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. althoughthe acoustics of wind instruments is complicated, a simple description in terms of open and closed tubes can in understanding the physical phenomena related to these instruments. for example, a flute can bedescribed 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 almostcompletely closed by the reed. throughout the problem, take the speed of sound in air to be 343 .part aconsider a pipe of length 80.0 open at both ends. what is the lowest frequency of the sound wave produced when you blow into the pipe?

f = 214.4 Hz

Explanation:

It is a process of standing waves two waves one that travels to the right and another to the left are added, it is assumed that the frequency of the wave does not change, the result is a stable wave.

In the process of reflection of the wave there are two possibilities if the end is closed a node is formed and if it is open a belly is formed (antinode or maximum)

Let's apply these reasoning to our problem. We assume that the wave initially travels to the right, as it enters through an open end at that point it has a belly. When it reaches the other end that is open we also have a belly. As we see we have bellies at the ends, for there to be a wave inside the tube there must be a node, for the lowest resonance there is only one node, which must be in the center of the tube.

The distance from a node to a maximum (belly) corresponds to ¼ wavelength, so in this case we have within ½ tube wavelength

    L = ½ λ

    λ = 2L

    λ = 2 0.80

    λ = 1.60 m

Having the wavelength and the speed of sound let's use

    v = λ f

    f = v / λ

    f = 343 / 1.60

    f = 214.4 Hz

  1.the balanced   chemical equation is as below

cu +   2 ag(no)₃ → cu(no₃)₂   +2 ag

2.the   much silver   produced   is   50.9 g

step 1: find the   moles of cu

moles =   mass÷   molar mass

  from periodic table the molar mass of cu = 63.5 g/mol

moles is therefore = 15.00 g÷ 63.5 g/mol = 0.236   moles

step 2: use the mole ratio to determine the   moles of ag

from equation above   cu: ag   is 1: 2

therefore the moles of ag   =0.236   moles   x 2/1 =0.472   moles

step 3: find the mass   of ag

mass = moles x molar mass

from periodic table   the molar mass of ag =107.87 g/mol

mass   = 0.472   moles x 107.87 g/mol = 50.9 g

3.   the   much   silver   collected worth is   1.797 oz

  convert   50.9 g   to oz

that is   1 gram = 0.0353 oz

          50.9 grams =?   oz

by cross multiplication

={(50.9 grams x 0.0353 oz) / 1   gram}   =1.797 oz

answer: the exterior temperature of commercial jet planes as they leave the stratosphere and return to the ground increases steadily.

explanation:

the lower most layer of atmosphere is troposphere which extends from 0 to 12 km above the surface of the earth. the temperature initially decreases as distance is increased from the earth's surface. commercial jets fly in lower part of the stratosphere. the stratosphere extends upto 50 km. the temperature in stratosphere starts rising again due to presence of ozone which traps the uv rays.

thus, when the planes leave the stratosphere and return to the ground, the exterior temperature increases steadily.