Download Physics 6B Practice Midterm Solutions

Physics 6B
Practice Midterm 2 Solutions
1.
A pair of speakers is hooked up to a stereo and placed 4m apart. The same sound is emitted by
both speakers and it has a frequency of 68Hz. If you stand between the speakers, at what distance from the
left speaker would the sound get quiet (destructive interference)? Use 340 m/s for the speed of sound.
x1
x2=4m-x1
Destructive interference will occur when the distance to one speaker is ½ wavelength
more than the distance to the other one. We can write a formula for this condition,
then rearrange to find the distance:
Now we need to find the wavelength:
Answer c)
2.
You are in the front row at a concert, standing 1m away from the speakers. The sound intensity level
is an earsplitting 120 db, so you decide to move away to a quieter position. How far away from the speaker do
you need to be so that the level is only 80 db?
I
 


10

log
The formula for intensity level in decibels is
 I0 
There is a great shortcut we can use in this problem. We need the level to decrease by 40db, so the Intensity
should decrease by a factor of 104. In general. For every 10db decrease in level, the intensity should be
divided by 10 (if the level increases, multiply instead).
The other relationship we need here is I 
1
r2
This says that the Intensity is inversely related to the square of the distance from the source of the sound.
So if the distance increases by a factor of 10, the intensity goes down by a factor of 100.
In this case, we need the intensity to change by a factor of 104 so the distance should increase by a factor of 102.
ANSWER: c) 100m
3. A loudspeaker playing a constant frequency tone is dropped off a cliff. As it accelerates downward, a person
standing at the bottom of the cliff will hear a sound of:
a) increasing frequency and decreasing amplitude
b) constant frequency and increasing amplitude
c) increasing frequency and increasing amplitude
d) decreasing frequency and constant amplitude
The sound source is getting closer to the person, so it will get louder (increasing amplitude).
Also, the source is accelerating toward the listener, so the frequency will be doppler-shifted upward more and
more as the speaker speeds up faster and faster..
Answer c)
4. Two strings on the same guitar (same length) are tuned so that string B is one octave higher frequency than
string A. Given that string A has 4 times the mass of string B, what is the ratio of the tensions in the strings?
(Hint: One octave higher frequency means the frequency is twice as high).
a) String A has 4 times the tension of String B
b) String A has 2 times the tension of String B
c) The tensions in the strings are equal
d) String A has half the tension of String B
The frequencies of the standing waves on a string are given by f  nv
2L
Since the lengths of the strings are the same, we see that the waves on string B must travel at twice the speed of
the waves on string A.
Ftension
v

Wave speed on a string is given by

2
Solve this for tension to get Ftension  v  
In this case, the mass/length of string A is 4 times as much as string B.
The tensions come out to be the same since the double speed (squared) gives a factor of 4 for string B.
Answer c).
5.
One of the harmonics on a string that is 1.30 m long has a frequency of 15.60 Hz. The next higher
harmonic has frequency 23.40 Hz. Find the fundamental frequency and the speed of the waves on the string.
The difference between any 2 successive harmonic frequencies on a particular string is always equal to the
fundamental frequency for that string. In this case that is (23.4-15.6) = 7.8Hz
To get the speed, use the formula
f1 
v
 v  7.8Hz 2  1.3m  20.28 m
s
2L
6. Charge q1 = -5.4µC is placed at the origin, and charge q2 = -2.2 µC is on the x-axis at x=1m. Where should a
charge q3 be placed between q1 and q2 so that the net force acting on it is zero?
Here is a diagram, with q3 in between. We just
need to find the forces on q3 due to the other
charges, and make them equal.
q1
F(q1 on q3)
x
q3 F(q2 on q3) q2
1-x
We are not given the sign of charge q3, but it
shouldn’t matter. The forces due to q1 and q2 will
be in opposite directions in either case.
kqq'
r2
kq q
Fq1  q3  12 3
x
kq2 q3
Fq2  q3 
1  x 2
Felec 
Setting these equal, and plugging in the given values:
kq1q3 kq2q3
q1
q2
5.4
2.2





