Physics

Question 1

In the figure above, an object O is placed a distance 2f from the centers of converging and diverging lenses both with focal length f. How does the height hc of the image created by the converging lens compare to the height hd of the image created by the diverging lens?

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Answer 1

hc = 3hd

Thin lens equation: (1/F) = (1/O) + (1/i)

converging at a convex shaped lens causes parallel light rays to convert at the focal length behind the lens, hence f is positive. For a diverging concave shaped lens, parallel light rays diverge, and f is negative.

In the converging case, 1/f = 1/2f + 1/i
solving for i: i=2f

In the diverging case, F=-f
1/-f = 1/2f + 1/i
i = -2/3f

Geometry implies that the ratio of the image height h and object height ho = the ratio of i and o

h/ho = i/o

in the converging height, the image height hc is:
hc/ho = 2f/2f = 1

in the diverging case, the image height hd =
hd/ho = 2/3f / 2f = ho/3

therefore, the relation between the heights is: hc = 3hd

Question 2

A 20 cm^3 marshmallow is placed in 250 cm^3 of hot chocolate. The densities of the marshmallow and hot chocolate are 0.5 g/mL and 1.1 g/mL respectively. Ignoring the effects of atmospheric pressure, what percentage of the marshmallow is above the surface of the hot chocolate?

Answer 2

55%

Wm = pm * g * Vm 
Fb = pf * g * Vd 

Wm=weight of marshmallow
Fb= bouyant force
Vd = volume of displaced hot chocolate

For floating object:
Wm = Fb
pM * vM = pF * Vd

Fraction of marshmallow below surface:
Vd/Vm = Pm/Pf

Fraction of marshammlow above surface
1- (Vd/Vm) = 1 - (Pm/Pf) = 1 - (0.5/1.1) = 0.55

Question 3

Ultrasound imaging at 1.8 MHz is used to evaluate a liver disease. The signal travels through a boundary between liver and fat tissue at an angle 30 degrees relative to the normal at the boundary. If the ultrasound signal has a wavelength of 0.8mm in fat tissue, what is the speed of sound in fat?

Answer 3

1440 m/s

v=lamda(f)

In fat tissue: 0.8mm = 8x10^-4 m
f = 1.8 mHz = 1.8 X 10^6 1/s

v = (8x10^-4 m) (1.8 X 10^6 1/s )
=1440 m/s

Question 4

A spring is extended 15 cm from its equilibrium point. If the spring constant k is 75 N/m, the magnitude and direction of the elastic force Fel are described by what force and direction?

Answer 4

1.1 x 10^1 N; oriented toward the equilibrium point

Fel = -kx
15 cm = 0.15 m
Fel - (75)(0.15) = -1.1 X 10^1, absolute value

Fel points toward the equilibrium point because Fel acts to oppose the extension and restore the stretched spring back to its shorter position. Therefore, extending the spring 15cm from its equilibrium point will generate an elastic force that has a magnitude of 1.1 x 10^1 N and is oriented TOWARD the equilibrium point.

Question 5

A 10 kg block is suspended in air by a single string that passes through a pulley and is attached on the other side to a 35 kg block that is on the verge of sliding on the ground.

What is the coefficient of static friction between the larger block and the ground? (The vertical component of the force the rope exerts on the large block is 87 N)

(402705)

Answer 5

0.19

The maximum magnitude of static friction Fs between two opposing surfaces is equal to the product of normal force and the coefficient of static friction, Fs = us * Fn

If a force applied to an object does not exceed Fs, the object will not move. If Fs is exceeded, the object will accelerate and static friction becomes kinetic friction.

In this question, the two blocks and pulley are in static equilibrium. For the smaller block, the tension T in the rope is directed vertically upward and equal to the weight Wsb of the smaller block:
T = Wsb = mg
T = 10 kg * 10 m/s^2 = 100N

On the left side of the pulley, T acts on the larger block and can be resolved into component forces orietned along the x and y axes. The horizontal component Tx equals:
Tx = Tcos(theta)
=100N * cos(60) = 50 N

And the vertical component Ty is:
Ty = Tsin(theta)
=100N * sin(60) = 87 N

Because the block is still in static equilibrium, Fs must be EQUAL to Tx:
Fs=us * Fn = 50

Furhtermore, in the y-direction the sum of Fn and Ty = the weight of the larger block:
Fn + Ty = Wlb

Solving for Fn and substituing into the equation for Fs gives: 
us * ( Wlb - Ty) = 50N 
us * [(35 kf * 10 m/s^2) - 87N] = 50 N 
us * 263N = 50N 
us = 50N/263N 
=0.19

Question 6

A swimmer enters a straight river at Point A and attempts to swim directly across to the opposite shore. However, the downstream velocity of the river is 0.5 m/s, which causes the swimmer to arrive at Point B on the opposite shore after 80 seconds of swimming. If the river is 30m wide, what is the minimum diagnol distance between points A and B?

Answer 6

50 m

Vector addition: driver + dswim = dcross
*magntiude of any vector d is equal to the sqaure root of the sum of the x and y components squared

If point A is set as the origin, the width of the river can be oriented along the y-axis. Therefore, the vector corresponding to the swimmer’s effortful crossing dswim depends only on the width of the river:
dswim = (0m, 30m)

The flow of the river acts to passively transport the swimmer down the river in the x-axis direction. As such, the swimmer’s motion may be expressed in terms of the velocity of river flow multiplied by the time it takes the swimmer to cross the river:
driver = [0.5 m/s * 80s], 0m)
driver = (40m, 0m)

Consequently,
dcross = [(0m + 40m], [30m, +0m]) = (40m, 30m)

Furthermore, the magntiude of dcross = 
sqaure root (40)^ + (30)^ 
=sqrt 1600 + 900 = sqrt 2500 = 50m

Question 7

A flask with height h and base area A is filled with water. If the pressure due to the weight of the water it 2atm halfway between the base and the top, what is the pressure due to the weight of the water at the base of the flask?

