Physics Final Study Guide

C) 3.6 N, repulsive

After connecting, the charges equalize.
Repulsive force is kq^2/r^2 = (9E9)(12E-6)^2/(.6)^2

Two identical small conducting spheres are separated by .6 m. The spheres carry different amounts of charge and each sphere experiences an attractive electric force of 10.8 N. The total charge on the two sphere’s is -24E-6 C. The two spheres are now connected by a slender conducting wire, which is then removed. The electric force on each sphere is closest to:
A) 5.4 N, attractive
B) zero
C) 3.6 N, repulsive
D) 3.6 N, attractive
E) 5.4 N, repulsive
A) 1/4 F
F1 = kq^2/r^2
F2 = k(q/2)^2/r^2
Two identical small charged spheres are a certain distance apart, and each one initially experiences an electrostatic force of magnitude F due to the other. With time, charge gradually leaks off of both spheres. When each of the spheres has lost half its initial charge, the magnitude of the electrostatic force will be
A) 1/4 F
B) 1/16 F
C) 1/8 F
D) 1/2 F
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A) F

Newton’s Third Law: For every action there is an equal and opposite reaction

A point charge Q is located a short distance from a point charge 3Q, and no other charges are present. If the electrical force on Q is F, what is the electrical force on 3Q?
A) F
B) 3F
C) F/sqrt(3)
D) sqrt(3)F
A) D/2

E = lambda / (2*pi*8.85E-12*r)
So, to double E, we need to cut r in half

At a distance D from a very long (essentially infinite) uniform line of charge, the electric field strength is 1000 N/C. At what distance from the line will the field strength to be 2000 N/C?
A) D/2
B) 2D
C) sqrt(2)D
D) D/4
E) D/sqrt(2)
A) is always perpendicular to the surface of the conductor
Under electrostatic conditions, the electric field just outside the surface of any charged conductor
A) is always perpendicular to the surface of the conductor
B) is perpendicular to the surface of the conductor only if it’s a sphere, a cylinder, or flat sheet
C) can have nonzero components perpendicular to and parallel to the surface
D) is always parallel to the surface
E) is always zero because the electric field is zero inside conductors
A) 6.36 Nm^2/C
Since there’s no charge on the cone, total flux = 0 (flux on bottom and sloped sides sum to 0).
phi(bottom) + phi(curved) = 0
-(4550)(pi*.0211)^2 + phi(curved) = 0
A cone is resting on a tabletop as shown in the figure with its face horizontal. A uniform electric field of magnitude 4550 N/C points vertically upward. How much electric flux passes through the sloping surface area of the cone? (figure not pictured)
A) 6.36 Nm^2/C
B) 10.4 Nm^2/C
C) 1.24 Nm^2/C
D) 25.5 Nm^2/C
E) 82.1 Nm^2/C
A) No work is required
A negative charge is moved from point A to point B along an equipotential surface. Which of the following statements must be true for this case?
A) No work is required to move the negative charge from point A to point B
B) Work is required to move the negative charge
C) The work done on the charge depends on the distance between A and B
D) The negative charge performs work in moving from point A to point B
E) Work is done in moving the negative charge from point A to point B
E) The potential is highest at the center of the sphere

Don’t mix this up witha conducting sphere where the potential is zero inside the sphere

A NONCONDUCTING sphere contains positive charge distributed uniformly through out its volume. Which statements about the potential due to this sphere are true? All potentials are measured relative to infinity.
A) The potential at the center of the sphere is the same as the potential at the surface.
B) The potential at the surface is higher than its center
C) The potential at the center is the same as the potential at infinity
D) The potential at the center of the sphere is zero.
E) The potential is highest at the center of the sphere
B) 11

phi(entire sphere) = q(enclosed)/8.85E-12

Phi(entire sphere) / 2 gives us the answer

The xy plane is “painted” with uniform surface charge density which is equal to 40 nC/m^2. Consider a spherical surface with a 4.0 cm radius that has a point in the xy plane at it’s center. What is the electric flux (in Nm^2/C) for that part of the spherical surface for which z>0? (8.85E-12)
A)14
B) 11
C) 17
D) 20
E) 23
C) x = .5 m, y = .5 m

Ex = -dv/dx = (-2y – 2x)
Ey = -dv/dy = (-2y – 1)

Set both equal to 0 and solve for x and y.

