrium), the horizontal components of the tension force must be equal and opposite. Rope 1 has a tension of 15N, giving (15N)(cos30)=13N. Thus, 13N=(F2)(cos6); F2 is 25.5N

An artificial leg designed for use by runners is spring-based, to mimic the compression required of a muscle during hard running. For safety reasons, it was determined that the leg should be able to absorb as much as 125 J of kinetic energy without compressing more than 10 cm, or the runner would be likely to stumble. What should the spring constant be?

A: 25000N/m When the leg "absorbs" kinetic energy, it's converted to elastic potential energy. While this process involves the loss of some energy as heat, we assume that it's a perfectly elastic process. Thus, the leg needs to hold ~ 125J of elastic PE. PE = (1/2) kx2. k = 2 KE/x2 k = (2)(125 J) / (0.10 m)2 k = (250 J) / (.01 m2) k = 25,000 N/m

A charged particle is accelerated from rest to 100m/s by an electric field. This particle has a charge-to-mass ratio of 0.17 C/kg. If this acceleration occurs in 20µs, how strong must the electric field be?

A: 3*10^7 N/C For, set F=ma equal to F=qE where E = ma/q. We can calculate acceleration by dividing the change in velocity by the time, which equals a = 5*10^6 m/s^2. m/q is the reciprocal of the charge to ratio (1/0.17) thus is 6! Multiplying it gives 3*10^7

Both the vertical displacement of her steps and the angle of her body from the vertical increase the energy required to realign the ground reaction forces. Recall that tanθ = sinθ/cosθ, so when θ is close to zero, so is tanθ. When θ is close to 90°, tanθ becomes arbitrarily large.

A: 37W The gravitational potential energy at the runner's height is: PE = (60 kg)(10 m/s2)(0.5 m) = 300 J Most of this energy is conserved as the runner hits the ground and his muscles capture the energy as spring potential energy, so the question only asks about the lost energy, amounting to 10%, or 30 J. Don't worry about the mechanics of which foot lands first and how much of the kinetic energy each one absorbs. The question stem doesn't give that kind of information, so there's nothing for it anyway. So, an additional 30 J is needed per stride, and a stride occurs every 0.8 s. Thus: P = (30 J)/(0.8 s) = 40 W

A slingshot propels a rubber ball horizontally from a 30 m platform at 15 m/s. Approximately how far from the base of the platform will the ball land?

A: 38m This can be solved by knowing that range = (initial horizontal velocity) * (time). First, use vertical components and the equation Δx = v0t + ½at2 to find time. 30 m = ½(10 m/s2)(t2) means that t2 = 6, so t ≈ 2.5 Then, use horizontal components to find range, which is equal to the product of initial horizontal velocity and time. Range = (15 m/s)(2.5 s) = 37.5 m.

What is the wavelength of the photons emitted by the 145Pm-m? (Note: 1 eV = 1.6 x 10-19 J)

A: 4.1 * 10^-12 m. The energy of the photon is given as 300Kev, which can be converted to joules using 1 eV = 1.6 * 10^-19J. The energy of a photon is related to E = hc/λ. We solve for the wavelength by plugging in.

An object is placed 40 cm to the left of a converging lens with a focal length of 20 cm. Where will the image appear?

A: 40cm to the right on the lens This question describes a converging lens and an object positioned more than one focal length away. In such a situation, a real and inverted image will form. Real images appear on the opposite side of the lens from the object, meaning that this image must fall on the right, not the left. To find an exact answer, use the formula 1/f = 1/di + 1/do. Plugging in 20 cm for f and 40 cm for do gives us a di value of 40 cm.

A number of astronauts want to use the swinging of a pendulum on a rover to measure the gravitational force on another planet. If a 1m pendulum with 0.1 kg mass at the end oscillates with a period 0.31s, what's acceleration due to gravity at the planet's surface in terms of earth's gravity?

A: 40g For a pendulum, period, length and gravitational acceleration all related by T = 2π√(L/g). By plugging in 0.1pi for period and 1m for length gives us 0.1 = 2π√(L/g). We solve that x = 400m/s^2, which is ~40times more than the approximate acceleration on earth

A small metal racecar (pictured as the square below) is positioned at the top of a curved ramp and released. The mass of the car (m) is 2 kg, the length of the track (L) is 15 m, and the track's height (h) is 3 m, while µk = 0.1. With what velocity will the racecar leave the track? Assume that the car's wheels slide instead of rotating.

A: 5.5 m/s The kinetic energy at the end of the track is equal to the initial potential energy (mgh) minus the energy lost to friction. This dissipated energy is the same as the work performed by the frictional force; thus, it can be written as µkmgL. In other words, ½ mv2 = mgh - µkmgL. Inserting given values shows us that ½ mv2 = (2 kg)(10 m/s2)(3 m) - (0.1)(2 kg)(10 m/s2)(15), so KE = 30 J. The final step is to recognize that velocity is thus equal to √30, which is closest to 5.5 m/s.

A foam life raft has a density of 491 kg/m^3. When placed in a pool of pure water, what percentage of the raft's volume will be above the surface of the liquid?

A: 50.9% We use its specific gravity to find the fraction of the raft that is underwater. Because the pure water has density of 1000kg/m^3, we know that raft's specific gravity is 0.491 (49.1% of the object will fall beneath the surface). Since the question asks for the amount that sticks above the water, 100%-49.1% = 50.9% If the problem doesn't tell use the liquid is water, specific gravity over 1 doesn't validate that it will reach the bottom of the tank.

If a pair of gamma rays emitted from the same annihilation event collide with sensors 0.33 nanoseconds apart, what is the minimum distance the annihilation could have occurred from the center of the machine?

