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
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the percent of markup <span> 62.3%
</span>$3.05 -> $4.95
<span>Difference = $1.90 </span>
<span>Percentage of increase = 1.90/3.05 = Approximately 0.6229 = 62.3% = D </span>
i think its undifined
maybe it is. Because you can't see something there
Answer:
x = 3.6
Step-by-step explanation:
3/5 = x/6
multiply both sides by 6
18/5 = x
as a decimal
x = 3.6
At 10 am, Car A is at 30 miles.
Then at 11 am, Car A is at 60 miles and Car B is at 40 miles
At 12 am, Car A is at 90 miles and Car B is at 80 miles
At 1 pm, Car A and B are tied for 120 miles
The very second after Car B passes Car A
Answer: 1:01 pm Car B Passes Car A
