Physics - MCAT Physical
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A pendulum with a mass of 405kg reaches a maximum height of 2.4m. What is its velocity at the bottommost point in its path?
A pendulum with a mass of 405kg reaches a maximum height of 2.4m. What is its velocity at the bottommost point in its path?
First solve for the potential energy of the pendulum at the height of 2.4m.
PE = mgh
PE = (405kg)(10m/s2)(2.4m) = 9720J
This must be equal to the maximum kinetic energy of the object.
KE = ½mv2
9720J = ½mv2
Plug in the mass of the object (405 kg) and solve for v.
9720J = ½(405kg)v2
v = 6.9m/s
First solve for the potential energy of the pendulum at the height of 2.4m.
PE = mgh
PE = (405kg)(10m/s2)(2.4m) = 9720J
This must be equal to the maximum kinetic energy of the object.
KE = ½mv2
9720J = ½mv2
Plug in the mass of the object (405 kg) and solve for v.
9720J = ½(405kg)v2
v = 6.9m/s
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An empty mining cart has a mass of
and is traveling down a track that has a slope of
to the horizontal. The cart is traveling at a rate of
when an operator notices a disturbance on the track ahead and locks the wheels of the cart. What is the speed of the cart after it has traveled
with the wheels locked?


An empty mining cart has a mass of and is traveling down a track that has a slope of
to the horizontal. The cart is traveling at a rate of
when an operator notices a disturbance on the track ahead and locks the wheels of the cart. What is the speed of the cart after it has traveled
with the wheels locked?
We need the equation for conservation of energy for this problem:

We can eliminate final potential energy if we set the final height to be zer. We are solving for final velocity, so let's rearrange for final kinetic energy.

Substituting our equations for each variable, we get:

Rearranging for final velocity we get:

If you derive this formula and are unsure of your work, simply check your units. Each term under the square root has units of
, which will ultimately give us units of
, which is what we want.
We have values for all variables except two: height and normal force.
Let's calculate the height. We know that the between the initial and final states, the cart has traveled 20 meters down a slope of 40 degrees. Therefore, we can calculate height with the formula:



Now we just need to find the normal force. The following diagram will help visualize this calculation.

If you are unsure whether to use sine or cosine, think about it practically. As the angle gets less and less, the normal force is going to get larger. This is characteristic of a cosine function.
Therefore, we can say that:

Now that we have all of our variables, it's time to plug and chug:


