Upload
others
View
3
Download
0
Embed Size (px)
Citation preview
King Fahd University of Petroleum and Minerals – Physics Department
Physics 102 Recitation – Term 172 – Spring 2018 – Section 7
Quiz # 1
Name: ID # _______ __
A transverse sinusoidal wave is travelling to the right (+ -axis) on a stretched
string. The general form of the wave is ( , ) = sin ( ± + ). The
amplitude of the wave is 0.800 cm, its wavelength is 0.550 m, and its phase
constant is zero. The speed of waves on the string is 24.0 m/s. What is the speed
of the particle of the string located at = 0.300 m when the time is 1.50 s?
SOLUTION: = 2 = = 2 / ( , ) = sin( − ) ( , ) = − cos( − ) ( , ) = −(2 / ) cos (2 / ) − (2 / ) ∴ = − 2 × 24.00.550 × 8.00 × 10 cos 2 × 0.3000.550 − 2 × 24.0 × 1.500.550 = − . /
King Fahd University of Petroleum and Minerals – Physics Department
Physics 102 Recitation – Term 172 – Spring 2018 – Section 8
Quiz # 1
Name: ID # _______ __
Two identical sinusoidal traveling waves, each with a frequency of 100 Hz, are moving
in the same direction along a stretched string. If they are interfering and the combined
wave has an amplitude that is 0.25 times that of the amplitude of each wave, calculate
the phase difference between the two waves in radians and in meters. The speed of
waves on the string is 25 m/s.
SOLUTION:
The amplitude of the resultant wave is : = 2 cos /2 .
Thus, the phase difference is : = 2 cos /2 = 2 cos 0.25 /2 = 2.89 rad
The wavelength is : = = = 0.25 m.
Therefore, the phase difference in meters is : ∆ = = 0.115 m.
King Fahd University of Petroleum and Minerals – Physics Department
Physics 102 Recitation – Term 172 – Spring 2018 – Section 9
Quiz # 1
Name: ID # _______ __
A standing wave is set up on a string of length 1.2 m that is fixed at its ends. The string vibrates
according to the equation: , = 0.50 sin 2.5 cos 40 , where and are in
meters and is in seconds.
(a) What is the speed of waves on this string?
(b) What is the harmonic number of the standing wave?
SOLUTION:
(a) = = . = 16 m/s.
(b) = ⟹ = = = = × . × . = 3
King Fahd University of Petroleum and Minerals – Physics Department
Physics 102 Recitation – Term 172 – Spring 2018 – Section 7
Quiz # 2
Name: ID #
The difference between two consecutive resonant frequencies in a tube closed at
one end is 80 Hz. What is the fifth harmonic frequency of this tube?
SOLUTION:
The resonant frequencies of the tube are given by: = , where n = 1, 3, 5, …
The difference between two consecutive resonant frequencies is: ∆ = − = − = .
The fifth harmonic frequency of the tube is = = = × 80 = 200 Hz.
King Fahd University of Petroleum and Minerals – Physics Department
Physics 102 Recitation – Term 172 – Spring 2018 – Section 8
Quiz # 2
Name: ID #
The intensity of sound is 3.20 × 10 W/m2 at a distance of 5.00 m from a
source. What is the sound level at a distance of 10.0 m from the source?
SOLUTION:
Recall that the intensity ( ) of sound at a distance from a source is: = . Let be the intensity at 5.00 m, and be the intensity at 10.0 m, then: = 4 × 100 ÷ 4 × 25 = 0.25
Thus, = 0.25 = 8.00 × 10 W/m .
