All MCAT Physical Resources
Example Questions
Example Question #1 : Doppler Effect
The source of a sound moves away from the listener. The listener has the impression that the source is __________.
lower in velocity
lower in pitch
higher in tone
higher in amplitude
lower in pitch
The formula for the Doppler effect is:
Only frequency of the sound is affected by the Doppler effect; velocity and amplitude remain unchanged. When the source is moving away from the observer the velocity of the source is added to the speed of light.
This increases the value of the denominator, decreasing the value of the observed frequency. Frequency corresponds to pitch or tone; a lower observed frequency will result in a lower observed pitch.
Example Question #1 : Doppler Effect
You are jogging on the sidewalk at a rate of . A police car behind you is patrolling at a rate of when it turns on its siren. If the siren has a frequency of , is the frequency you perceive higher or lower than the frequency emitted?
Neither higher nor lower because the doppler effect only affects sounds at lower frequencies
Higher because the siren is traveling the same direction as you
Lower because the overall distance between you and the siren is increasing
Higher because the overall distance between you and the siren is decreasing
Lower because you are running slower than the police car
Higher because the overall distance between you and the siren is decreasing
The doppler effect states that if two objects are moving closer together, perceived frequencies for emitted waves will be higher. If you are jogging away from the car at , but the car is traveling at , the overall distance between you and the siren is decreasing. You will hear a higher frequency than what the siren is emitting.
The numerator is subtracted when the observer moves away from the source. The denominator is subtracted when the source moves toward the observer. In this problem both terms would be subtracted, but the denominator would be decreased by more than the numerator. This would result in a fraction greater than one, and an overall increase in the final frequency.
Example Question #1 : Sound
A star emits visible color at . If the star is moving away from a stationary observer, which of the following cannot be the wavelength observed?
The Doppler effect accounts for observed frequency versus actual frequency emitted by a sound or light source. The equation for the Doppler effect is:
The numerator terms are summed when the observer moves toward the source, and the denominator terms are summed when the source moves away from the observer. In this question, we know that the source moves away from the observer; therefore, the observed frequency will be less than the actual frequency because the denominator in the equation will increase.
The question, however, asks about wavelength. If the frequency decreases, then the wavelength must increase according to the equation:
Velocity will remain constant (the speed of light). Any change in frequency will cause a change in wavelength via their direct relationship in the given equation.
Example Question #1 : Doppler Effect
Two cars approach each other at when one car starts to beep its horn at a frequency of 475Hz. What is the wavelength of the horn as heard by the other driver?
The Doppler equation is:
.
Because the cars are approaching each other, the frquency heard will be increased. This fundamental knowledge allows you to determine the signs of the equation. The top of the fraction will be addition and the bottom will be subtraction to make a coefficient greater than 1.
Use the given values to solve:
Now that we know the frequency, we can solve for the wavelength:
Example Question #81 : Mcat Physical Sciences
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 .
Example Question #1 : Power
Two children are playing on an icy lake. Child 1 weighs 50kg, and child 2 weighs 38kg. Child 1 has a backpack that weighs 10kg, and child 2 has a backpack that weighs 5kg.
Over the course of the afternoon, they collide many times. Four collisions are described below.
Collision 1:
Child 1 starts from the top of a ramp, and after going down, reaches the lake surface while going 5m/s and subsequently slides into a stationary child 2. They remain linked together after the collision.
Collision 2:
Child 1 and child 2 are sliding in the same direction. Child 2, moving at 10m/s, slides into child 1, moving at 2m/s.
Collision 3:
The two children collide while traveling in opposite directions at 10m/s each.
Collision 4:
The two children push off from one another’s back, and begin moving in exactly opposite directions. Child 2 moves with a velocity of +8m/s.
After child 1 comes to a full stop in the above collisions, she decides she wants to keep going. If she sits on a sled and moves across the lake surface by pushing off the lake surface with her feet, which scenario below will have her producing the most power? Assume the sled has a mass of 10kg.
She goes from 2m/s to 4m/s in 1s
She goes from 2m/s to 5m/s in 4s
She goes from 2m/s to 4m/s in 2s
She goes from 2m/s to 5m/s in 6s
She can't produce net power unless there is an external input of energy
She goes from 2m/s to 4m/s in 1s
Power is equal to work/time. The work that child 1 is performing is also equal to the change in kinetic energy. When she goes from 2m/s to 4m/s, she is increasing her kinetic energy by mv2.
This is the higest number of all the given scencarios. To arrive at this conclusion, you must compare calculations for the other given scenarios.
Example Question #1 : Power
A student holds a 10kg stack of books 1.5m above the ground for 20s. How much power was used to hold the books during this time?
0W
300W
147W
15W
7.35W
0W
Power is given by . Although the student exerts a force on the books during this time, they are not moved at all, so the displacement d is 0, and thus the power is also 0.
Example Question #13 : Work, Energy, And Power
A 100-Watt lightbulb is 3m away from a window with height 0.5m and width 0.5m. At what rate does the bulb's energy exit the window? Assume the light spreads out evenly in all directions.
1.77J/s
3.54J/s
0.88J/s
8.33J/s
2.65J/s
0.88J/s
First we can find the intensity of light in Watts per square meter, when the light reaches the window, by using .
Assuming the light spreads out evenly in all directions, it has spread out over the area of a sphere of radius 3m by the time it reaches the window.
Multiplying this intensity times the area of the window gives the power (or energy per second) leaving the window.
Example Question #14 : Work, Energy, And Power
Which of the following CANNOT be a unit for power?
In physics, power is often given the unit, W, or watts. In mechanics, a watt often represents the amount of work done over a unit of time, or Joules per second.
In electrostatics a watt is expressed as the amount of current times the voltage, or as the current squared times the resistance.
A Newton-meter is a measure of energy or work, and can be written as a Joule, and is therefore the correct answer because it is not a measure of power.
Example Question #11 : Work, Energy, And Power
An elevator lifts 100kg a distance of ten meters in a time of four seconds. What is the power of the elevator?
The power of an elevator can be determined using the equation , with being the work done by the elevator, and being the time in which energy is transferred. Using this, we can plug in the known values and solve for the power.
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