A: 5cm Gamma rays travel at the speed of light, so both of the rays involved here must move at 3 x 108 m/s. Since the two rays arrived 0.33 nanoseconds apart, one ray must have traveled 10 cm farther than the other: (0.33 x 10-9 s)(3 x 108 m/s) = 1 x 10-1m = 10 cm. According to the passage, the apparatus in question is circular, with the sensors positioned along the outer perimeter of the circle. This may be difficult to visualize, so when in doubt, draw it out! Suppose the radius of the machine is 2 m, as shown below. If the annihilation event occurs 5 cm from the center, and if the axis of gamma ray emission goes directly through the center, then one gamma ray will travel 95 cm to reach the sensors on the perimeter of the circle. The other gamma ray, which is propelled in the opposite direction, will have 105 cm to travel to reach the sensors on the other side. This represents a difference of 10 cm, which would explain the 0.33 ns time disparity outlined in the question. It turns out that this is the closest you can get to the center and still have one ray travel 10 cm farther.

The distance between the ears for an average adult male is 20 cm. If a wave with a frequency of 17,000 Hz hits both the right and left ear and has a phase difference of 5π radians between the two, what is the difference in distance between the origin of the sound and each ear (velocitysound in air = 343 m/s)?

A: 5cm If the phase difference between the sound wave hitting the right and the left ear is 5π, then the difference in distance from the origin of the wave to the right and the left ear is 2½ wavelengths, as (5п radians) / (2π radians / λ) = 2 ½ λ. Thus, the wavelength is needed in order to determine the difference in distance. The wavelength can be found by using the frequency and velocity in the following equation: v = λf. Rearranging for wavelength gives: λ = v/f. Plugging in velocity and frequency gives: λ = 343 m/s / 17000 Hz. In order to estimate, round 343 to 340 and use scientific notation: λ = 3.40 x 102 / 1.7 x 104 = 2 x 10-2 m = 2 cm Thus, the difference in distance = 2.5λ = (2.5)(2 cm) = 5 cm.

Compared to the moon, Earth is about 100 times more massive and has a radius about four times larger. By what factor is the gravity on Earth greater than that experienced on the surface of the moon?

A: 6 Fgrav = GMm/r2. Since Earth has a mass that is 100 times that of the moon and a radius that is four times the moon's radius, it will have an Fgrav that is (100)/(42) = (100/16) times greater. This fraction is closest to 6

Radiation oncologists use machines called gamma knives to perform radiosurgery. Lead vests are used to protect non-cancerous areas and medical personnel from radiation exposure. A typical gamma knife uses a photon beam with a frequency of 1.6 x 1019 Hz. Suppose a stray photon hits a lead atom and ejects an electron and a lower-frequency photon of 1.599 x 1019 Hz. What is the maximum kinetic energy of the ejected electron? (Lead has a work function of 4.14 eV, Planck's constant is 6.626 x 10-34 J∙s, and 1 eV = 1.6 x 10-19 J.)

A: 6.0 x 10-18 J The photons used for radiation therapy are so high-energy that they cannot be absorbed by an atom all at once. Instead, they display a scattering behavior where a collision produces a second, lower-energy photon that moves off at an angle. Because of the law of conservation of energy, the energy of the initial photos must equal the sum of the energies of the emitted photon and the electron. We use E=hf to find the energies required to solve. First, we must find the frequency the electron, which is equal to the difference between the two photon frequencies. 1.6 x 1019 Hz - 1.599 x 1019 Hz = 1016 Hz E = hf = (6.6 x 10-34 J∙s)(10 16 Hz) = 6.6 x 10-18 J This is NOT equal to the kinetic energy of the electron as some energy is required simply to liberate the electron from lead's poisonous grasp. The work function represents this amount of energy. We must first convert lead's work function to joules then subtract the work function from our calculated energy to get the kinetic energy: Wpb = (4eV)(1.6*10^-19J) = 6.4 * 10^-19 J KE = Etotal - Wpb = 6.6 x 10-18 J - 6.4 x 10-19 J = 6 x 10-18 J

The energy required to move one mole of electrons through a potential difference of 1V is closest to

A: 6.022*10^23 eV The definition of eV: the amount of energy needed to move *one electron* (or any particle with an elementary charge) through *one-volt potential*. Thus, moving an entire mole of electrons in a similar manner must require one mole of electron-volts.

When a downward force is applied at a point 0.6m to the left of a fulcrum, equilibrium is obtained by placing a mass of 10^-7kg at a point 0.4m to the right of the fulcrum. What is the magnitude of the downward force?

A: 6.5*10^-7N 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. At equilibrium the left and right forces point vertically downward with the beam horizontal, hence θ=90 degree. At equilibrium, the torque magnitudes are equal so: Fleft*Lleft = Fright*Lright. The force at Lright = 0.4m and weight is mg = (10^-7kg)(9.8m/s^2)=9.8*10^-7N. The unknown force is at Lleft=0.6m thus, we find the answer by plugging in

A small, flat plane cut entirely from diamond is submerged in a tank of freshwater. A ray of visible light travels through this water (n=1.34) to contact the water-diamond interface (n=2.4). If the incoming angle is 21 degree, what's the angle of reflection with respect to the normal (assuming no total inflection)?

A: 69 degree For a ray contacting a flat surface, the angle of incidence equals the angle of reflection. But since this question asks for the angle with respect to normal, 90-21 = 69 degree

Consecutive resonances occur at wavelengths of 0.8m and 4.8m 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..)

A: 6m Since n increases from n to n+2 for two consecutive wavelengths one may write 4L=n(8m) and 4L=(n+2)(4.8m). Thus, putting b=3, L = 6m

Experiment 1 is repeated and it is observed that when the drivers applies the brakes, the air-filled balloon moves until the string forms a 30° angle with the vertical. What is the acceleration of the car while braking?

A: 6m/s^2 First, we isolate the forces acting on the balloon. These forces are gravity and the tension in the thread, which has vertical and horizontal components. Since the string pulls upward on the balloon, the vertical component of the tension will offset the force of gravity; the horizontal component of the tension comes from the deceleration. We see that the forces form a triangle, with the string as the hypothenuse and the components of tension as the legs. We use the tangent of the 30 angle, as it is equal to the ratio of the horizontal component of the force (ma_h) decided by the vertical component (mg). The mass of the ballon cancels, giving tan30=sin30/cos30. 10/√3 = 5.8 m/s!

