Ask question

Ask question

All categories

11111nata11111 [884]

4 years ago

8

A helium-filled pipe which is closed at both ends is 10 cm long and has a little hole on one side. Blowing into that hole will c

reate standing waves in the pipe. What is the wavelength of the first overtone?

1 answer:

Anna11 [10]4 years ago

6 0

Answer:

0.133m

Explanation:

Since the pipe is closed at one end and opened at the other end (the end where a hole is created), the pipe in question is a closed pipe.

Overtones are defined as the multiple integrals of their fundamental frequency.

If fundamental frequency of the pipe is Fo, the first overtone for closed pipe F1 = 3fo

Since Fo = V/4L

F1 = 3(V/4L)

Given the length of the pipe L = 10cm = 0.1m

Speed of sound wave V = 330m/s

F1 = 3(330)/4(0.1)

F1 = 990/0.4

F1 = 2475Hertz

Using the relationship to get the wavelength

V = F1λ

λ = v/F1

λ= 330/2475

λ = 0.133m

You might be interested in

What is your favorite place to live

Gekata [30.6K]

Florida. The beaches are great

3 0

3 years ago

Read 2 more answers

In an experiment involving pendulums, you want to see how changing the mass of the bob affects the period (amount of time) of a

Alborosie

Answer:

The longer the length of string, the farther the pendulum falls; and therefore, the longer the period, or back and forth swing of the pendulum. The greater the amplitude, or angle, the farther the pendulum falls; and therefore, the longer the period.

Explanation:

3 0

3 years ago

Which of the following is correct? *

Natalija [7]

Answer:

The last one

1m = 100 cm

Explanation:

If you do not trust me look it up

5 0

3 years ago

Three forces act on an object. Two of the forces have the magnitudes 56 N and 27 N, and make angles 53° and 156°, respectively,

Ludmilka [50]

Answer:

Explanation:

Given

Two forces $F_1$ and $F_2$ at an angle of $\theta _1$ $\theta _2$

$F_1=56\ N$

$F_2=27\ N$

$\theta _1=53^{\circ}$

$\theta _2=156^{\circ}$

As resultant force is zero therefore horizontal component as well as vertical component of force is zero

$\sum F_x=F_1\cos \theta _1+F_2\cos \theta _2+F_3\cos \theta _3$

$\sum F_x=56\cos 53+27\cos 156+F_3\cos \theta$

$F_3\cos \theta_3=-9.035\ N----1$

$\sum F_y=F_1\sin \theta _1+F_2\sin \theta _2+F_3\sin \theta _3$

$F_3\sin \theta _3=55.704\ N----2$

squaring and adding 1 and 2

$F_3^2(\cos ^2\theta _3+\sin^2\theta _3)=86.583+3102.93$

$F_3=56.47\ N$

Divide 1 and 2 to get $\theta _3$

$\frac{\sin \theta_3 }{\cos \theta_3 }=\frac{-55.704}{9.305}$

$\tan \theta_3=-6.16$

$\theta _3=180-80.78$

$\theta _3=99.22^{\circ}$

6 0

3 years ago

Which statement is true? A) The work done to lift an object 6 meters is greater than the gravitational potential energy it gains

inna [77]

The work required to raise an object to a height is equal to the gravitational potential energy the object gains. <em>(C)</em>

4 0

4 years ago