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Parker-NCSU Physics Test Bank-Feedback & Solutions (Homework)

James Finch

Physics - College, section 1, Fall 2019

Instructor: Dr. Friendly

Current Score : 11 / 11

Due : Monday, January 28, 2030 00:00 EST

Last Saved : n/a Saving...  ()

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11/11 (100.0%)

  • Instructions

    The NCSU Physics Test Bank is a free resource, available to be used with any textbook (or no textbook at all!).

    The NCSU Physics Test Bank is a collection of multiple choice calculus-based introductory physics questions based on actual exam questions used at North Carolina State University. Each question has unique Socratic-style feedback for each individual answer choice, authored by NC State Physics professor George Parker. This specialized feedback is designed to help students overcome common misconceptions and problem solving errors.

    Every question also has a detailed solution available to the student at the instructor's discretion, useful for student review and self-study. Available as an Additional Resource at no extra charge, these questions are ideally suited as extra practice problems, supplements to homework assignments, or practice examinations. This demo assignment allows many submissions, more than recommended in a normal homework situation.

    The answer key and solutions will display after the first submission for demonstration purposes. Instructors can configure these to display after the due date or after a specified number of submissions.

Assignment Submission

For this assignment, you submit answers by question parts. The number of submissions remaining for each question part only changes if you submit or change the answer.

Assignment Scoring

Your last submission is used for your score.

1. 1/1 points  |  Previous Answers WACalcPhysMC 4.1.004. My Notes
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1/1
 
A boat crossing a river is shown at two different times. The current is 6 mi/h relative to the bank. The boat's velocity relative to the water is 8 mi/h, and it is directed so that the keel remains perpendicular to the current. What is the magnitude of the boat's velocity relative to the bank?
Displays two images of the boat with its keel perpendicular to the current shown with a vector parallel to the bank of the river and directed to the right. The boat closer to the upper bank is further to the right than the boat closer to the lower bank.
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The velocity of the boat relative to the water is perpendicular to the current and it forms one side of a vector right triangle with the velocity of the water relative to the bank the other orthogonal side. The vector forming the hypotenuse is the desired velocity.


Solution or Explanation

  • Part 1

    The velocity of the boat relative to the bank of the river is given by
    vboat,bank = vboat,water + vwater,bank
    The velocity of the boat relative to the water is perpendicular to the current and it forms one side of a vector right triangle with the velocity of the water relative to the bank the other orthogonal side. The vector forming the hypotenuse is the desired velocity.
    Displays a right triangle made up of vectors. The vertical leg pointed up is labeled v sub boat, water. The horizontal leg at the top of the vertical leg and pointed right is labeled v sub water, bank. The hypotenuse pointed up and to the right is labeled v sub boat, bank.
    The magnitude of the hypotenuse is given by
    vboat,bank = 
    vboat,water2 + vwater,bank2
     
     = 
    (8 mi/h)2 + (6 mi/h)2
     
     = 10 mi/h
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2. 1/1 points  |  Previous Answers WACalcPhysMC 6.1.001. My Notes
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1/1
 
A conical pendulum consists of a 0.5 kg mass attached to a string and rotating in a horizontal circle of radius 0.2 meter. The string makes an angle of 37 degrees to the vertical, as shown, and T is the tension in the string. What is the speed of the mass?
A conical pendulum has a mass rotating in a horizontal circle on the end of a string that makes an angle of 37 degrees with the vertical axis.
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The net force in the vertical direction is zero, which gives the tension in terms of the known weight. The radial force due to tension equals the mass times the centripetal acceleration ar = v2/r. Solving for the speed v gives the answer.


Solution or Explanation

  • Part 1

    A free-body diagram of the mass is shown below. The tension T is directed towards the suspension point and the weight mg is directed vertically downward. The instantaneous acceleration lies in the plane of these two forces and points towards the center of the horizontal circle.
    Displays the free-body diagram for the mass with the tension T pointing 37 degrees from the vertical and to the right and the weight of the mass pointing down.