Suppose that a moderately hard surface such as the hard-packed earth described in the passage is found to decelerate a person making a feet-first landing to a stop in an average of 0.09 seconds. What is the person's deceleration in terms of g (where 1 g = 9.81 m/s2) when falling from a height of two meters?

A: 7.10g A person is falling from 2m height then hitting the ground; from the time of impact (where he hits at some unknown velocity) to the end of the collision (where v=0m/s), it takes 0.09s. To find the magnitude of this acceleration, we need to know the velocity at impact. For this, we use vf^2 = vi^2 + 2ad. Thus, vf^2 = 0 + 2(10)(2m) = 40m^2/s^2. Thus, vf ~ 6.4. Now, acceleration is calculated by (change in velocity)/(time) = (6.4m/s)/(0.09s) = 64m/s^2. Because out answer must be written in terms of g, estimating g as 10m/s^2, 64m/s^2 equals to 7.1g

Given the data in the passage, if a hospital requires at least 250g of 145Nd to complete a standard cardiac image, how long will it be before it must replace a 2000g sample?

A: 75 minutes The decay of a substance calculated by knowing its half-life. The passage states that ~93% of the decay occurs in the first 100 minutes, indicating that about 7% of the samples are left (close to 4 half-lives) because 0.5^4 = 0.0625 = 6.25%. Therefore, one half-life is 25 minutes. The hospital sample is 2000g and must be replaced once it reaches 250/2000; 1/8 of its original radioactivity. Using the shortcut of the amount remaining (1/2)^n where n is the number of half-lives, we calculate 1/8 = (1/2)^3; thus, 3 half-lives pass before the sample drops to 250g. Given that each half life is 25 minutes, it takes 75 minutes

A positively-charged particle with a mass of 1.67*10^-27 and a charge 1.6*10^-19 is traveling at 1*10^4m/s in 0.5T magnetic field. What magnitude of magnetic force will this particle experience?

A: 8*10^-16N To solve this, consider F=qvB = (1.6 × 10-19 C)(1 × 104 m/s)(0.50 T), or 0.8 × 10-15 N. This value is equivalent to 8 × 10-16 N.

A cylinder fitted with a piston exists in a high-pressure chamber (3 atm) with an initial volume of 1 L. If a sufficient quantity of a hydrocarbon material is combusted inside the cylinder to produce 1 kJ of energy, and if the volume of the chamber then increases to 1.5 L, what percent of the fuel's energy was lost to friction and heat?

A: 85% The energy used for expansion must come from the energy of combustion minus the amount lost to friction. Since W=PV =(3atm)(0.5L), it's 1.5L*atm, which can be converted to ~150J. As the fuel is theoretically capable of producing 1000J energy, the amount lost to the environment must be 1-(150/1000) ~ 85%

Depth (cm) & Pressure (N/m^2) : 5 & 250 10 & 450 15 & 650 The table above gives pressure measured at many depths below the surface of a liquid in a container. A second liquid (whose density is twice that of the first one) is pouted into a second container. Similar pressure measurements are taken for the second liquid at many depths below the surface of the second liquid. What is the pressure at a depth of 10cm for the second liquid?

A: 850 N/m^2 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 200N/m^2 per 5cm of depth), the data imply an extra pressure of 50N/m^2 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.

When an infrared ray transitions from air (n = 1) to ethanol (n = 1.36), which wave characteristic is LEAST likely to change?

A: Frequency (Wavelength and velocity change!) Velocity of light changes when moving between media of different refractive indices; however, frequency always remains constant (what changes in E=hf is the wavelength)!

If the transducer measures a wavelength of 616 μm and a time to return of 104 μs for the waves hitting the uterus, approximately how far from the skin is the uterus?

A: 8cm For this question, we need to calculate a distance from the wavelength and timespan presented. The passage tells us that the frequency of the ultrasound is 2.5 MHz = 2.5 x 106 Hz. We are told by the question stem that wavelength (λ) = 616 μm, or approximately 6 x 10-4 m. Using the equation for wave velocity, v = λf, we can calculate: v = (2.5 x 106 Hz)(6 x 10-4 m) = 15 x 102 m/s Then, using distance = v x t allows us to calculate: d = (15 x 102 m/s)(104 x 10-6 sec) = 1560 x 10-4 m = 0.15 m = 15 cm. Before we grab choice D, recall that D is the distance to the uterus and back. Dividing by two gives us the distance from the skin to the uterus. 15/2 = 7.5 cm, or about 8 cm.

If the red LED used in this experiment was designed to operate at 30 mW and there was a voltage drop of 3.2 V when the LED was lit, what was the current flowing in the circuit?

A: 9.4 * 10^-3 amps The electrical power P is given by P = IV. 30mW = 3.0*10^-2 W. Thus, I = P/V = (3.0*10^-2)/(3.2V) = 1*10^-2.

A billiard ball is submerged in a small tank of water at depth of 60cm. The thank is located in colorado where its ambient pressure is 0.83atm. What's the absolute pressure experience by the ball?

A: 90245 Pa The question asks for absolute pressure, which includes both gauge and atmospheric pressure. Gauge pressure P = ρgy, where ρ denotes the density of the fluid and y refers to the depth of the object. Thus, P = (1000kg/m^3)(10m/s^2)(0.6m) = 6000 Pa. Then we must add atmosphere pressure, which is 0.83atm. Since 1 atm equals 101,500 Pa, 0.83 atm is equivalent to 84245 Pa, providing the final answer of 90245 Pa.

*A 10-kg sled rests on a 30° ramp with a coefficient of static friction of 0.5. An upward force is applied to the sled, parallel to the slope of the ramp, in incremental values until the sled begins to accelerate up the ramp. Approximately what minimum force is required to perform this action?