(402715)

Answer 7

3,000 mm Hg

P=pgh

In this question, a flask with variable cross-sectional area is filled with water. Because the water is not flowing, it forms a fluid column that generates hydrostatic pressure. The hydrostatic pressure at half the depth of the total flask is 2atm, which is twice the atmospheric pressure experienced by objects on the Earth’s surface at sea level:
P=pg(h/2) = 2atm

Therefore, the hydrostatic pressure experienced at the bottom of the flask is
P=pgh = 4atm

Converting the units from atm to mmHg gives:
P = 4atm * (760 mmHG / 1 atm) = 3000 mmHg

Question 8

When standing barefoot on a floor, body heat is transferred into the floor based on the thermal conductivity k of the flooring material. The heat flow rate per unit area H is defined by:
H=k(deltaT) / L

Where deltaT is the difference in temperature over a distance L. The tiled part of a floor in a room at room temperature feels colder than the carpeted part because:

Answer 8

the thermal conductivity of the tile is HIGHER than the thermal conductivity of the carpet.

Heat flows from the warmer body to the cooler objects. However, tile is a better thermal conductor than carpet and therefore has a higher k. Consequently, heat transfer from the body to the tile is more rapid than the heat transfer to the carpet such that the person perceives the tile being colder than the carpet.

Question 9

An observer perceives a sound to be lower in frequency when the sound source is moving away from the observer. This shift in frequency of the perceived sound occurs because:

A) the sound waveform velocity decreases as the distance between source and observer increases
B) sound waveforms are reflected off environmental objects nearby
C) successive sound waveforms are emitted further away from the observer
D) the sound wave intensity decreases as the distance between source and observer increases

Answer 9

C) successive sound waveforms are emitted further away from the observer

Doppler effect: fo = v+/- vs (or vo) / lamba

The motion of the observer or waveform source relative to each other leads to a shift in only the apparent, perceived velocity of the sound waveform as successive waveform crests and troughs are perceived closer together or farther apart. When the source or observer is moving away from the other, the distance between successive waveforms increases and the perceived waveform velocity decreased.

PERCEIVED VELOCITY, NOT ACTUAL VELOCITY (which is why the answer is not A)

Question 10

A marble rolls down a slope at a velocity of 10 m/s. If the KE of the marble is 2J, what is the mass of the marble?

Answer 10

40 g

KE = 1/2m*v^2
2J = 1/2 m * (10)^2
m=0.04 kg = 40 g

Question 11

A large number of monoatomic ideal gas molecules are stored within a closed container at a temperature of 200k. What is the average translational kinetic energy of the individual gas molecules? (Boltzmann’s constant k = 1.381 x 10^-23 J/K)

Answer 11

4.2 X 10^-21 J

KE = 3/2kT

= 3/2 (1.381 x 10^-23 ) (200K)
=4.2 x 10^-21 J

Question 12

Two identical balls of equal mass m1 and m2 are placed at rest at the top of separate hills. How do the velocities v1 and v2 of the balls comapre, measured after each has rolled down to the bottom of its respective hill?

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Answer 12

v1 = sqrt2/2 * v2

Total energy: U = KE + PE

Conservation of energy requires that an object’s total energy observed between two points remains the same, but contributions from potential energy may convert to kinetic energy or vice versa within mechanical systems:
1/2mVa^2 + mghA = 1/2mVb^2 + mghB

Based on the conservation of energy, all the potential energy stored within the balls at the top of the hills is converted into kinetic energy as the balls reach the bottom of the hills such that:
m1gh = 1/2m1v1^2 and m2g(2h) = 1/2m2v2^2
Rearrange equation of these eqautions to solve for the respective velocities

V1 = sqrt 2gh ; V2 = sqrt 4gh = 2 * sqrt gh

Ratio: v1/v2 gives answer

Question 13

Blood flows through a vessel in the leg at an avg velocity of 10 cm/s. How much glucose passes by a segment of the blood vessel each second if the radius of the blood vessel is 0.20cm and each 1.0cm^3 of blood contains 1.0 mg of dissolved glucose?

Answer 13

1.3 X 10^0 mg/s

Q=Av

For a noncompressible ideal fluid with constant density, the mass flow rate Mf quantifies the amount of mass flowing past a point within the conduit per unit time:
Mf = Qp

A= pi * r^2
= pi * (0.20)^2 = 0.126 cm^2

Q = 0.126 cm^2 * 10cm/s = 1.26 cm^3/s

Furthermore, the mass flow rate of glucose Mglu at a segment of the artery is equal to the product of volumetric flow rate and the blood glucose mass concentration:
Mglu = 1.26 cm^3/s * 1 mg/cm^3 = 1.26 mg/s

Question 14

Four charged particles of equal mass are initially fixed in place along a straight line. If all the charged particles are released at the same time, which particle would initially accelerate most rapidly?

Particle A: +4q
Particle B: +3 q
Particle C: -3 q
Particle D: -2q

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Answer 14

Particle B

F=kq1q2 / r^2

Particle B experiences 3 forces: repulsive from A, and two attractive from C and D. Each of the forces acting on particle B is oriented along the same line towards particles C and D (to the right). Furthermore, FAB has the largest magnitude in this setup because the charges of particles A and B are relatively large and the distance between particles A and B is relatively small.

Question 15

Compared to other diffraction techniques, the advantage of XR diffraction is that it enables:

Answer 15

determination of 3D molecular structure

Question 16

Which of the following will occur if a corrective lens is placed in front of an eye that is unable to form a clear image of a nearby object?

A) the focal length will shift away from the eye’s lens
B) the focal length will shift closer to the eye’s lens
C) spherical aberration will increase
D) optical power will decrease

Answer 16

B) the focal length will shift closer to the eye’s lens

Hyperopia/farsightedness

Uncorrected: eye too short or optical power too low, light focused behind focal plane of retina

Corrected: Converging lens increases optical power, light now focused on focal plane of retina

Question 17

The suspension of mass M within a 2-pulley system requires a force F. If pulley B is removed from the system, what is the ratio of the original value of the force with two pulleys to the new value of the force with one pulley?

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Answer 17

1:2

Removing Pulley B eliminates the mechanical advantage of the pulley system: Fnew = 2 * Foriginal

Question 18

The figure shows a board placed on top of a fulcrum that is balanced by three masses. If the mass of M1 and M2 are 60 kg and 30 kg, what is the mass of M3?