In a certain region, the electric potential due to a charge distribution is given by the equation V(x,y) = 2xy – x^2 – y, where x and y are measured in meters and V is in volts. At which point is the electric field equal to 0? [Hint: if the electric field is 0 at a point, each component must be zero at this point] A) x = 1 m, y = .5 m
B) x = 1 m, y = 1 m
C) x = .5 m, y = .5 m
D) x = .5 m, y = 1 m
E) x = 0 m, y = 0 m
D) -2.8
phi = q(enc)/8.85E-12 = 15 +(-40) /8.85E-12
Two charges 15 pC and -40 pC are inside a cube with sides that are .40 m in length. Determine the net electric flux for teh surface of the cube in Nm^2/C. (8.85E-12)
A) + 2.8
B) -1.1
C) 1.1
D) -2.8
E) -.47
D) 24

d = 3i + 4j
Va – Vb = E.d = (4i + 3j)(3i +4j) = 24

Points A [at (2,3) m] and B [at (5,7) m] are in a region where the electric field is uniform and given by E = 4i + 3j N/C. What is the potential difference Va-Vb in volts? [Hint: it might be helpful to think of the separation between the points as a vector] A) 33
B) 27
C) 30
D) 24
E) 11
C) 0

E is conservative (unless created by magnetic fields, in which case it wouldn’t be constant), so its line integral on any clsoed loop is always 0.

A CONSTANT electric field E pointing along the x axis is integrated along the circumference of a circle in the xy plane, over an entire loop, in the clockwise direction. The radius of the circle is r. Find the result of this line integral.
A) E(2*pi*r)
B) E*pi*r^2
C) 0
D) -E(2*pi*r)
B) energy
The electron-volt is a unit of
A) charge
B) energy
C) electric field
D) voltage
E) electric force
E) 96 kV

Va = k(4E-6)/.3 + k(-6E-6)/.5
Vb = k(-6E-6)/.3 + k(4E-6)/.5

Va-Vb

A +4.0E-6 C point charge and a -4.0E-6 C point charge are placed as shown in the figure. What is the potential difference Va-Vb between the points A and B? (k = .25(pi)(8.85E-12) = 8.99E9 Nm^2/C^2)
A) 96 V
B) 0 V
C) 48 V
D) 48 kV
E) 96 kV
E) A is the brightest, B and C are equal but not as bright as A
More current goes through the path with bulb A as there is less total resistance in that path. More current = more brightness. Since the current does not separate between B and C, they have the same brightness
In the circuit shown in the figure, all the lightbulbs are identical. Which of the following is the correct ranking of the brightness of the bulbs? (A is alone and B and C are in series with eachother. A is parallel with B and C)
A) B and C have equal brightness, and A is the dimmest
B) All three bulbs have the same brightness
C) A is the brightest, C is dimmest, B is in between
D) A and B have equal brightness, and C is the dimmest
E) A is the brightest, B and C are equal but not as bright as A
B) Charge on C2 increases
Two capacitors, C1 and C2, are connected in series across a battery. With the battery still connected, a dielectric is now inserted between the plates of capacitor C1. What happens to the charge on capacitor C2?
A) Charge on C2 decreases
B) The charge on C2 increases
C) the charge on C2 either inreases or decreases depending on the value of the dielectric constant
D) The charge on C2 remains the same
E) The charge on C2 either increases or decreases depending on whetehr C2 is greater or less than C1
E) mvsin(theta)/(qB)
The centripetal force is given by m*v(perpendicular)^2/r, where v(perpendicular) = vsin(theta)
Thus, r = mvsin(theta)/(qB)
A charged particle of mass m and charge q, and moving with a velocity v enters a region of magnetic field B, making an angle (theta) with the magnetic field (0
E) +19 V
Use loop rule to set up that Epsilon – i1r1 – i2r2 = 0
A multiloop circuit is shown in the figure. Some circuit quantities are not labled. It is not necessary to solve the entire circuit. The emf (Epsilon) is closest to: (figure not shown)
A) -3 V
B) -10 V
C) -1 V
D) +3 V
E) +19 V
E) 120 ohms