A: 95N The block will begin to accelerate once all opposing forces have been overcome; in this case, these forces are static friction (opposes motion) and gravity (opposes upward movement). Since the applied force needs only to minimally surpass equilibrium, we can set our net force equation as: Fnet = Fapplied - (Fg + Fs) = 0 N. This simplifies to Fapplied = Fg + Fs. Gravitational force parallel to the plane is equal to mg sin 30º = (10 kg)(10 m/s2)(0.5) = 50 N. The maximum force exerted by static friction is given by µmg cos 30º, since normal force is equal and opposite to the y-component of the gravitational force. Thus, Fs = (µmg cos 30º) = (0.5)(10 kg)(10 m/s2)(0.866) = 43.3 N. Fapplied is then equivalent to 50 N + 43.3 N, or 93.3 N.

If a given channel in a nerve membrane shows a preference for transporting larger solutes, which of the following would increase the likelihood of a given particle being transported by the channel?

A: Adding an electron to a Cl atom to create a Cl- ion The question states that the channel favors larger solutes. Adding electrons to make a negatively charged ion increase the radius of the particle, making it larger. Thus, going from a Cl to Cl- would increase the likelihood of it being transported by this channel. - Removing electrons to make positively charged ion would decrease radius, making the particle smaller, less likely to be transported by this channel. - Likewise, cleaving a large molecule into smaller pieces would also decrease the likelihood of this channel transmitting the particle.

At the exact same instant, one of the gorillas drops a 15-kg table directly downward, another throws a 20-kg refrigerator horizontally with a velocity of 10 m/s, and a third drops a 1-kg lunchbox straight down. If air resistance can be neglected, which object will hit the ground last?

A: All three objects will reach the ground at the same time. Time in flight does not depend on mass; in other words, if two objects of different masses are simultaneously dropped from the same height, they will hit the ground at the same time. Additionally, though the refrigerator is thrown horizontally while the other two objects are dropped, all projectiles begin with a vertical velocity of zero.

In order for the image produced by the eyepiece to be erect, where must the image focused by the objective lens fall?

A: At a distance from the eyepiece smaller than the eyepiece's focal length In other words, the image is "upright" and virtual. To create a virtual image with a converging lens, the object distance must be smaller than the focal distance of the lens in question. This means that the object must fall within the focal length of the eyepiece lens - If the image produced by the objective falls outside the focal length of the eyepiece, a real/inverted image would form

Light inside the thin glass tube of a laproscopic surgical device strikes the edge of the glass tube and is entirely reflected back into the tube, with none of the light exiting to the surrounding medium. Which of the following must be true? A. θincident = 90º B. θincident = 0º C. θincident ≥ θcritical D. θrefracted = θincident

A: C The question describes total internal reflection, when a light ray bounces inside a medium with a higher refractive index than the surrounding medium. For a light ray to totally reflect, rather than exit and refract (bend), the light ray must strike the edge of the glass tube at an angle equal or greater than the critical angle.

Sound waves reaching the ultrasonic transducer propagate through brain tissue via which of the following mechanisms?

A: Collision of adjacent particles within the tissue Sound waves in liquids and gases (and most in solids) are examples of longitudinal waves. Sound waves propagate due to vibratory changes in pressure or collision of adjacent particles within an elastic medium. Longitudinal pressure waves, including the waves generated by the PA effect, are created by local regions of compression and rarefaction, which occurs partly because of elastic collision between adjacent particles within the medium. The energy carried by the oscillating wave alternates in form between the potential energy of the compressed material and kinetic energy of the displacement velocity of particles of the medium. - "Vibration of particles perpendicular to the direction of wave motion in the tissue" is a description of a transverse wave

Myopia is a condition of the eye where the crystalline lens focuses the light rays to a position between the lens and the retina. To correct this, a thin lens is placed in front of the eye to help focus light on the retina.

A: Diverging Because a diverging lens will spread out the light before it reaches the lens of the eye which will cause light to focus on a point closer to the retina

Which of the following describes the direction of motion of alpha, beta, and gamma rays in the presence of an external magnetic field?

A: Gamma rays travel straight; alpha and beta rays are bent in opposite directions (Picked: "They all travel straight"- because magnetic field does work only on a moving charge) 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 experiences forces of opposite direction. Gamma rays possess zero charge and thus experience no force from a magnetic field. Alpha and beta particles possess charges of opposite sign and experience forces in opposite directions

In a follow-up experiment, two identical gurneys are placed side-by-side on a ramp with their wheels locked to eliminate spinning. Gurney 1 has a dummy placed on it to give it a total mass of 200 kg, while Gurney 2 is loaded with a dummy that makes it only 50 kg overall. If the ramp has a coefficient of friction of μs, which gurney is more likely to slide down the ramp?

A: Gurney 1 and Gurney 2 are equally likely to slide. In order for the gurney to slide down the ramp, the force pulling it downward (mgsinƟ) must be greater than the static frictional force (μsFN = μsmgcosƟ). The net force on each gurney is thus Fnet = (mgsinƟ) - (μsmgcosƟ). Since net force is equal to the product of mass and acceleration, we can rewrite this equation as ma = (mgsinƟ) - (μsmgcosƟ), where an acceleration greater than 0 means the gurney will slide down the ramp. The mass of the gurney is present in all terms and can be canceled, meaning that it is not a factor in whether the gurney slides. Thus, both gurneys have an equal likelihood of slipping down the ramp, regardless of the fact that they have different total masses. - Heavier gurney does experience a larger force of gravity; however, its mass also contributes to a larger static friction force (and vice versa with a smaller gurney).

Which circuit elements store energy? I. Capacitors II. Resistors III. Batteries

A: I and III The capacitor charges up and stores energy in the electric field between the places. The battery is the source of energy for the circuit.