x2
1  x 2 x2 1  x 2 x 2 1  x 2
2.2  x2  5.4  1  x 2  1.48  x  2.32  1  x   3.8x  2.32  x  0.61m
7. An object with a charge of – 3.6µC and a mass of 12g experiences an upward electric force, due to a uniform
electric field, equal in magnitude to its weight. Find the direction and magnitude of the electric field.
Electric force is simply charge times electric field.
Felec
E
mg
 qE  mg  E 
q
0.012kg 9.8 sm2 
6
3.6  10 C
Felec
 33,000 CN
This is the magnitude of the field. The direction must be downward to give an upward
force on a negative charge. One way to remember this is that the direction of the field is
always the direction of the force on a positive charge. Since our charge is negative, it
must be the opposite direction.
mg
8. Two conducting spheres have net charges q1=+8 µC and q2=-2 µC. The spheres touch and some charge is
transferred. How many electrons are transferred, and to which sphere?
BEFORE
AFTER
q1
q2
q1
q2
+8 µC
-2 µC
+3 µC
+3 µC
When the spheres touch, charges are transferred until both spheres have equal charge.
The total charge before they touch is +6µC, so the charge on each sphere after they touch must be +3µC.
So there must be a total of 5µC of charge transferred.
We need to divide by the charge on the electron to find the # of electrons transferred:
5  106 C
 3.1  1013 electrons
19
C
1.6  10 electron
Since sphere 1 gets less positive, it must be gaining the negative electrons. Conversely, sphere 2 is getting more
positive, so it is losing electrons. Thus the electrons must be transferred to sphere 1.
9. During a lightning strike, electrons are transferred from the bottom of a thundercloud to the ground. During this
process, the electrons:
a) gain potential energy as they move toward a higher potential
b) lose potential energy as they move toward a lower potential
c) gain potential energy as they move toward a lower potential
d) lose potential energy as they move toward a higher potential
Since the electron is moving downward, the ground must be positively
charged, and the bottom of the cloud must be negative.
- - - - e-
So the electron is moving toward a higher (positive) potential.
Because its own charge is negative, however, the potential energy of the
electron is getting more negative as it moves toward the ground.
So it is losing potential energy.
+ + + + + + + + +
10. During a lightning strike, electrons are transferred from the bottom of a thundercloud to the ground. This
occurs due to dielectric breakdown of the air, when the electric field is greater than 3x106 V/m. The distance from
the ground to the cloud is 1000m. Find the magnitude of the potential difference between the cloud and the
ground.
Assuming a constant electric field, we can use the formula V=Ed


V  3 106 Vm 1000m  3 109Volts
- - - - 1000m
+ + + + + + + + +
11. An electric dipole consists of two equal charges, +q and –q, a distance d apart. Find the total electric potential
at a point that is a distance of d/2 to the right of the positive charge, as shown.
kq
r
In this formula we include the sign of the charge to tell us the
sign of the potential. All we need to do is find the potential
due to each charge, and then add them together.
The electric potential due to a point charge is given by:  
left 
k  q 
3d
2
right 
total 
k q 
1d
2
k  q  k q   2 kq
kq 4 kq



2

3d
1d
3
d
d
3 d
2
2
-q
d
+q
d/2
12. How much charge is on each plate of the capacitors in the circuit shown? The battery has voltage 12V, and
the capacitances are: C1=3µF, C2=2µF, C3=4µF.
First we need to find the equivalent capacitance for the circuit.
C2 and C3 are in parallel, so we just add them together.
12V
C2
Next we see that C1 is in series with the (C2+C3) combination, so we
need to use the reciprocal formula.
1
1
1
1
1
 
