Kiren
I haven't touched a physics book in a few years but from what I can remember

F = m*a

p = m*v

Ek = 1/2(m*v^2)

Does F or E = mass times velocity squared? Is it derived?

I do know E = m*c^2 and c is the speed of light (so you could substitue v, but I'm guessing that would be only for near light speeds)

Hopefully someone can clear this up for me
newton
wow. you join this forum at 5:04 am, and ask me to answer this 'question' on a physics board in a private message at 5:12 am?

HAHAHAHAHA! nice try.

i'm not your teacher, unless you're interested in conspiracy theory.
Kiren
i'm out of school, taking a break. i for the life of me can't remember if this is right or not and was hoping someone could help me out
yesitdid
QUOTE (Kiren+Apr 6 2006, 05:07 AM)
I haven't touched a physics book in a few years but from what I can remember

F = m*a

p = m*v

Ek = 1/2(m*v^2)

Does F or E = mass times velocity squared? Is it derived?

I do know E = m*c^2 and c is the speed of light (so you could substitue v, but I'm guessing that would be only for near light speeds)

Hopefully someone can clear this up for me

No, unless you are speaking about the E=mc^2 which is Einstein's energy equivalent of a given mass and IIRC is derived from the relatavitic effects of a mass approaching lightspeed , E is not equal to mv^2

You note that momentum is p=mv

If you recall calculus then you will see that the derivative of E=1/2(mv^2) wrt to velocity is equal to mv
Limon
How can v be equal to v²? If the mass of an object changes as it does in ballistic pendulums a change in v can't satisfy both equations. Newton wins but we pretend that The Law of Conservation of Energy does not lose (that magic, unmeasured heat).
Precursor562
E=MC^2 is the potential energy found in mass. Hence it was the equation used to determine the energy that would be released from a certain mass of uranium or plutonium. It actually best explains the energy that would be released when equal quantities of matter and its antimatter combine.

To determine the energy required to move objects at certain velocities is an equation that is a bit larger. I can't remember off hand

E = (m - x)/sqrt(V^2-C^2)

Something like that. There is no "x" I just put it there in place of what really goes there that I can't remember does.

As for the equation

N=M*V
is either
N=kg*m/s or N=kg*m/s^2

The difference is the force in Newtons to move an object of mass at a specific velocity and the force in Newtons to move an object of mass up to a specific velocity(acceleration). Yes the force that will move an object from 0 to 5 mph will keep it there.

socratus
And that about a very simple question:
What is the specific formula of the electron ( not a symbol)?
========
Commentaries:
==========
1.
e-
2.
electrons donot have formulas
but have charge which is expressed by e and
unit is coulombs. e=1.6*10^-19.
or for mass comparing with proton
(mass of electron=(1/1836) *mass of proton).
3.
The formula for what? the speed? the densiity?
Be more specific.
If you just mean the symbol it is (e-)
4.
(A very stupid question indeed!)
What does one really mean by the formula for an electron?
If u mean symbol, then the answer could be e-
5.
Time-independent Schrodinger wave equation
in one dimension for an electron is:

d^2 Ō/ dx^2 = - (2m/h^2)[E - U(x)] Ō(x)
6.
Not simple, you are actually right,
expecting even an electron to have a formula.
But the very behaviour of a simple electron can
constitute a formula occupying miles in paperwork.
If one day we work out that formula,
even to some approximation,
then a true window called science has opened to us.
Limon
A simple Quiz
The ballistic pendulum turns linear motion (the projectile) into circular motion (the pendulum block or cup plus projectile). If the pendulum is laid on a horizontal plane the same change in motion will occur except now the cup and projectile combination will be moving around the center post of the frictionless plane. This motion is identical to a spinning cylinder, and could be used to conduct the cylinder and spheres experiment. That is: All the motion would be given back to the projectile. Momentum is conserved when the projectile comes in; wouldn’t momentum be conserved as the projectile goes out? If it is not; it is the first experiment to not conserve momentum.
Lets put some numbers to this: A 1 kg projectile moving 5 m/sec collides with a 4 kg object at rest. The 5 kg combination then gives all of its motion back to the one kg object by using the cylinder and spheres experiment. What is the speed of the one kilogram projectile.
Precursor562
If a 1kg solid mass hit a 4kg solid mass at 5 m/s. Leaving all other forces out the 1 kg solid mass would stop and the 4kg solid mass would then move at 1/4th the speed.
Where force in newtons(N) equals mass times the change in velocity.
N=kg*dV
N=1*5
N=5

The 1kg mass would strike the 4kg mass with a force of 5 newtons.

N=kg*dV
N/kg=dV
5/4=1.25m/s

1/4 of 5 is 1.25.

In reality however the 1kg mass will "bounce" in that the force (kinetic energy) would reverberate off the far end of the 4kg mass. So the 4 kg mass will begin to move at 1.25m/s but the 1kg object will begin to travel the other way at 3/4ths the speed at which it came. 3.75m/s.

If the other way around you would end up with the 4kg mass reducing in speed by 1/4 and the 1 kg mass increasing to speed equal to the speed of the 4kg mass prior to impact.

N=kg*dV
N=4*1.25
N=5

N=kg*dV
N/kg=dV
5/1=5

Where the 1kg object had a starting velocity of zero than an increase of 5m/s will give it a velocity of 5m/s.

Limon
“Lets put some numbers to this: A 1 kg projectile moving 5 m/sec collides with a 4 kg object at rest. The 5 kg combination then gives all of its motion back to the one kg object by using the cylinder and spheres experiment. What is the speed of the one kilogram projectile.”

Let us assume that the collision is inelastic and that the motion is contained in the combination of the two objects. That gives us 1 m/sec for the 5 kg combination. All this motion is given back (by the cylinder and spheres experiment) to the one kilogram. Does your last sentence state that the final velocity of the one kilogram object should approach 5 m/sec?

Precursor562
Yeah I didn't take into consideration the whole pendulum, cylinder and spheres stuff. All I talked about were round masses in frictionless space.
Guest_john
A net force of 5 newtons causes an object to accelerate at a rate of 25m/s2 what is the mass of the object ?
Farsight
Kiren: forget the equations. Kinetic Energy is a way of writing down "stopping distance" if you apply a constant deceleration force. Momentum is a way of writing down "stopping time".

All you've really got is a moving mass. It doesn't have any momentum or energy, in that these are expressions rather than true properties. It's just a mass, moving. That's all there realy is out there in the world that we study with Physics.

Imagine if there was you and a cannonball. And the cannonball was moving at 10m/s. You could use the equations and say "the cannonball has this much kinetic energy". But now imagine it's you moving instead of the cannonball. Does the cannonball "have" any kinetic energy? Because it hasn't moved an inch, and it hasn't changed one iota. The answer is no.

The equations are valid. But what they represent aren't what you think.
AlphaNumeric
F = mv^2 cannot be right, no matter what you're attempting to descrive because the units don't match. Force has units of kgm/s^2, but mv^2 has units of kgm^2/s^2. If the units don't match then the equation is physically meaningless.

You're saying apples = pears.
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