A scientist uses an ultrasound device mounted to a vehicle to measure fluid flow underground. The device makes use of the Doppler effect to track fluid movement in the water table. Which of the following scenarios is most likely to produce a readable Doppler shift? I. The fluid is flowing at a velocity twice that of the sound-emitting device, in the same direction as the device is moving. II. The fluid is flowing at the same velocity and in the same direction as the sound-emitting device is moving. III. The fluid is not moving at all.

A: I only The doppler effect is used to analyze *moving* objects/fluids. A doppler shift is registered only if the fluid is moving relative to the source of the sound (device). More specifically, at least some component of the fluid's velocity must exist in the same directional plane as the wave's velocity (otherwise, the device will register the fluid as not moving at all), and this component must be different from the velocity of the sound source.

Which of the following statements are true? I. The blood velocity and pressure are higher in arteries than in capillaries. II. Blood vessels do NOT follow Bernoulli's principle if viscosity of blood is taken into account. III. For an incompressible fluid undergoing laminar flow through a pipe with both narrow and wide sections, the fluid velocity is higher in the narrow areas than in wider areas. IV. For an incompressible fluid undergoing laminar flow through a pipe with both narrow and wide sections, the fluid pressure is higher in narrow areas than in wide areas.

A: I, II, and III Recall that the resistance of capillaries to blood flow is much higher than that of arteries. This causes the capillaries to exhibit lower pressure than arteries. Bernoulli's principle has two underlying assumptions: incompressibility and non-viscosity of the fluid. These are not true of blood (or any real fluid), so blood vessels don't follow Bernoulli's principle. From Bernoulli's principle, the narrower sections have lower pressure and higher velocity than non-occluded areas

An experimenter proposed employing a material with an index of refraction strongly dependent upon the wavelength of transmitted light for use in an objective lens. How would the use of such a material impact the performance of a microscope?

A: Images formed by the eyepiece from transmitted light containing multiple wavelengths will be unfocused If transmitted light rays refracted differently depending on their wavelengths, then the effective focal length of the lens would differ for each wavelength of light. As a result, for any light containing rays of multiple wavelengths, the rays would focus at multiple points, resulting in an unfocused image; this is known as a "chromatic aberration", in which optical instrument fails to converge light rays from a source to a single point - Monochromatic light is composed of light of a single wavelength. Parallel rays of such light would converge to a single point

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), t would

A: Increase by a factor of 6 t of 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

Making which of the following changes to a circuit element will increase the capacitance of the capacitor described in the passage

A: Increasing the area of the capacitor plates Capacitance depends on geometric factors only, and in parallel plates, C is proportional to the plate area and inversely proportional to the separation distance of the plates

Visible light travels more slowly through an optically dense medium than through a vacuum. A possible explanation for this could be that the light

A: Is absorbed and re-emitted by the atomic structure of the optically dense medium

Which of the following best describes the movement of an electron after it is ejected from the cathode?

A: It accelerates toward the anode A charged particle accelerates in an electric field. The electron starts with a velocity that increases as it approaches the anode through the vacuum

A circular segment of gold wire is positioned in an external magnetic field of strength B that points out of the page. If the strength of the field is increased to 2B, what happens to the wire?

A: It experiences a clockwise current This is related to Lenz's law which states that a change in magnetic field will always generate a current that *counteracts* the change. Here, the external field is becoming stronger, or more heavily positioned out of the page. To resist this change, it will induce a current that promotes a field pointing the opposite direction, into the page. The right-hand rule tells us that this current must travel clockwise.

What's the magnitude of the detected sound frequency shift from 170Hz during the projectile flight described in the passage?

A: It falls to zero, then increases 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 170Hz, only the absolute value matters

After a block began to slide, how did its speed vary with time? (Assume that the tension and kinetic friction on the block were constant in magnitude)

A: It increased linearly with time The coefficient of kinetic friction is always lower than that of static friction. Thus, there is a net accelerating force on the block once it starts to slide. A constant force on a mass produces a constant acceleration, causing the velocity of the block to increase linearly with time

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?

A: It will be approximately equal The near equality of the photon energy and the work function means that little kinetic energy will be left for the electron. This initial kinetic energy is small compared to the 50eV it will gain from the potential difference between electrodes

An object has a mass of 100 kg on Earth. If Jupiter has a mass that is about 320 times the mass of Earth and a radius that is about 11 times Earth's radius, what will be the approximate mass of the object if moved to to Jupiter?

A: It will not change The mass of an object will never change when it is moved to a different area. This is different than weight, which will change if the gravitational field is altered

Which of the following best describes the velocity profile of laminar flow from left to right in a stationary tube?

A: Laminal flow is due to shear forces (friction) between the fluid and the solid surface of the tube. This results in layers having a gradient of velocities where the flow is the fastest in the middle of the tube (where friction is low) and slowest near the surface (where friction is high)

While standing in a packed bus, a child in the back thinks it is funny to nudge the person in front of him. That person stumbles forward a bit, then returns to his previous position. However, the damage is done; when the passenger fell forward, he nudges the person in front of him, and in this way, a wave of stumbling proceeds to the front of the bus. This wave could best be characterized as a:

A: Longitudinal wave This wave's oscillation is forward-backwards in amplitude while its overall direction of propagation is forward. Because the amplitude of oscillation is parallel to the direction of the wave's motion, it is a longitudinal wave. A transverse wave has an amplitude perpendicular to its overall direction of propagation.

A "loop-the-loop" on an amusement park ride has a radius of R, while the cars on the ride travel with a velocity of v. If a man of mass M were able to step on a scale at the instant when he is at the bottom of this loop, how much would he seem to weigh?

A: Mg + Mv2/R If the man were not moving in a circular pattern, his weight would simply be Mg. Here, however, he is also subject to centripetal force. Remember that apparent weight can be conceptualized as the normal force exerted on an object. In this case, the normal force is that which points directly upward from the bottom of the loop. This is the combination of centripetal force (Mv2/r) and gravitational force (Mg).