 
 Ceq  2F
Ceq C1 C2  C3 3 2  4
C1
C3
The total charge for the system is given by Q=CV. Qtotal=24µC
Now we need to work backward to find the charge on each capacitor.
First look at capacitor C1. There are no parallel paths connected to it,
so all the charge must land on its plates. Q1=24µC
12V
Next we notice that since the other 2 capacitors are in parallel, the
charge must split up so that the total adds up to 24µC.
Option 1
C2+C3
Option 2
12V
The voltage across C2 and C3
must be equal because they are in
parallel. Using V=Q/C we get
V1=8volts. This leaves 4volts for
the others. Now using Q=CV
This adds up to 24, and Q2 is half
again we get Q2=8 and Q3=16
as much as Q3.
Since C2 has half the
capacitance of C3, it will only get
half the charge. So we need
Q2=8µF and Q3=16µF.
Ceq=2µF
C1
13.
A parallel-plate capacitor is initially charged by a battery with voltage V. The battery is disconnected,
and a dielectric with constant k is inserted between the plates. What happens to the energy stored in the
capacitor?
When the battery is disconnected, the charges that have built up on the plates have nowhere to go, so the charge
on the capacitor will remain constant when the dielectric material is inserted. To calculate the amount of charge
we could use the formula Q=CV.
Now what happens when the dielectric is inserted?
The capacitance is increased by a factor of k (the dielectric constant).
So what does that do to the energy?
(1)U cap  12 CV 2
We have 3 versions of the potential energy formula for capacitors:
Let’s use version (3). We already know that the charge (Q) is not
going to change.
(2)U cap  12 QV
(3)U cap  12
Q2
C
Since the capacitance is increased by a factor of k, the energy
must decrease by that factor (the C is in the denominator of the
formula).
So the answer is c) The stored energy decreases by a factor of k
Note: if the battery had remained connected, the energy would have increased by a factor of k instead. Think
about why this happens.
14. How much power is dissipated in each of the resistors in the circuit shown? The battery has voltage 12V, and
the resistances are: R1=4Ω, R2=3Ω, R3=6Ω.
First we need to find the equivalent resistance for the circuit.
12V
R2
R2 and R3 are in parallel, so we use the reciprocal trick.
1
1
1 1 1


   Req  2
Req R2 R3 3 6
Next we see that R1 is in series with the (R2 and R3) combination, so
we just add them together, For a total of 6Ω.
V 12V
 2A
Now we can use Ohm’s law to find the current: I  
R 6
All of this current must go through R1, since there is no other path
parallel to it. We can use the formula for power dissipated by a
resistor: P=I2R = IV = V2/R. In this case the first one works fine:
P1  2 A2 4  16W
R1
R3
12V
The current will split when it gets to the R2/R3 parallel combination:
Option 1
Option 2
Since R2 has half the resistance The voltage across R2 and R3
of R3, it will get twice the current. must be equal because they are in 12V
So we get .I 2  43 A and I3  23 A
parallel. Using V=IR we get
V1=8volts. This leaves 4volts for
This adds up to 2, and I2 is half
the others.
as much as I3.
Now using P=V2/R we get:
Now use P=I2R again:
2
4V 2 16
P2  43 3  16
W

5
.
3
W
P

 3 W  5.3W
2
3
3

2
P3  23  6  83 W  2.7W
2Ω
P3 
4V 2
6
 16
W  2.7W
6
6Ω
R1
15.
In the circuit below all the bulbs have the same resistance.
When the switch is closed, what happens to bulb A?
A
First realize that since we assume that the resistance stays constant, we
know that bulbs get brighter when the current increases.
C
B
Next, see from the diagram that all the current must go through bulb A.
So now think about what happens when the switch is closed. The path
through bulb B is closed, creating a parallel path to bulb C. This has the
effect of reducing the equivalent resistance (adding a parallel path will
always reduce the overall resistance).
Since the resistance is reduced, the current must increase. (V=I/R)
So the current through bulb A increases, and bulb A gets brighter.
Bonus question – what happens to bulb C?
Bulb C gets dimmer because it only gets half of the total current now.
16. Resistor 1 is a solid cylinder with length L and diameter D. Resistor 2 is made of the same material, but it has
length 2L and diameter 2D. Compared to the resistance of resistor 1, resistance 2 is:
Use the formula for resistance: R 
L
A
The resistivity is the same for each resistor.
The length is twice as large.
The area is 4 times as big, because area depends on the diameter squared
Now compare the 2 resistances:
R1 
R2 
L
A
 2 L 
4A
 12
L
A
 12 R1
So the answer is b) half as large
17. Consider the three electric charges shown below. Charge B is equidistant from charges A and C. List the
charges in order of the magnitude of the force they experience, from smallest to largest.
Each charge will feel 2 forces from the other charges.
Remember – same sign repels, opposites attract.
FBonA  
kqq
kqq
kqq
; FConB   2  Fnet , B  2 2
2
r
r
r
kqq
kqq
3 kqq

;
F



F


BonC
net
,
C
4 r2
r2
2r 2
FAonB  
FAonC
kqq
kqq
5 kqq
;
F



F


ConA
net
,
A
4 r2
r2
2r 2
-q
A
r
+q
B
r
+q
C