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Answer 18

17 kg

Torque = r * Fsin(theta)
Sum of torques = 0
T1 = T2+ T3

60 kg(2m) = 30kg*2m + M3*3.5m
M3= 17kg

Question 19

cos (60)

Answer 19

1.2

Question 20

A 1,200 kf car is traveling at 10 m/s. What is the minimum distance at which an automatic braking system needs to activate to avoid hitting a stationary object? (the braking system can decelerate the car at 2m/s^2)

Answer 20

25 m

V^2 = V0^2 +2a(deltaX)
deltax = V^2 - V0^2 / 2a
= (0)^2 - (10)^2 / -4
=-100/-4 = 25 m

Question 21

Researchers collide a 1,000kg car into a stationary concrete barrier with a mass of 100,000 kg. What best explains why the barrier remains stationary when it is struck by the car?

Answer 21

The inertia of the barrier is much greater than the inertia of the car

Inertia is proportional to Fs and to mass

Question 22

A steel tow cable is used to pull a car at a constant velocity toward a barrier during a crash test. What information about the test is needed to calculate the tension in the cable?

I. Friction force on the car’s wheels
II. Mass of the car
III. Velocity of the car

Answer 22

I only

SumFy = 0 = Fg - Fw
Sum Fx = 0 = T-Ff

T depends only on Ff when the vehicle is being pulled at a constant velocity.

Question 23

A seatbelt holds a 70kg crash test dummy in the car seat during a collision that slows the car from 90 km/hr to 0 km/hr. What is the approximate magnitude of the work done by the seat belt on the crash test dummy?

Answer 23

20,000 J

W = Change in KE = KEf - KEi
=0 - 1/2mv^2
=-1/2(70)(90km/hr)^2

90 km/hr (1000m /km) (1hr / 3600 s) = 25 m/s

W = =-1/2(70)(25 m/s)^2 = -21,875 kg * m^2 / s^2 ~ -20,000 J

Question 24

A monoatomic ideal gas is initially sealed within a container featuring a piston fixed in place at one end. The piston is then unlocked and allowed to move under constant pressure. How does the initial heat capacity of the gas at constant volume Cv compare with the heat capacity of the gas at constant pressure Cp and why?

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Answer 24

Cp will be greater than Cv, because the gas can expend energy by doing work on the piston

Up = Q-W

Cv is less than Cp because some heat energy transferred into a gas stored at constant pressure may be converted to pressure-volume work during the expansion or compression of the gas. Therefore, allowing a prevoiusly fixed piston to move freely (to maintain constant pressure) will increase the molar heat capacity of a monoatomic ideal gas.

Question 25

In a graph of force versus distance, the area under the curve

Answer 25

is equal to work associated with the displacement

because the Fd component of the equation W=Fd is equal to a y-axis value (a force) multiplied by an x-axis value (a distance)

Question 26

Ideal fluids are used to model the behavior of fluids both in motion and at rest. Which one of the following assumptions does not apply to an ideal fluid?

A) The direction of the flow is equal at all points within a moving fluid
B) Fluid pressure is not influenced by fluid velocity
C) frictional forces bewteen fluid molecules are negligible
D) the fluid is incompressible

Answer 26

B) Fluid pressure is not influenced by fluid velocity

Question 27

Ideal fluid viscosity

Answer 27

No viscosity - no tendency to resist flow (no friction between fluid molecules)

Question 28

Non-ideal fluid viscosity

Answer 28

viscous! tendency to resist flow (friction present between fluid molecules)

Question 29

ideal fluid flow

Answer 29

laminar flow - smooth flow in layers, fluid elements travel together in straight lines

Question 30

non-ideal fluid flow

Answer 30

turbulent flow - disrupted layers of flow . fluid elements can rotate and swirl

Question 31

ideal fluid compressibility

Answer 31

incompressible - uniform density . The density of the fluid is modified by neither external forces nor its own weight when oriented in a fluid column

Question 32

non-ideal fluid compressibility

Answer 32

compressible - variable density

Question 33

The spherical aberration of a converging lens can be corrected by reducing the thickness of the lens periphery. The spherical aberattion of the lens is caused because light rays emerging from the lens:

A) are insufficiently refracted at the lens periphery
B) converge at the lens focal point uniformly
C) are excessively refracted at the lens periphery
D) are scattered due to chromatic dispersion

Answer 33

C) are excessively refracted at the lens periphery

An ideal lens generates an image of an object at a single focal point, regardless of the locations at which light rays enter and exit the lens. Spherical aberration describes the phenomenon by which real lenses’ perfectly rounded surfaces do not produce an image at a single point, but rather at a series of focal points.

Spherical aberration is most pronounced among rats entering and exiting the lens periphery. Furhtermore, correcting spherical aberration in converging lenses requires using an aspherical lens in which the thickness of the lens periphery is decreased relative to a perfectly rounded lens.

because reducing the thickness of the lens periphery will lead to less refraction of light, light rays exiting the periphery of a spherical converging lens can be said to refract excessively, converging on a focal point that is too close to the lens.

Question 34

Suppose that a simple circuit comprising one voltage source and one metallic resistor yields current I. If the resistor were replaced with another resistor that is identical except that is has 75% lower conductivity, then I would:

Answer 34

Decrease by a factor of 4

conductivity = 1/resistivity

Conduct new = conduct old / 4
Because conduct and resist are inversely proportional, the resistivity of the new resistor is increased by a factor of 4 compared to the original
Rnew = 4Rold

Ohm’s LawL V=IR, I=V/R

Inew = Iold /4 
4Inew = Iold

Question 35

A concave mirror has a 4-m radius of curvature. This mirror will focus distant objects at a location that is approximately:

Answer 35

2 m in front of the mirror

f = r/2 
f = 4/2 =2 

Point of focus is positioned in front of a concave mirror and behind a convex mirror

Question 36

A cube with a side length of 5cm floats on the surface of water. If 4/5 of the cube volume floats above the surface of the water, what is the density of the cube? Density of water = 1000 kg/m^3

Answer 36

200 kg /m^3

pobject/pfluid = vfluid/vobject

1/5 = pobject/pfluid = vfluid/vobject

Given that pfluid is 1,000, pobject=
pobject/1,000 = 1/5 = 200

Question 37

Energy order of wavelengths

Answer 37

Radio, Microwaves, IR, Visible (ROY G BIV), UV, X rays, Gamma

Increasing energy and frequency, decreasing wavelength

Question 38

Which of the following changes will increase the rate at which a mammalian organism loses body heat to the environment?