The typical range of human hearing spans from 20 Hz to 20,000 Hz, and the speed of sound in air is approximately 340 m/s. With this information in mind, an ultrasound signal

A: Must have a wavelength lower than 0.017m The prefix "ultra" means "higher," as we can discern using our knowledge that ultraviolet light is light with a frequency higher than that of visible light. Ultraviolet sound, then, likely also has a frequency higher than the typical range heard by a human. No answer choice states this directly, so we must calculate the wavelength that this corresponds to. To do so, we can use the simple equation v = λf. 340 m/s = λ (20000 Hz) 3.4 x 102 m/s = λ (2 x 104 Hz) λ = (3.4 x 102 m/s) / (2 x 104 Hz) λ = 1.7 x 10-2 m = 0.017 m

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?

A: Number of electrons ejected 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 frequency of each photon directed at the cathode is increased, the speed of the ejected electrons increase. The only effect the photon frequency has on the ejected electron is on its kinetic energy. *(Photon energy)=(cathode work function)+(electron kinetic energy)*. The number of electrons ejected (= CURRENT) depends on the number of incident photons

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?

A: Part of the sound energy is reflected by the solid Sound wave are reflected/absorbed when incident on materials just as light waves are. In doing so the waves are reduced in their forward intensity. When a wall is placed between a source of sound and listener, some of the sound energy is transmitted to the listener, but some is reflected and sent backward toward the source/absorbed in the material. The listener hears less sound, but at the same frequency and wavelength as the unimpeded sound

Which of the following describes the direction of the magnetic force on an ion moving in an artery past a flowmeter?

A: Perpendicular to both the direction of v and the direction of B 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.

What information about an axon is required to calculate the current associated with an NCV pulse?

A: Potential, resistance per unit length, and length This is a question about Ohm's law, I=V/R. To determine R, the resistance per unit length, the length, and the potential V are needed.

Changing which of the following will change the focal length of the convex mirror?

A: Radius of curvature of the mirror (Picked: "focal length of the lens at 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 won't affect the focal length of the mirror

A stationary charge lies to the right of a current-carrying loop of wire as depicted in the figure below. While the current is flowing, the large will

A: Remain motionless The force of a magnetic field (generated by the wire) on a charge is found with the equation F=qvBsin(θ). Here, the charge is motionless so v = 0. Therefore, F =0 and with no force pushing on it, the charge will remain stationary

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

A: Remain the same 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

At a given temperature, the resistance of a wire to direct current depends only on the

A: Resistivity, length, and cross sectional area

Refractive myopia is corrected by the use of a diverging lens of appropriate optical power. When compared to a healthy eye, a myopic, but otherwise normal, eye's near point is most likely

A: Smaller than that of a healthy eye A diverging lens decreases the optical power of an instrument by increasing its effective focal length. Such a lens is corrective for the myopic eye where its optical power exceeds that required for the myopic eye, where its optical power exceeds that required for the axial length of the eye. As a result of this increased optical power, the myopic near point should be smaller than found in a healthy eye. This is due to the myopic eye's capacity to focus light of great divergence on the retina

The brain is encased in the skull, surrounded by cerebrospinal fluid (CSF). If the brain has a specific gravity of 1.03 and the specific gravity of CSF is 1.07, what portion of the skull will the brain hit first, immediately after a person is struck in the face?

A: The back of the skull first, because the brain is less dense than the CSF The question gives us the specific gravities of the objects inside the skull. Since the brain has a lower specific gravity, it will float in the CSF (CSF is more dense). The CSF will "slosh" to the front of the skull as the head moves backward, causing the brain to hit the back of the skull first

The students untie both balloons and poke an identical pinhole in each, allowing the gas to escape each balloon. Which of the following best describes the movements of the balloons and the gas molecules as they escaped the balloon? (Note: aside from the identity of their contents, all properties of the balloons and the gases are identical.)

A: The balloons will move at the same speeds, and the helium molecules will move faster If the balloons exist under the same conditions, then the gas molecules that escape will have the same temeperature and thus, the same kinetic energy. This kinetic energy will act as thrust, pushing the balloons around the car. KEair = KEHe However, remember that KE = 1/2 mv2. 1/2 mairvair2 = 1/2 mHevHe2 Thus, if the gas molecules have the same kinetic energy, the He particles (which have less mass) must be moving faster.

The surface temperature of the sun (5770 K) is much greater than that of the earth (298 K). As a result, solar photons have a much higher energy than Earth photons. Given that information, how does the frequency and velocity of an Earth photon differ from that of a solar photon?

A: The frequency of the earth photon would be lower; the speed of the photons would be identical Recall Planck's law: E=hf. The greater the energy of a photon, the higher its frequency. Therefore, earth photons will have a lower frequency than solar photons. The speed of photons is the speed of light and does not depend on the energy of the photons. Thus, the speed of the photons are identical

Unpolarized light with an intensity of 200kW/m^2 shines on a series of vertical polarizers as shown below. The intensity of light incident on the eye after passing through both polarizers is recorded as the "reference intensity". If the second polarizer is then rotated by 90 degree, how much will the intensity of light reaching the eye change compared to the reference intensity?

A: The intensity changes by 100kW/m^2. The intensity of unpolarized light will be cut in half as it passes through a single polarizing filter, leaving an intensity of 100kW/m^2. Originally the second polarizer is oriented in the same direction as the first where its intensity unchanged as the light reaches it. making our reference intensity equal to 100. Then turning the second polarizer by 90 degree will cause the filters to be perpendicular, blocking 100% of the light. The new intensity will be recorded as 0, meaning that the intensity changed by 100kW/m^2 compared to the reference intensity.

Two charges in a medium with a dielectric constant ε experience an attractive electrostatic force, F. How would this force change if the dielectric constant were increased to 2ε and the distance were doubled?

