PhysicsBowl 2009
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Instruction
- Questions: The test is composed of 50 questions; however, students answer only 40 questions.
Division 1 students will answer only questions 1 – 40. Do not answer questions 41 – 50.
Division 2 students will answer only questions 11 – 50. Do not answer questions 1 – 10. - Calculator: A hand-held calculator may be used. Any memory must be cleared of data and programs. Calculators may not be shared.
- Formulas and constants: Only the formulas and constants provided with the contest may be used.
- Time limit: 45 minutes.
Approximately how many seconds is it until the PhysicsBowl takes place in the year $\textit{2109}$ ?
$\textbf{(A) }$ $10^2 \qquad$ $\textbf{(B) }$ $10^7 \qquad$ $\textbf{(C) }$ $10^8 \qquad$ $\textbf{(D) }$ $10^9 \qquad$ $\textbf{(E) }$ $10^{12}$
$\textbf{D}$
$100\ \text{yr}=100\ \text{yr}\times\dfrac{365\ \text{dy}}{1\ \text{yr}}\times\dfrac{24\ \text{hr}}{1\ \text{dy}}\times\dfrac{60\ \text{min}}{1\ \text{hr}}\times\dfrac{60\ \text{s}}{1\ \text{min}}=3.15\times10^9\ \text{s}$
A room has a floor area of $25\ \text{m}^2$. What is this area written in $\text{cm}^2$?
$\textbf{(A) }$ 25000000$ \qquad$ $\textbf{(B) }$ 250000$ \qquad$ $\textbf{(C) }$ 2500$ \qquad$ $\textbf{(D) }$ 0.25$ \qquad$ $\textbf{(E) }$ 0.0025
$\textbf{B}$
$25\ \text{m}^2=25\ \text{(100 cm)}^2=250000\ \text{cm}^2$
Which one of the following quantities can have its unit expressed as $\dfrac{\text{kg}\cdot\text{m}}{\text{s}^2}$ ?
$\textbf{(A) }$ Force$ \qquad$ $\textbf{(B) }$ Power$ \qquad$ $\textbf{(C) }$ Energy$ \qquad$ $\textbf{(D) }$ Pressure$ \qquad$ $\textbf{(E) }$ Linear Momentum
$\textbf{A}$
By Newton’s Second Law $F=ma$, the unit of force is $\text{N}=(\text{kg})\cdot(\text{m/s}^2)=\dfrac{\text{kg}\cdot\text{m}}{\text{s}^2}$.
The length measurement $L=0.01230\ \text{cm}$ has how many significant digits?
$\textbf{(A) }$ 6$ \qquad$ $\textbf{(B) }$ 5$ \qquad$ $\textbf{(C) }$ 4$ \qquad$ $\textbf{(D) }$ 3$ \qquad$ $\textbf{(E) }$ 2
$\textbf{C}$
In a measurement, all non-zero values are significant, as well as any zeroes at the end of a measurement after the decimal. Therefore, the significant digits are 1230.
At an instant of time, a block of mass 0.50 kg has a position of 0.3 m, a speed of 4.0 m/s, and an acceleration of $1.0\ \text{m/s}^2$. What is the block’s kinetic energy (in Joules) at this instant?
$\textbf{(A) }$ 1.0$ \qquad$ $\textbf{(B) }$ 1.5$ \qquad$ $\textbf{(C) }$ 2.0$ \qquad$ $\textbf{(D) }$ 4.0$ \qquad$ $\textbf{(E) }$ 8.0
$\textbf{D}$
The kinetic energy $KE=\dfrac12mv^2=\dfrac12(0.5)(4)^2=4.0\ \text{J}$.
For a standing wave mode on a string fixed at both ends, adjacent antinodes are separated by a distance of 20 cm. Waves travel on this string at a speed of 1200 cm/s. At what frequency is the string vibrated to produce this standing wave?
$\textbf{(A) }$ 120 Hz$ \qquad$ $\textbf{(B) }$ 60 Hz$ \qquad$ $\textbf{(C) }$ 40 Hz$ \qquad$ $\textbf{(D) }$ 30 Hz$ \qquad$ $\textbf{(E) }$ 20 Hz
$\textbf{D}$
Since the distance between adjacent antinodes is 20 cm, the wavelength is $\lambda=40\ \text{cm}$. The frequency of the wave is $f=\dfrac{v}{\lambda}=\dfrac{1200}{40}=30\ \text{Hz}$.
The equivalent resistance of the circuit shown with resistances $R_1=4.00\ \Omega$, $R_2=3.00\ \Omega$, and $R_3=2.00\ \Omega$ is
$\textbf{(A) }$ $0.111\ \Omega \qquad$ $\textbf{(B) }$ $0.923\ \Omega \qquad$ $\textbf{(C) }$ $1.08\ \Omega \qquad$ $\textbf{(D) }$ $3.00\ \Omega \qquad$ $\textbf{(E) }$ $9.00\ \Omega$
$\textbf{D}$
Three resistors are connected in parallel, sop the equivalent resistance is $R_{eq}=1/\left(\dfrac1{R_1}+\dfrac1{R_2}+\dfrac1{R_3}\right)=1/\left(\dfrac14+\dfrac13+\dfrac12\right)=0.923\ \Omega$.
Consider the motion of an object given by the position vs. time graph shown. For what time(s) is the speed of the object greatest?
