Electromagnetics, Waves, and Optics - Physics
Card 0 of 148
Suppose that car A and car B are both traveling in the same direction. Car A is going 
and is sounding a horn with a frequency of 
. If car B is traveling at a speed of 
in front of car A, what frequency of sound does car B hear?
Note: The speed of sound in air is 
.
Suppose that car A and car B are both traveling in the same direction. Car A is going
Note: The speed of sound in air is
The important concept being tested in this question is the Doppler effect. When a source of sound waves (or any other wave) is emitted from a source and is moving relative to some observer, the actual frequency of the wave will be different for the observer.
For starters, we'll need to use the Doppler equation.
The trickiest part about this question is to decide whether to add or subtract in both the numerator and denominator. To find the right sign orientation, it's helpful to do a quick thought experiment. Imagine that only one of them is moving and the other is stationary. Decide how that will affect the observed frequency; will it increase it or decrease it? Then repeat for the other one.
First, let's consider the detector (aka observer). We're told that car B, the detector, is traveling ahead of car A and also in the same direction. This means that car B is driving away from car A. So in this situation, is the frequency that car B hears expected to increase or decrease? The answer is that it will decrease. Since car B is traveling away from car A, each successive wave will take longer to reach car B. Hence, we will use subtraction in the numerator because that will make the observed frequency smaller.
Now let's apply this same logic to the denominator, which deals with the source of the sound waves. We know that car A is traveling in the same direction as car B and is behind. This means that car A is traveling toward car B. So from this perspective, each successive wave is expected to get closer together, thus making the time between each wave smaller and the frequency bigger. In the denominator, will adding or subtracting make the observed frequency bigger? The answer is subtraction. By making a smaller number in the denominator, the entire fraction becomes larger.
Keeping this information in mind, we'll need to use subtraction in both the numerator and denominator. Once we plug in the values given in the question stem, we have everything we need to solve for the answer.
The important concept being tested in this question is the Doppler effect. When a source of sound waves (or any other wave) is emitted from a source and is moving relative to some observer, the actual frequency of the wave will be different for the observer.
For starters, we'll need to use the Doppler equation.
The trickiest part about this question is to decide whether to add or subtract in both the numerator and denominator. To find the right sign orientation, it's helpful to do a quick thought experiment. Imagine that only one of them is moving and the other is stationary. Decide how that will affect the observed frequency; will it increase it or decrease it? Then repeat for the other one.
First, let's consider the detector (aka observer). We're told that car B, the detector, is traveling ahead of car A and also in the same direction. This means that car B is driving away from car A. So in this situation, is the frequency that car B hears expected to increase or decrease? The answer is that it will decrease. Since car B is traveling away from car A, each successive wave will take longer to reach car B. Hence, we will use subtraction in the numerator because that will make the observed frequency smaller.
Now let's apply this same logic to the denominator, which deals with the source of the sound waves. We know that car A is traveling in the same direction as car B and is behind. This means that car A is traveling toward car B. So from this perspective, each successive wave is expected to get closer together, thus making the time between each wave smaller and the frequency bigger. In the denominator, will adding or subtracting make the observed frequency bigger? The answer is subtraction. By making a smaller number in the denominator, the entire fraction becomes larger.
Keeping this information in mind, we'll need to use subtraction in both the numerator and denominator. Once we plug in the values given in the question stem, we have everything we need to solve for the answer.
Compare your answer with the correct one above
On planet Borg, a male insect is flying toward his mate at 
while buzzing at 
. The female insect is stationary and is buzzing at 
. What is the speed of the sound on planet Borg?
On planet Borg, a male insect is flying toward his mate at
To begin, we'll start with the Doppler Effect equation:
Rearrange the equation by solving for 
to get:
Finally we plug in our known values and solve for 
:
To begin, we'll start with the Doppler Effect equation:
Rearrange the equation by solving for
Finally we plug in our known values and solve for
Compare your answer with the correct one above
When a ray of light is reflected off of the surface of another medium, some of the energy from the light is transferred into this medium. As a result, which of the following is true about the reflected light ray?
When a ray of light is reflected off of the surface of another medium, some of the energy from the light is transferred into this medium. As a result, which of the following is true about the reflected light ray?