P = v^2/R

(R ends up being 121 ohms but 120 is the closest value given)

A light bulb is connected to a 110-V source. What is the resistance of the bulb if it is a 100-W bulb?
A) 8.0 milli-ohms
B) 100 ohms
C) 6.0 milli-ohms
D) 240 ohms
E) 120 ohms
C) 10E-6 F

*A cap is discharged in 5 time constants, thus

5(RC) = 50E-3 s… solve for C to get 1E-5 F –> 10E-6 F

A defibrillator is essentially a discharging RC circuit, with the heart muscle serving as the resistor. The resistance of the heart muscle is 1000 ohms. What capacitance value is used if the discharge through the heart is completed in 50 ms?
A) 100E-6 F
B) 2E-6 F
C) 10E-6 F
D) 20E-6 F
E) 200E-6 F
E) 600 V

E = vB for no deflection, but also E = V/d (where v = velocity and V = potential difference).

vB = V/d –> V = vBd

The figure shows a velocity selector that can be used to measure the speed of a charged particle. A beam of particles is directed along the axis of the instrument. A parallel plate capacitor sets up an electric field E, which is oriented perpendicular to a uniform magnetic field B. IF the plates are separated by 2.0 mm and the value of the magnetic field is .60 T, what voltage between the plates will allow particles of speed 5.0E5 m/s to pass straight through without deflection?
A) 94 V
B) 190 V
C) 1900 V
D) 3800 V
E) 600 V
D) at none of these speeds
When electric current is flowing in a metal, the electrons are moving
A) at the speed of sound in the metal
B) at the speed of light
C) at the speed of sound in air
D) at none of these speeds
E) at nearly the speed of light
A) 1.7E-6

Capacitors act the opposite as resistors for equivalence

Three capactiors are connected as shown in the figure. What is the equivalent capacitance between points A and B? (6E-6||4E-6)s(2E-6)
A) 1.7E-6 F
B) 8.0E-6 F
C) 12E-6 F
D) 4.0E-6 F
E) 7.1E-6 F
B) toward the west
A vertical wire carries a current vertically upward in a region where the magnetic field vector points toward the north. What is the direction of the magnetic force on this current due to the field?
A) toward the east
B) toward the west
C) toward the south
D) toward the north
E) downward
C) zero
Magnetic forces never do work. ya dingus.
A proton (charge e) is shot with velocity v in a region of uniform magnetic field B. The angle between the velocity and the magnetic field is initially theta. How much work is done by the magnetic force F on the proton, when the displacement of the proton is d?
A) An essential piece of information is missing
B) e(v x B)
C) zero
D) F.d
E) Fdcos(theta)
D) the current
Two wires made of the same material have the same length but different diameters. They are connected in series to a battery. The quantity that is the same for the wires is
A) the end to end potential difference
B) the electron drift velocity
C) the electric field
D) the current
E) the current density
D) f