A) vasoconstriction of the organism’s superficial blood vessels
B) vasoconstriction of the organisms pulmonary blood vessels
C) replacement of the organism’s lean muscle mass with superficial fat
D) increase in the organism’s rate of pulmonary ventilation

Answer 38

D) increase in the organism’s rate of pulmonary ventilation

Heat transfer to the environment through ventilation: conductive heat transfer to inhaled air is followed by convective heat transfer through exhalation. Therefore, increasing the rate of respiration increases the rate of heat loss to the environment.

Question 39

Lung: body heat transferred to inhaled air

Answer 39

conduction

Question 40

warmed air exhaled into environment

Answer 40

convection

Question 41

superficial blood vessels transfer heat to the ambient air

Answer 41

conduction

Question 42

conduction

Answer 42

thermal energy passes between two objects through physical contact

Question 43

convection

Answer 43

the movement of a fluid, such as blood or air, serves to transport heat from warmer regions to cooler regions. Heat transfer between two objects or regions occurs until the temperature equalizes

Question 44

A gram-negative bacterium moves along a flat surface. What is the maximum displacement of the bacterium during a 1hr period of the bacterium moves at a constant speed of 25 um/s?

Answer 44

9 X 10^-2 m

displacement: 25 um/s * 1hr
= 2.5 X 10^-5 m/s * 3600 s
=9 X 10^-2 m

Question 45

um to m

Answer 45

1 X 10^-6 m = 1 um

Question 46

Which of the following procedures best describes the use of diffractive phenomena in separating the individual components of an electromagnetic wavefront?

A) single-slit diffraction of monochromatic light
B) double-slit diffraction of polychromatic light
C) grated diffraction of polychromatic light
D) grated diffraction of monochromatic light

Answer 46

C) grated diffraction of polychromatic light

A diffraction grate is a diffractive array composed of numerous slits with near-zero width similar to the wavelength of incident light. Compared to other diffractive phenomena, the band pattern caused by grated diffraction is characterized by relatively sharp and narrow peaks. This occurs because constructive and destructive interference occurs not only between waves emerging from adjacenet slits, but also between waves emerging from slits separated from each other by other slits.

The sharp, narrow peaks of the band pattern are useful in analyzing the properties of the light that enters the slit.

Consequently, transmitting polychromatic light compoased of light waves with different wavelengths through a diffraction grate will produce a series of multicolored arrays that enables an analysis of wavefront composition.

Question 47

Thin-film interference is observed when polychromatic light in incident on an interface formed when one semitransparent medium is layered on top of a second semitransparent medium. Which best explains why thin-film interference generates a multicolored array when two semitransparent fluids are useful?

A) the density of the top fluid varies along the fluid interface
B) the thickness of the top fluid varies along the fluid interface
C) the osmolarity of the top fluid varies along the fluid interface
D) the two semitransparent fluids decrease the number of reflection events at the fluid interface

Answer 47

B) the thickness of the top fluid varies along the fluid interface

Thin-film interference is influenced by the thickness of the film because light waves of a specific wavelength / color that interfere constructively or destructively at one film thickness may no longer interefere in the same fashion at a different thickness. Therefore, surface tension and other effects that logically distort the surface of a fluid may also cause multicolored arrays to appear on the surface as light that emerges from adjacent segments of the thin film varies in color composition

Question 48

Ultrasonic shock wave devices may be used in medical settings to disrupt disease structures located deep in the body. Among other variables, the device user can control the frequency of emitted waveforms. The adjustment of waveform frequency:

Answer 48

enables high-amplitude resonance within diverse target tissues.

Question 49

A circuit is constructed from two voltage sources and three resistors. If the potential of each voltage source is 10V and the resistance of each resistor is 5ohms, what is the current exiting Point A into R3?

402361

Answer 49

1.3 C/s

Kirchoff’s junction rule: Ienter = Iexit
Loop rule: Vn = I1R1 + I2R2 + … InRn

In this question, current flows through a complicated circuit with two voltage sources and three resistors. The circuit contains two loops, the first is composed of V1 and R1 and R3, the second is V2 and resistors R2 and R3.

One current unique to each loop (Ia and Ib) enters Point A and another current shared by both loops exits Point A (Iexit)
Iexit = Ia + Ib
Vloop 1 = IaR1 + IexitR3
Vloop 2 = IbR2 + IexitR3

Since R1=R2=R3, these variables can simply be replaced with R
V loop1 = (Ia+Iexit)R
V loop2 = (Ib + Iexit)R

The equation for Iexit can be substituted into each of these equations, yielding
V loop1 = (Ia + Ia + Ib)R = (2Ia + Ib)R
Vloop2 = (Ib + Ia + Ib)R = (Ia + 2Ib)R

Substiting the values of the voltage sources and resistors gives:
10=10Ia + 5Ib
10 = 5Ia + 10Ib

Multiplying the Vloop1 equation by 2 gives:
20 = 20Ia + 10Ib

Substituing the V loop 2 equation from the equation above:
20 - 10 = (20Ia + 10Ib ) - 10Ia + 5Ib
10 = 15Ia
Ia = 10/15 = 2/3

Substituing this value into the equation for Vloop 1 yields:
10 = 10 * 2/3 + 5Ib
Ib = 2/3

From the equation for Iexit:
I exit = 2/3 + 2/3 = 4/3 A

Question 50

relationship between fluid pressure and increase in fluid velocity

Answer 50

Venturri effect: the pressure of an ideal fluid in motion will decrease as fluid velocity increases

*this is not linear

Question 51

The absolute temperature of ideal gas molecules stored in a container is directly proportional to the:

A) quantity of gas molecules
B) intermolecular forces between gas molecules
C) average kinetic energy of gas molecules
D) maximum velocity of gas molecules

Answer 51

C) average kinetic energy of gas molecules

T propotional to Sum KE each molecules / quantity of molecules

Question 52

Light from an incandescent lightbulb passes through an ideal linear polarizer. The intensity of the light is 100 lm before passing through the filter, but 50 lm immediately after. Which of the following explains this discrepancy?

A) Refraction occurs as light enters the polarizer
B) Light waves with magnetic fields oriented perpendicular to the axis of polarization are absorbed
C) Light waves with electric fields oriented perpendicular to the axis of polarization are absorbed
D) Reflection occurs before light enters the polarizer

Answer 52

C) Light waves with electric fields oriented perpendicular to the axis of polarization are absorbed

A linear polarization filter allows for the transmission of electromagnetic radiation oriented parallel to the axis of polarization but inhibits the passage of radiation oriented perpendicular to this axis.