Therefore, the sound level at a distance of 10.0 m is: = 10 log = 10 × log 8.00 × 10 10 = 89 dB
King Fahd University of Petroleum and Minerals – Physics Department
Physics 102 Recitation – Term 172 – Spring 2018 – Section 9
Quiz # 2
Name: ID #
A stationary speaker sends sound waves of frequency 1000 Hz toward a car
moving away at a speed of 35 m/s. What is the frequency of the waves reflected
back to the speaker? [speed of sound in air = 343 m/s]
SOLUTION:
= speed of sound, = speed of car
From the speaker to the car:
Source = speaker, frequency =
Observer = car, frequency = = −
From the car to the speaker:
Source = car, frequency =
Observer = speaker, frequency = = + = − + = −+ = 1000 × 343 − 35343 + 35 = 815 Hz
King Fahd University of Petroleum and Minerals – Physics Department
Physics 102 Recitation – Term 172 – Spring 2018 – Section 7
Quiz # 3
Name: ID #
An ice cube, of mass 20 g and initial temperature 0 °C, is placed in an aluminum
container whose initial temperature is 80 oC. The system comes to an equilibrium
temperature of 30 °C. What is the mass of the container? [cAl = 900 J/kg.K]
SOLUTION: For ice:
Melting: = = 20 × 10 × 333 × 10 = 6660 J Heating: = ∆ = 20 × 10 × 4190 × 30 = 2514 J For aluminum:
Cooling: = ∆ = × 900 × −50 = − 45000
Conservation of energy: + + = 0 6660 + 2514 − 45000 = 0
Thus: mAl = 0.204 kg
King Fahd University of Petroleum and Minerals – Physics Department
Physics 102 Recitation – Term 172 – Spring 2018 – Section 8
Quiz # 3
Name: ID #
A metal rod has a length of 7.50 m at 15 °C and a length of 7.60 m at 95 °C. What
is the temperature of the rod when its length is 7.45 m?
SOLUTION:
In the first change (15 → 95 °C): = ∆ ∆ = 7.60 − 7.507.50 × (95 − 15) = 1.67 × 10 (℃)
In the second change (15 °C → ??): ∆ = ∆ = 7.45 − 7.501.67 × 10 × 7.50 = − 40 C = + ∆ = 15 − 40 = −25 ℃
King Fahd University of Petroleum and Minerals – Physics Department
Physics 102 Recitation – Term 172 – Spring 2018 – Section 9
Quiz # 3
Name: ID #
The figure below shows a steel bar 20.0 cm long welded end to end to a copper
bar 40.0 cm long. The two bars have the same cross sectional area. The free end
of steel is maintained at 100 oC, and the free end of copper is maintained at 0 oC.
Find the temperature (Tx) at the junction between the two bars, assuming steady
state. [Thermal conductivity: ksteel = 50.2 W/m.K, kcopper = 385 W/m.K]
SOLUTION: In steady state: ( ) = ( ) ( ) = (100 − )
= 100 − 3.835 = 100 −
Thus: Tx = 20.7 oC
King Fahd University of Petroleum and Minerals – Physics Department
Physics 102 Recitation – Term 172 – Spring 2018 – Section 7
Quiz # 4
Name: ID #
An ideal gas, initially occupying a volume of 0.400 m3 at a pressure of 1.00 atm,
expands isothermally to double its initial volume. How much heat is absorbed by
the gas in this process?
SOLUTION:
Isothermal process: = constant; thus = constant; therefore ∆ = 0.
From the 1st law of thermodynamics: ∆ = − = 0.
Thus: = = ln = ln = 0.400 × 1.01 × 10 × ln 2 = 28 kJ
King Fahd University of Petroleum and Minerals – Physics Department
Physics 102 Recitation – Term 172 – Spring 2018 – Section 8
Quiz # 4
Name: ID #
An ideal monatomic gas, initially occupying a volume of 1.00 L at a pressure of
1.00 atm, absorbs 500 J of heat and expands isobarically. What is the final volume
of the gas?
SOLUTION: For an ideal monatomic gas: = 3 /2, = 5 /2. Isobaric process: = ∆ = 5 /2 ∆ .
Thus: ∆ = 2 /5. = ∆ = ∆ = 2 /5
Thus: ∆ = = × × . × = 1.98 × 10 m = 1.98 L.
Therefore: = 2.98 L.
