Hi, there is a chance you could get 6000 point ! :) However, you must know somet

Question

Hi, there is a chance you could get 6000 point ! :)

However, you must know something before solving these problems.

Yes, I can't bet 6000 points here on the one post, so I gonna give more links under the pictures.

just go to the link when you finish to solve problems.

each post have different problems but every problems connect each others.

This post shows you the first page of my papers

Each post have different page of the papers.

you know that each post could have bet 300 points, so I gonna post 20 posts.

the number of papers which you have to solve is 5 pages (include this page) and it means actual number of posts which you have to consider is 5 posts

when it reaches over 6 posts with solving, then I gonna repeat the picture.

Then just repeat your answer with each post on each posts.

I gonna check your username if you posted every solution with clear explanation.

And I gonna rate you right away.

Don't worry I have never avoided to rate before.

i always check that and give right away with no problem.

If you can't see the link, it means that I'm making links and upload those as soon as possible.

Thanks for reading and hope you solve this as soon as possible :)

Okay, I start with first page here :)

while you're reading this, wait for links :)

I gonna post other link on comment :) check the comment :)

The speed of sound in air. Once again we return to the wave equation, but this time it is to get an idea of what determines the speed of sound in air. The situation we imagine is a mass that accelerates to the left or right due to a net force. Here the "mass" is just a small amount of air. Since the volume of air can change, we should he specific and say that by "amount" we mean a specified mass or a certain number of molecules but again, you should keep in mind that we assume that the changes in volume and pressure will be very small compared to the total volumes and pressures themselves. Be careful to distinguish between total quantities and the changes in those quantities. The situation is shown in the following diagram: We imagine a small cylinder of air, with small mass m, all of w hich is part of a sound w ave moving to the right. Let us also call the direction to the right the "positive" x direction and ignore all motion that is not in the x direction. There is pressure on both sides of this air. and a difference in pressure will cause it to accelerate Furthermore, the molecules on the left-hand side of this little mass may have a different velocity than the molecules on the right-hand side, causing the little mass to change in volume. IMPORTANT: The symbols needed for this problem can be very confusing! In order for us to understand each other, we all need to use the same symbols and since there are lots of P's and Y's in this problem, we must work to avoid confusing them with each other. In order to get full credit for this problem I must be able to understand your answ ers, so print neatly and use these symbols:

Explanation / Answer

1)

The general formula for the speed of sound in a gas is

v2 = dp/d?

where dp/d? is the derivative of pressure p with respect to density ?.

The presentation below is a derivation of this relationship. This derivation is more pedestrian than the usual derivation and does not require so many feats of mental agility.

Consider a tube of gas with constant cross-sectional area A. Let u be the fluid velocity, p the pressure and ? the density. The momentum equation for this flow is:

?du/dt = -?p/?x

which says that the force on a parcel is equal to the negative of the pressure gradient. The rate of change of the velocity du/dt is the instantaneous rate of change ?u/?t plus the advection term u?u/?x. When the left is divided by density and the terms are replaced the result is:

?u/?t + u?u/?x + (1/?)?p/?x = 0

These may be moved toward linearization by assuming u is small and ignoring products of small terms. The form of the equations is then

?u/?t + (1/?)?p/?x = 0

??/?t + ??u/?x = 0

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