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
This is a geometric sequence since there is a common ratio between each term. In this case, multiplying the previous term in the sequence by 2 gives the next term.
Answer:
Random Sample
Step-by-step explanation:
Answer:
1.)4188.79
2.) 7238.23
3.)1.02
4.)4189
5.)170 cm³
Step-by-step explanation:
1.) 4πr2 * 5
2.) 4πr2 * 12
3.) V = 4/3(PI*r3). *4.5
4.) V = 4/3 π r ^3 * 10
5.) Given:
Cylindrical container: height = 18 cm ; diameter = 6 cm.
3 balls each have a radius of 3 cm.
Volume of a cylinder = π r² h
V = 3.14 * (3cm)² * 18 cm
V = 508.68 cm³
Volume of rubber ball = 4/3 π r³
V = 4/3 * 3.14 * (3cm)³
V = 113.04 cm³
113.04 cm³ * 3 balls = 339.12 cm³
508.68 cm³ - 339.12 cm³ = 169.56 cm³ or 170 cm³
There is 170 cm³ free space in the container.
Answer:
1 = x and -2 = y
I am fairly happy with this answer