We need the equation for conservation of energy for this problem:
We can eliminate final potential energy if we set the final height to be zer. We are solving for final velocity, so let's rearrange for final kinetic energy.
Substituting our equations for each variable, we get:
Rearranging for final velocity we get:
If you derive this formula and are unsure of your work, simply check your units. Each term under the square root has units of , which will ultimately give us units of
, which is what we want.
We have values for all variables except two: height and normal force.
Let's calculate the height. We know that the between the initial and final states, the cart has traveled 20 meters down a slope of 40 degrees. Therefore, we can calculate height with the formula:
Now we just need to find the normal force. The following diagram will help visualize this calculation.
If you are unsure whether to use sine or cosine, think about it practically. As the angle gets less and less, the normal force is going to get larger. This is characteristic of a cosine function.
Therefore, we can say that:
Now that we have all of our variables, it's time to plug and chug:
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Which of the following forces is not conservative?
Which of the following forces is not conservative?
Conservative forces are forces that do not lose energy to heat, sound, or light. Of these answers, energy is completely conserved and transferred from kinetic energy to potential energy, or vice versa. Gravitational forces, electrostatic forces, and elastic forces all work by providing a potential that will work in the same direction as the motion of an object or particle, allowing kinetic and potential energy to interconvert. Frictional forces lose energy as heat when sliding across a surface, and the more force (the more rough the surface), the more energy that is lost.
Conservative forces are forces that do not lose energy to heat, sound, or light. Of these answers, energy is completely conserved and transferred from kinetic energy to potential energy, or vice versa. Gravitational forces, electrostatic forces, and elastic forces all work by providing a potential that will work in the same direction as the motion of an object or particle, allowing kinetic and potential energy to interconvert. Frictional forces lose energy as heat when sliding across a surface, and the more force (the more rough the surface), the more energy that is lost.
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Which of the following is not a conservative force?
Which of the following is not a conservative force?
Friction is a non-conservative force, meaning that the work it does depends on the path taken by the object. For example, moving a brick in a long zig-zag across the table will generate more heat from friction than moving it in a straight line across the table.
Electric and gravitational forces are conservative. This can be tested by knowing a constant equation to calculate the energy associated with these forces; such equations are applicable regardless of path. No such equation exists for frictional energy.
Friction is a non-conservative force, meaning that the work it does depends on the path taken by the object. For example, moving a brick in a long zig-zag across the table will generate more heat from friction than moving it in a straight line across the table.
Electric and gravitational forces are conservative. This can be tested by knowing a constant equation to calculate the energy associated with these forces; such equations are applicable regardless of path. No such equation exists for frictional energy.
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A 2kg lead ball is loaded into a spring cannon and the cannon is set at a 45º angle on a platform. The spring has a spring constant of 100N/m and the ball and spring system is compressed by 1m before launch. While the ball is in flight air resistance can be neglected, and the ball finishes its flight by landing at a cushion placed some distance away from the cannon.
Find the horizontal component of velocity once the ball has left the cannon.
A 2kg lead ball is loaded into a spring cannon and the cannon is set at a 45º angle on a platform. The spring has a spring constant of 100N/m and the ball and spring system is compressed by 1m before launch. While the ball is in flight air resistance can be neglected, and the ball finishes its flight by landing at a cushion placed some distance away from the cannon.
Find the horizontal component of velocity once the ball has left the cannon.
This asks us to understand the vector components of velocity. Remember that the final velocity is the hypotenuse of a triangle (solved to be 10m/s in the previous problem), and that by knowing the hypotenuse value we can solve for the horizontal component by using cosine.
vx = (10m/s)(cos(45o)) = 7.1m/s
This asks us to understand the vector components of velocity. Remember that the final velocity is the hypotenuse of a triangle (solved to be 10m/s in the previous problem), and that by knowing the hypotenuse value we can solve for the horizontal component by using cosine.
vx = (10m/s)(cos(45o)) = 7.1m/s
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A 2kg lead ball is loaded into a spring cannon and the cannon is set at a 45º angle on a platform. The spring has a spring constant of 100N/m and the ball and spring system is compressed by 1m before launch. While the ball is in flight air resistance can be neglected, and the ball finishes its flight by landing at a cushion placed some distance away from the cannon.
What is the initial vertical component of velocity of the ball?
A 2kg lead ball is loaded into a spring cannon and the cannon is set at a 45º angle on a platform. The spring has a spring constant of 100N/m and the ball and spring system is compressed by 1m before launch. While the ball is in flight air resistance can be neglected, and the ball finishes its flight by landing at a cushion placed some distance away from the cannon.
What is the initial vertical component of velocity of the ball?
This question also asks us to understand the vector components of velocity. Remember that the final velocity is the hypotenuse of a triangle, and that by knowing the hypotenuse value (solved as 10m/s in a previous problem) we can solve for the vertical component using sine.
vy = (10m/s)(sin(45o)) = 7.1m/s
This question also asks us to understand the vector components of velocity. Remember that the final velocity is the hypotenuse of a triangle, and that by knowing the hypotenuse value (solved as 10m/s in a previous problem) we can solve for the vertical component using sine.
vy = (10m/s)(sin(45o)) = 7.1m/s
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A 2kg lead ball is loaded into a spring cannon and the cannon is set at a 45º angle on a platform. The spring has a spring constant of 100N/m and the ball and spring system is compressed by 1m before launch. While the ball is in flight air resistance can be neglected, and the ball finishes its flight by landing at a cushion placed some distance away from the cannon.
What is the horizontal acceleration of the ball during its flight?
A 2kg lead ball is loaded into a spring cannon and the cannon is set at a 45º angle on a platform. The spring has a spring constant of 100N/m and the ball and spring system is compressed by 1m before launch. While the ball is in flight air resistance can be neglected, and the ball finishes its flight by landing at a cushion placed some distance away from the cannon.
What is the horizontal acceleration of the ball during its flight?
Without an additional force acting in the horizontal direction during flight (we are told we can neglect air resistance), there is no acceleration. Remember that Newton's second law, F = ma, requires that a force act on an object to produce acceleration. Here, we have no additional force and thus no acceleration.
Without an additional force acting in the horizontal direction during flight (we are told we can neglect air resistance), there is no acceleration. Remember that Newton's second law, F = ma, requires that a force act on an object to produce acceleration. Here, we have no additional force and thus no acceleration.
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Which factors increase the maximum velocity of a pendulum?
Which factors increase the maximum velocity of a pendulum?
Both the length of the pendulum's string and the angle of displacement affect the maximum velocity of the pendulum. Increasing the length of the pendulum's string and increasing the angle of displacement both increase the distance the pendulum must travel in a single period, increasing its potential energy at its maximum height, and therefore the maximum velocity at its lowest point.