Light waves with magnetic fields oriented perpendicular to this axis of polarization have electric fields parellel to the axis. Such waves are transmitted through the polarizer and are not absorbed.

Question 53

A battery is connected to a circuit joining it to a lightbulb as shown. Increasing the internal resistance of the battery will

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Answer 53

Decrease the terminal voltage observed

EMF: V = EMF - IR

Question 54

A scientist studying capillary fluid exchange in a laboratory environment can most effectively increase the net fluid filtration out of the capilarries by doing which of the following?

A) increasing the speed of blood flow
B) increasing interstitial hydrostatic pressure
C) replacing blood with a less viscous fluid
D) decreasing blood osmotic pressure

Answer 54

D) decreasing blood osmotic pressure

Sterling equation: Jv = K [(Pc - Pif) - (OsC - Osif]
Therefore, only a decrease in the osmotic pressure within the capillary (eg, reduced concentration of plasma proteins or salts, would result in a furhter net positive Jv, increasing the movement of fluid out of the capillary

Question 55

An 8F parallel plate capacitor generates a voltage of 5 V when fully charged. How much charge is stored by the capacitor?

Answer 55

40 C

Q = CV = 8*5

Question 56

A cable with a tension of 45 N is used to suspend a 5kg mass M at rest against a wall. A horizontal force also pushes the mass against the wall. What is the magnitude and direction of the force of friction between the mass and the wall?

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Answer 56

5N , upward

Static equilibrium, sum of forces in x and y directions must both equal 0

Fg = mg = 5x10 = 50N

Sum y forces = (+45N + Ff) + (-50N) = 0
Ff - 5N = 0
Ff = +5 N

Therefore, the frictional force acting between the object and the wall must be 5N and oriented in an upward y direction

Question 57

Electric current units and equation

Answer 57

I = Q/t = V/R

also related to P=IV

Question 58

Blood flows 5cm from Point A to Point B within an artery supplying the brain. The systolic blood pressure at Point A is 8kPA. What is the systolic blood pressure when the blood reaches Point B? Assume the blood velocity and vessel diameter are constant over this distance. The density of blood is 1200 kg/m^3 and acceleration = 10 m/s^2.

Answer 58

7.4 kPA

Bernoullis equation: Pa + pghA + 1.2pvA^2 = PB + pghB + 1.2pvB^2

Because blood velociy and density are constant, the velocity terms can be elimiinated, and the equation can be rearranged to solve for the unknown pressure Pb

Pb = Pa + pg(ha - hb)
= 8000 PA + 1200(10) * (-0.05m)
=7400 kg/ms^2 = 7.4 kPA

Question 59

What graph would illustrate how the intensity of sound varies as a detector is moved further and further from a stationary source projecting sound in all directions. Ignore the doppler effect

Answer 59

Decreases, not linearly

I proportional to 1/r^2

Question 60

Two convex lenses are placed in series such that the image produced by first lens becomes the object of the second lens. If each lens is capable of producing a three-fold magnified image, what is the maximum magnification that this two-lens system can provide?

Answer 60

9.0

Mtotal = M1 * M2
placed in series, Mtotal = 3*3 = 9

Question 61

A sample of water vapor undergoes deposition and then melting. What happens to its entropy after each phase change?

Answer 61

Decreased with deposition, increasing with melting

Deposition = gas to solid 
melting = solid to liquid

Question 62

In the presence of an external magnetic field, a proton initially traveling upward is deflected into the path labeled as A in the figure. An electron with the same initial velocity as the proton travels through the magnetic field. What would be the expected path of the electron?

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Answer 62

D

Lorentz force: F=qvB

In the figure, the path of the proton is bent to the left in the presence of the external magnetic field. Consequently, the Lorentz force initially point to the left for positive charges in the magnetic field. However, because an electron is negatively charged, the trajectory of an electron would curve in the opposite direction, which is toward the right.

Although the Lorentz forces on the proton and electron have identical magnitudes, their paths are determined by the acceleration a imparted on each particle. Newton’s second law of motion implies that acceleration is inversely proportional to particle mass m:
a= F/m = qvB/m
Because the mass and inertia of an electron are much smaller than those of a proton, the electron experiences greater acceleration. Therefore, its path will have a greater curvature.

Question 63

An ideal fluid flows through a tube at an initial speed of 4 m/s. What will be the spped if the cross-sectional area decreases from 6m^2 to 2m^2?

Answer 63

12 m/s

Q1=Q2
A1V1 = A2V2
64 = 2v2
V2 = 12

Question 64

A 30 year old woman has an average bone tissue density of 1.8 g/cm^3 with an estimated bone tissue volume of 5,000 cm^3. Assuming her bone tissue volume remains the same, what will her bone tissue density be if she is expected to lose 1.5 kg of bone mass by age 70 due to osteoperosis?

Answer 64

1.5 g/cm^3

density = m/v

Mass age 30 = 1.8 (5000) = 9000 g

9000 g - 1500 g = 7500 g

density age 70 = 7500 g / 5000 cm^3 = 1.5

Question 65

In which of the following processes does energy convert from chemical energy to thermal and kinetic energy?

A) digesting a piece of bread
B) toasting a piece of bread
C) running a combustion engine
D) charging a cell phone with an external battery

Answer 65

C) running a combustion engine

Combustion is a chemical process through which the chemical energy stored in the molecular bonds of reactants is released by creating new bonds with less energy. The combustion event produces a flash of light and sound. The temperature of the gases increases (thermal) and the gases expand to move the engine’s components (kinetic)

A) chemical –> thermal. no KE produced
B) electric –> thermal + chemical
D) chemical –> electric –> chemical

Question 66

Laser Doppler vibrometers detect the Doppler shift of a laser beam reflected off an individual’s chest to measure respiratory rate. The calculated displacement of the chest during a breath using a 500-nm laser is shown below. If the laser source is stationary, at which point (s) would the observed frequency of the reflected last be 6 X 10^14 Hz?

Answer 66

I and III only

At maxima and minimua, v=0, no frequency shift

F= C/lamba = 3x10^8 / 500 x 10^-9 m = 6 X 10^14 Hz?

