albertlee

Thx Dark, the information is relavant and helpful......... Anyway, I have also problems with the direction of forces................. Can anyone tell me the basic concepts of the direction of force? for eg, how can an object move diagonally? or move in an indefined route, since it makes even harder to measure the distance and to say the direction of the force......... Aprciate for furthur helps!

Ok, just for sure, is my way of doing correct in my previous message?

Thx Dark........But the equation from Wolfson confused me: k.e. = 1/2mv^2 since m=1kg, v=av*2=2m/s Can I work it as k.e.=1/2(1)(2^2)? But what I wonder is where does this equation take the distance into acount? Any help? thx

How to find out the power by using average velocity? given info: object:1kg, average velocity:1m/s, distance:10m Aprciate for the responds

Any body?

Ok I got it.... Thx for the responds....

So is the rate of drag increases faster than thrust? just this simple.....

Yes, dissipated energy............. Can anyone explain more detailed about what is dissipated energy? and how it refers to my question?

So, Swansot, Do you also mean that the drag increases the force more faster than the thrust?

When a car is driving from left to right in the direction of force(right) using kinetic energy, what energy does the K.E. transfer to? I was thinking that it would be potential energy, but since energy is the capacity to do work, the such energy cannot do work due to likely the gavity, therefore it is just the displacement anyway.............. Any help?

Any idea?

So, Skye, so what is the name of the energy of the air? Skye, are u think about this equation: k.e. = 1/2mv^2?

Skye, transfer of kinetic energy to what?......

This is how i think........... The drag always equal to the the thrust according to Newton's first law, therefore it is a constant speed...... The reason why the plane increases its speed is because that the plane exerts more force on the thrust, hence the drag increases, therefore the constant speed increases.......... Again, this is just how i think..................but i dont think it is just this simple........ Any suggestions?

So......Any suggestions?

I think the "v^2" must come up with a different way of calculation......

Is it really sure the term "v^2" does come from v^2 = u^2 + 2as in the drag equation? since v^2 means initial velocity squared and 2 times acceleration times distance.......which just meaning nothing.............. Any ideas?

Ok Let me say it more clearly: Why the drag equation use the term "v^2" from the simultaneous equations of constant acceleration?Since, The term has no physical meaning...... Therefore, the drag equation I knew must be after the cancelation....Hence we get the odd term "v^2" which has no physical meaning............. So What is the drag equation before cancelation? many thanks for furthur responds.

Thx wolfson, but......sorry, it does only tell about what is the square of velocity as a concept in drag equation .......But I want to know about the significance of v^2, since you cannot feel it such as force, acceleration, etc. Or in another saying, what is the expansion of drag equation? since it must be after cancelation which finally shows an odd term like v^2........ Any ideas?

Thx Dave.....Now I know the calculation of v^2, but I am still confused why the drag equation use this term?, since it is just a "part" of equation of displacement, which is quite meaningless....... Any help?

Ok...... I might think it is very hard to explain the drag equation very completely in a concept..........It must be at a very high level in physics.......Since i am just too curious to know, so i ask.......... I think no one here would know what is the concept behind the velocity^2.......Anyway, thx to every one who answers my questions..........

Ok......, thx anyway to wolfson, but although i see the expansion of v^2, i still cannot realize the "concept" behind it................since the term 'v" means the final velovity right? And the final velocity = acceleration*time + initial velovity, where initial velovity is almost always 0, because in 0 second, there is no velocity......which means: v^2 = (at)^2............. What i really need is the "concept" of the squared final velocity.......... Secondly, I was also asking what does the term "area" refer to? Please help.....thx

Ok.....Thanks to everyone for the effort on answering my question............ Except that i dont what is Wolfson talking about in his last message of my thread........ The end of my thread...................

Thx YT for relevant information..........But simply i just want to know what is the "meaning" of squared velocity....or in another saying, why is it squared velovity? and secondly, The term "area" in the equation is for what? eg, area of the airplane, air, or etc............... Any help?

Ok............ So why dont the non-metals apply to "metallic bonding" if they can do so, since if it takes less energy? Can any one answer to that? Cheers