Both the length of the pendulum's string and the angle of displacement affect the maximum velocity of the pendulum. Increasing the length of the pendulum's string and increasing the angle of displacement both increase the distance the pendulum must travel in a single period, increasing its potential energy at its maximum height, and therefore the maximum velocity at its lowest point.
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Which of the following would not cause a decrease in the pressure of a gas in a sealed container?
Which of the following would not cause a decrease in the pressure of a gas in a sealed container?
A decrease in pressure means a decrease in gas particle collisions. The only option that would not cause a decrease in collisions is adding moles of a different gas. Even though different molecules are added, there will be greater pressure as particle collisions will be more frequent.
Reducing temperature slows the gas particles, thus decreasing the frequency of collisions. Similarly, increasing the volume of the container and removing particles will cause a decrease in collisions, and subsequent pressure.
A decrease in pressure means a decrease in gas particle collisions. The only option that would not cause a decrease in collisions is adding moles of a different gas. Even though different molecules are added, there will be greater pressure as particle collisions will be more frequent.
Reducing temperature slows the gas particles, thus decreasing the frequency of collisions. Similarly, increasing the volume of the container and removing particles will cause a decrease in collisions, and subsequent pressure.
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Which of the following factors does not explain why measurements of real gases deviate from ideal values?
Which of the following factors does not explain why measurements of real gases deviate from ideal values?
Measurements of real gases deviate from ideal gas predictions because intermolecular forces and the volume of the particles themselves are not taken into consideration for ideal gases. The volume of the space between particles is considered for ideal gases and does not contribute to deviation from ideal gas behavior.
Attraction between molecules causes real pressure to be slightly less than ideal pressure, while the volume of gas particles causes real volume to be slightly greater than ideal volume.
Measurements of real gases deviate from ideal gas predictions because intermolecular forces and the volume of the particles themselves are not taken into consideration for ideal gases. The volume of the space between particles is considered for ideal gases and does not contribute to deviation from ideal gas behavior.
Attraction between molecules causes real pressure to be slightly less than ideal pressure, while the volume of gas particles causes real volume to be slightly greater than ideal volume.
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Which of the following best describes the effect of the Doppler shift on the appearance of stars moving towards Earth?
Which of the following best describes the effect of the Doppler shift on the appearance of stars moving towards Earth?
The Doppler shift equation for light is
, where f is the source frequency, f' is the observed frequency, v is the relative velocity between source and observer, and c is the speed of light.
When the source and observer are moving closer together, v is positive, so the observed frequency is greater than the source frequency. Greater frequency also implies shorter wavelength, so visible light is shifted towards the blue end of the spectrum.
The Doppler shift equation for light is , where f is the source frequency, f' is the observed frequency, v is the relative velocity between source and observer, and c is the speed of light.
When the source and observer are moving closer together, v is positive, so the observed frequency is greater than the source frequency. Greater frequency also implies shorter wavelength, so visible light is shifted towards the blue end of the spectrum.
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Each of the following types of waves experience polarization except __________.
Each of the following types of waves experience polarization except __________.
Polarization is the property that allows tansverse waves to oscillate in multiple orientations. A transverse wave can oscillate, for example, in either the xy-plane or the yz-plane.
Sound waves are longitudinal, and thus do no experience polarization as medium is displaced in one direction only. A longitudinal wave will travel in only one dimension via compression and rarefraction.
Polarization is the property that allows tansverse waves to oscillate in multiple orientations. A transverse wave can oscillate, for example, in either the xy-plane or the yz-plane.
Sound waves are longitudinal, and thus do no experience polarization as medium is displaced in one direction only. A longitudinal wave will travel in only one dimension via compression and rarefraction.
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Which of these is an example of a longitudinal wave?
Which of these is an example of a longitudinal wave?
Longitudinal waves transmit energy by compressing and rarefacting the medium in the same direction as they are traveling. Sounds waves are longitudinal waves and travel by compressing the air through which they travel, causing vibration.
Light, X-rays, and microwaves are all examples of electromagnetic waves; even if you cannot recall if they are longitudinal or transverse, they are all members of the same phenomenon and will have the same type of wave transmission. Transverse waves are generated by oscillation within a plane perpendicular to the direction of motion. Oscillating a rope is a transverse wave, as it is not compressing in the direction of motion.
Longitudinal waves transmit energy by compressing and rarefacting the medium in the same direction as they are traveling. Sounds waves are longitudinal waves and travel by compressing the air through which they travel, causing vibration.
Light, X-rays, and microwaves are all examples of electromagnetic waves; even if you cannot recall if they are longitudinal or transverse, they are all members of the same phenomenon and will have the same type of wave transmission. Transverse waves are generated by oscillation within a plane perpendicular to the direction of motion. Oscillating a rope is a transverse wave, as it is not compressing in the direction of motion.
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A 2kg lead ball is loaded into a spring cannon and the cannon is set at a 45º angle on a platform. The spring has a spring constant of 100N/m and the ball and spring system is compressed by 1m before launch. While the ball is in flight air resistance can be neglected, and the ball finishes its flight by landing at a cushion placed some distance away from the cannon.
How much energy is stored in the spring before the ball is launched?
A 2kg lead ball is loaded into a spring cannon and the cannon is set at a 45º angle on a platform. The spring has a spring constant of 100N/m and the ball and spring system is compressed by 1m before launch. While the ball is in flight air resistance can be neglected, and the ball finishes its flight by landing at a cushion placed some distance away from the cannon.
How much energy is stored in the spring before the ball is launched?
In order to determine how much energy is stored, we first need to understand what type of energy we want to consider. A spring stores potential energy; the potential energy of the spring is maximized at maximal displacement from its resting state. In order to compute the potential energy stored, we need both the spring constant (100N/m) and the displacement from resting (1m).
PEs = ½k(Δx)2 = ½(100N/m)(1m)2 = 50J
In order to determine how much energy is stored, we first need to understand what type of energy we want to consider. A spring stores potential energy; the potential energy of the spring is maximized at maximal displacement from its resting state. In order to compute the potential energy stored, we need both the spring constant (100N/m) and the displacement from resting (1m).
PEs = ½k(Δx)2 = ½(100N/m)(1m)2 = 50J
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What is the period of a pendulum that has a string length of 9.8m?
What is the period of a pendulum that has a string length of 9.8m?
The key to answering this question is to recall the following important formula for a simple pendulum:
.

