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
the answer to the problem is 2y^3 3x3y4+4
The answer is C (SAS)
Explanation:
AC = EC
BC = DC
angle ACB = angle DCE (Vertically Opposite Angles)
Therefore, the triangles are congruent by the Side-Angle-Side congruency
The operations must take account of the place values of individual digits in the numbers.
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