Frequency is indpendent of velocity in the cone

A charged particle is moving with speed v perpendicular to a uniform magnetic field. A second identical charged particle is moving with speed 2v perpendicular to the same magnetic field. If the frequency of revolution of the first particle is f, the frequency of revolution of the second particle is
A) f/2
B) f/4
C) 4f
D) f
E) 2f
B) in parallel across the source
Maximizes the equivalent capacitance, so more charge can be stored by both capacitors and hence stored energy is maximized
Suppose you have two capacitors and want to use them to store the maximum amount of energy by connecting them across a voltage source. You should connect them
A) in series across the source
B) in parallel across the source
C) it doesn’t matter because the stored energy is the same either way.
B) 2 times the diameter of the original wire
Just memorize it I guess

r^2 so you only need to multiply by 2 to get 4

A wire of resistivity p must be placed in a circuit by a wire of the same material but 4 times as long. If, however, the resistance of the new wire is to be the same as the resistance of the original wire, the diameter of the new wire must be
A) 1/4 the diameter of the original wire
B) 2 times the diameter of the original wire
C) 4 times the diameter of the original wire
D) The same diameter
E) 1/2 the diameter of the original wire
B) 2Uo

Uo = Q^2/2Co
New C = (8.85E-12)(A)/(2d)

–> U = Q^2/2C = Q^2/(2(Co/2)) = Q^2/Co = 2Uo

An ideal parallel-plate capacitor consists of a set of two parallel plates of area A separated by a very small distance d. When the capacitor plates carry charges +Q and -Q, the capacitor stores energy Uo. If the separation between the plates is doubled, how much electrical energy is stored in the capacitor?
A) Uo
B) 2Uo
C) 4Uo
D) Uo/2
E) Uo/4
B) 208 T/s, decreasing

The changing B field produces an EMF which opposes the battery.

|EMF| = dphi(B)/d(t) = A dB/dt
V(battery) – |EMF| = iR
4 – .12^2(dB/dt) = .1(10)
dB/dt = 208 T/s

Must be decreasing to set direction of EMF correct by Lens’ law

The circuit shown below is in a uniform magnetic field that is into the page. The current in the circuit is .10 A. At what rate is the magnitude of the magnetic field changing? Is it increasing or decreasing?
A) 420 T/s, increasing
B) 208 T/s, decreasing
C) 140 T/s, increasing
D) 140 T/s, decreasing
E) 208 T/s, increasing
C) no induced current

Since the B field is in the plane of the loop, the flux of B over the loop is ALWAYS 0. No change in flux = no change in induced emf.

In the figure, a straight wire caries a steady current I perpendicular to the plane of the page. A bar is in contact with a pair of circular rails, and rotates about the straight wire. The direction of the induced current through the resistor R is: (This is the problem with the two curved wires connected by a resistor R that look like part of a circle)
A) from b to a
B) from a to b
C) No induced current through the resistor
D) .32 V

|EMF| = N(dphi/dt) = N(BA – 0)/dt

A ten-loop coil having an area of .23 m^2 and a very large resistance is in a .047-T uniform magnetic field oriented so that maximum flux goes through the coil. The coil is then rotated so that the flux through it goes to zero in .34 s. What is the magnitude of the average emf induced in the coil during the .34 s?
A) .032 V
B) .0032 V
C) 0 V
D) .32 V
E) 1.0 V
C) 2.5E-6
Integral of B.dl = Mu(o)*I = Mu(o)*2
A long straight wire carrying 2.0 A current enters a room through a window 1.5 m high and 1.0 m wide. The line integral of the magnetic field of the wire evaluated around the window frame has the value (in T.m)
A) 2.5E-7
B) none of these
C) 2.5E-6
D) 3.8E-6
E) 3.0E-7
E) 4*Epsilon
|EMF| = N (dphi/dt) = NA (dB/dt).
Doubling both N and dB/dt multiplies the whole thing by 4
A coil of wire containing N turns is in an external magnetic field that is perpendicular to the plane of the coil and is steadily changing. Under these circumstances, an emf is induced in the coil. If the rate of change in the magnetic field and the number of turns in the coil are now doubled (nothing else changes), what will be the induced emf in the coil?
A) Epsilon/4
B) Epsilon
C) 2*Epsilon
D) Epsilon/2
E) 4*Epsilon
A) 3.7E-5 J/m^3

Energy Density = B^2/(2*Mu(o))

Where B = Mu(o)*i/(2*pi*r)