The key to answering this question is to recall the following important formula for a simple pendulum: .
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Two charges of
coulombs are a distance
apart from each other. Which of the following would reduce the force exerted between the charges by a factor of 4?
Two charges of coulombs are a distance
apart from each other. Which of the following would reduce the force exerted between the charges by a factor of 4?
Given Coulomb's Law electrostatic forces: 
We can see that distance and force are inversly related. Also distance is squared, so if we increase the distance by 2, the force between the two charges will be reduced by a factor of four.
Given Coulomb's Law electrostatic forces:
We can see that distance and force are inversly related. Also distance is squared, so if we increase the distance by 2, the force between the two charges will be reduced by a factor of four.
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How many electrons would it require to generate a 5 N attractive force on a +3 µC charge from 0.1 m away.
How many electrons would it require to generate a 5 N attractive force on a +3 µC charge from 0.1 m away.
To find the answer we must first solve for q, using Coulomb's Law. Then after finding q = 1.85 x 10–6 C, we must divide by 1.6 x 10–19 C to get the amount of electrons required to make such a charge.
To find the answer we must first solve for q, using Coulomb's Law. Then after finding q = 1.85 x 10–6 C, we must divide by 1.6 x 10–19 C to get the amount of electrons required to make such a charge.
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A sound source with a frequency of 790Hz moves away from a stationary observer at a rate of 15m/s. What frequency does the observer hear?
The speed of sound is 340m/s.
A sound source with a frequency of 790Hz moves away from a stationary observer at a rate of 15m/s. What frequency does the observer hear?
The speed of sound is 340m/s.
In this scenario the Doppler effect is described by the following equation.