A: The new force would be 0.125 times the original value Coulomb's law is typically given as, F=k(Qq)/r^2 where "r" represents the distance between the two charges. From this relationship, we can see that doubling "r" reduces the force by a factor of 4. However, it is slightly trickier to assess the role of the dielectric constant. Technically, though, Coulomb's constant "k" is equal to 1/4piε , so doubling ε halves the force generated. The product of one-fourth and one-half is one-eighth, or a change to the original force by a factor of 0.125.

An object floats in water with 4/5 of its volume submerged. If the object is then placed in a type of oil with a density half that of water, which of the following is true about the object placed in the oil?

A: The object will sink to the bottom When dealing with the principles of floating, buoyancy, and density (as is the case of an object floating with a certain portion of it submerged), we can use a shortcut: Vfluid / Vobj = ρ obj / ρ fluid , where V fluid is the volume of displaced fluid (here, it's water). V fluid / V obj = 4/5, so ρ obj / ρwater = 4/5. The question tells us that ρwater = 2ρoil. This means that 4/5 =ρobj / 2ρoil . Therefore, ρobj / ρoil = 8/5. Since the ratio of densities is larger than 1, we know that the object is more dense than oil, and it will sink to the bottom.

Measurements are made in two arteries with the same diameter and the same flow rate. One artery is in the head and the other is in the leg. In a person standing upright, which of the following statements is true?

A: The pressure of the blood in the leg is greater than the pressure of the blood in the head The gravitational potential energy of the blood in the head is greater than the gravitational potential energy of the blood in the leg. As a result the blood in the leg will have a greater pressure due to the conversion of the gravitational potential energy into the kinetic energy of molecular motion. The increased pressure is the result of the increased number of collisions between the molecules in the blood of the leg and the surface of the blood vessel

A 10kg crate is set on a cargo ramp with a 30 degree incline. if the crate remains in place, which of the following statements must be true?

A: The ramp's coefficient of static friction must be at least 0.6 Static friction opposes movement when an object is at rest, where Fs = µsFN. The normal force is FN = mgcos30 = (10 kg)(10 m/s2)(cos 30) = 86.6 N. In contrast, the component of gravity that acts in *parallel* to the ramp is equal to mgsin30, or (10 kg)(10 m/s2)(sin 30) = 50 N. We can place these values into our equation for static friction to determine the required coefficient. 50 N = µs(86.6 N), so µs = (50) / (86.6) or about 0.58.

An ultrasound examination could show the motion of a fetus. In order to image this motion, the ultrasound examination devices require what minimal information?

A: The speed of of the sound, and the frequencies of the sound waves emitted and observed (I picked: "The speed of the sound and the moving object and the frequencies of the sound waves emitted and observed) The doppler effect is used with ultrasound waves to provide fetal images. The doppler effect relates the frequency of the ultrasound wave as detected by a moving detector the the frequency of the wave when the source is stationary, the speed of the source and the detector.

A hoop and a sphere, each of mass M, are rolled down a frictionless ramp. At the bottom of the ramp, which object will have the greater translational velocity, and why?

A: The sphere, because its moment of inertia is lower. An object's moment of inertia contributes to its rotational kinetic energy. Since a higher moment of inertia correlates to a higher rotational KE, it also means that translational KE must be lower, as the sum of these two values must equal the potential energy at the top of the incline. For the sphere, more of its mass is concentrated towards the center than for the hoop. As a result, the sphere's moment of inertia will be lower.

A ball attached to a string on a tabletop is swung in a circle with a velocity of v. The string is threaded through a small hole in the table such that it can be pulled on to be shortened. If the string is loosened and allowed to extend to three times its original length, how will the ball's velocity change?

A: The velocity of the ball will decrease to 1/3 of its previous value. Don't think of this problem in terms of centripetal acceleration. The initial and final values of the ball's angular momentum should be identical where L=mvR in which R is the length of the string.

While evaluating the timing of pitches, the researchers noticed a couple of high-performing athletes who were able to generate higher velocities by increasing the length of their acceleration phase. How did the power production for these athletes compare to those originally mentioned?

A: There may have been no change in power output between the two groups The power output is P = W/t. If the velocity is increased, then the work also increased, but the time spent in the acceleration phase increased as well. Thus, there should be no significant difference in power output between the two groups. Q2. If researchers failed to take into account the effect of air resistance on the pitch, how would it impact their measurements of the efficiency of energy transfer from the arm to the baseball? A: It would be lower than the actual efficiency as there was a higher initial velocity at release. (Thought it would be higher efficiency due to no air resistance) Air resistance would decrease the velocity of the ball as it travels from the mound to home plate (where the velocity was recorded). Thus, the measured velocity should be lower than the velocity at release. The decreased velocity results in a decreased calculation energy for the baseball, leading to a decreased calculation of efficiency for the energy transfer.

During an experiment involving the effect of roller coaster riding on neck muscles, all participants reported a feeling of weightlessness in the loop segment of the roller coaster ride. At which of the following positions did this feeling occur?

A: Top of the loop The feeling of weightlessness occurs when the normal force is minimal. At the bottom of the loop, the normal force points toward the center of the loop, while the gravitational force points downward. The net force will be the centripetal force directed toward the center of the circle. At the bottom of the loop, the normal force is largest because in order for the net force to be directed inward, the normal force should exceed the downward gravitational force. At the top of the loop, both the gravitational force and the normal force point downward. At the top of the loop, the normal force is the least because gravity is downward and therefore, there is no need for a large normal force in order to stay in a circular path.

A circuit is comprised of a 12-volt battery with three light bulbs with the same resistance. If two of the light bulbs are wired in parallel and are in series with the third light bulb, what happens when one of the parallel light bulbs is unscrewed?

A: Two of the light bulbs remain lit Even after unscrewing one of the parallel light bulbs, there is still a circuit for current to flow from the positive terminal to the negative terminal of the battery through the two remaining light bulbs.