For this question, we're told that light is reflected off of the surface of another medium. As a result, the light loses some of its energy, which is transferred into the medium where it reflected.
Because the light has lost some of its energy, we need to determine how the wavelength and frequency of the wave will be affected. To do this, we can recall the equation for the energy of a wave.
Where 
is Planck's constant and 
represents the frequency. We can see that when energy is decreased, the frequency also decreases. Thus, we can eliminate two of the answer choices.
To see how the wavelength changes, it's important to recall the relationship that wavelength and frequency have with each other.
Where 
is the speed of light and 
is the wavelength. Since the light is still traveling in the same medium, its speed will not change. Thus, as the frequency of the wave decreases, the wavelength has to increase. We can alternatively show how all of these variables are related as follows.
So, in summary, when the energy of light decreases, the frequency will decrease and the wavelength will increase.
For this question, we're told that light is reflected off of the surface of another medium. As a result, the light loses some of its energy, which is transferred into the medium where it reflected.
Because the light has lost some of its energy, we need to determine how the wavelength and frequency of the wave will be affected. To do this, we can recall the equation for the energy of a wave.
Where
To see how the wavelength changes, it's important to recall the relationship that wavelength and frequency have with each other.
Where
So, in summary, when the energy of light decreases, the frequency will decrease and the wavelength will increase.
Compare your answer with the correct one above
A 
string under tension is oscillated at the 5th harmonic. What is the wavelength of this oscillation?
A
Each harmonic has a wavelength that behaves according to the equation
, where 
is the wavelength and 
is the harmonic.
Since this is the 5th harmonic, 
and 
. Solving for wavelength gives 
.
Each harmonic has a wavelength that behaves according to the equation
Since this is the 5th harmonic,
Compare your answer with the correct one above
Suppose that the fifth harmonic of a standing wave contained within a pipe closed at both ends has a wavelength of 
. What is the length of the pipe?
Suppose that the fifth harmonic of a standing wave contained within a pipe closed at both ends has a wavelength of
For this question, we're told that a standing wave is contained within a pipe closed at both ends. We're also given the wavelength for the fifth harmonic, and are asked to find the length of the pipe.
The first step to solve this problem is to use the equation for a pipe closed at both ends.
Since we're told which harmonic the wave is on, as well as its wavelength, we have everything we need to solve for the length of the pipe.
For this question, we're told that a standing wave is contained within a pipe closed at both ends. We're also given the wavelength for the fifth harmonic, and are asked to find the length of the pipe.
The first step to solve this problem is to use the equation for a pipe closed at both ends.
Since we're told which harmonic the wave is on, as well as its wavelength, we have everything we need to solve for the length of the pipe.
Compare your answer with the correct one above
A pipe has a length of 
. If the pipe is open at both ends, what is the frequency of the third harmonic assuming that the velocity in which sound moves through the air is 
.
A pipe has a length of
Since the pipe is open at both ends, we can use the equation 
where f is the frequency we are solving for, 
is the number of the harmonic, 
is the velocity at which sound moves through air, and L is the length of the pipe.
We know from the question that:
When we plug in these values into the frequency equation, we get 
as the answer.
Since the pipe is open at both ends, we can use the equation
We know from the question that:
When we plug in these values into the frequency equation, we get
Compare your answer with the correct one above
A convex lens connected to a projector projects an image onto a board of an object that is 
away from the lens. If the object is 
tall and the inverted image is 
tall, what is the focal length of the projector lens?
A convex lens connected to a projector projects an image onto a board of an object that is
To solve this one will require two equations. First, to find focal length, you will need the lens equation.
But we are only given the object distance, not the image distance. So to find that we need to know the relationship between magnification and object and image distances.
Plug in the quantities we know.
(remember it was inverted)
. Then solve for 
which is 
. So the projector would sit that far from the board.
Now finally we can find the focal length of this lens.
Plugging in:
Solving for 
gives a focal length of 
or 
To solve this one will require two equations. First, to find focal length, you will need the lens equation.
But we are only given the object distance, not the image distance. So to find that we need to know the relationship between magnification and object and image distances.
Plug in the quantities we know.
Now finally we can find the focal length of this lens.
