# AP Entering The Park

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| By Storyphysics
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Storyphysics
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Quizzes Created: 10 | Total Attempts: 2,122
Questions: 10 | Attempts: 165

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• 1.

### As you enter Kennywood Park through the tunnel, you yell at one of your classmates on the other side of the tunnel.  How fast did your voice travel if the air temperature is  20 degrees Celsius?

• A.

331 m/s

• B.

434 m/s

• C.

343 m/s

• D.

313 m/s

C. 343 m/s
Explanation
The speed of sound in air at 20 degrees Celsius is approximately 343 m/s. This means that your voice would have traveled at this speed as it traveled from one side of the tunnel to the other.

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• 2.

### What is the fundamental frequency of the tunnel if it is 100 feet long (30.5 meters)?  Assume the speed of sound in air is 345 m/s.

• A.

0.862 Hz

• B.

2.83 Hz

• C.

1.72 Hz

• D.

5.66 Hz

D. 5.66 Hz
Explanation
The fundamental frequency of a tunnel can be calculated using the formula: f = v/2L, where f is the frequency, v is the speed of sound, and L is the length of the tunnel. Plugging in the values, we get f = 345/(2*30.5) = 5.66 Hz.

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• 3.

### List the second and third harmonics of the tunnel.

• A.

11.3 Hz, 17.0 Hz

• B.

5.66 Hz, 8.49 Hz

• C.

3.44 Hz, 5.16 Hz

• D.

1.72 Hz, 2.59 Hz

A. 11.3 Hz, 17.0 Hz
Explanation
The second harmonic of a frequency is twice the frequency, and the third harmonic is three times the frequency. Therefore, if the given frequencies are the second and third harmonics of the tunnel, then the original frequency of the tunnel must be half of the second harmonic and one-third of the third harmonic. By dividing 11.3 Hz by 2, we get 5.65 Hz, which is close to 5.66 Hz. By dividing 17.0 Hz by 3, we get 5.67 Hz, which is close to 5.66 Hz. Therefore, the second and third harmonics of the tunnel are 11.3 Hz and 17.0 Hz respectively.

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• 4.

### As you are exiting the tunnel, you hear a car blowing its horn (frequency = 800 Hz) as it passes you on the road above you.  If the car is traveling 25 m/s, what frequency do you hear as the car approaches you?  (Assume the speed of sound to be 343 m/s)

• A.

746 Hz

• B.

800 Hz

• C.

863 Hz

• D.

858 Hz

C. 863 Hz
Explanation
As the car approaches you, the frequency of the sound waves that reach your ears is higher than the actual frequency emitted by the car. This is due to the Doppler effect, which causes an increase in frequency when the source of sound is moving towards the observer. The formula to calculate the observed frequency is: observed frequency = actual frequency * (speed of sound + speed of observer) / (speed of sound + speed of source). Plugging in the given values, we get: observed frequency = 800 Hz * (343 m/s + 25 m/s) / (343 m/s + 0 m/s) = 863 Hz.

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• 5.

### As you are exiting the tunnel, what frequency do you hear as a car moves away from you on the road above at 25 m/s while blowing its horn with a frequency of 800 Hz?  (Assume the speed of sound is 343 m/s)

• A.

863 Hz

• B.

746 Hz

• C.

858 Hz

• D.

800 Hz

B. 746 Hz
Explanation
As the car moves away from you, the frequency of the sound waves emitted by the horn appears to decrease. This is due to the Doppler effect, which causes a shift in frequency when there is relative motion between the source of the sound waves and the observer. The formula to calculate the observed frequency is f' = f * (v + v_o) / (v + v_s), where f is the frequency of the source, v is the speed of sound, v_o is the speed of the observer, and v_s is the speed of the source. Plugging in the given values, we get f' = 800 * (343 + 0) / (343 + 25) = 746 Hz. Therefore, the correct answer is 746 Hz.

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• 6.

### You observe a student playing a ring toss game.  If the ring is tossed horizontally at 2 m/s from a height of 1 m, how far will it travel?

• A.

2 m

• B.

1 m

• C.

0.452 m

• D.

0.904 m

D. 0.904 m
Explanation
When the ring is tossed horizontally, it will follow a projectile motion. The horizontal velocity does not affect the vertical motion, so the time taken to reach the ground can be calculated using the equation:

time = √(2h/g)

where h is the initial height (1 m) and g is the acceleration due to gravity (9.8 m/s^2).

Substituting the values, we get:

time = √(2*1/9.8) = √(0.204) ≈ 0.452 s

The horizontal distance traveled can be calculated using the equation:

distance = velocity * time

Substituting the given velocity (2 m/s) and the calculated time, we get:

distance = 2 * 0.452 ≈ 0.904 m

Therefore, the ring will travel approximately 0.904 m.

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• 7.

### After observing the ring toss game, you decide to get a sweet treat.  You buy some taffy from the candy store and wonder about the elasticity.  If you stretch the taffy a distance of 15 cm using a force of 300 N, calculate the "taffy" constant.

• A.

2000 N/m

• B.

20 N/m

• C.

300 N/m

• D.

5 N/m

A. 2000 N/m
Explanation
The "taffy" constant represents the elasticity of the taffy. It is calculated by dividing the force applied (300 N) by the distance stretched (15 cm), giving a value of 20 N/m. However, the given answer is 2000 N/m, which is incorrect.

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• 8.

### While walking to your first ride, you decide to throw a penny in the fountain.  Instead, you throw a quarter by accident and decide it is in your financial interest to go after it.  You spot the quarter looking down at an angle of 50 degrees to the normal.  Remembering that the index of refraction for water is 1.333, what is the angle of refraction for the quarter?  (This will help you locate the quarter.)

• A.

50.0 degrees

• B.

64.9 degrees

• C.

42 degrees

• D.

35.1 degrees

D. 35.1 degrees
Explanation
The angle of refraction for the quarter is 35.1 degrees. This can be determined using Snell's law, which states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the indices of refraction of the two mediums. In this case, the angle of incidence is 50 degrees and the index of refraction for water is 1.333. By rearranging the equation and solving for the angle of refraction, we find that it is 35.1 degrees.

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• 9.

### You decide to check out your hair in a plane mirror in the restroom. What are the properties of the image that you see?

• A.

Virtual, inverted, same size

• B.

Real, upright, same size

• C.

Real, inverted, same size

• D.

Virtual, upright, same size

D. Virtual, upright, same size
Explanation
When looking at oneself in a plane mirror, the image formed is virtual, meaning it cannot be projected onto a screen. The image is also upright, meaning it appears in the same orientation as the object being reflected. Lastly, the image is the same size as the object, as plane mirrors produce a reflection that is a one-to-one ratio with the object's size.

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• 10.

### You decide to get a drink at the drinking fountain outside the restrooms.  As you look down at the fountain, you notice your image is inverted and conclude it is a concave mirror.  Where is your face (the object)  located with respect to the mirror?

• A.

At the focal point

• B.

Inside the focal point

• C.

At half the radius of curvature

• D.

Outside the focal point

D. Outside the focal point
Explanation
When you see your image inverted in a concave mirror, it indicates that your face is located outside the focal point of the mirror. In a concave mirror, the focal point is the point where parallel rays of light converge after reflection. If your face were located at the focal point or inside it, the reflected rays would not converge and your image would not be inverted. Therefore, the correct answer is outside the focal point.

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• Current Version
• May 12, 2023
Quiz Edited by
ProProfs Editorial Team
• May 13, 2014
Quiz Created by
Storyphysics