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
ankoles [38]
3 years ago
12

The physics of wind instruments is based on the concept of standing waves. When the player blows into the mouthpiece, the column

of air inside the instrument vibrates, and standing waves are produced. Although the acoustics of wind instruments is complicated, a simple description in terms of open and closed tubes can help in understanding the physical phenomena related to these instruments. For example, a flute can be described as an open-open pipe because a flutist covers the mouthpiece of the flute only partially. Meanwhile, a clarinet can be described as an open-closed pipe because the mouthpiece of the clarinet is almost completely closed by the reed.
A. Consider a pipe of length 80.0 cm open at both ends. What is the lowest frequency f of the sound wave produced when you blow into the pipe?
B. A hole is now drilled through the side of the pipe and air is blown again into the pipe through the same opening. The fundamental frequency of the sound wave generated in the pipe is now_________ .
a. the same as before
b. lower than before
c. higher than before
Physics
1 answer:
Fed [463]3 years ago
7 0

Answer:

1. 214.38hz

2. higher than before.

Explanation:

1. the lowest frequency f of the sound wave produced

f = v/2L

l = 80cm = 0.8m

v = 343 metre per second, this is the sound speed of air

f = 343/2*0.8

the lowest frequency = 214.38hz

2. Since there is ahle n the pipe, this would shorten the length of air column that is able to oscillate in this pipe. we use the same formula as we used in question 1 here.

Vf ∝1/L

so we conclude that, c. it is higher than before

You might be interested in
A car battery with a 12 V emf and an internal resistance of 0.11 Ω is being charged with a current of 56 A. What are (a) the pot
denpristay [2]

Answer:

Part a)

V = 18.16 V

Part b)

P_r = 345 Watt

Part c)

P = 672 Watt

Part d)

V = 5.84 V

Part e)

P_r = 345 Watt

Explanation:

Part a)

When battery is in charging mode

then the potential difference at the terminal of the cell is more than its EMF and it is given as

\Delta V = E + i r

here we have

E = 12 V

i = 56 A

r = 0.11

now we have

\Delta V = 12 + (0.11)(56) = 18.16 V

Part b)

Rate of energy dissipation inside the battery is the energy across internal resistance

so it is given as

P_r = i^2 r

P_r = 56^2 (0.11)

P_r = 345 W

Part c)

Rate of energy conversion into EMF is given as

P_{emf} = i E

P_{emf} = (56)(12)

P_{emf} = 672 Watt

Now battery is giving current to other circuit so now it is discharging

now we have

Part d)

V = E - i r

V = 12 - (56)(0.11)

V = 12 - 6.16 = 5.84 V

Part e)

now the rate of energy dissipation is given as

P_r = i^2 r

P_r = 56^2 (0.11)

P_r = 345 W

7 0
3 years ago
A small box of mass m1 is sitting on a board of mass m2 and length L. The board rests on a frictionless horizontal surface. The
Nadusha1986 [10]

Answer:

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is \left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}

Explanation:

The Newton’s second law states that the net force on an object is the product of mass of the object and final acceleration of the object. The expression of newton’s second law is,

\sum {F = ma}

Here, is the sum of all the forces on the object, mm is mass of the object, and aa is the acceleration of the object.

The expression for static friction over a horizontal surface is,

F_{\rm{f}}} \leq {\mu _{\rm{s}}}mg

Here, {\mu _{\rm{s}}} is the coefficient of static friction, mm is mass of the object, and g is the acceleration due to gravity.

Use the expression of static friction and solve for maximum static friction for box of mass {m_1}

Substitute  for in the expression of maximum static friction {F_{\rm{f}}} = {\mu _{\rm{s}}}mg

{F_{\rm{f}}} = {\mu _{\rm{s}}}{m_1}g

Use the Newton’s second law for small box and solve for minimum acceleration aa to pull the box out.

Substitute  for , [/tex]{m_1}[/tex] for in the equation .

{F_{\rm{f}}} = {m_1}a

Substitute {\mu _{\rm{s}}}{m_1}g for {F_{\rm{f}}} in the equation {F_{\rm{f}}} = {m_1}a

{\mu _{\rm{s}}}{m_1}g = {m_1}a

Rearrange for a.

a = {\mu _{\rm{s}}}g

The minimum acceleration of the system of two masses at which box starts sliding can be calculated by equating the pseudo force on the mass with the maximum static friction force.