$\textbf{(A) }$ At all times from $t=0.0\ \text{s}\rightarrow t=2.0\ \text{s} \qquad\newline$
$\textbf{(B) }$ At time $t=3.0\ \text{s} \qquad\newline$
$\textbf{(C) }$ At time $t=4.0\ \text{s} \qquad\newline$
$\textbf{(D) }$ At all times from $t=5.0\ \text{s}\rightarrow t=7.0\ \text{s} \qquad\newline$
$\textbf{(E) }$ At time $t=8.5\ \text{s}$
$\textbf{C}$
Speed is the slope of the tangent line is the position vs. time graph. To achieve the greatest speed, the tangent line should be as vertical as possible, which happens at time $t=4.0\ \text{s}$.
The free fall trajectory of an object thrown horizontally from the top of a building is shown as the dashed line in the figure. Which sets of arrows best correspond to the directions of the velocity and of the acceleration for the object at the point labeled P on the trajectory?
$\textbf{A}$
The velocity is tangent to the trajectory pointing bottom right. The gravitational acceleration always points to the ground.
A car with mass $M$ initially travels to the East with speed $4V$. A truck initially travels to the West with speed $3V$. After the vehicles collide, they move together to the West with common speed $2V$. What is the mass of the truck?
$\textbf{(A) }$ $2M \qquad$ $\textbf{(B) }$ $3M \qquad$ $\textbf{(C) }$ $4M \qquad$ $\textbf{(D) }$ $5M \qquad$ $\textbf{(E) }$ $6M$
$\textbf{E}$
According to the conservation of linear momentum, $M_2(3V)-M(4V)=(M_2+M)(2V)$, so we get $M_2=6M$.
What is the orientation of the Earth, Sun, and Moon during a total lunar eclipse?
$\textbf{(A) }$ The Sun is between the Earth and Moon.$ \qquad\newline$
$\textbf{(B) }$ The Earth is between the Sun and Moon.$ \qquad\newline$
$\textbf{(C) }$ The Moon is between the Sun and Earth.$ \qquad\newline$
$\textbf{(D) }$ The Earth, Moon, and Sun make a right triangle.$ \qquad\newline$
$\textbf{(E) }$ The Earth is above the Sun and Moon.
$\textbf{B}$
During a lunar eclipse, sunlight is blocked from the Moon by the Earth. Hence, the Earth is between the Sun and the Moon.
A 5.0 kg solid sphere is in free fall near the surface of the Earth. What is the magnitude of the gravitational force acting $\textit{on the Earth by the solid sphere}$? The Earth’s mass is $5.98\times10^{24}\ \text{kg}$.
$\textbf{(A) }$ 0 N$ \qquad\newline$
$\textbf{(B) }$ 5 N$ \qquad\newline$
$\textbf{(C) }$ It is immeasurably small, but not zero.$ \qquad\newline$
$\textbf{(D) }$ 50 N$ \qquad\newline$
$\textbf{(E) }$ $5.98\times10^{25}\ \text{N}$
$\textbf{D}$
By Newton’s Third Law, the gravitational force acting on the sphere by the earth $G=mg=(5)(10)=50\ \text{N}$ is equal to the force exerted on the earth by the sphere.
Questions 13 – 14 deal with the following information:
In the figure to the right, a box moves with speed 5.00 m/s at the bottom of a rough, fixed inclined plane. The box slides with constant acceleration to the top of the incline as it is being pushed directly to the left with a constant force of $F=240\ \text{N}$. The box, of mass $m=20.0\ \text{kg}$, has a speed of 2.50 m/s when it reaches the top of the incline.
What is the magnitude of the acceleration of the box as it slides up the incline?
$\textbf{(A) }$ $12.0\ \text{m/s}^2 \qquad$ $\textbf{(B) }$ $10.0\ \text{m/s}^2 \qquad$ $\textbf{(C) }$ $5.88\ \text{m/s}^2 \qquad$ $\textbf{(D) }$ $1.88\ \text{m/s}^2 \qquad$ $\textbf{(E) }$ $0.938\ \text{m/s}^2$
$\textbf{E}$
The mass moves with a constant acceleration from the initial speed $v_i=5.00\ \text{m/s}$ to the final speed $v_f=2.50\ \text{m/s}$. According to $v_f^2-v_i^2=2ax$, we have $a=\dfrac{v_f^2-v_i^2}{2x}=\dfrac{2.5^2-5^2}{2(10)}=-0.9375\ \text{m/s}^2$.
How much work is done by the applied force, $F$ , to the box?
$\textbf{(A) }$ 2400 J$ \qquad$ $\textbf{(B) }$ 1920 J$ \qquad$ $\textbf{(C) }$ 1200 J$ \qquad$ $\textbf{(D) }$ 988.5 J$ \qquad$ $\textbf{(E) }$ -187.5 J
$\textbf{B}$
The work done on the mass is $W=\vec{F}\cdot\vec{s}$. The force points to the left, and the horizontal component of the displacement is $x=8.0\ \text{m}$. Thus we get $W=Fx=(240)(8)=1920\ \text{J}$.
An ideal gas in a closed container of volume 0.6 L is at a temperature of 100$^\circ\text{C}$. If the pressure of the gas is 5.2 atm , how many moles of gas are in the container?