  • Part 2

    Newton's second law for the vertical y direction states that the net force must be zero since there is no motion in that direction.
    Fnet,y = Tcos37° − mg
     = may
     = m(0)
     = 0
    The tension is then
    T = 
    mg
    cos37°

  • Part 3

    Newton's second law in the x direction which will be the radial direction of the motion states that the net force is the mass times the radial acceleration.
    Fnet,r = mar
     = 
    mv2
    r
    This net radial force is Tsin37°, so the radial equation becomes
    Tsin 37° = 
    mv2
    r
    Using our previous result
    T = 
    mg
    cos37°
    and substituting this value of T gives
    mg
    cos 37°
    sin 37°    		 = 
    mv2
    r
     
    mgtan 37° = 
    mv2
    r
     
    v = 
    rmgtan 37°
    m
    The mass cancels out giving
    v = 
    rgtan 37°
     
     = 
    (0.20 m)(9.8 m/s2)tan 37°
     
     = 1.2 m/s
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3. 1/1 points  |  Previous Answers WACalcPhysMC 8.4.002. My Notes
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1/1
 
A box of mass 20 kg slides from rest at the top of the incline shown to the bottom where it is 3 m vertically below and 12 m horizontally to the right of its initial position. If the block's speed at the bottom is 6 m/s what was the work done on the box by friction?
A box slides down an incline with friction present.
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Since friction acts on the block, its mechanical energy decreases during the fall.

From the work-kinetic energy theorem and the definition of mechanical energy E = K + U, it follows that the work done on the block by friction Wf equals the change in mechanical energy: Wf = E2 - E1 where E2 is the final mechanical energy and E1 is the initial mechanical energy.


Solution or Explanation

  • Step 1

    Since friction acts on the block, its mechanical energy decreases during the fall.

    From the work-kinetic energy theorem and the definition of mechanical energy E = K + U, it follows that the work done on the block by friction Wf equals the change in mechanical energy E2 - E1 where E2 is the final mechanical energy and E1 is the initial mechanical energy.

    As usual, change means final value E2 minus initial value E1, not the other way around.

  • Step 2

    The initial kinetic energy is zero since the block starts from rest.

    The initial potential energy is
    U1 = mgy1
     = (20 kg)(9.8 m/s2)(3 m)
     = 588 J
    taking the zero of potential energy to be at the bottom of the ramp.

    The initial value of the mechanical energy is then
    E1 = K1 + U1
     = 0 + 588 J
     = 588 J
    The final potential energy is
    U2 = mgy2
     = (20 kg)(9.8 m/s2)(0 m)
     = 0 J
    The final kinetic energy is
    K2 = (1/2)mv22
     = (1/2)(20 kg)(6 m/s)2
     = 360 J
    The final mechanical energy is then
    E2 = K2 + U2
     = 360 J + 0
     = 360 J
    The work done by friction follows:
    Wf = E2 − E1
     = 360 J − 588 J
     = −230 J
    The sign is negative since friction always opposes motion.
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4. 1/1 points  |  Previous Answers WACalcPhysMC 16.4.001. My Notes
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A four meter long string is under tension that results in a wave speed of 20 m/s. A rectangular shaped pulse is created at the center and moving to the left. What is the least time you will have to wait before the string has exactly the same appearance as it did initially? That is, the string will again look as it does in the sketch below.
Displays a rectangular pulse in the center of a 4 m long string. The pulse is above the string. The string is fixed at both ends. Above the pulse is a vector pointing left and labeled 20 m/s.
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The distance traveled back to the center after two reflections divided by the wave speed gives the minimum time to reproduce the starting wave pulse.


Solution or Explanation

  • Part 1

    The distance traveled back to the center after two reflections divided by the wave speed gives the minimum time to reproduce the starting wave pulse. The first reflection at the left end inverts the pulse. The second reflection at the right end returns the pulse to its initial upright shape. The time required is
    t = 
    distance
    wave speed
     
     = 
    (2 + 4 + 2 ) m
    20 m/s
     
     = 
    8 m
    20 m/s
     
     = 0.4 s
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5. 1/1 points  |  Previous Answers WACalcPhysMC 23.3.001. My Notes
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Three point charges lie on the x axis as shown. The +40 µC charge is 3 cm to the left of the +10 µC charge while the +20 µC charge is 2 cm to the right of the +10 µC charge. What is the magnitude of the net force on the +10 µC charge?
Three point charges lie on the x axis
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Coulomb's law and the superposition principle give the net force on the designated charge.


Solution or Explanation

  • Step 1

    Coulomb's law and the superposition principle give the net force on the designated charge.

    Numbering the charges 1, 2, and 3, going from left to right, the magnitude of the force on #2 due to #1 is
    F21 = k
    q1q2
    r2
     
     = (9.0 × 109 N · m2/C2)
    (40 × 10−6 C)(10 × 10−6 C)
    (3 × 10−2)2
     
     = 4000 N
    By inspection of the figure, we see that this force is to the right or in the positive x direction since two like charges repel.