Plugging in:
Solving for
Compare your answer with the correct one above
For most people, the nearest distance that objects can be located away from the eye and still seen clearly is 
. This is referred to as the near point; if the object comes any closer, the object cannot be seen clearly. Suppose that a person who needs glasses cannot see objects clearly if they are closer than 
from the eye; that is to say, their near point is 
. A lens with what refractive power is needed in order to correct this person's vision to bring their near point to 
?
For most people, the nearest distance that objects can be located away from the eye and still seen clearly is
For this question, we're given the definition of near point. We're told what the near point is in the average person, and also the near point for a certain person who can't see well and needs glasses. We're asked to find the refractive power of a lens that will bring this individual's near point to the average, healthy value.
As was stated in the question stem, the near point is the closest distance of an object from the eye where that object can still be seen clearly. In the question, we're told that the normal value for this is 
. Moreover, a person with a near point of 
means that the object will need to be twice as far away, and no closer, to be seen clearly. Thus, in order to correct for this, a lens will be needed.
The idea is to be able to make the individual see things clearly when objects are located 
away. To accomplish this, a lens will need to diffract the light coming from an object 
away. This diffracted light will then need to form an image 
away, which is where this particular individual's near point is.
With this information in hand, we can use the lens-makers equation to solve for the refractive power of the lens.
We know that the object will be located a distance of 
away. Moreover, the image will need to form at a distance of 
away. However, since the image is forming on the same side of the lens that the object is located, the image will be virtual. Thus, the value used in the equation will be 
.
Also, remember that to find refractive power, we'll need to have our units be in meters.
Furthermore, recall that refractive power is equal to the inverse of focal length.
Hence, the power of the refractive lens will need to be 
diopters to correct this person's near point.
For this question, we're given the definition of near point. We're told what the near point is in the average person, and also the near point for a certain person who can't see well and needs glasses. We're asked to find the refractive power of a lens that will bring this individual's near point to the average, healthy value.
As was stated in the question stem, the near point is the closest distance of an object from the eye where that object can still be seen clearly. In the question, we're told that the normal value for this is
The idea is to be able to make the individual see things clearly when objects are located
With this information in hand, we can use the lens-makers equation to solve for the refractive power of the lens.
We know that the object will be located a distance of
Also, remember that to find refractive power, we'll need to have our units be in meters.
Furthermore, recall that refractive power is equal to the inverse of focal length.
Hence, the power of the refractive lens will need to be
Compare your answer with the correct one above
An object located 
away from a convex lens whose focal point is 
produces an image located at what distance away from the lens?
An object located
For this question, we'll need to use the thin lens equation in order to determine the image distance from the lens.
Because this is a convex lens, the refracted light will form an image on the side of the lens where light is expected to go (real image). Thus, we know that the focal length should be a positive value. Moreover, the object distance is also positive.
Rearranging the above equation, we can isolate the term for image distance.
Then we can plug in the values given to us in the question stem.
And finally we take the inverse of this value to arrive at the correct answer.
For this question, we'll need to use the thin lens equation in order to determine the image distance from the lens.
Because this is a convex lens, the refracted light will form an image on the side of the lens where light is expected to go (real image). Thus, we know that the focal length should be a positive value. Moreover, the object distance is also positive.
Rearranging the above equation, we can isolate the term for image distance.
Then we can plug in the values given to us in the question stem.
And finally we take the inverse of this value to arrive at the correct answer.
Compare your answer with the correct one above
A light ray is traveling through air hits a transparent material at an angle of 
from the normal. It is then refracted at 
. What is the speed of light in the material?
A light ray is traveling through air hits a transparent material at an angle of
This problem requires Snell's Law and the corresponding equation:
We know that the index of refraction of air is:
We also know that:
and 
Now we can plug in these values into the Snell's Law equation to find the index of refraction for the transparent material.
Finally, we need to calculate the speed of light moving through this transparent material now that we know the index of refraction for it. To do that, we need to use this equation:
Where 
is the speed of light and 
is the index of refraction. We plug in our known values and get:
This problem requires Snell's Law and the corresponding equation:
We know that the index of refraction of air is:
We also know that:
Now we can plug in these values into the Snell's Law equation to find the index of refraction for the transparent material.
Finally, we need to calculate the speed of light moving through this transparent material now that we know the index of refraction for it. To do that, we need to use this equation:
Where
Compare your answer with the correct one above
A single string of wire has a resistance of 
. If the wire is connected to a 
power source, what is the strength of the magnetic field 
away from the wire?
