(1) The harmonic number for the mode of oscillation is 3.
(2) The pitch (frequency) of the sound is 579.55 Hz
(3) The level of the water inside the vertical pipe is 0.1 m.
<h3>The harmonic number</h3>
The harmonic number for the mode of oscillation illustrated for the closed pipe is 3.
<h3>Frequency of the wave</h3>
The pitch (frequency) of the sound is calculated from third harmonic formula;
f = 3v/4L
where;
- v is speed of sound
- L is length of the pipe
f = (3 x 340) / (4 x 0.44)
f = 579.55 Hz
<h3>level of the water</h3>
wave equation for first harmonic of a closed pipe is given as
f = v/(4L)
251.1 = 340/(4L)
4L = 340/251.1
4L = 1.35
L = 1.35/4
L = 0.34 m
level of water = 0.44 m - 0.34 m = 0.1 m
Thus, the level of the water inside the vertical pipe is 0.1 m.
Learn more about harmonics of closed pipes here: brainly.com/question/27248821
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Answer:
Explanation:
From this exercise, our knowable variables are <u>hight and initial velocity </u>
To find how much time does the <u>ball strike the ground</u>, we need to know that the final position of the ball is y=0ft
Solving for t using quadratic formula
or
<u><em>Since time can't be negative the answer is t=6.96s</em></u>
Answer:
Explanation:
Let the critical angle be C .
sinC = 1 / μ where μ is index of refraction .
sinC = 1 /1.2
= .833
C = 56°
Then angle of refraction r = 90 - 56 = 34 ( see the image in attached file )
sin i / sinr = 1.2 , i is angle of incidence
sini = 1.2 x sinr = 1.2 x sin 34 = .67
i = 42°.
Initial velocity = Vo= 25 m/s
Final velocity = V = x
Acceleration= a = 6 m/s^2
time= t = 4 seconds
Appy the equation:
V = Vo + at
Replacing:
V = 25 + 6(4) = 25 + 24 = 49 m/s