King Fahd University of Petroleum and Minerals – Physics Department
Physics 102 Recitation – Term 172 – Spring 2018 – Section 9
Quiz # 4
Name: ID #
An ideal monatomic gas ( = 2.00), initially occupying a volume of 1.00 L at
300 K, expands adiabatically to double its initial volume. How much work is done
by the gas in this process?
SOLUTION: For an ideal monatomic gas: = 3 /2, = 5 /2, = / = 5/3. Adiabatic process: = 0, and = .
= = 12 / × 300 = 189 K
From the 1st law of thermodynamics: ∆ = − .
Thus: = −∆ = − ∆ = − 3 /2 − = −2.00 × 3 × 8.312 189 − 300 = 2.77 kJ
King Fahd University of Petroleum and Minerals – Physics Department
Physics 102 Recitation – Term 172 – Spring 2018 – Section 7
Quiz # 5
Name: ID #
An ideal diatomic gas is the working substance in an engine that operates on the
cycle shown in the figure below, where V3 = 4V1. What is the engine’s efficiency?
SOLUTION:
Consider process 2→3:
Thus: 3 0.431
Consider process 4→1:
Thus: 0.144 ∆ 5 /2 ∆ 2.5 ∆ 2.5 ∆ ∴ 2.5 4 0.144 0.431 2.87 ( ⟹ ) ∆ 5 /2 ∆ 2.5 ∆ 2.5 ∆ ∴ 2.5 3 5.00 ( ⟹ ) 0 | | 1 | | 1 2.875.00 0.426
King Fahd University of Petroleum and Minerals – Physics Department
Physics 102 Recitation – Term 172 – Spring 2018 – Section 8
Quiz # 5
Name: ID #
An ideal diatomic gas is the working substance in an engine that operates on the
cycle shown in the figure below. Processes BC and DA are reversible and
adiabatic. What is the efficiency of this engine?
SOLUTION:
Consider process BC:
Thus: / / /32 / 2 23.78
Consider process DA:
Thus: / / /32 / 11.89 ∆ 7 /2 ∆ 3.5 ∆ 3.5 ∆ ∴ 3.5 2 3.5 (+ ⟹ ) ∆ 7 /2 ∆ 3.5 ∆ 3.5 ∆ ∴ 3.5 11.89 23.78 1.30 ( ⟹ ) 0 | | 1 | | 1 1.33.5 0.629
King Fahd University of Petroleum and Minerals – Physics Department
Physics 102 Recitation – Term 172 – Spring 2018 – Section 9
Quiz # 5
Name: ID #
An ideal diatomic gas is the working substance in an engine that operates on the
cycle shown in the figure below. What is the engine’s efficiency?
SOLUTION: ln 3 ln 2 3 ln 2 ( ⟹ ) ∆ 5 /2 ∆ 2.5 3 5 ( ⟹ ) ln ln 1/2 ln 2 ( ⟹ ) ∆ 5 /2 ∆ 2.5 3 5 ( ⟹ ) 3 ln 2 5 ln 2 5 | | 1 | | 1 ln 2 53 ln 2 5 0.196
King Fahd University of Petroleum and Minerals – Physics Department
Physics 102 Recitation – Term 172 – Spring 2018 – Section 7
Quiz # 6
Name: ID #
Two small identical conducting spheres, initially uncharged are separated by a
distance of 0.50 m. Find the number of electrons that must be transferred from
one sphere to the other in order to produce an attractive force of 2.0 × 104 N
between the spheres.
SOLUTION: The magnitude of the charge on each sphere is the same; let’s call it Q. = = ⟹ = = 2.0 × 109.0 × 10 × 0.50 = 7.5 × 10 C
=
Thus: = = . × . × = 47 × 10 electrons
King Fahd University of Petroleum and Minerals – Physics Department
Physics 102 Recitation – Term 172 – Spring 2018 – Section 8
Quiz # 6
Name: ID #
Two point charges are placed on the x-axis as follows: q1 = – 6.00 × 10-6 C at the origin,
and q2 = + 2.00 × 10-6 C at x = 10.0 cm. A third positive point charge q3 is to be located
somewhere, on the x axis, such that the net electrostatic force on it due to q1 and q2 is
zero. What is the x-coordinate of q3?