What is the energy density in the magnetic field 25 cm from a long straight wire carrying a current of 12 A?
A) 3.7E-5 J/m^3
B) 1.2E-4 J/m^3
C) 3.6E-4 J/m^3
D) 7.3E-5 J/m^3
E) The density cannot be determined without knowing the voluem
C) Rotate about z-axis

C is the only possibility that doesn’t cause a change in the fluxx across the turns of the solenoid

As shown in the figure, a coil of wire is placed in the xy-plane, centered on the z-axis. A solenoid is centered on the z-axis and is carrying a steady dc current. Which one of the following actions will NOT result in an induced emf in the coil?
A) Rotate the coil about the y-axis
B) Rotate about the x-axis
C) Rotate about the z-axis
D) Move the coil toward point P
E) Change the current in the solenoid
A) .234 s
Remember: i = I(o)*e^(-t/T), where T = L/R
The current in an RL circuit drops from 1.00 A to 14.0 mA in the first second following removal of the battery from the circuit. Find the time constant of the circuit
A) .234 s
B) 14.0 ms
C) 2.73 s
D) 1.00 s
E) 2.190 s
D) -4E-6 T i – 4E-6 T k

B = -i*B(loop) – k*B(wire)
(Directions determined by right hand rule)
B(loop) = Mu(o)*10/(2*pi*.5)
B(wire) = (Mu(o)/2)*(q(.5)^2/(.5^2+.5^2)^3/2

A long straight very thin wire on the y-axis carries a 10-A current in the positive y-direction. A circular loop .5 m in raidus, also of very thin wire and lying on the yz-plane, carries a 9.0-A current, as shown. Point P is on the positive x-axis, at a distance of .50 m from the center of the loop. What is the magnetic field vector at point P due to these two currents?
(Picture not shown)
A) zero
B) -8E-6 T k
C) -4E-6 T i – 8E-6 T k
D) -4E-6 T i – 4E-6 T k
E) +4E-6 T i – 4.0E-6 T k
A) (1/2)pi*k*R^4

Just memorize it.

A circular loop with radius R is in a magnetic field that is perpendicular to it. Assume that the normal to the loop is in teh same direction as the field. The strength of the field, as a function of radial distance from the center r, is given by B = kr^2 (where k is a constant). What is the flux over the loop?
A) (1/2)pi*k*R^4
B) pi*k*R^3
C) (2/3)pi*k*R^3
D) r*k*R
C) 10,000 A/s
Epsilon = L (di/dt)
1000 = (100E-3)(di/dt)
At what rate would the current in a 100-mH inductor have to change to induce an emf of 1000 V in the inductor?
A) 100 A/s
B) 10 A/s
C) 10,000 A/s
D) 1 A/s
E) 1000 A/s
B) from b to a

The loop sees a magnetic field pointed into the paper, decreasing. To counteract this change in flux, the loop creates an emf in the clockwise direction (b to a)

In the figure, two parallel wires carry currents of magnitude I in opposite directions. A rectangular loop is midway between the wires. The current I is decreasing with time. The induced current through resistor R is: (picture not shown)
A) from a to b
B) from b to a
C) there is no induced current through the resistor
D) the induced current will be oscillating back and forth
C) counterclockwise in the east and clockwise in the west
The B fields shown are increasing, applying lens’ law gives the answer
A long straight wire lies on a horizontal table and carries an ever-increasing current toward the north. Two coils of wire lie flat on the table, one on either side of the wire. When viewed from above, the direction of the induced current in these coils is
A) conterclockwise in both coils
B) clockwise in both coils
C) counterclockwise in the east coil and clockwise in the west coil
D) clockwise in the east coil and counterclockwise in the west coil
B) clockwise

Loop 2 experiences an increasing magnetic field pointed to the left, applying lenz’ law forces current to move clockwise

A closed, circular loop has a counter-clockwise current flowing through it as viewed by a person on the right. If a second closed circular loop with the same radius approaches this loop with constant velocity along a common axis as shown, in what direction will a current flow in the approaching loop as viewed by the person on the right? (figure not shown but it’s the one with the eye looking at two rings)
A) counter-clockwise
B) clockwise
C) no current will be induced because velocity is constant
D) no current will be induced because of Lenz’s law
.53 m^2

T = Mu*B*Sin(theta)
T = (IA)*B*sin(theta)
1.1 = 4A(.6)sin(60)
A = .53 m^2

Note: Theta is the perpendicular – the plane of the loop’s angle given = 60 degrees.

