(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
#SPJ1
Answer:
d)energy
Explanation:
Waves can transfer energy over distance without moving matter the entire distance. For example, an ocean wave can travel many kilometers without the water itself moving many kilometers. The water moves up and down—a motion known as a disturbance. It is the disturbance that travels in a wave, transferring energy.
Question
A banked highway is designed for traffic moving at v 8 km/h. The radius of the curve = 330 m. 50% Part (a) Write an equation for the tangent of the highway's angle of banking. Give your equation in terms of the radius of curvature r, the intended speed of the turn v, and the acceleration due to gravity g
Part (b) what is the angle of banking of the highway? Give your answer in degrees
Answer:
a. Equation of Tangent
tan(θ) = v²/rg
b. Angle of the banking highway
θ = 0.087°
Explanation:
Given
Radius of the curve = r = 330m
Acceleration of gravity = g = 9.8m/s²
Velocity = v = 8km/h = 8 * 1000/3600
v = 2.22 m/s
a . Write an equation for the tangent of the highway's angle of banking
The Angle is calculated by
tan(θ) = v²/rg
θ = tan-1(v²/rg)
b.
Part (b) what is the angle of banking of the highway? Give your answer in degrees
θ = tan-1(v²/rg)
Substituting the values of v,g and r
θ = tan-1(2.22²/(330 * 9.8)
θ = tan-1(0.001523933209647)
θ = 0.087314873580116°
θ = 0.087°