Question 67

A 100g block of iron at 31C is submerged in 50g of water at 25C in a thermally insulted container. If the temp of the system reaches equilibrium at 26C, what is the specific heat of iron? Note: the specific heat ofwater is 4.2J/g*C

Answer 67

0.42 J/g*C

Heat gained by water:
q=mcdeltat
=50 (4.2) (1) = 210 J

this is equal to heat lost by iron
qiron = 100 (ciron) (-5)
therefore qiron = -210 J
solve for c, c = 0.42

Question 68

A single 4-uF capacitor is added in series with the set of parallel capacitors shown above. What is the ratio of the new equivalent capacitance to the orginal equivalent capacitance?

401294

Answer 68

1:2

C parallel = 2uf + 2uf = 4uf

C series: I/C series = 1/C1 + 1/C2 + 1/C3 + …

When the 4uf capacitor is added in series with the given capacitor circuit, the equivalent capacitance of the new circuit is:
1/Ceq = 1/4uf + 1/4uf = 2/4 = 1/2
Ceq=2

Therefore, the equivalent capacitance decreased to 2 from 4

Question 69

The researchers want to measure the friction properties of the walking track using a block and a force gague. They apply a force to the block, causing it to slide at a constant velocity along the track. What information is required to calculate the coefficient of kinetic friction for the track surface?

A) mass of block, magnitude of force, contact area between block and track
B) mass of block, magnitude of force, desnity of block
C) mass of block, magnitude of force
D) mass of block, velocity of block, contact area between block and track

Answer 69

C) mass of block, magnitude of force

uk = Fa / mg

Question 70

During the weight test, a subject lifts the object 15 cm off the table, holds it motionless for a moment, and then drops it. What is the velocity of the object immediately before it lands on the table?

403234

Answer 70

1.7 m/s

KE = 1/2mv^2 
V = sqrt 2KE / m 

In this question, the 3kg object is dropped from a height of 15cm. KE initial is zero and E is equal to PE initial.
E= 3 (10) (0.15m) = 4.5 J
therefore KEfinal must equal 4.5J

V = sqrt 2(4.5) / 3 ~ sqrt 3 = 1.7 m/s

Question 71

The students compress a 40g golf ball a distance of 80cm against a spring with a spring constant k =100 N/m. After it is released from the spring, what is the velocity of the ball as it enters the air-filled tube?

Answer 71

40 m/s

Conservation of energy: 1/2kx^2 = 1/2mv^2
v= sqrt k/m * x
= sqrt 100m50 * 0.8 m
= 40 m/s

Question 72

Which graph best illustrates how the magnitude of the golf ball’s acceleration varies with its velocity in the air-filled region of the tube?

403220

Answer 72

increases, not linear because of formular given in passage, F=1/2CdpAv^2
a is directly proportional to sqaure of the ball’s velocity

Newton’s second law: f=ma, a=f/m

Question 73

In the first experiment, which of the following best describes the vertical motion of the gold ball through the water when the drag force, the bouyant force, and the weight of the ball sum to zero?

A) it accelerates downward
B) it accelerates upward
C) it moves with constant speed downward
D) it moves with constant speed upward

Answer 73

C) it moves with constant speed downward

Fnet = Fb + Fd - W = 0
When the net force is zero, Newton’s first law implies that the golf ball must move at a constant velocity (zero acceleration). Therefore, because the ball was initially dropped downward, it continues to move downward at a constant velocity in the water.

Question 74

frictional force acting on a mass sliding down an incline in dynamic equilibrium

402441

Answer 74

equal in magnitude to the weight of the object multiplied by the sine of the incline angle , equal and opposite to the component gravitational force that acts parallel to the surface of the incline F

mgsin(theta)

Question 75

What information about the LED circuit is required to calculate which resistance value should be used?

A) Source voltage and LED maximum forward current
B) wavelength of light, LED forward voltage, and LED maximum forward current
C) source voltage, LED forward voltage, and LED maximum forward current
D) LED forward voltage and LED maximum forward current

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Answer 75

C) source voltage, LED forward voltage, and LED maximum forward current

Ohm’s Law: V=IR
R= V/I

In this question, the resistor and the LED are connected in a series circuit, so the same current flows through each element. Thus, the LED maximum forward current If = current through the resistor.
By conservation of energy, the voltage rise from the voltage source Vs = the sum of the voltage drops across the other elements in the circuit, the resistor Vr, and the LED Vf
Consequently: Vr = Vs - Vf
and Ohm’s law implies that
R = (Vs - Vf )/ IF

Therefore, the two voltages and the current are needed to calculate the resistance

Question 76

The ECG and PPG signals show a time delay between the heart contraction and the inflow of blood into the finger due to the pulse wave velocity. What other information is needed to calculate the average pulse wave velocity?

A) distance from the heart to the finger
B) average blood pressure
C) relative blood volume in the finger
D) average acceleration of the blood

Answer 76

A) distance from the heart to the finger

V = deltax / deltat

Question 77

A charged particle is accelerated in a uniform electric field. When its velocity is 2 m/s, its electric potential energy is 100J and its KE is 10J. What is the particle’s potential energy when its velocity reaches 4m/s?

Answer 77

70 J

E initial = PE initial + KE initial = 110J

KEfinal = 1/2m(2v)^2 (velocity increases by a factor of 2, from 2 to 4 m/s)
Therefore KE increases by a factor of 4 , 40 J

110-40 = 70 J

Question 78

One horsepower is the amount of power required to lift a 75kg mass a vertical distance of 1 m in 1 s. What is 2ph equivalent to in watts?

Answer 78

1500 W

W = change in PE = mgdeltaH

P = W/t = mgdeltaH /t

1 hp = 75(10)(1) / 1 = 750 therefore 2 hp = 1500 W

Question 79

Uniform magnetic fields created by advanced medical imaging devices most likely come from the:

A) nuclear decay of device materials
B) electric forces generated by charged particles
C) motion of electrons along a circular pathway
D) storage of electrical charges at small distances

Answer 79

C) motion of electrons along a circular pathway

Question 80

An object is launched with a one-time burst of propulsion away from the surface of the moon. After the burst, which best describes the changes that occur as the object moves away from the moon’s surface?

A) the object mass is dissipated as heat
B) potential energy is converted into KE
C) KE is converted into PE
D) total mechanical energy is not conserved

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Answer 80

C) KE is converted into PE

Question 81

A star orbiting a black hole in the clockwise direction begins to slowly spiral inward due to a counterclockwise drag force. When the star’s radial distance from the black hole is 1.0 X 10^6 km, a drag froce of 250N acts perpendicular to the orbit radius. What is the torque on the star?