A single string of wire has a resistance of
So this is all about the magnetic field strength around a current carrying wire.
The equation for this is:
But you must use Ohm's Law 
in order to find the current in the wire.
Since the wire has 
of resistance and the voltage through the wire is 
, that means the current in the wire is 
.
Being sure to change 
into 
, plug everything in and get the answer, which is
So this is all about the magnetic field strength around a current carrying wire.
The equation for this is:
But you must use Ohm's Law
Since the wire has
Being sure to change
Compare your answer with the correct one above
Suppose that a magnetic field is oriented such that it is pointing directly to the left, as in the picture shown below. If a positively charged particle were to begin traveling through this magnetic field to the right, in which direction would the particle's trajectory begin to curve?
Suppose that a magnetic field is oriented such that it is pointing directly to the left, as in the picture shown below. If a positively charged particle were to begin traveling through this magnetic field to the right, in which direction would the particle's trajectory begin to curve?
In order to answer this question, it's important to understand the factors that determine the magnetic force experienced by a charge. We can begin by writing out the equation for magnetic force.
As shown in the above equation, the magnetic force is directly proportional to the particle's charge, its velocity, and the strength of the magnetic field itself. But, for the purposes of the this question, the most important factor is the angle of the particle's velocity with respect to the magnetic field.
Notice that if theta is equal to zero, then the sine of theta will be equal to zero as well. This, in turn, will cause the magnetic force to also be zero. This is also true if we were to define theta as 
.
Since the particle is moving in a direction that is parallel to the magnetic field lines but in the opposite direction, we have a situation in which theta is equal to 
. This means that the magnetic force on the particle is zero. As a result, the particle will continue to move through the magnetic field without changing its direction.
In order to answer this question, it's important to understand the factors that determine the magnetic force experienced by a charge. We can begin by writing out the equation for magnetic force.
As shown in the above equation, the magnetic force is directly proportional to the particle's charge, its velocity, and the strength of the magnetic field itself. But, for the purposes of the this question, the most important factor is the angle of the particle's velocity with respect to the magnetic field.
Notice that if theta is equal to zero, then the sine of theta will be equal to zero as well. This, in turn, will cause the magnetic force to also be zero. This is also true if we were to define theta as
Since the particle is moving in a direction that is parallel to the magnetic field lines but in the opposite direction, we have a situation in which theta is equal to
Compare your answer with the correct one above
What is the main difference between electrical and gravitational forces?
What is the main difference between electrical and gravitational forces?
Electric forces can be attractive or repulsive because charges may be positive or negative. In the case for gravitational forces, there are only attractive forces because mass is always positive.
Electric forces can be attractive or repulsive because charges may be positive or negative. In the case for gravitational forces, there are only attractive forces because mass is always positive.
Compare your answer with the correct one above
When the magnitude of two interacting charges is increased by a factor of 2, the electrical forces between these charges is __________.
When the magnitude of two interacting charges is increased by a factor of 2, the electrical forces between these charges is __________.
In Coloumb's law, an increase in both interacting charges will cause an increase in the magnitude of the electrical force between them. Specifically if the magnitude of both interacting charges is doubled, this will quadruple the electrical force.
In Coloumb's law, an increase in both interacting charges will cause an increase in the magnitude of the electrical force between them. Specifically if the magnitude of both interacting charges is doubled, this will quadruple the electrical force.
Compare your answer with the correct one above
Three equal charges are at three of the corners of a square of side d. A fourth charge of equal magnitude is at the center of the square as shown in Figure above. Which of the arrows shown represents the net force acting on the charge at the center of the square?
Three equal charges are at three of the corners of a square of side d. A fourth charge of equal magnitude is at the center of the square as shown in Figure above. Which of the arrows shown represents the net force acting on the charge at the center of the square?
Because of the principles of superposition, each electric force that acts from the charges at the corners on to the charge at the center can be broken into components. Since all the charges are positive, all the forces will be repulsive. The forces acting from the top left and bottom right corners will cancel, leaving only the repulsive force coming from the bottom left corner.
Because of the principles of superposition, each electric force that acts from the charges at the corners on to the charge at the center can be broken into components. Since all the charges are positive, all the forces will be repulsive. The forces acting from the top left and bottom right corners will cancel, leaving only the repulsive force coming from the bottom left corner.
