The temperature of the air is 20.0 degrees Celsius when a tourist drops a coin into a wishing well. She hears the splash 5.30 s after drops the coin. If air resistance is ignored, how deep is the well?
time to fall:
h = .5 g t^2
so t = sqrt (2 g h)
time for sound to rise = h/c
5.3 = h/c + sqrt (2 g h)
sqrt (2 g h) = 5.3 - h/c
2 g h = 5.3^3 - 10.6 (h/c) + h^2/c^2
look up c at 20 deg C
and then solve the quadratic
120
100
To determine the depth of the well, we can use the formula for the distance traveled by an object in free fall:
d = 0.5 * g * t^2
Where:
d = depth of the well
g = acceleration due to gravity (approximately 9.8 m/s^2 on Earth)
t = time
First, let's convert the temperature from Celsius to Kelvin, as it is needed for the speed of sound calculation. The conversion formula is:
K = °C + 273.15
So, 20.0 degrees Celsius is equal to (20.0 + 273.15) = 293.15 Kelvin.
Next, we need to calculate the speed of sound, as this will give us the time it takes for the sound of the splash to reach the tourist. The speed of sound can be approximated using the formula:
v = 331.4 * √(T)
Where:
v = speed of sound in m/s
T = temperature in Kelvin
Substituting the temperature value we calculated earlier:
v = 331.4 * √(293.15)
Now, we can calculate the time taken for the sound to reach the tourist using the formula:
t_sound = distance / speed of sound
Since the distance is the same as the depth of the well, we can rewrite this as:
t_sound = d / v
Finally, we substitute the given values:
5.30 s = d / [331.4 * √(293.15)]
Now, we can solve for the depth (d) of the well by rearranging the equation:
d = 5.30 s * [331.4 * √(293.15)]
Evaluating this expression will give us the solution for the depth of the well in meters.