A rectangular loop of wire carrying a 4.0-A current is placed in a magnetic field of .60 T. The magnitude of the torque acting on this wire when the plane of the loop makes a 30 degree angle with the field is measured to be 1.1 Nm. What is the area of this loop?
A) .53 m^2
B) .20 m^2
C) .80 m^2
D) .4 m^2
E) .26 m^2
B) 1.36 µH

V = IX where X = 2*pi*f*L

Solve for L

The inductor in a radio receiver carries a current of amplitude 200 mA when a voltage of amplitude 2.40 V is across it at a frequency of 1400 kHz. What is the value of the inductance?
A) 1.43 µH
B) 1.36 µH
C) 9.20 µH
D) 4.42 µH
E) 1.97 µH
A) 458

Vs = Vp (Ns/Np)

In a transformer, how many turns are necessary in a 110-V primary if the 24-V secondary has 100 turns?
A) 458
B) 240
C) 110
D) 22
E) 4
D) 14

Is = Ip(Np/Ns)

^different than Voltage transformations

When a current of 2.0 A flows in the 100-turn primary of an ideal transformer, this causes 14 A to flow in the secondary. How many turns are in the secondary?
A) 700
B) 356
C) 114
D) 14
E) 4
D) f0/2

Just memorize it.

In a series LRC circuit, the frequency at which the circuit is at resonance is f0. If you double the resistance, the inductance, the capacitance, and the voltage amplitude of the ac source, what is the new resonance frequency?
A) 4 f0
B) 2 f0
C) f0
D) f0/2
E) f0/4
E) The current amplitude is a maximum

Remember lab

When an LRC series circuit is at resonance, which one of the following statements about that circuit is accurate? (There may be more than one correct choice.)
A) The impedance has its maximum value.
B) The reactance of the inductor is zero.
C) The reactance of the capacitor is zero.
D) The reactance due to the inductor and capacitor has its maximum value.
E) The current amplitude is a maximum.
C) It does not change
How does the average power dissipated in the resistor change as the frequency in the ac power supply decreases?
A) It decreases.
B) It increases.
C) It does not change.
D) It increases or decreases depending on the sign of the phase angle.
C) A capacitor
An ac source of period T and maximum voltage V is connected to a single unknown ideal element that is either a resistor, an inductor, or a capacitor. At time t = 0 the voltage is zero. At time t = T/4 the current in the unknown element is equal to zero, and at time t = T/2 the current is I = -Imax, whereImax is the current amplitude. What is the unknown element?
A) a resistor
B) an inductor
C) a capacitor
D) an inductor or a capacitor
x (position)
v (velocity)
k (spring constant)
m (mass)
Block-Spring System vs LC Oscillator… What does each value correlate to?

q
i
1/C
L

w = sqrt (k/m)
What is the angular frequency w for a block-spring system?
U(E) = q^2/(2C)
What is the formula for energy stored in the electric field of a capacitor?
X = 2*pi*f*L
Impedance of an inductor?
X = 1 / (2*pi*f*C)
Impedance of a capacitor?
B) 5.20 MHz

I = V/Xc

Calculated f may be in kHz, but the answer is in MHz for some reason.