$\textbf{(A) }$ 0.0048$ \qquad$ $\textbf{(B) }$ 0.018$ \qquad$ $\textbf{(C) }$ 0.49$ \qquad$ $\textbf{(D) }$ 1.83$ \qquad$ $\textbf{(E) }$ 490
$\textbf{C}$
By the ideal gas equation $PV=nRT$, we have the mole number of molecules $n=\dfrac{PV}{RT}=\dfrac{\left(2.5\times1.013\times10^5\right)\left(0.6\times10^{-3}\right)}{(8.31)(100+273)}=0.49\ \text{mol}$.
What condition must be met in order to use the rotational kinematics equation $\Delta\theta=\omega_0t+\dfrac12\alpha t^2$?
$\textbf{(A) }$ The angular acceleration is constant.$ \qquad\newline$
$\textbf{(B) }$ The angular velocity is constant.$ \qquad\newline$
$\textbf{(C) }$ The linear acceleration is zero.$ \qquad\newline$
$\textbf{(D) }$ The angular acceleration is zero.$ \qquad\newline$
$\textbf{(E) }$ There is no restriction on the use of this equation.
$\textbf{A}$
Just like the equation $\Delta x=v_0t+\dfrac12at^2$ is only valid when the linear acceleration is a constant, $\Delta\theta=\omega_0t+\dfrac12\alpha t^2$ is valid only when the angular acceleration is a constant.
A toy car moves 3.0 m to the North in one second. The car then moves at 9.0 m/s due South for two seconds. What is the average speed of the car for this three second trip?
$\textbf{(A) }$ 4.0 m/s$ \qquad$ $\textbf{(B) }$ 5.0 m/s$ \qquad$ $\textbf{(C) }$ 6.0 m/s$ \qquad$ $\textbf{(D) }$ 7.0 m/s$ \qquad$ $\textbf{(E) }$ 12.0 m/s
$\textbf{D}$
The car moves 0.3 m to the north and then $x=vt=(9)(2)=18\ \text{m}$ to the south. The total distance is $3+18=21\ \text{m}$, so the average speed is $21/3=7.0\ \text{m/s}$.
A wire has a conventional current $I$ directed to the right. At the instant shown in the figure, an electron has a velocity directed to the left. The magnetic force on the electron at this instant is
$\textbf{(A) }$ zero.$ \qquad\newline$
$\textbf{(B) }$ directed out of the plane of the page.$ \qquad\newline$
$\textbf{(C) }$ directed into the plane of the page.$ \qquad\newline$
$\textbf{(D) }$ directed toward the top of the page.$ \qquad\newline$
$\textbf{(E) }$ directed toward the bottom of the page.
$\textbf{E}$
The magnetic field associated with the current in the wire is directed out of the plane of the page at the location of the electron (Right-hand rule… right thumb along current, fingers wrap in the sense of the field). Using the right hand rule for the force on a charged particle in a field… right fingers point to the left (velocity), the fingers are curled toward the field (out of paper) and the right thumb points toward the top of the paper… except that since the charge is negative (electron), the hand is flipped 180 degrees giving the final direction of the force to be toward the bottom of the page.
A scientist claims to be investigating “The transfer of energy that results from the bulk motion of a fluid.” Which of the following terms best describes the energy transfer method that this scientist is studying?
$\textbf{(A) }$ radiation$ \qquad$ $\textbf{(B) }$ convection$ \qquad$ $\textbf{(C) }$ conduction$ \qquad$ $\textbf{(D) }$ latent heat$ \qquad$ $\textbf{(E) }$ specific heat
$\textbf{B}$
Energy transfer from the bulk motion of a fluid is best related to the term convection.
A projectile launched from the ground landed a horizontal distance of 120.0 m from its launch point. The projectile was in the air for a time of 4.00 seconds. If the projectile landed at the same vertical position from which it was launched, what was the launch speed of the projectile? Ignore air resistance.
$\textbf{(A) }$ 22.4 m/s$ \qquad$ $\textbf{(B) }$ 30.0m/s$ \qquad$ $\textbf{(C) }$ 36.1 m/s$ \qquad$ $\textbf{(D) }$ 42.4 m/s$ \qquad$ $\textbf{(E) }$ 50.0 m/s
$\textbf{C}$
The horizontal component of the velocity is a constant in 4.00 s, so $v_x=120/4=30\ \text{m/s}$. In the vertical direction, the projectile experiences a free fall for 4/2=2 s, so the vertical component of the velocity when reaching the ground is $v_y=\dfrac12gt^2=\dfrac12(10)(2)^2=20\ \text{m/s}$. By the Pythagorean Theorem, $v=\sqrt{v_x^2+v_y^2}=\sqrt{30^2+20^2}=36.1\ \text{m/s}$.
A 20.0 kg box remains at rest on a horizontal surface while a person pushes directly to the right on the box with a force of 60 N . The coefficient of kinetic friction between the box and the surface is $\mu_k=0.20$. The coefficient of static friction between the box and the surface is $\mu_s=0.60$. What is the magnitude of the force of friction acting on the box during the push?
$\textbf{(A) }$ 200 N$ \qquad$ $\textbf{(B) }$ 120 N$ \qquad$ $\textbf{(C) }$ 60 N$ \qquad$ $\textbf{(D) }$ 40 N$ \qquad$ $\textbf{(E) }$ 0 N
$\textbf{C}$
Since the box remains at rest, the force of friction is equal to the force applied by the person.