  • Step 2

    Similarly, the magnitude of the force on #2 due to #3 is
    F23 = k
    q2q3
    r2
     
     = (9.0 × 109 N · m2/C2)
    (10 × 10−6 C)(20 × 10−6 C)
    (2 × 10−2)2
     
     = 4500 N
    Since charges 2 and 3 repel, this force is to the left or in the negative x direction.

  • Step 3

    The net force on #2 is the vector sum of the forces on it due to #1 and #3. Since these forces are in opposite directions, the magnitude of their sum is the difference in their magnitudes or 500 N. This can be written in vector form as
    F2 = F21 + F23
     = (4000 N)i hat + (−4500 N)i hat
     = (−500 N)i hat
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6. 1/1 points  |  Previous Answers WACalcPhysMC 24.3.001. My Notes
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An insulating sphere of radius R has a total positive charge Q distributed uniformly throughout its volume. Which figure below best represents the variation of the magnitude of the electric field as a function of radius r, where r = 0 at the sphere's center?
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Using Gauss's law, the magnitude of the electric field can be determined as a function of r. The spherical symmetry of the charge means the field outside the sphere is that of a point charge at the origin.


Solution or Explanation

  • Step 1

    Using Gauss's law, the magnitude of the electric field can be determined as a function of r. The spherical symmetry of the charge means the field outside the sphere is that of a point charge at the origin. Spherical symmetry also means that the field is zero at the origin. For 0 < r < R the field increases from zero to its maximum at r = R. For r > R, the point charge field decreases as 1/r2.

  • Step 2

    The dependence of the field on r inside the charge follows from Gauss's law and spherical symmetry. Gauss's law states that the electric flux through any closed surface equals the net charge inside that surface divided by
    ε0
    . Applying this law to a spherical surface of radius r with 0 < r < R gives (ρ is the uniform charge density of the sphere)
    E · da
    		 = 
    qin
    ε0
     
    (4πr2)E = 
    1
    ε0
    4πr3/3
    ρ
    (4πr2)E = 
    1
    ε0
    4πr3/3
    Q
    4πR3/3
     
    E = 
    Q
    4πR2
    r
    R
    This shows that the field is linear in r inside the charge. The maximum value of the field occurs at the surface of the charge (r = R).
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7. 1/1 points  |  Previous Answers WACalcPhysMC 25.3.001. My Notes
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Two point charges (+10 µC each) are 6 m apart on the x axis as shown. Calculate the magnitude of the electric potential at the point P, which is half way between them. Take the zero of electrical potential at infinity.
Two point charges lie on the x axis
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Adding the potential at P due to each charge gives the total potential there (superposition).


Solution or Explanation

  • Part 1

    Adding the potential at P due to each charge gives the total potential there (superposition). Thus,
    V = V1 + V2
    The potential of the first charge is
    V1 = k
    q1
    r1
     
     = (9.0 × 109 N·m2/C2)
    (10 × 10−6 C)
    3 m
     
     = 3.0 × 104 V
    Since r2 = r1 and q2 = q1, the potential of the second charge at P is the same as that of the first. This gives
    V = 2V1
     = 2 × 3.0 × 104 V
     = 6.0 × 104 V
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8. 1/1 points  |  Previous Answers WACalcPhysMC 28.2.001. My Notes
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If R = 30 Ω, what is the equivalent resistance between points A and B?
A resistor circuit is shown with 2 resistors in parallel and their combination in series with a third resistor; the series combination is in parallel with a fourth resistor.
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Two resistors in parallel are combined and their equivalent resistor is in series with one of resistance R. The equivalent resistance of that combination is then in parallel with the last resistor of resistance R.


Solution or Explanation

  • Part 1

    Two resistors in parallel are combined and their equivalent resistor is in series with one of resistance R. The equivalent resistance of that combination is then in parallel with the last resistor of resistance R.

