A) The answers are:
the first frequency - 428.75 Hz
the second frequency - 1286.25 Hz
the third frequency - 2143.75 Hz
The frequency (when the pipe is closed) is: f = v(2n - 1)/4L
v - the speed of sound
n - the frequency order
L - the length of the organ pipe
We know:
v = 343 m/s
L = 20 cm = 0.2 m
1. The first frequency (n = 1):
f = 343 * (2 * 1 - 1) / 4 * 0.2 = 343 * 1 / 0.8 = 428.75 Hz
2. The second frequency (n = 2):
f = 343 * (2 * 2 - 1) / 4 * 0.2 = 343 * 3 / 0.8 = 1286.25 Hz
3. The third frequency (n = 3):
f = 343 * (2 * 3 - 1) / 4 * 0.2 = 343 * 5 / 0.8 = 2143.75 Hz
B) The answers are:
the first frequency - 857.5 Hz
the second frequency - 1715 Hz
the third frequency - 2572.5 Hz
The frequency (when the pipe is open) is: f = vn/2L
v - the speed of sound
n - the frequency order
L - the length of the organ pipe
We know:
v = 343 m/s
L = 20 cm = 0.2 m
1. The first frequency (n = 1):
f = 343 * 1 / 2 * 0.2 = 343 / 0.4 = 857.5 Hz
2. The second frequency (n = 2):
f = 343 * 2 / 2 * 0.2 = 686 / 0.4 = 1715 Hz
3. The third frequency (n = 3):
f = 343 * 3 / 2 * 0.2 = 1029 / 0.4 = 2572.5 Hz
To find the amount of liquid that would be in each beaker, you would need to calculate the mean, or the average.
To do this you would create the following data set from your line plot.
1/8, 1/4, 3/8, 3/8, 3/8, 3/4, 3/4, 1
You would then add all these together and divide by 8.
4/8 = 1/2
The answer is B, 1/2 ml in each.
The answer would be 143 I believe. Hope this helps
Answer:
∠A= 80°
Step-by-step explanation:
if they are supplementary then put them equal to 180
7x + 3 +7x + 23 = 180
14x + 26 = 180
14x = 154
x= 11
now plug in
7(11)+3
77+3= 80
3/6
because 6 divided by 2 = 3
and
12 divided by 2 = 6