A point object is connected to the end of a long string of negligible mass and the system swings as a simple pendulum with period $T$. What is the period of the pendulum if the string is made to have one-quarter of its original length?
$\textbf{(A) }$ $4T \qquad$ $\textbf{(B) }$ $2T \qquad$ $\textbf{(C) }$ $T \qquad$ $\textbf{(D) }$ $\dfrac12T \qquad$ $\textbf{(E) }$ $\dfrac14T$
$\textbf{D}$
The period $T=2\pi\sqrt{\dfrac{L}{g}}$. By shortening the length of the string to $\dfrac14L$, the new period is $\dfrac12T$.
A person rubs a neutral comb through their hair and the comb becomes negatively charged. Which of the following is the best explanation for this phenomenon?
$\textbf{(A) }$ The hair gains protons from the comb.$ \qquad\newline$
$\textbf{(B) }$ The hair gains protons from the comb while giving electrons to the comb.$ \qquad\newline$
$\textbf{(C) }$ The hair loses electrons to the comb.$ \qquad\newline$
$\textbf{(D) }$ The comb loses protons to the person’s hand holding the comb.$ \qquad\newline$
$\textbf{(E) }$ The comb loses protons to the person’s hand while also gaining electrons from the hair.
$\textbf{C}$
Protons are confined to the nuclei of the atoms and are not mobile, so the charging is done via electron transfer. That the comb becomes negative means that electrons were accepted by the comb from the hair.
A point mass moves along a horizontal circular path of radius 8.0 m with a constant kinetic energy of 128 J. What is the magnitude of the net force acting on the mass as it moves?
$\textbf{(A) }$ 64 N$ \qquad$ $\textbf{(B) }$ 32 N$ \qquad$ $\textbf{(C) }$ 16 N$ \qquad$ $\textbf{(D) }$ 8 N$ \qquad$ $\textbf{(E) }$ 0 N
$\textbf{B}$
Now we know the kinetic energy $KE=\dfrac12mv^2=128\ \text{J}$. The net force acting on the mass provides the centripetal force of the circular motion, thus $F_{net}=\dfrac{mv^2}{r}=\dfrac{2KE}{r}=\dfrac{2(128)}{8}=32\ \text{N}$.
Which of the following types of electromagnetic radiation has the largest magnitude of momentum per photon?
$\textbf{(A) }$ FM radio$ \qquad\newline$ $\textbf{(B) }$ Microwaves$ \qquad\newline$ $\textbf{(C) }$ Violet light$ \qquad\newline$ $\textbf{(D) }$ Infrared light$ \qquad\newline$ $\textbf{(E) }$ Gamma rays
$\textbf{E}$
The momentum of a photon $p=\dfrac{E}{c}=\dfrac{h\nu}{c}$. So the photon with highest frequency $\nu$ has the largest momentum.
For the circuit shown, when a shorting wire (no resistance) connects the points labeled A and B, which of the numbered light bulbs become brighter? Assume that all four bulbs are identical and have resistance $R$.
$\textbf{(A) }$ Bulb 1 only$ \qquad\newline$
$\textbf{(B) }$ Bulb 2 only$ \qquad\newline$
$\textbf{(C) }$ Bulb 3 only$ \qquad\newline$
$\textbf{(D) }$ Bulbs 1 and 3 only$ \qquad\newline$
$\textbf{(E) }$ Bulbs 1, 2, and 3
$\textbf{D}$
By adding the shorting wire, bulb 4 is now bypassed. This reduces the overall resistance of the circuit. By decreasing the resistance, the current in the circuit increases. Using Ohm’s Law, bulb 1 now has more current and therefore there is a larger power associated with it and it burns more brightly. Bulb 2, however, dims using Kirchhoff’s Loop Rule… the battery voltage is unchanged while bulb 1 has a larger voltage… meaning that the voltage across bulb 2 must decrease and it therefore is dimmer. Finally, since there is a bit more current in the circuit and bulb 2 becomes dimmer, more of the current will be directed to bulb 3, thereby makes it brighter.
A proton moves straight up the plane of this page into a region that has a magnetic field directed to the right. If the particle is undeflected as it passes through this region, in what direction must there be a component of electric field? Ignore gravity.
$\textbf{(A) }$ To the left$ \qquad\newline$
$\textbf{(B) }$ Into the page$ \qquad\newline$
$\textbf{(C) }$ Out of the page$ \qquad\newline$
$\textbf{(D) }$ Down the page $ \qquad\newline$
$\textbf{(E) }$ To the right
$\textbf{C}$
By the right-hand rule, there is a magnetic force on the proton directed into the plane. (Fingers point up the page, curl them to the right… thumb points inward). Since the proton is undeflected, there must be an electric force of equal size directed out of the plane of the page.
White light shines through ideal filters to produce cyan light. If an ideal pigment appears yellow in white light, what color does the pigment appear in the cyan light?
$\textbf{(A) }$ magenta$ \qquad$ $\textbf{(B) }$ yellow$ \qquad$ $\textbf{(C) }$ blue$ \qquad$ $\textbf{(D) }$ red$ \qquad$ $\textbf{(E) }$ green
$\textbf{E}$
Cyan = blue + green, so the red light is blocked by the filter. Yellow = red + green, so the pigment will absorb blue light. When cyan light shines on the yellow pigment, blue light will be absorbed and green light reflected. So the pigment appears green.