402701

Answer 81

2.5 X 10^11 Nm in the counterclockwise direction

T=rFsin(theta)
=rFsin(90)
T= R* F

In this question, a star orbits a black hole in a counterclockwise direction at a radial distance of 1.0 X 10^6 km. The translational force applied on the star acts in a counterclockwise direction (opposition to the clockwise orbit of the star), perpendicular to the radial vector. Consequently, theta is 90, and torque magnitude is:
T = 1.0 X 10^6 km * 250 N
=2.5 X 10^11

Question 82

A train is traveling west at a velocity of 25 m/s. Another train is traveling east directly toward the west-bound train at a velocity of 15 m/s. The west-bound train blows its whistle with a frequency of 600 Hz when the two trains are 1000 m apart and then blows its whistle again 10 seconds later. For passengers on the east-bound train, how will the perceived frequency of the first whistle compare with the perceived frequency of the second whistle? Note: Use 350 m/s for the speed of sound in air

Answer 82

It will be approximately identical to the frequency of the second whistle

The velocities of the two trains have not changed

Question 83

Blood flow in the aorta becomes turbulent at a Reynolds number of 6,000. If the diameter of the aorta is approximately 2 cm, what is the avg velocity of blood at this Reynolds number?

402658

Answer 83

85 cm/s

Vcritical = (n * Re) / (p * d)

= 0.003 (viscosity of blood) * 6000 / 1060 (density of blood) * 0.02
=90 ~ 85

Question 84

The diameter of a segment of an artery is reduced by a factor of two due to an obstruction. Assume that the flow is incompresible and laminar, and therefore follows the continuity equation. Compared to an unobstructed segment of the artery, the velocity of blood in the obstructed segment of the artery is:

402659

Answer 84

4 times as large

A1V1=A2V2

A=pi * (d/2)^2 = pi/4 * d^2

v2= (d1 / 0.5 * d1) ^2 * v1 = 4V1

Question 85

A fluid is stirred until it becomes turbulent. After the stirring stops, the fluid gradually returns to its original static state. The turbulent flow does not continue indefinitely because:

A) the fluid is incompressible
B) the KE in the flow is transferred to gravitational potential energy
C) the energy put into the flow by stirring is transferred by convection
D) the energy put into the flow by stirring is dissipated by the fluid’s viscosity

402660

Answer 85

D) the energy put into the flow by stirring is dissipated by the fluid’s viscosity

Shear force is proportional to viscosity n * delta V/delta Y

Once the stirring stops , the only force acting on the flow is the viscous shearing force resulting from the velocity gradients present within the fluid. Consequently, the KE of the flow is dissipated by frictional shearing forces such that the velocity decreases and the fluid gradually returns to a steady state.

Question 86

The blood leaving the aorta reaches the circle of Willis, a collection of arteries that supplies blood to the brain. Assume that it takes 600ms for blood to reach the circle of Willis and that its average velocity is reduced to 61 cm/s. What is the average accelerated experienced by the blood?

Answer 86

-67.0 cm/s^2

61-101 / 600 ms
=-40/600ms ( * 1X10^-3 ms/s)
=-67.0 cm/s^2

Question 87

Is entropy a state or path function

Answer 87

state

Question 88

Is work a state or path function

Answer 88

Path

Question 89

Adiabatic processes

Answer 89

no heat exchange between the system and the environment

Question 90

The compressor exerts a pressure of 400 PA to compress nitrogen gas from 3L to 1L. If no heat transfer occurs, how much will the internal energy of the gas change? Note: 1 m^3 = 1000L, and 1 J = 1 PA*m^3

401248

Answer 90

0.8 J

W=P(deltaV)

Delta U = Q + W , though no direct heat transfer, Q = 0, Delta U = W

Change in volume 3-1 = 2L, must be converted to m^3
2 L * (1m^3 / 10^3 L) = 0.002 m^3

W= 400 Pa * 0.002 m^3 = 0.8J

Question 91

Which of the following best describes the type of heat transfer that occurs in the external cooler of the Hampson-Linde cycle apparatus?

A) Conduction occurs as dry ice loses heat to the coil of the external cooler
B) Convection occurs as the coil of the external cooler loses heat to dry ice
C) Conduction occurs as the coil of the external cooler gains heat from nitrogen gas
D) Convection occurs as nitrogen gas gains heat from the coil of the external cooler?

401249

Answer 91

C) Conduction occurs as the coil of the external cooler gains heat from nitrogen gas

As the nitrogen gas flows through the coil of the external cooler, it transfers energy obtained during the compression stage to different sections of the coil. Therefore, the heat transfer results from the physical contact of the nitrogen gas and the coil and is primarily a conductive transfer of heat.

Question 92

A tissue sample at 275 K is submerged in 2kg of liquid nitrogen at 70K for cyropreservation. The final temperature of the nitrogen is 75K. What is the heat capacity of the sample in J/K?

401250

Answer 92

100 J/K

q = 2 (2) (5K) = 20 kJ
*specific heat is given in Table 1, c = 2

Because nitrogen reaches a final temperature at 75K, the system must have reached thermal equilibrium. At equilibrium, temperature is uniform throughout the system, and therefore the final temperature of the tissue sample is also 75K. Given that the tissue’s initial temperature was 275K, deltaT = -200 K. The heat lost by the tissue was transferred to the nitrogen, q=-20kg. The heat capacity of the tissue is:
C= q/deltaT
= -20kj/-200K = 0.1 kj/K
* 1000J = 100 J/K

Question 93

Approximately how much heat in kJ was released from the combustion of the 1g sample?