If the voltage amplitude across an 8.50-nF capacitor is equal to 12.0 V when the current amplitude through it is 3.33 mA, the frequency is closest to:
A) 32.6 MHz
B) 5.20 MHz
C) 32.6 kHz
D) 5.20 kHz
E) 32.6 Hz
(a) the source voltage leads the current
(b) 700 volts
An ac series circuit consists of a voltage source, an 880-Ω resistor, and an inductance L. (There is no capacitance in the circuit.) The current amplitude is 0.60 A, and the phase angle between the source voltage and the current has a magnitude of 41°
(a) Does the source voltage lag or lead the current?
(b) What is the voltage amplitude of the source?
A) 98 V

Z = sqrt(R^2 + (Xl – Xc)^2)

A series ac circuit consists of an inductor having a reactance of 80 Ω and an inductance of 190 mH, a 40-Ω resistor, a capacitor whose reactance is 100 Ω, and an ac source. The rms current in the circuit is measured to be 2.2 A. What is the rms voltage of the source?
A) 98 V
B) 96 V
C) 93 V
D) 91 V
E) 88 V
A) 14 Hz

Xc = 1/(2*pi*f*C)
Xl = 2*pi*L*f

Xc = 2(Xl) and solve for f

A series circuit consists of ac source, a 90-Ω resistor, a 0.80-H inductor, and an 80-μF capacitor. The frequency of the source is adjusted so that the capacitive reactance is equal to twice the inductive reactance. What is the frequency of the source?
A) 14 Hz
B) 13 Hz
C) 16 Hz
D) 17 Hz
E) 19 Hz
A) .32 J
U = .5*L*i^2
*remember that the given current is in rms and must be converted into the peak current by multiplying by sqrt(2) to get 1.838 A
An ac circuit is shown in the figure. The rms current in the circuit is 1.3 A. What is the peak magnetic energy in the inductance?
A) 0.32 J
B) 0.16 J
C) 0.48 J
D) 0.64 J
E) 0.80 J
E) 19.7 mH

tan(phase angle) = (Xl – Xc)/R

The phase angle of an LRC series circuit with a capacitive reactance of 40 Ω, a resistor of 100 Ω and a certain inductor at 1000 Hz is 40.0°. What is the value of the inductance in this circuit?
A) 11.8 mH
B) 124 mH
C) 212 mH
D) 61.9 mH
E) 19.7 mH
P(avg) = I^2rms*R = Vrms*Irms*cos(phi)
where phi is the phase angle
Average Power P(avg) formula?
cos(phi)
What is the power factor of a circuit?
A) 930 kW

Just use P(avg) = I^2rms*R and then multiply by cos(39)

For an RLC ac circuit, the rms current is 10 A. If the impedance is 12 kΩ when the voltage leads the current by 39°, find the average power of the circuit.
A) 930 kW
B) 47 kW
C) 93 kW
D) 190 kW
zero
If we place a closed Gaussian surface around a magnet, what is the net magnetic flux through the surface?
zero
If we place a closed Gaussian surface around the north pole of a magnet, what is the net magnetic flux through the surface?
They form rings concentric with the plates
In a circular parallel plate capacitor with a changing electric field, how are the induced magnetic field lines oriented?
From the positive plate to the negative plate
When a capacitor is being charged, what is the direction of the displacement current?
There is no displacement current
After the charging of a capacitor is complete, what is the direction of the displacement current?
1.1E-5 T

B = (Mu(o)*i)/(2*pi*r) where i = (epsilon.o)(area)(change in E field/change in time)

A 0.70-m radius cylindrical region contains a uniform electric field that is parallel to the axis and is increasing at the rate 5.0 x 1012 V/m·s. The magnetic field at a point 1.2 m from the axis has a magnitude of
50 V/(m*s)

i = (epsilon.o)(area)(change in E field/change in time)

A 1.2-m radius cylindrical region contains a uniform electric field along the cylinder axis. It is increasing uniformly with time. To obtain a total displacement current of 2.0 x 10-9 A through a cross section of the region, the magnitude of the electric field should change at a rate of
C) Perpendicular to one another and perpendicular to the direction of wave propagation
In an electromagnetic wave, the electric and magnetic fields are oriented such that they are
A) parallel to one another and perpendicular to the direction of wave propagation.
B) parallel to one another and parallel to the direction of wave propagation.
C) perpendicular to one another and perpendicular to the direction of wave propagation.
D) perpendicular to one another and parallel to the direction of wave propagation.
C) -z-direction.

|E x B|.