The diagram shows the lowest four energy levels for an electron in a hypothetical atom. The electron is excited to the -1 eV level of the atom and transitions to the lowest energy state by emitting only two photons. Which of the following energies could not belong to either of the photons?
$\textbf{(A) }$ 2 eV$ \qquad$ $\textbf{(B) }$ 4 eV$ \qquad$ $\textbf{(C) }$ 5 eV$ \qquad$ $\textbf{(D) }$ 6 eV$ \qquad$ $\textbf{(E) }$ 9 eV
$\textbf{B}$
To transit to the -12 eV state with only two photon emitted, the two-step transition could be $-1\ \text{eV}\rightarrow-3\ \text{eV}\rightarrow-12\ \text{eV}$, which emitted photons of energy of 2 eV and 9 eV, or $-1\ \text{eV}\rightarrow-7\ \text{eV}\rightarrow-12\ \text{eV}$, which emitted photons of energy of 6 eV and 5 eV. 4 eV is impossible.
In terms of the seven fundamental SI units in the MKS system, the Ohm is written as
$\textbf{(A) }$ $\dfrac{\text{kg}\cdot\text{m}^2}{\text{A}^2\cdot\text{s}^3} \qquad$ $\textbf{(B) }$ $\dfrac{\text{kg}\cdot\text{m}^2\cdot\text{s}}{\text{C}^2} \qquad$ $\textbf{(C) }$ $\dfrac{\text{kg}\cdot\text{m}}{\text{C}\cdot\text{s}} \qquad$ $\textbf{(D) }$ $\dfrac{\text{kg}\cdot\text{m}^2}{\text{A}\cdot\text{s}^2} \qquad$ $\textbf{(E) }$ $\dfrac{\text{kg}\cdot\text{s}^2}{\text{A}^2\cdot\text{m}^2}$
$\textbf{A}$
By Ohm’s law $R=\dfrac{U}{I}$, we have $\Omega=\dfrac{\text{V}}{\text{I}}=\dfrac{\text{J/C}}{A}=\dfrac{\text{N}\cdot\text{m}}{\text{A}\cdot\text{A}\cdot\text{s}}=\dfrac{\text{kg}\cdot\text{m/s}^2\cdot\text{m}}{\text{A}^2\cdot\text{s}}=\dfrac{\text{kg}\cdot\text{m}^2}{\text{A}^2\cdot\text{s}^3}$.
A car moves to the right along a one-dimensional track for total time $T$ in two parts.
Part One: The car maintains constant non-zero speed $V$ for the first 3/4 of the total time.$\newline$
Part Two: The car accelerates uniformly to rest during the last 1/4 of the total time.
What is the ratio of the distance traveled during Part One of the trip to the distance traveled during Part Two of the trip?
$\textbf{(A) }$ 6:1$ \qquad\newline$
$\textbf{(B) }$ 3:2$ \qquad\newline$
$\textbf{(C) }$ The values of $V$ and $T$ are required to answer the question.$ \qquad\newline$
$\textbf{(D) }$ 4:3$ \qquad\newline$
$\textbf{(E) }$ 8:3
$\textbf{A}$
The distance traveled in Part One is $d_1=\dfrac34VT$. The distance traveled in Part Two is half the distance if it remains a constant speed V, thus $d_2=\dfrac12\cdot\dfrac14VT=\dfrac18VT$. The ratio of the two trip is $d_1:d_2=6:1$.
A cube of unknown material and uniform density floats in a container of water with $60\%$ of its volume submerged. If this same cube were placed in a container of oil with density $\rho_{oil}=800\ \text{kg/m}^3$, what portion of the cube’s volume would be submerged while floating?
$\textbf{(A) }$ $33\% \qquad$ $\textbf{(B) }$ $50\% \qquad$ $\textbf{(C) }$ $58\% \qquad$ $\textbf{(D) }$ $67\% \qquad$ $\textbf{(E) }$ $75\%$
$\textbf{E}$
By placing the cube in water and drawing the free body diagram, we have only gravitational force and buoyancy, thus $\rho_{water}gV_{displace}=\rho_{cube}V_{cube}g$, so $\dfrac{\rho_{cube}}{\rho_{water}}=\dfrac{V_{displace}}{V_{cube}}=0.6$. When placing the cube in oil, we have $\dfrac{V_{displace}}{V_{cube}}=\dfrac{\rho_{cube}}{\rho_{oil}}=\dfrac{600}{800}=0.75$. $75\%$ of the cube is submerged in the oil.
An ideal gas undergoes an isobaric expansion followed by an isochoric cooling. Which of the following statements must be true after the completion of these processes?
$\textbf{(A) }$ The final pressure is less than the original pressure.$ \qquad\newline$
$\textbf{(B) }$ The final volume is less than the original volume.$ \qquad\newline$
$\textbf{(C) }$ The final temperature is less than the original temperature.$ \qquad\newline$
$\textbf{(D) }$ The total quantity of heat, $Q$, associated with these processes is positive.$ \qquad\newline$
$\textbf{(E) }$ The internal energy of the gas is unchanged.
$\textbf{A}$
The PV diagram shows qualitatively what the processes of interest look like. Since the first process is constant pressure, the constant volume cooling reduces the pressure from the initial value, making (A) true. The volume clearly increases, making (B) wrong. The temperature could end up greater, less than, or the same as it started depending on the exact nature of the processes. The same is true of the total heat associated with the processes. Finally, the internal energy is not changed ONLY if the initial and final temperatures are exactly the same.