SOLUTION: Since q1 and q2 are of opposite signs, the point of equilibrium (zero force) must be: (1) on the
x-axis, (2) outside the region between q1 and q2, (3) closer to the weaker charge.
Thus, q3 must be placed at a distance d to the right of q2.
Let the distance between q1 and q2 be L, then: = ∶ + = ⟹ + =
Thus: 1 + = ± ⟹ = −1 ± = 0.732
Therefore, = .. = 13.7 cm, and the x-coordinate of q3 is x = + 23.7 cm.
King Fahd University of Petroleum and Minerals – Physics Department
Physics 102 Recitation – Term 172 – Spring 2018 – Section 9
Quiz # 6
Name: ID #
The figure below shows an equilateral triangle ABC. A point charge q = + 1.0 μC is located at
each of the three corners. A point charge Q is placed at the midpoint between B and C.
Determine the magnitude and sign of Q so that the net force on the charge at A is zero.
SOLUTION: Note the following:
(1) Q must be negative.
(2) The forces on A due to B and C are equal in magnitude ( = ), and each of them
makes an angle of 60o with the horizontal.
(3) The distance d between Q and A is such that: = − = .
(4) The force on A due to Q must be equal to the sum of the y-components of the forces
due to B and C.
Thus: = + = 2 = 2 sin 60 = 2 √32 ⟹ = √3 = √3 × 34 × 1.0 = 1.3 μC
Therefore: Q = – 1.3 μC.
King Fahd University of Petroleum and Minerals – Physics Department
Physics 102 Recitation – Term 172 – Spring 2018 – Section 7
Quiz # 7
Name: ID #
A neutral conducting spherical shell has an outer radius of 2.0 m. A point charge
is placed at the center of the shell. If the charge induced on the inner surface of
the shell is + 5.0 μC, what are the magnitude and direction (inward/outward) of
the electric field at a distance of 3.0 m from the center of the sphere?
SOLUTION:
The shell is neutral, thus: shell = 0. = + ⟹ = − = 0 − 5.0 = − 5.0 μC.
The requested point is outside the shell; thus, the only charge that will contribute
to the electric field is the charge on the outer surface of the shell.
Since that charge is negative, the electric field is radially inward. = = × × . × . = / .
King Fahd University of Petroleum and Minerals – Physics Department
Physics 102 Recitation – Term 172 – Spring 2018 – Section 8
Quiz # 7
Name: ID #
Two infinite parallel lines of charge are shown in the figure below, and are separated
by 5.0 cm. Their uniform linear charge densities are = + 2.0 × 10 C/m and = − 3.0 × 10 C/m. Find the net electric field at a point that is 1.0 cm to the left
of . Express your answer in unit vector notation.
SOLUTION: The electric fields due to the lines are: = − ̂ = − × × × . × . × ̂ = − 3.6 × 10 ̂ (N/C) = + ̂ = + × × × . × . × ̂ = + 9.0 × 10 ̂ (N/C)
The net electric field is: = + = − 3.6 × 10 ̂ + 9.0 × 10 ̂ = − 2.7 × 10 ̂ (N/C)
King Fahd University of Petroleum and Minerals – Physics Department
Physics 102 Recitation – Term 172 – Spring 2018 – Section 9
Quiz # 7
Name: ID #
A cube of edge length 2.00 m is placed in a region in which the electric field is given
by = 20.0 − 28.0 (N/C). Determine the net charge contained within the cube.
SOLUTION: The area of each face is 4.0 m2.
Since the electric field is in the z direction, there will be electric flux only through
the top ( ) and bottom ( ) faces.
At the top (z = 2.0 m): Φ = . = − 36.0 . + 4.00 = − 144 N. m /C.