    The two resistors in the upper right portion of the figure are in parallel (same potential difference). The equivalent resistor representing them has a resistance Req1.
    A resistor circuit is shown with 2 resistors in series and their combination in parallel with a third resistor.
    1
    Req1
    		 = 
    1
    R
     + 
    1
    R
     
     = 
    2
    R
     
    Req1 = 
    1
    2
    R

  • Part 2

    This equivalent resistor is in series with the remaining resistor R in the upper branch of the circuit. These two combine to give an equivalent resistor with a resistance Req2
    A resistor circuit is shown with 2 resistors in parallel.
    Req2 = R + 
    1
    2
    R
     = 
    3
    2
    R

  • Part 3

    Finally, this last equivalent resistor is in parallel with the resistor with resistance R in the lower branch of the circuit. Combining them gives the single equivalent resistor for all four resistors with resistance Req3.
    A resistor circuit is shown with 1 resistor.
    1
    Req3
    		 = 
    1
    R
     + 
    1
    3
    2
    R
     
     = 
    5
    3R
     
    Req3 = 
    3
    5
    R
     = 
    3
    5
    (30 Ω)
     = 18 Ω
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9. 1/1 points  |  Previous Answers WACalcPhysMC 29.2.001. My Notes
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An electron enters a region having a uniform magnetic field directed into the screen as shown. What is the path of the electron (assume zero gravity)?
An electron enters a magnetic field directed into the screen
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The magnetic force on the electron is perpendicular to its velocity and it pulls the electron in a circular arc downward towards the bottom of the screen.


Solution or Explanation

  • Part 1

    The magnetic force on the electron is perpendicular to its velocity and it pulls the electron in a circular arc downward towards the bottom of the screen.

    The magnetic force on an electron, a point charge with q = -e, is
    F = qv × B
     = (−e)v × B
    The electron enters the field with a velocity to the right. The cross product
    v × B
     is then directed to the top of the screen. The negative charge on the electron gives a force initially towards the bottom of the screen. As the electron moves in a circular arc, the magnetic force remains perpendicular to the velocity (think of a mass whirled in a circle on the end of a string with tension supplying the centripetal force). Of course, when it leaves the field again it travels in a straight line (zero magnetic force).
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10. 1/1 points  |  Previous Answers WACalcPhysMC 30.1.001. My Notes
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Two long straight wires are parallel and carry currents of 20 A to the left and 10 A to the left as shown. A point P is 0.20 m from each wire and in the same plane as the wires. Compute the magnitude of the total magnetic field at P.
Displays two parallel horizontal wires with current flowing to the left. The top wire has a 20 A current and the bottom wire has a 10 A current. A point P is shown that is 0.2 m from each wire.
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The magnetic field at P is the vector sum of the fields each infinite straight wire produces at P. The right hand rule is used to determine the directions of each field.


Solution or Explanation

  • Step 1

    The magnetic field at P is the vector sum of the fields each infinite straight wire produces at P. The right hand rule is used to determine the directions of each field.

    Using the right hand rule, we see that the 20 A wire produces a field at P that is out of the screen (with the thumb of right hand in the direction of the current, the fingers of that hand circulate around the thumb in the direction of the field). The 10 A wire produces a field into the screen. Since the field of the 20 A wire is the larger of the two fields, the net field is out of the screen with a magnitude equal to the difference in the magnitudes of the two fields.

    The magnitude of the field of the 20 A wire is
    B1 = 
    μ0i1
    2πr1
     
     = 
    (4π × 10−7 T · m/A)(20 A)
    2π(0.20 m)
     
     = 2.0 × 10−5 T

  • Step 2

    Similarly, the field of the 10 A wire is
    B2 = 
    μ0i2
    2πr2
     
     = 
    (4π × 10−7 T · m/A)(10 A)
    2π(0.20 m)
     
     = 1.0 × 10−5 T

  • Step 3

    The magnitude of the total field at P is then
    B = B2 − B1
     = 2.0 × 10−5 T − 1.0 × 10−5 T
     = 1.0 × 10−5 T
    This field is directed out of the screen.
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11. 1/1 points  |  Previous Answers WACalcPhysMC 42.3.001. My Notes
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Bohr's formula for the energy of hydrogen can be written as
En = −
13.6 eV
n2
Here the energy is expressed in electron-volts and n is an integer (n = 1, 2,
  
). According to the formula, what is the minimum energy that must be supplied to excite the electron in hydrogen to a higher energy state if it is initially in its ground state?
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The energy to raise the electron to the next allowed energy state is the difference between the energies of these states calculated from the formula.


Solution or Explanation

  • Part 1

    The energy to raise the electron to the next allowed energy state is the difference between the energies of these states calculated from the formula.

    To go from the n = 1 state to the n = 2 state requires an energy
    ΔE = E2 − E1
     = −
    13.6 eV
    22
     + 
    13.6 eV
    12
     
     = 0.75(13.6 eV)
     = 10.2 eV
    This is the minimum energy required to excite a hydrogen atom out of its ground state (n = 1).
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