The diagram below shows the path taken by a monochromatic light ray traveling through three media. The symbols $v_1$, $\lambda_1$, and $f_1$ represent the speed, wavelength, and frequency of the light in Medium 1, respectively. Which of the following relationships for the light in the three media is true?
$\textbf{(A) }$ $\lambda_1<\lambda_3<\lambda_2 \qquad\newline$
$\textbf{(B) }$ $v_2<v_3<v_1 \qquad\newline$
$\textbf{(C) }$ $f_2<f_1<f_3 \qquad\newline$
$\textbf{(D) }$ $v_3<v_1<v_2 \qquad\newline$
$\textbf{(E) }$ $\lambda_2<\lambda_1<\lambda_3$
$\textbf{E}$
According to Snell’s Law at the interface between Medium 1 and 2, we have $n_1\sin\theta_1=n_2\sin\theta_2$. Apparently $\theta_1>\theta_2$, so the index of refraction $n_1<n_2$. Similarly, we get $n_1\sin\theta_1=n_2\sin\theta_2=n_3\sin\theta_3$ and $\theta_3>\theta_1>\theta_2$, so $n_3<n<1<n_2$. The speed of light in media is expressed as $v=\dfrac{c}{n}$, so $v_3>v_1>v_2$. The frequency of light does not change in different media, so $f_1=f_2=f_3$. By $v=\lambda f$, we have $\lambda_3>\lambda_1>\lambda_2$.
A piece of an ideal fluid is marked as it moves along a horizontal streamline through a pipe, as shown in the figure. In Region I, the speed of the fluid on the streamline is $V$. The cylindrical, horizontal pipe narrows so that the radius of the pipe in Region II is half of what it was in Region I. What is the speed of the marked fluid when it is in Region II?
$\textbf{(A) }$ $4V \qquad$ $\textbf{(B) }$ $2V \qquad$ $\textbf{(C) }$ $V \qquad$ $\textbf{(D) }$ $\dfrac12V \qquad$ $\textbf{(E) }$ $\dfrac14V$
$\textbf{A}$
The continuity equation determines that the total fluid moving through Region I must be equal to the fluid through Region II. Hence, we have $\pi R_1^2V_1=\pi R_2^2V_2$. Since the radius $R_2=\dfrac12R_1$, we have $V_2=4V_1=4V$.
For the $RC$ circuit shown, the resistance is $R=10.0\ \Omega$, the capacitance is $C=5.0\ \text{F}$ and the battery has voltage $\xi= 12\ \text{V}$. The capacitor is initially uncharged when the switch S is closed at time $t=0$. At some time later, the current in the circuit is 0.50 A. What is the magnitude of the voltage across the capacitor at that moment?
$\textbf{(A) }$ 0 V$ \qquad$ $\textbf{(B) }$ 5 V$ \qquad$ $\textbf{(C) }$ 6 V$ \qquad$ $\textbf{(D) }$ 7 V$ \qquad$ $\textbf{(E) }$ 12 V
$\textbf{D}$
By Kirchhoff’s Loop Rule, the sum of voltage across the capacitor and resistor is equal to the voltage of the battery. When the current is 0.50 A, the voltage across the resistor is $V_R=IR=(0.5)(10)=5\ \text{V}$, so the voltage across the capacitor $V_C=\xi-V_R=12-5=7\ \text{V}$.
For the figure shown, the variable resistance in the circuit on the left is increased at a constant rate. What is the direction of the magnetic field at the point P at the center of the left-hand circuit and in what direction is the conventional current through the resistor in the right-hand circuit?
$\textbf{A}$
Conventional current in the left-hand circuit is directed clockwise. By a right-hand rule, the magnetic field interior to the loop with current is therefore INTO the page. Since magnetic field lines form closed loops, outside the current-carrying loop, the field has a component directed OUT of the plane of the page. Since the resistance in the left-hand circuit is increasing, there is a decrease in the current and hence a decrease in the magnetic field strength. By Lenz’s Law in the right-hand circuit, there is an induced current in the wires to generate a field directed out of the plane of the page. This means that there is a counterclockwise current in the wire and the current is from B to A through the resistor in the diagram.
A small object of mass $M$ and charge $Q$ is connected to an insulating massless string in a vacuum on Earth. A uniform electric field exists throughout the region of the vacuum as indicated. The mass remains in static equilibrium at an angle of $\theta$ with the vertical as shown in the figure. When the string is cut, which of the illustrated paths best indicates the trajectory of the mass?
$\textbf{B}$
The object is at rest when the gravitational force, the electric force and the tension in the string exerted. So the sum the gravitational force and electric force must be in the opposite direction of the tension as shown in choice B. When the string is cut, the tension is lost, thus the object will move in the direction of the sum of gravitational force and electric force in a straight line, as shown in choice B.
For the compound machine comprised of the inclined plane and pulley shown, a 75.0 N force is required to slide a box with mass M = 10.0 kg up the incline at a constant speed. What is the efficiency of this compound machine?
