1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Radda [10]
4 years ago
5

You are playing a violin, where the fundamental frequency of one of the strings is 440 Hz, as you are standing in front of the o

pening of a long tube that is closed at the other end. As you play, you notice that the first time you hear an echo from the tube is when the sound from the 440 Hz string is in its fourth harmonic. Assuming you are playing this string on earth, what must be the length of the tube? What is the next higher harmonic number for the same string on the violin when you hear the next higher harmonic echo from the tube?
Physics
1 answer:
Natalka [10]4 years ago
6 0

Answer:

a)   L = 440 cm

Explanation:

In the open tube on one side and cowbell on the other, we have a maximum in the open part and a node in the closed part, therefore the resonance frequencies are

             λ₁ = 4L             fundamental

             λ₃ = 4L / 3       third harmonic

             λ₅ = 4L / 5       five harmonic

             

The violin string is a fixed cure in its two extracts, so both are nodes, their length from resonance wave are

              λ₁ = 2L                    fundamental

             λ₂ = 2L / 2              second harmonic

             λ₃ = 2L / 3              third harmonic

             λ₄= 2L / 4               fourth harmonic

They indicate that resonance occurs in the fourth harmonic, let's look for the frequency

              v =λ f

for the fundamental

              v = λ₀ f₀

              V = 2L f₀

for the fourth harmonica

              v = λ₄  f ’

              v = L / 2  f'

             2L f₀ = L / 2 f ’

             f ’= 4 f₀

             f ’= 4 440

             f ’= 1760 Hz

for this frequency it has the resonance with the tube

           f ’= 4L

           L = f ’/ 4

           L = 1760/4

           L = 440 cm

b) let's find the frequency of the next harmonic in the tube

             λ₃ = 4L / 3

             λ₃ = 4 400/3

             λ₃ = 586.6 cm

            v = λf

            f = v / λlam₃

            f₃3 = 340 / 586.6

            f3 = 0.579

as the minimum frequency on the violin is 440 Beam there is no way to reach this value, therefore there are no higher resonances

You might be interested in
A 52-kg snow skier is at the top of a 245-m-high hill. After she has gone down a vertical distance of 112 m, what is her total e
Assoli18 [71]

Answer:

  a. 125 kJ

Explanation:

Her total energy is the same as the potential energy she had at the top of the hill:

  PE = mgh

  = (52 kg)(9.8 m/s^2)(245 m) = 124,852 J

  ≈ 125 kJ . . . . matches choice A

_____

After skiing down 112 m, some of her initial energy is converted to kinetic energy, and some remains as potential energy. We assume the ski slope is essentially frictionless, and air resistance is negligible.

7 0
3 years ago
A bar-magnet with magnetic moment 2.5 Am^2 is placed in a homogeneous magnetic field (of 0.1 T that is directed along the z-axis
Angelina_Jolie [31]

Answer:

1)

Force on bar magnet  = 0

Torque on bar magnet = 0

2)

Force on bar magnet  = 0

Torque on bar magnet = 0.177 Nm

3)

Force on bar magnet  = 0

Torque on bar magnet = 0.25 Nm

Explanation:

Part 1)

net force on bar magnet in uniform magnetic field is always zero

Torque on bar magnet is given as

\tau = MBsin\theta

when bar magnet is inclined along z axis along magnetic field

then we will have

\tau = MBsin0 = 0

Part 2)

net force on bar magnet in uniform magnetic field is always zero

Torque on bar magnet is given as

\tau = MBsin\theta

when bar magnet is pointing 45 degree with z axis then we will have

\tau = MBsin45

\tau = (2.5)(0.1)sin45

\tau = 0.177 Nm

Part 3)

net force on bar magnet in uniform magnetic field is always zero

Torque on bar magnet is given as

\tau = MBsin\theta

when bar magnet is pointing 90 degree with z axis then we will have

\tau = MBsin90

\tau = (2.5)(0.1)sin90

\tau = 0.25 Nm

8 0
4 years ago
Sasha did an experiment to study the solubility of two substances. She poured 100 mL of water at 20 °C into each of two beakers
chubhunter [2.5K]

Answer:

b is the higher solubility then A

4 0
3 years ago
A star emits electromagnetic radiation which closely resembles the spectrum of a blackbody. The three star system named Albireo
aksik [14]

Answer:

A = 13000K has a maximum at lam = 1,9984 10⁻⁷ m = 199.84 nm , this star is visually separated from the other two by its constant emission spectrum and is not affected by the other two.

we have a fluctuation of the intensity emitted by the stars.   Consequently by this fluctuation the amateur astronomer can conclude that this system is made up of two stars.

Explanation:

The radiation of a black body is characterized by its temperature, with Wien's law of displacement we can find the maximum wavelength emitted by each star.

                  λ T = 2,898 10⁻³

therefore the emission the star of A = 13000K has a maximum at lam = 1,9984 10⁻⁷ m = 199.84 nm

The emission of the premiere is in the ultraviolet light range, as this star is visually separated from the other two by its constant emission spectrum and is not affected by the other two.

The burst with A = 4300K ​​has a bad emission maximum = 6.7395 10⁻⁷ m = 673.95 nm, which corresponds to an emission in the visible in the orange range, giving a blackbody spectrum of this range, but since the emission is formed by two stars, we see that when the two are placed one in front of the other the intensity of the emission must increase significantly and when they are placed next to each other the intensity reaches its minimum, consequently we have a fluctuation of the intensity emitted by the stars.

Consequently by this fluctuation the amateur astronomer can conclude that this system is made up of two stars.

7 0
3 years ago
Complete the following:
masha68 [24]

When light is incident parallel to the principal axis and then strikes a lens, the light will refract through the focal point on the opposite side of the lens.

To find the answer, we have to know about the rules followed by drawing ray-diagram.

<h3>What are the rules obeyed by light rays?</h3>
  • If the incident ray is parallel to the principal axis, the refracted ray will pass through the opposite side's focus.
  • The refracted ray becomes parallel to the major axis if the incident ray passes through the focus.
  • The refracted ray follows the same path if the incident light passes through the center of the curve.

Thus, we can conclude that, when light is incident parallel to the principal axis and then strikes a lens, the light will refract through the focal point on the opposite side of the lens.

Learn more about refraction by a lens here:

brainly.com/question/13095658

#SPJ1

8 0
2 years ago
Other questions:
  • A closed loop conductor that forms a circle with a radius of 1.8 m is located in a uniform but changing magnetic field. If the m
    14·1 answer
  • Could anyone help me figure out d and e? How can you find the work of the friction when the force of friction is not given?
    12·1 answer
  • Non-moving electrical charges are often called
    13·2 answers
  • If 34.7 g of O2 reacts with iron to form 79.34 g of iron oxide, how much iron was used in the reaction?
    6·1 answer
  • 1. Water boils at 100°C at sea level. If the water in this experiment did not boil at 100°C, what could be the reason?
    11·2 answers
  • Reflected light from a thin film of oil gives constructive interference for light with a wavelength inside the film of λfilm. By
    5·1 answer
  • A large rocket has a mass of 2.00×10⁶ kg at takeoff, and its engines produce a thrust of 3.50×10⁷ N. Find its initial accelerati
    14·1 answer
  • Beyond what point must an object be squeezed for it to become a black hole
    13·1 answer
  • Which of the following has no energy?"<br><br> Amoving ball<br> Awrecking ball<br> Non of the above
    8·1 answer
  • Does frequency affect wavelength?
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!