The pseudo force acts on in the direction opposite to the motion of the board and the static friction force on this mass acts in the direction opposite to the pseudo force. If these two forces are cancelled each other (balanced), then the box starts sliding.

Use the Newton’s second law for the system of box and the board.

Substitute for for in the equation .

{F_{\min }} = \left( {{m_1} + {m_2}} \right)a

Substitute for in the above equation .

{F_{\min }} = \left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is \left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g

There is no friction between the board and the surface. So, the force required to accelerate the system with the minimum acceleration to slide the box over the board is equal to total mass of the board and box multiplied by the acceleration of the system.

5 0
3 years ago
A train whistle has a sound intensity level of 70. dB, and a library has a sound intensity level of about 40. dB. How many times
kodGreya [7K]

Answer:

The sound intensity of train is 1000 times greater than that of the library.

Explanation:

We have expression for sound intensity level,

            L=10log_{10}\left ( \frac{I}{I_0}\right )

A train whistle has a sound intensity level of 70 dB

We have

           70=10log_{10}\left ( \frac{I_1}{I_0}\right )

A library has a sound intensity level of about 40 dB

We also have

           40=10log_{10}\left ( \frac{I_2}{I_0}\right )

Dividing both equations

           \frac{70}{40}=\frac{10log_{10}\left ( \frac{I_1}{I_0}\right )}{10log_{10}\left ( \frac{I_2}{I_0}\right )}\\\\\frac{7}{4}=\frac{log_{10}\left ( \frac{I_1}{I_0}\right )}{log_{10}\left ( \frac{I_2}{I_0}\right )}\\\\10^7\frac{I_2}{I_0}=10^4\frac{I_1}{I_0}\\\\\frac{I_1}{I_2}=10^3=1000

The sound intensity of train is 1000 times greater than that of the library.

3 0
3 years ago
If two charged objects each have 2.5 C of charge on them and are located 100 m apart, how strong is the electrostatic force betw
LiRa [457]

Answer:

5.619×10⁶ N

Explanation:

Applying,

F = kqq'/r²................... Equation 1

Where F = electrostatic force between the charges, k = coulomb's constant, q = first charge, q' = second charge, r = distance btween the charges

From the questiion,

Given: q = 2.5 C, q' = 2.5 C, r = 100 m

Constant: 8.99×10⁹ Nm²/C²

Substitute these values into equation 1

F = (2.5×2.5×8.99×10⁹)/100²

F = 56.19×10⁵

F = 5.619×10⁶ N

4 0
3 years ago
After a collision between two different massed objects; the larger objects accelerate at a faster rate than the smaller object?
Nitella [24]

Answer: Things continue doing what they are doing unless a force is applied to it. Objects have a natural tendency to resist change. This is INERTIA. Heavier objects (objects with more mass) are more difficult to move and stop. Heavier objects (greater mass) resist change more than lighter objects, so true

Explanation:

Pushing a bicycle or a Cadillac, or stopping them once moving. The more massive the object (more inertia) the harder it is to start or stop. The Cadillac has more of a tendency to stay stationary (or continue moving), and resist a change in motion than a bicycle.

6 0
3 years ago
Other questions:
  • Kyle lays a mirror flat on the floor and aims a laser at the mirror. The laser beam reflects off the mirror and strikes an adjac
    6·1 answer
  • Two cars are driving at the same constant speed on a straight road, with car 1 in front of car 2. car 1 suddenly starts to brake
    12·1 answer
  • Which of the following foods contains large amounts of protein? chicken, beans, fish, dairy, berries
    15·2 answers
  • Two electrons are at rest and separated by a distance of 4.32 × 10-10 m. When they are released they accelerate away from each o
    6·1 answer
  • Newly discovered planet has twice the mass and three times the radius of the earth. What is the free-fall acceleration at its su
    14·2 answers
  • What is the fundamental frequency of a 0.003 kg steel piano wire of length 1.3 m and under a tension of 2030 N? Answer in units
    9·1 answer
  • A mass of 3.0 kg rests on a smooth surface inclined 34° above the horizontal. It is kept from sliding down the plane by a spring
    5·1 answer
  • A car traveling at 90 m/s can stop in a distance of 110 m. What is the magnitude of the cars acceleration as it slows down?
    7·1 answer
  • 9. If 100 J of work is done in 20 second the power is ..
    9·2 answers
  • What is the acceleration of Karla's I Phone being thrown from Mr. Higley's classroom at 0m/s if it hits the wall 1.2 seconds lat
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!