$\textbf{(A) }$ $\dfrac13 \qquad$ $\textbf{(B) }$ $\dfrac38 \qquad$ $\textbf{(C) }$ $\dfrac12 \qquad$ $\textbf{(D) }$ $\dfrac23 \qquad$ $\textbf{(E) }$ $\dfrac34$
$\textbf{D}$
Efficiency can be computed as “what you get” divided by “what you paid for”. Here, we got the mass to rise to the top of the incline of height $H$. We paid for it by applying a force over the length of the hypotenuse $L$ and putting work into the system. Writing this mathematically, we have $$\eta=\dfrac{W_{out}}{W_{in}}=\dfrac{mgH}{FL}=\dfrac{mg}{F}\sin30^\circ=\dfrac{(10)(10)}{75}\cdot\dfrac12=\dfrac23$$.
A uniform solid cylinder of mass $M=2.00\ \text{kg}$ and radius $R=10.0\ \text{cm}$ is connected about an axis through the center of the cylinder to a horizontal spring with spring constant 4.00 N/m. The cylinder is pulled back, stretching the spring 1.00 m from equilibrium. When released, the cylinder rolls without slipping. What is the speed of the center of the cylinder when it returns to equilibrium?
$\textbf{(A) }$ 0.577 m/s$ \qquad$ $\textbf{(B) }$ 1.00 m/s$ \qquad$ $\textbf{(C) }$ 1.15 m/s$ \qquad$ $\textbf{(D) }$ 1.22 m/s$ \qquad$ $\textbf{(E) }$ 1.41 m/s
$\textbf{C}$
By the conservation of energy, the potential energy of the spring is converted into the translational and rotational kinetic energy of the cylinder, which is, $$\dfrac12k\Delta x^2=\dfrac12Mv^2+\dfrac12I\omega^2$$ The moment of inertia $I=\dfrac12MR^2$. Since the cylinder rolls without slipping, we have $v=\omega R$. So the equation of conservation of energy can be rewrote as $$\dfrac12k\Delta x^2=\dfrac12Mv^2+\dfrac12\cdot\dfrac12MR^2\omega^2=\dfrac12Mv^2+\dfrac12\cdot\dfrac12Mv^2$$ Thus we get $v=\sqrt{\dfrac{2k}{3M}}\Delta x=\sqrt{\dfrac{2(4)}{3(2)}}(1)=1.15\ \text{m/s}$.
For the circuits shown below, which will take the least time to charge the capacitor(s) to $90\%$ of the full charge? All batteries are ideal and identical with emf $\xi$, all resistors are identical with resistance $R$, and all capacitors are identical with capacitance $C$ and are initially uncharged.
$\textbf{E}$
The circuit with the smallest time constant $\tau=\text{resistance}\times\text{capacitance}$ will charge to $90\%$ the fastest. The time constants are $RC$, $RC$, $2RC$, $RC$, and $\dfrac12RC$ for the circuits given.
A uniform, solid disk with mass $M$ and radius $R$ is rotating on a fixed, frictionless platform with constant angular speed $\omega_0$ about a fixed axis through its center. A second uniform solid disk of mass $2M$ and radius $R/2$ is placed from rest directly on top of the first disk so that the centers of the disks line up. When equilibrium is established, the disks are spinning at the same rate. What is the angular speed of the disks at equilibrium?
$\textbf{(A) }$ $\dfrac14\omega_0 \qquad$ $\textbf{(B) }$ $\dfrac13\omega_0 \qquad$ $\textbf{(C) }$ $\dfrac12\omega_0 \qquad$ $\textbf{(D) }$ $\dfrac23\omega_0 \qquad$ $\textbf{(E) }$ $\sqrt{\dfrac23}\omega_0$
$\textbf{D}$
The angular momentum is conserved, thus we have $I_1\omega_0=(I_1+I_2)\omega$, where $I_1=\dfrac12MR^2$ is the momentum of inertia of the bottom disk and $I_2=\dfrac12(2M)\left(\dfrac12R\right)^2=\dfrac14MR^2$ is for the top disk. So $$\omega=\dfrac{I_1}{I_1+I_2}\omega_0=\dfrac23\omega_0$$
A real object is located in front of a convex lens at a distance greater than the focal length of the lens. What type of image is formed and what is true of the image’s size compared to that of the object?
$\textbf{B}$
When the object distance $d_o$ is greater than the focal length $f$, by $\dfrac1{d_o}+\dfrac1{d_i}=\dfrac1f$, the image distance $d_i=\dfrac{d_of}{d_o-f}>0$. so the image is real. The absolute value of magnification $|M|=\left|\dfrac{d_i}{d_o}\right|=\dfrac{f}{d_o-f}$ might be greater, equal, or less than 1, depending on the value of object distance and focal length.
In an ideal $LC$ circuit, what is the time difference between all of the energy in the circuit being stored in the inductor and all of the energy being stored in the capacitor?
$\textbf{(A) }$ No time difference$ \qquad\newline$
$\textbf{(B) }$ One-eighth of a period of oscillation$ \qquad\newline$
$\textbf{(C) }$ One-quarter of a period of oscillation$ \qquad\newline$
$\textbf{(D) }$ One-half of a period of oscillation$ \qquad\newline$
$\textbf{(E) }$ After one full period of oscillation has passed
$\textbf{C}$
The $LC$ circuit has charge oscillate as a simple harmonic oscillator. Consequently, it is one-quarter of a period for energy to switch from the inductor to the capacitor, just like the transition of kinetic and potential energy in a spring.
