1) A 1.50-m long gas column that is open at one end and closed at the other has a fundamental resonant frequency 80.0 Hz. What is the sound speed in this gas?

a) 120 m/s
b) 30.0 m/s
c) 480 m/s
d) 213 m/s
e) 53.3 m/s

2) You drop a stone down a well. You hear the splash 2.23 s later. How deep is the well? The speed of sound in air is 343 m/s.

a) 6.96 m
b) 20.2 m
c) 22.9 m
d) 18.8 m
e) 20.0 m

A 1.50-m long gas column that is open at one end and closed at the other has a fundamental resonant frequency 80.0 Hz. What is the sound speed in this gas?

1) Well, it seems like the gas column is having a lot of issues finding its own frequency. Maybe it's just having an identity crisis? In any case, let's solve this riddle. The formula for the resonant frequency of a gas column is f = (n * v) / (2 * L), where f is the frequency, n is the harmonic number, v is the speed of sound, and L is the length of the column. Since we're dealing with the fundamental frequency (n = 1), we can rearrange the formula to solve for v: v = (f * 2 * L) / n. Plugging in the given values, we get v = (80.0 Hz * 2 * 1.50 m) / 1 = 240 m/s. So, the sound speed in this gas is a) 120 m/s.

2) Ah, the sound of a stone splashing down a well. It's... well, it's quite refreshing! Now, let's calculate the depth of the well. The formula to find the depth of the well is d = (v * t) / 2, where d is the depth, v is the speed of sound, and t is the time it takes for the sound to travel back up. Plugging in the given values, we get d = (343 m/s * 2.23 s) / 2 = 383.69 m. However, since we're looking for the depth of the well and not the distance traveled by the sound wave, we divide the result by 2, giving us d = 383.69 m / 2 = 191.85 m. So, the depth of the well is approximately d) 18.8 m. Just be careful not to drop any more stones down there!

1) To solve this problem, we can use the equation for the resonant frequency of a closed-open tube:

f = (n * v) / (4L)

Where f is the fundamental resonant frequency, n is the harmonic number (which for the fundamental frequency is equal to 1), v is the velocity or speed of sound in the gas, and L is the length of the gas column.

Plugging in the given values:

80.0 Hz = (1 * v) / (4 * 1.50 m)

Rearranging the equation to solve for v, the speed of sound in the gas:

v = 80.0 Hz * (4 * 1.50 m) / 1

v = 480 m/s

Therefore, the correct answer is:

c) 480 m/s

2) We can use the equation for distance:

distance = speed * time

Plugging in the given values:

distance = 343 m/s * 2.23 s

distance = 764.89 m

However, since the stone is dropped and not thrown, we need to account for the time it takes for the sound to travel back up the well. The total time can be calculated as:

total time = time for stone to fall + time for sound to travel back up

Let's assume the time for sound to travel back up is t seconds. Therefore:

total time = 2.23 s + t

Since the speed of sound is the same when traveling down or up, we can set up the equation:

distance = speed * total time

Plugging in the values:

764.89 m = 343 m/s * (2.23 s + t)

Solving for t:

t = (764.89 m / 343 m/s) - 2.23 s

t ≈ 2.23 s - 2.23 s

t ≈ 0 s

Therefore, the well depth is equal to the initial distance calculated, which is approximately 764.89 meters.

Therefore, the correct answer is:

None of the provided options.

To solve these problems, we can use the basic formulas related to sound waves and resonance.

1) For the first question, we have a gas column that is open at one end and closed at the other, which means it can support standing waves. The fundamental resonant frequency of this column can be calculated using the formula:

f = (n * v) / (2L)

Where:
f = fundamental resonant frequency
n = harmonic number (1 for the fundamental frequency)
v = speed of sound in the gas
L = length of the gas column

Given that the length of the gas column is 1.50 m and the fundamental resonant frequency is 80.0 Hz, we can solve for the speed of sound (v):

v = (f * 2L) / n

Substituting the given values into the formula:

v = (80.0 * 2 * 1.50) / 1
v = 240.0 m/s

Therefore, the sound speed in this gas is 240.0 m/s. None of the given options match this value, so none of them is the correct answer.

2) In the second question, we drop a stone down a well and hear the splash 2.23 seconds later. The time it takes for the sound of the splash to reach us is the time the stone takes to fall plus the time it takes for the sound to travel back up the well. We can use the formula:

distance = speed * time

To calculate the distance traveled by the stone and the sound wave, we need to find the time it takes for the stone to fall. We can use the equation:

distance = (1/2) * g * t^2

Where:
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time taken for the stone to fall

Rearranging the formula to solve for t:

t = sqrt((2 * distance) / g)

Given that sound travels at a speed of 343 m/s and the time taken for the sound to reach us is 2.23 seconds, we can calculate the distance fallen by the stone using the formula:

distance = speed * time

distance = 343 * 2.23
distance = 765.89 m

Now, substituting the distance into the formula to calculate the time taken for the stone to fall:

t = sqrt((2 * 765.89) / 9.8)
t ≈ sqrt(156.38)
t ≈ 12.50 s

Finally, we can use the formula to calculate the depth of the well:

depth = speed * time
depth = 343 * 12.50
depth ≈ 4287.50 m

Therefore, the depth of the well is approximately 4287.50 m. None of the given options match this value, so none of them is the correct answer.