At the bottom (z = 0): Φ = . = + 20.0 . − 4.00 = − 80.0 N. m /C.
Thus, the net flux is Φ = Φ + Φ = −224 N. m /C.
Gauss’ law: Φ = .
Therefore: = Φ = −224 × 8.85 × 10 = − 1.98 × 10 C ≈ −2 nC .
King Fahd University of Petroleum and Minerals – Physics Department
Physics 102 Recitation – Term 172 – Spring 2018 – Section 7
Quiz # 8
Name: ID #
In the rectangle shown below, the sides have lengths of 2.0 cm and 4.0 cm. The charges
are 1.0μC and 4.0μC. How much work is required to a move a third
charge 3.0μC from A to B along the diagonal of the rectangle?
SOLUTION: 9 10 1.0 100.04 4.0 100.02 1575kV
9 10 1.0 100.02 4.0 100.04 450kV ∆ 3.0 10 450 1575 10 3.375J
King Fahd University of Petroleum and Minerals – Physics Department
Physics 102 Recitation – Term 172 – Spring 2018 – Section 8
Quiz # 8
Name: ID #
Two isolated concentric conducting thin spherical shells have radii = 0.50 m and = 1.0 m; uniform charges = + 3.0 μC and = − 4.0 μC. With V = 0 at infinity,
what is the electric potential at a point that is 0.80 m from the center?
SOLUTION: = + = 9 × 10 3.00.80 − 4.01.0 × 10 = − 2250 V
King Fahd University of Petroleum and Minerals – Physics Department
Physics 102 Recitation – Term 172 – Spring 2018 – Section 9
Quiz # 8
Name: ID #
In the figure below, two parallel charged plates are separated by distance 5.00mm.
The plate potentials are 50.0V and 100V. An electron is released from
rest at the midpoint between the plates. Which plate will it strike? With what speed will
it strike that plate?
SOLUTION: The electron will move toward the higher potential, i.e. to plate 1.
The electric field is uniform, so the potential varies linearly in the region between
the plates.
Thus, the potential at the midpoint is – 75.0 V.
Apply the conservation of mechanical energy: ⟹ { 0} 12
2 2 1.60 109.11 10 50.0 75.0 8.78 10
Therefore, 2.96 10 m/s.
King Fahd University of Petroleum and Minerals – Physics Department
Physics 102 Recitation – Term 172 – Spring 2018 – Section 7
Quiz # 9
Name: ID #
In the circuit shown below, each capacitor has a capacitance of 3.0μF and 12V.
How much total energy is stored by this combination of capacitors?
SOLUTION: and are connected in parallel: 6.0μF
and are connected in series: 2.0μF
The stored energy is : 2.0 10 144 0.144mJ
King Fahd University of Petroleum and Minerals – Physics Department
Physics 102 Recitation – Term 172 – Spring 2018 – Section 8
Quiz # 9
Name: ID #
In the circuit shown below, each capacitor has a capacitance of 3.0μF and 12V.
How much total energy is stored by this combination of capacitors?
SOLUTION:
and are connected in series: 1.5μF
and are connected in parallel: 4.5μF
The stored energy is : 4.5 10 144 0.324mJ
King Fahd University of Petroleum and Minerals – Physics Department
Physics 102 Recitation – Term 172 – Spring 2018 – Section 9
Quiz # 9
Name: ID #
A 15.0 μF capacitor is charged using a 12.0 V battery. The battery is removed, and an
insulator of dielectric constant κ = 5.00 is inserted filling the space between the plates of
the capacitor. Calculate the energy stored in the capacitor after insertion of the insulator.
SOLUTION: =
This charge is maintained after removing the battery.
After inserting the dielectric, the stored energy is = 2 = 2 = 12 1 = 2 = 15.0 × 10 × 1442 × 5.00 = 0.216 mJ
King Fahd University of Petroleum and Minerals – Physics Department
Physics 102 Recitation – Term 172 – Spring 2018 – Section 7
Quiz # 10
Name: ID #
A potential difference of 0.500 V is applied to a conducting wire of length 3.00 m and
resistivity of 9.68 × 10-8 Ω.m. Calculate the magnitude of the current density in the wire.