Compare your answer with the correct one above
An electron traveling along the +x-axis enters an electric field that is directed vertically down, i.e., along the negative y-axis. What will be the direction of the electric force acting on the electron after entering the electric field?
An electron traveling along the +x-axis enters an electric field that is directed vertically down, i.e., along the negative y-axis. What will be the direction of the electric force acting on the electron after entering the electric field?
Positive charges in an electric field will experience an electric force that is in the same direction as the electric field. If the charge is negative, the force will be in the opposite direction of the electric field. Since we are talking about an electron moving in an electric field that points in the negative y-direction, the electron will feel a force that points in the positive y-direction, or upwards.
Positive charges in an electric field will experience an electric force that is in the same direction as the electric field. If the charge is negative, the force will be in the opposite direction of the electric field. Since we are talking about an electron moving in an electric field that points in the negative y-direction, the electron will feel a force that points in the positive y-direction, or upwards.
Compare your answer with the correct one above
A charged rod carrying a negative charge is brought near two spheres that are in contact with each other but insulated from the ground. If the two spheres are then separated, what kind of charge will be on the spheres?
A charged rod carrying a negative charge is brought near two spheres that are in contact with each other but insulated from the ground. If the two spheres are then separated, what kind of charge will be on the spheres?
When the negatively charged rod is brought near one of the two spheres, the presents of the negative charge will induce a flow of charge in the spheres such that regions farthest away from the charged rod will become most negative and regions near the rod will become most positive. This is called charge by induction.
When the negatively charged rod is brought near one of the two spheres, the presents of the negative charge will induce a flow of charge in the spheres such that regions farthest away from the charged rod will become most negative and regions near the rod will become most positive. This is called charge by induction.
Compare your answer with the correct one above
By what method will a positively charged rod produce a negative charge on a conducting sphere that is placed on an insulating surface?
By what method will a positively charged rod produce a negative charge on a conducting sphere that is placed on an insulating surface?
Charge by induction happens when a charged object is brought in the vicinity of a neutral object. The presents of the charged object will cause the free charges in the neutral object to shift such that the neutral object becomes polarized. When the charged object is positive, this will induce a negative charge on a neutral object.
Charge by induction happens when a charged object is brought in the vicinity of a neutral object. The presents of the charged object will cause the free charges in the neutral object to shift such that the neutral object becomes polarized. When the charged object is positive, this will induce a negative charge on a neutral object.
Compare your answer with the correct one above
A charged particle traveling along the +x axis enters an electric field directed vertically upward along the +y-axis. If the charged particle experiences a force downward because of this field, what is the sign of the charge on this particle?
A charged particle traveling along the +x axis enters an electric field directed vertically upward along the +y-axis. If the charged particle experiences a force downward because of this field, what is the sign of the charge on this particle?
Positive charges in an electric field will experience an electric force that is in the same direction as the electric field. If the charge is negative, the force will be in the opposite direction of the electric field. Since the charged particle experiences a force which is opposite to the electric field, the sign of the charge must be negative.
Positive charges in an electric field will experience an electric force that is in the same direction as the electric field. If the charge is negative, the force will be in the opposite direction of the electric field. Since the charged particle experiences a force which is opposite to the electric field, the sign of the charge must be negative.
Compare your answer with the correct one above
A conductor is placed in an electric field under electrostatic conditions. Which of the following statements is correct for this situation?
A conductor is placed in an electric field under electrostatic conditions. Which of the following statements is correct for this situation?
A conductor is defined as a object free to move charges. In particular, valence electrons, which are the outer most electron in each atom and the most free to move, travel inside the conductor until the net electric field inside the conductor is zero. These electrons will move until this condition has been met. Because of the presents of charged particles at the surface and the condition that they are no longer moving, any electric field at the surface must be perpendicular to that surface.
A conductor is defined as a object free to move charges. In particular, valence electrons, which are the outer most electron in each atom and the most free to move, travel inside the conductor until the net electric field inside the conductor is zero. These electrons will move until this condition has been met. Because of the presents of charged particles at the surface and the condition that they are no longer moving, any electric field at the surface must be perpendicular to that surface.
Compare your answer with the correct one above