Download Physics 212 Spring 2009 Exam 2 Version C

Physics 212 Spring 2009 Exam 2 Version C (832725)
Question
1 2 3 4 5 6 7 8 9 10
Instructions
Be sure to answer every question. Follow the rules shown on the first page for filling in the Scantron form. Each problem is
worth 10% of the exam. When you are finished, check with Dr. Mike or his TA to be sure you have finished the scantron
correctly.
1.
Question Details Electric Current and Resistance (2) [742620]
Five resistors are made of a material with the same value of resistivity. Which has the most resitance?
*v*10*A cylindrical resistor with length 3 cm and radius 0.01 cm.
*w*05*A cylindrical resistor with length 1 cm and radius 0.01 cm.
*y*02*A cubic resistor with side length 4 cm.
*z*02*A cubic resistor with side length 2 cm.
*x*05*A cylindrical resistor with length 2 cm and radius 0.02 cm.
Solution or Explanation
Use the equation
R = L/A
where L is the length of the resistor, A is its area and
2.
is its resistivity.
Question Details Magnetic Force (4) [623658]
A proton is moving through a magnetic field whose magnitude is 7.80 10-4 T. The proton experiences only a magnetic force
and has an acceleration of magnitude 3.80 1012 m/s2. At a certain instant, it has a speed of 67.0 106 m/s. Determine the
angle (less than 90°) between the proton's velocity and the magnetic field.
y*02*37.20°
x*02*40.63°
v*10*49.37°
w*05*24.46°
z*06*0.86°
Solution or Explanation
The angle can be found from the equation,
F = |q|vB sin ,
where |q| = 1.602x10-19 C, v = 67.0
106 m/s and B = 10-4 T.
The force is equal to
F = m a = (1.67x10-27 kg) (3.80
3.
1012 m/s2).
Question Details Equivalent Circuits (4) [567357]
In the figure below, the resistors all have a value of 5
is the current across the resistor R1?
except R1, which has a value of 15 . The battery is a 9 V battery. What
y*03*0.26 A
w*04*1.80 A
v*10*0.60 A
z*01*0.00 A
x*06*1.50 A
Solution or Explanation
Because the resistor, R1, is in parallel with the battery, we see that
V1 = V = 9 V.
Then, Ohm's law gives us
I1 = V1 / R1 = (9 V) / (15 ) = 0.60 A.
4.
Question Details Capacitance (3) [742619]
How many electrons are on the negative plate of a 37 F capacitor with a 8 V potential difference across the plates?
v*10*1.85 1015 electrons
x*04*1.85 1027 electrons
z*01*4.63 10-6 electrons
y*03*2.96 10-4 electrons
w*04*2.89 1013 electrons
Solution or Explanation
By definition, the capitance of a capacitor is
C = Q/ V.
Thus,
Q = - C V = - (37 F) (8 V) = - 0.000296 C.
Each electron has -1.602 x 10-19 C. Thus,
n = Q / -e = (- 0.000296 C) / (- 1.602 x 10-19 C) = 1.85
1015 electrons.
5.
Question Details Electric Current and Resistance (3) [742622]
How many electrons per second must a 1.5 V battery produce in order to generate a current of 41 A?
*v*10*2.56 1020 electrons/second
*y*02*3.76 1041 electrons/second
*w*04*9.36 1018 electrons/second
*x*03*2.28 1017 electrons/second
*z*01*I would need to know the resistance in the circuit to calculate this.
Solution or Explanation
Remember that by definition, the charge per unit time that passes through a given cross-sectional area is the current. Thus, the
amount of chrage produced by the battery in one second is
q = - I * (1 sec) = - 41 C.
The charge of a single electron is e = -1.602 x 10-19 C. Thus, we have the number of electrons produced by the battery.
n = q / -e = - 41 C / (- 1.602 x 10-19 C) = 2.56
6.
1020 electrons/second.
Question Details Mass Spectrometer (3) [750382]
A negative ion with only one extra electron moves through the magnetic field (B = 0.36 mT) of a mass spectrometer. The ion
had an initial speed of 200 m/s. If the radius of the circle is 1.94 cm, what is the mass of the ion?
x*02*6.73 1016 kg
y*01*6.73 1021 kg
w*05*5.60 10-22 kg
v*10*5.60 10-27 kg
z*01*I would need to know what kind of ion it is to determine this.
Solution or Explanation
The mass can be found from the equation,
R = (m v) / (|q| B),
where R = 1.94 cm, v = 200 m/s, B = 0.36 mT, and |q| = 1.602 x 10-19 C. Remember to convert your units.
7.
Question Details Capacitors (4) [742609]
An isolated spherical conductor of radius 15 cm is charged to 2 kV and has a capacitance of 16.69 pF. What would the
capacitance be if the sphere is charged to 7 kV?
z*02*1.19 pF
y*02*233.66 pF
w*05*58.42 pF
x*05*4.77 pF
v*10*16.69 pF
Solution or Explanation
The capacitance does not change unless the shape, size or dielectric of the capacitor changes. In this case it was the voltage
that changed. Since the charge changes proportional to the voltage, the capacitance deos not change.
8.
Question Details Capacitors (3) [742612]
A 4.0-µF and an 8.0-µF capacitor are connected in series across a 25-V battery. Find the total charge stored on the two
capacitors.
z*05*9.38 µC
v*10*66.67 µC
y*04*0.11 µC
x*02*0.48 µC
w*05*300.00 µC
Solution or Explanation
The equivalent capacitance for two capacitors in SERIES is given by the equation
1 / Ceq = 1 / C1 + 1 / C2 and the total charge on both capacitors is then Qeq = Ceq V.
9.
Question Details Resistors (4) [742618]
Which of the following equivalent resistances CANNOT be made with one 1 and two 3 resistors? (Hint: Try ALL of the
combinations of parallel and series. There are 4 unique combinations each having a different equivalent resistance.)
v*10*1.67
z*03*0.60
y*05*3.75
w*01*7.00
x*05*2.50
Solution or Explanation
Put the resistors is series to get 7.00 .
Put the 3
resistors in parallel and this combination in series with the 1
Put one of the 3
3.75 .
resistors in parallel with the 1
resistor to get 2.50 .
resistor and put this combination in series with the other 3
resistor to get
Put all of the resistors in parallel to get 0.60 .
10.
Question Details Equivalent Circuits (3) [567355]
A 28 F and a 69 F capacitor are connected in series with each other and a 5 V battery. What is the voltage across each
capacitor?
v*10*3.56 V and 1.44 V
w*04*5.00 V and 5.00 V
y*02*17.32 V and 7.03 V
x*02*140.00 V and 345.00 V
z*05*2.53 V and 2.47 V
Solution or Explanation
For resistors in series, the charge is the same and the voltages add. Thus,
Q1 = Q2 = Qeq = Ceq Veq = (1 / C1 + 1 / C2)-1 Veq = 9.96
10-5 C.
Thus,
V1 = Qeq / C1 = 3.56 V
and
V2 = Qeq / C2 = 1.44 V
Assignment Details
Name (AID): Physics 212 Spring 2009 Exam 2 Version C (832725)
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