If the magnetic field of an electromagnetic wave is in the +x-direction and the electric field of the wave is in the +y-direction, the wave is traveling in the
A) xy-plane.
B) +z-direction.
C) -z-direction.
D) -x-direction.
E) -y-direction.
A) equally divided between the electric and magnetic fields
The energy per unit volume in an electromagnetic wave is
A) equally divided between the electric and magnetic fields.
B) mostly in the electric field.
C) mostly in the magnetic field.
D) all in the electric field.
E) all in the magnetic field.
D) F/2
When an electromagnetic wave falls on a white, perfectly reflecting surface, it exerts a force F on that surface. If the surface is now painted a perfectly absorbing black, what will be the force that the same wave will exert on the surface?
A) 4F
B) 2F
C) F
D) F/2
E) F/4
B) its speed and wavelength decrease, but it’s frequency stays the same
When light goes from one material into another material having a HIGHER index of refraction
A) its speed, wavelength, and frequency all decrease.
B) its speed and wavelength decrease, but its frequency stays the same.
C) its speed decreases but its wavelength and frequency both increase.
D) its speed decreases but its frequency and wavelength stay the same.
E) its speed increases, its wavelength decreases, and its frequency stays the same.
A) 53°

I = Io*cos(theta)^2

if Io is initially unpolarized, I = Io/2.

Unpolarized light is incident upon two polarization filters that do not have their transmission axes aligned. If 18% of the light passes through this combination of filters, what is the angle between the transmission axes of the filters?
A) 53°
B) 73°
C) 85°
D) 80°
E) 90° and 140°

The angles should be exactly 90° from either of the other polarizers

The following are positioned in sequence: A source of a beam of natural light of intensity I0; three ideal polarizers A, B, and C; and an observer. Polarizer axis angles are measured clockwise from the vertical, from the perspective of the observer. The axis angle of polarizer A is set at 0° (vertical), and the axis angle of polarizer C is set at 50°. Polarizer B is set so that the beam intensity is zero at the observer. Which of the following pairs of angles are possible axis angle settings of polarizer B?
A) 40° and 90°
B) 40° and 130°
C) 40° and 140°
D) 90° and 130°
E) 90° and 140°
D) f
A convex lens has focal length f. If an object is located extremely far from the lens (at infinity), the image formed is located what distance from the lens?
A) infinity
B) 2f
C) between f and 2f
D) f
E) between the lens and f
E) 180 cm

m = -i/p
*since the image is bigger, the location of the image must be negative
(1/30) + (1/-45) = 2/r

A man’s face is 30 cm in front of a concave spherical shaving mirror. If the image is erect and 1.5 times as large as his face, what is the radius of curvature of the mirror?
A) 40 cm
B) 60 cm
C) 100 cm
D) 140 cm
E) 180 cm
A) 2.2 cm

The image is in a negative orientation because it’s formed behind the mirror (virtual)

The image of a plant is 4.0 cm from a concave spherical mirror having a radius of curvature of 10 cm. Where is the plant relative to the mirror?
A) 2.2 cm in front of the mirror
B) 4.4 cm in front of the mirror
C) 9.0 cm in front of the mirror
D) 1.0 cm in front of the mirror
E) 20 cm in front of the mirror
C) 4.5 mm
In a double-slit experiment, the slit separation is 2.0 mm, and two wavelengths, 750 nm and 900 nm, illuminate the slits simultaneously. A screen is placed 2.0 m from the slits. At what distance from the central maximum on the screen will a bright fringe from one pattern first coincide with a bright fringe from the other?
A) 1.5 mm
B) 3.0 mm
C) 4.5 mm
D) 6.0 mm
E) 9.0 mm
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