The density of a human body can be calculated from its weight in air, W_air, and its weight while submersed in water, W_water. The density of a human body is proportional to

A: W_air/(W_air - W_w) According to Archimedes' principle, the ratio of the density of an object to the density of the fluid it is submersed in its equal to the ratio of the weight of the object in air to the difference of submersed weight and in air.

Which changes are experienced by visible light as it moves from Medium 1 (n = 1.16) to Medium 2 (n = 1.68)?

A: Wavelength decreases while frequency remains constant We know that velocity decrease from v= λf. The frequency of light does not change when light transitions between media; thus, it must be wavelength that drops to yield the predicted decrease in speed.

A perfectly round, transparent bucket is filled with equal volumes of distilled water (n = 1.33) and benzene (n = 1.50), which immediately separate into two layers. If light is then shone into this beaker from multiple directions, which of these statements is accurate?

A: When a ray initially moving through the water hits the benzene at a 76° angle to the normal, its angle of refraction will be smaller than 76° because light travels faster in water than in benzene.

When does a runner output the most additional energy to keep the ground reaction forces most nearly vertical and through her body's center of mass?

A: When she takes high, bouncing strides and leans her upper half into her run Both the vertical displacement of her steps and the angle of her body from the vertical increase the energy required to realign the ground reaction forces. Recall that tanθ = sinθ/cosθ, so when θ is close to zero, so is tanθ. When θ is close to 90°, tanθ becomes arbitrarily large.

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 packed on the Moon where the gravitation force is 1/6 that of Earth, the ice in the same soft drink would float

A: With 9/10 submerged The floating ice cube implies that its weight is balanced by the buoyant force on it: Wice = mg = r_fluid*Vsubmerged*g Because both the weight and buoyant force are proportional to g, the numerical value of g is irrelevant to the volume of the ice cube submerged

A physician hypothesizes that the PA (photoacoustic) effect could be generated by modulating the intensity of radiation delivered by a continuous light source, rather than by illumination with pulsed light. Is this physician correct?

A: Yes, because the magnitude of thermal expansion and contraction will vary with the intensity of incident radiation (I picked: "No, consistent thermal expansion will take place rather than cyclic expansion and contraction") The physical basis of the PA: An object absorbs EM radiation energy, the absorbed energy is converted into heat, and the temperature of the object increase. As soon as the temperature increases, thermal expansion takes place, followed by contraction, generating the compression and rarefaction needed for compressional (sound) waves to be generated in and travel through the medium. However, steady thermal expansion, caused by the delivery of constant energy at a constant rate, does not generate rarefaction/compression. Thus, the heating source must be time-variant. This can be accomplished by delivering energy in a pulsatile (on-off) fashion or simply by modulating the amount of energy delivered over time. In the latter instance, the magnitude of thermal expansion and contraction will still vary, generating detectable pressure waves and ultimately a PA signal.

Would it be feasible to utilize the Doppler effect to visualize the movement of a planet in a solar system?

A: Yes, but light would have to be used instead of sound, since sound waves are longitudinal This question is referring to the simple fact that outer space is a vacuum. Transverse waves (like electromagnetic waves including visible light) can travel through a vacuum, but longitudinal waves cannot as they require a compressible medium to propagate.

The acceleration of the gurney during the initial 6 seconds of Experiment 1 (Fapplied linearly increases) is

A: Zero Thought: The answer is "rising a F_applied increases" because F = ma; since F_applied is increasing, acceleration must increase too? The passage states that the force applied was raised until the gurney moved (i.e static friction was overcome); this implies that the gurney has not moved yet. Thus, F_net = 0N and a = 0.

The capacitance of a nerve membrane can be increased by

A: decreasing the width of the membrane Capacitance is coulombs per volt. Thus, decreasing the charge decreases the capacitance. Increasing voltage also decrease the capacitance. Also, capacitance is directly proportional to area but inversely proportional to the distance between the two sides of the capacitor. Therefore, decreasing the width of the membrane actually increase capacitance.

The energy released when an atom splits comes from

A: mutual repulsion of the fragments The kinetic energy of the fission particles is produced by the repulsive Coulomb force between the particles.

A small metal ball (m = 0.5 kg) is launched off a spring at a 30° angle to the ground. Friction and air resistance are negligible in this case. If the ball has a horizontal velocity of 17.3 m/s after one second of flight, what was its initial kinetic energy?

A= 100J Initial kinetic energy can be quickly calculated if we know the total initial velocity of the ball. Luckily, we are given horizontal velocity, which remains constant throughout a projectile arc as long as air resistance can be ignored. For that reason, we know that the initial vx is also equal to 17.3 m/s. In other words, vtotal(cos 30°) = 17.3; cos 30° = √3/2, or about 0.866. Finishing that calculation tells us that the total initial velocity of the ball was 20 m/s. KE = ½ (0.5 kg)(20 m/s)2 = 100 J.

Which of these particles will experience a force when subjected to a 6.5T magnetic field? I. A neutron traveling at 6 × 102 m/s II. A stationary proton III. An electron moving at 4 × 104 m/s perpendicular to the field

III only. A particle must possess both velocity and a charge (whether positive or negative) to be affected by a magnetic field. Magnetic fields always exert a force on a moving particle that is perpendicular to both the particle's velocity and field. Given the equation for work W=Fdcos θ, work performed by magnetic field is 0J.

During trial 5, the wire was heated from 293K to 673K while V was held constant at 28V. How did the current through the circuit change during this time?

It decreased from 4.6A to 2A (Picked "remains constant at 2A" due to V=IR) Remember the linear relationship between resistivity and temperature. If the temperature is increased by a factor of 673/293 ~ 2.3, the resistance R will increase by the same amount. From Ohm's law, current and resistance are inversely proportional, thus current will decrease by a factor of 2.3.

Which image best illustrates the electric field lines between the inside of an axon and the surrounding extracellular solution?

The electric field line exit from positive and enter negative charges. The axon is negatively charged and the extracellular solution has a 0 net potential. Therefore, it flows from extracellular fluid to axon interior.