SOLUTION: = = 1 × = × = = 0.5009.68 × 10 × 3.00 = 1.72 × 10 A/m
King Fahd University of Petroleum and Minerals – Physics Department
Physics 102 Recitation – Term 172 – Spring 2018 – Section 8
Quiz # 10
Name: ID #
A conducting wire is 25.0 m long, has a radius of 0.200 mm, and is made of a material
with resistivity of 1.50 × 10-6 Ω.m. If it carries a current of 0.300 A, how much power
is dissipated in it?
SOLUTION: = = = 1.50 × 10 × 25.0 × 4.00 × 10 = 298 Ω = = 0.0900 × 298 = 26.9 W
King Fahd University of Petroleum and Minerals – Physics Department
Physics 102 Recitation – Term 172 – Spring 2018 – Section 9
Quiz # 10
Name: ID #
A conducting wire is 25.0 m long, has a radius of 0.200 mm, and is at 20.0 oC. If the
temperature is increased to 340 oC, what is the resistance of the wire? Ignore the change
in the dimensions of the wire. [ρo = 1.50 × 10-6 Ω.m, α = 0.400 × 10-3 ( oC )-1]
SOLUTION: = 1 + ∆ = 1.50 × 10 1 + 0.400 × 10 × 320 = 1.692 × 10 Ω. m = = = 1.692 × 10 × 25.0 × 4.00 × 10 = 337 Ω
King Fahd University of Petroleum and Minerals – Physics Department
Physics 102 Recitation – Term 172 – Spring 2018 – Section 7
Quiz # 11
Name: ID #
Three resistors are connected as shown in the figure. The potential difference between
points A and B is VA – VB = + 26 V. How much current flows through the 2.0-Ω resistor?
SOLUTION:
The equivalent resistance of the 2.0 Ω and 4.0 Ω resistors (connected in parallel) is
(4×2)/(4+2) = 4/3 Ω.
The total equivalent resistance is (4/3) + 3 = 13/3 Ω.
Thus, the current flowing the 3 Ω resistor is 26/(13/3) = 6 A.
Now, consider the path going from A through the 3 Ω resistor, the 2 Ω resistor,
then to B: − 3 × 6 − 2 =
Therefore: = = = 4 A
King Fahd University of Petroleum and Minerals – Physics Department
Physics 102 Recitation – Term 172 – Spring 2018 – Section 8
Quiz # 11
Name: ID #
In the circuit shown below, determine the power dissipated by the 40-Ω resistor?
SOLUTION: Reduce the circuit to a single-loop:
The 10 Ω and 40 Ω resistors are connected in parallel, their equivalent resistance is
(40×10)/(40+10) = 8 Ω.
The total equivalent resistance is 8 + 2 = 10 Ω.
Thus, the current from the battery is i = 18/10 = 1.8 A.
Now, consider the left loop, and call ix the current through the 40 Ω resistor:
Start from the battery and go clockwise
+18 – (40 ix) – (2 × 1.8) = 0.
Therefore, ix = 0.36 A.
The power dissipated in it is P = (0.36)2 × 40 = 5.2 W
King Fahd University of Petroleum and Minerals – Physics Department
Physics 102 Recitation – Term 172 – Spring 2018 – Section 9
Quiz # 11
Name: ID #
Five resistors, each of resistance 6.0 Ω, are connected as shown in the diagram.
The potential difference between points A and B is 32 V. What is the current in
resistor R?
SOLUTION: Number the resistors as shown below:
R23 = 12 Ω
R234 = (12×6)/(12+6) = 4 Ω
Req = 6 + 4 + 6 = 16 Ω
The current through resistor 1 is 32/16 = 2 A.
Now, consider the path that goes from A to B through R, and let the current through R be i : − 6 × 2 − 6 − 6 × 2 =
Thus: = = = 1.3 A