Using the values from the problem, we know that vo is zero and vf is 15m/s. v is 340m/s and fs is 790Hz.

In this scenario the Doppler effect is described by the following equation.
Using the values from the problem, we know that vo is zero and vf is 15m/s. v is 340m/s and fs is 790Hz.
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A fire truck emits an 880Hz siren. As the truck approaches an obeserver on the sidewalk, he perceives the pitch to be 950Hz. Approximately what pitch does he hear after the truck passes and is moving away? Assume the truck's velocity remains constant, and that the velocity of sound in air is 340m/s.
A fire truck emits an 880Hz siren. As the truck approaches an obeserver on the sidewalk, he perceives the pitch to be 950Hz. Approximately what pitch does he hear after the truck passes and is moving away? Assume the truck's velocity remains constant, and that the velocity of sound in air is 340m/s.
The equation for Doppler effect is
, where the + sign applies when the source and observer are moving farther apart, and the - sign applies when they are moving closer together. In these equations, v is the speed of sound, 340m/s,
is the frequency of sound emitted by the source,
is the freqency perceived by the observer, and
is the relative velocity between the source and observer.
We can apply this equation to the first part of the motion, as the truck moves closer to the observer, to solve for the velocity of the truck.


Now we can plug this velocity into the equation again for when the truck moves farther away from the observer and solve for
.

The equation for Doppler effect is , where the + sign applies when the source and observer are moving farther apart, and the - sign applies when they are moving closer together. In these equations, v is the speed of sound, 340m/s,
is the frequency of sound emitted by the source,
is the freqency perceived by the observer, and
is the relative velocity between the source and observer.
We can apply this equation to the first part of the motion, as the truck moves closer to the observer, to solve for the velocity of the truck.
Now we can plug this velocity into the equation again for when the truck moves farther away from the observer and solve for .
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A fire truck emitting a siren at
moves at
towards a jogger. The jogger is moving at
towards the fire truck. Take the speed of sound to be
.
At what frequency does the jogger perceive the siren?
A fire truck emitting a siren at moves at
towards a jogger. The jogger is moving at
towards the fire truck. Take the speed of sound to be
.
At what frequency does the jogger perceive the siren?
In order to solve this problem we must know how to utilize the Doppler formula.

is the velocity of the observer and
is the velocity of the source. Notice that the frequency must increase as the observer and source move closer, and therefore the plus sign is used in the numerator and the minus sign is used in the denominator. Had the jogger been moving away from the fire truck, the subtraction function would be used in both the top and bottom.
In this case we see that the source is the fire truck, moving at
, and the observer is the jogger, moving at
. By plugging these numbers into the formula and
for
, we find the perceived frequency or
to be
.



In order to solve this problem we must know how to utilize the Doppler formula.
is the velocity of the observer and
is the velocity of the source. Notice that the frequency must increase as the observer and source move closer, and therefore the plus sign is used in the numerator and the minus sign is used in the denominator. Had the jogger been moving away from the fire truck, the subtraction function would be used in both the top and bottom.
In this case we see that the source is the fire truck, moving at , and the observer is the jogger, moving at
. By plugging these numbers into the formula and
for
, we find the perceived frequency or
to be
.
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