In neutron decay, the process is written as $n\rightarrow p+e^-+\bar{\nu}$, where $\bar{\nu}$ is an antineutrino necessary to have conservation of momentum and energy. Who postulated the existence of neutrinos?
$\textbf{(A) }$ Pauli $ \qquad$ $\textbf{(B) }$ Dirac$ \qquad$ $\textbf{(C) }$ de Broglie$ \qquad$ $\textbf{(D) }$ Heisenberg$ \qquad$ $\textbf{(E) }$ Einstein
$\textbf{A}$
It was Pauli who conjectured the existence of the neutrino (Fermi later worked out the theory of beta decay and named the particle).
Monochromatic light is incident on a slide containing six infinitely thin, equally spaced slits as shown in the figure. The resulting interference pattern on a distant screen reveals a place of total destructive interference. Of the following, what must be the phase difference (in radians) between the electric fields from each of the adjacent slits for this location on the screen?
$\textbf{(A) }$ $0 \qquad$ $\textbf{(B) }$ $\dfrac{\pi}{12} \qquad$ $\textbf{(C) }$ $\dfrac{\pi}6 \qquad$ $\textbf{(D) }$ $\dfrac{\pi}3 \qquad$ $\textbf{(E) }$ $\dfrac{\pi}2$
$\textbf{D}$
In order for the 6 electric fields to completely cancel, one can draw a unit circle $(2\pi\ \text{rad})$ and divide it equally into 6 pieces. Doing so gives an angle of $\pi/3$ for each piece. This is the phase difference for each of the fields so that when added together (in a phasor diagram), the total field will be zero. It can be expressed as $\cos0+\cos\pi/3+\cos2\pi/3+\cos3\pi/3+\cos4\pi/3+\cos5\pi/3=0$.
A spatially uniform electric field is constrained within the circular region of radius $R$ as shown. The field is directed out of the plane of the page and its strength is increasing uniformly in time. What is the direction of the force on the proton in the figure if the proton is moving to the right at the instant shown? Ignore gravity.
$\textbf{(A) }$ No Force$ \qquad$ $\textbf{(B) }$ Up$ \qquad$ $\textbf{(C) }$ Down$ \qquad$ $\textbf{(D) }$ Out of the page$ \qquad$ $\textbf{(E) }$ Into the page
$\textbf{E}$
The time-changing electric field induces a magnetic field. Since the electric field is out of the page and increasing in time, the time-changing field acts like a current (a displacement current) out of the plane of the page. The magnetic field associated with a current out of the plane of the page is directed in a counterclockwise manner around the circular region. At the location of the proton, there is a magnetic force therefore directed into the page (fingers point to the right, curl down the page, the right thumb point into the page).
Questions 48 – 49 deal with the following information:
An electromagnetic wave has an electric field given by the expression (in Cartesian coordinates): $\vec{E}(x,t)=6.0\cos\left(1.14\times10^7x-3.43\times10^{15}t\right)\hat{z}$
What is the direction of the energy flow for the wave?
$\textbf{(A) }$ $-x \qquad$ $\textbf{(B) }$ $+x \qquad$ $\textbf{(C) }$ $-y \qquad$ $\textbf{(D) }$ $+y \qquad$ $\textbf{(E) }$ $+z$
$\textbf{B}$
A traveling wave has the form $\cos(kr\pm\omega t)$ where the direction of wave travel would be $\mp r$. For the field given, this means that the wave is traveling in the $+x$ direction.
What is the direction of the magnetic field at time $t=0$ and position $x=0$?
$\textbf{(A) }$ $-x \qquad$ $\textbf{(B) }$ $+x \qquad$ $\textbf{(C) }$ $-y \qquad$ $\textbf{(D) }$ $+y \qquad$ $\textbf{(E) }$ $+z$
$\textbf{C}$
The direction of energy flow is computed by the Poynting Vector which is related to the cross product of the electric and magnetic fields $\vec{S}=\vec{E}\times\vec{H}$. At the origin at time $t=0$, the electric field has value $6.0\hat{z}$. We must therefore solve the problem $\hat{x}= \hat{z}\times?$. The magnetic field is perpendicular to the electric field and from the rules of cross products, $\hat{z}\times(-\hat{y})=\hat{x}$. The magnetic field is directed along $-y$.
What is the magnitude of the linear momentum (in units of $\text{kg}\cdot\text{m/s}$) of an electron moving in a straight line if it has $3.2\times10^{-13}\ \text{J}$ of kinetic energy?
$\textbf{(A) }$ $0 \qquad$ $\textbf{(B) }$ $2.6\times10^{-22} \qquad$ $\textbf{(C) }$ $7.6\times10^{-22} \qquad$ $\textbf{(D) }$ $1.3\times10^{-21} \qquad$ $\textbf{(E) }$ $1.9\times10^{-12}$
$\textbf{D}$
Using $E^2=c^2p^2+m_0^2c^4$ and $KE=E-m_0c^2$, the linear momentum $$p=\sqrt{\dfrac{E^2-m_0^2c^4}{c^2}}=\sqrt{\dfrac{\left(KE+m_0c^2\right)^2-m_0^2c^4}{c^2}}=\sqrt{\left(\dfrac{KE}{c}\right)^2+2KE\cdot m_0}=1.3\times10^{-21}\ \text{kg}\cdot\text{m/s}$$ where the rest mass of electron is $m_0=9.1\times10^{-31}\ \text{kg}$.