401166

Answer 93

20 kJ

q=mcAT
In the experiment described in the passage, the heat released from the combustion reaction is absorbed by the surrounding water, m=500g. The description of a gram calorie (cal) is the definition of water’s specific heat: c=1 cal/g * degree C). The max temperature change deltaT = 10 (Figure 1) is due to the total heat released from the combustion reaction.
q = 500(1)(10) = 5000 cal

Use conversion in the passage
q=5000 cal (4.2 kJ / 1000 cal) = 21 kJ

Question 94

Fatty tissue has a lower electrical conductivity than lean non-fat tissue. This property allows body fat to be estimated by modeling the tissue between two points as a single resistor. For a constant current entering the body, relatively high body fat content would be indicated by:

A) an increased current leaving the body
B) a decreased current leaving the body
C) a lower voltage difference across the body
D) a higher voltage difference across the body

401072

Answer 94

D) a higher voltage difference across the body

V=IR

Higher body fat, higher resistance, increased voltage
less conductive = more resistance

Question 95

An individual that weighs 700N in air has an apparent weight of 40N when submerged in water. What is the volume of the displaced water?

401073

Answer 95

66 L

Wapparent = Wair - Fb
Fb = Wair - Wapparent
=700 N - 40 N = 660 N

Archimedes principle: Fb=pVg
V=Fb/pg
=660 N / (1)(10m/s^2) = 66 L

Question 96

A bone sample (specific graviety = 3) completely immersed in isopropyl alcohol (specific gravity = 0.8) is subject to a bouyant force of 4N. What bouyant force would the same bone sample experience when immersed in water?

401074

Answer 96

5N

FB1/FB2 = p1Vg/p2Vg
FB, water = 1/0,8 * 4N = 5N

Question 97

The hydrostatic weighing container above is filled with water to a depth of 2m. If point A is 0.7 m above the bottom of the container, what is the pressure due to weight of the water at point A? Note: the density of water is 1,000 kg/m^3.

401075

Answer 97

1.3 X 10^4 N/m^2

P=pgh
h=2-0.7m = 1.3 m
P = 1000 * 10 * 1.3 = 13,000 N/M^2

Question 98

To test the integrity of the capacitor, researchers maintained a voltage of 50mV across Cm and Rm for several minutes. During this time, how much energy was stored in the capacitor?

401065

Answer 98

2.5 X 10^-9 J

U=1/2CV^2

The 50Mv source is connected to the parallel circuit consisting of Cm and Rm. Because these two components are in parallel, they have the same V across them. Therefore, the V across Cm (and the value of U) only depends on the 50 mV source, and is independent of Rm.

C is given as 2 uF, and V is given as 50mV. Using 1uF = 10^-6 F and 1mv = 10^-3 V
U=1/2 (2 X 10^-6 F) (5 X 10^-2 V) ^2 = 2.5 X 10^-9 J

Question 99

If the voltage acorss Cm and Rm is 2V, how much power is dissipated from Rm?

401067

Answer 99

20 mW

P=IV=I^2 * R = V^2/R
P= 2V^2 / 200 ohms = 0.02 W

1 W = 1000 mW , 0.02 W = 20 mW

Question 100

A voltmeter is placed across RL. If a charge of 0.1 C flows into the input node every 2s, what will be the reading on the voltmeter?

401068

Answer 100

50 V

I = Q/t
=0.1C / 2s
=0.05 A

V=IR
=0.05 A (1000 ohms) = 50 V

Question 101

An anmeter is added to the circuit as shown. The voltage across the capacitor Cm is kept constant while the frequency is increased from 200 Hz to 400 Hz. Through this change, the value shown by the anmeter:

401069

Answer 101

Will increase by 100%

I=V/(R/2) = 2VR
Reading doubles

Question 102

For equally sized trocar-cannula systems, incisions made for blunt-tripped trocars are smaller than thsoe made by sharp-tipped trocars, forcing the skin to stretch more around the cannula. This option better prevents slipping after the cannula is inserted because:

A) it decreases the contact surface area
B) it decreases the static friction
C) it increases the normal force
D) it increases the coefficient of static friction

Answer 102

C) it increases the normal force

Less stretch = lower normal force and lower friction

Question 103

To remove the cannula at the end of a laraposcopy, the surgeon pulls on the cannula with increasing force. How does the frictional force vary with the applied pulling force before and after the cannula begins to slide?

401017

Answer 103

check Uworld answers,

increasing applied force (static friction = applied force), peak at threshhold of motion, then decreases, then straight line when kinetic friction is constant

Question 104

After the load leaves the ramp, it slides on a horizontal table with a coefficient of kinetic friction uk of 0.1 and a coefficient of static friction us of 0.3. If the direction of the load’s velocity is positive, what is the acceleration of the load?

401018

Answer 104

-1m/s^2

Fk =ukN = ukmg

A=Fnet/m (Newton’s 2nd law)

Because the direction of the load’s velocity is positive and friction acts in the direction opposite the force, the net force and acceleration are both negative (fnet = -Fk). The acceleration of the load is calculated as the negative of the kinetic friction divided by the object’s mass:
a=-ukmg / g = -ukg
=-(0.1)(10 m/s^2) = -1m/s^2

Question 105

When the angle of inclination of the ramp increases, kinetic friction:

A) decreases because the weight component perpendicular to the ramp decreases
B) decreases because the weight component parallel to the ramp decreases
C) increases because the weight component perpendicular to the ramp increases
D) increases because the weight component parallel to the ramp increases

401019

Answer 105

A) decreases because the weight component perpendicular to the ramp decreases

Fk=ukN
The normal force is equal and opposite to this perpendicular component : N=Wcos(theta)
Fk = uk W cos (theta)

As theta increases, the value of cosine decreases, and both the perpendicular component of the weight and normal force also decrease. Therefore, because kinetic friction is proportional to the normal force, kinetic friction must decrease.

Question 106

Which of the following expressions gives the magnitude of the work done by the frictional force F is the synthetic skin and the load slide down the entire length L of the ramp and then are pushed back up to their original position?

A) 0
B) F(L)
C) F (2L)
D) F(3L)

401020

Answer 106

C) F (2L)

W=change E
W= -FL + -FL = Wdown + Wup
Absolute value of W = 2L
(As the load slides down the ramp of length L, the frictional force F acts in the opposite direction, up the ramp. When the load slides up the ramp, F again opposes the motion)

The work done by a conservative force (gravity, electrostatic) depends on the net displacement (not distance) the force is applied. Because there is no net displacement in this case, the work done by the gravity would be zero. However, friction is NOT a conservative force.

Question 107

When the load slides down the ramp during the experiment, the magnitude of the kinetic friction force between the ramp and the synthetic skin depends on:

I. the area of the contact surface
II. the mass of the load
III. the speed of the synthetic skin

401021

Answer 107

II only

uk depends only on the mass of the load and the normal force